TITLE: Let $p(x)$ have a zero $a\in \mathbb{Q}$ ...
QUESTION [1 upvotes]: "Let $p(x)$ have a zero $a\in \mathbb{Q}$..."
Where $p(x)$ is a polynomial.
I came about this part of a statement and I was not entirely sure what it meant. Although, I assumed that it meant that there exists an element $a\in \mathbb{Q}$ such that if I substitute it into the polynomial $p(x)$ it would give $zero$. But I'm not entirely sure.
If someone can clarify this for me, I would appreciate it. Thanks

REPLY [2 votes]: You're correct. That sentence, in English, means that there exists a rational number such that the polynomial evaluated at that rational number yields $0$.

REPLY [2 votes]: Yes, you are entirely correct. It means that there is some $a\in \mathbb Q$ such that $p(a) = 0$.
But you can also think of it this way: $(x - a)$ divides $p(x)$. That is $$p(x) = (x-a)q(x)$$ where $q(x)$ whose degree is one less than the degree of $p(x)$.


TITLE: Value of $ \sum_{t \in \mathbb{F}_{p^n}} \zeta ^{Tr (t)} $
QUESTION [0 upvotes]: Let $\zeta = e^{2\pi i/p } $. Then is it true that $ \sum_{t \in \mathbb{F}_{p^n}} \zeta ^{Tr (t)} = 0  $, where $Tr$ is the map $ Tr:\mathbb{F}_{p^n} \rightarrow \mathbb{F}_{p}$ defined by $Tr(a) = a+a^p+a^{p^2}+......a^{p^{n-1}}$? If so, why? I know that $Tr$ is onto.

REPLY [2 votes]: Look at $$\sum_{l=1}^{p-1} \sum_{t \in \mathbb{F}_{p^n}} \zeta^{l\ Tr (t)}$$
Equivalently show that since $Tr$ is a surjective group homomorphism then $|Tr^{-1}(t)| = p^{n-1}$.


TITLE: Can the isoperimetric dimension of a d-generated group attain any value?
QUESTION [3 upvotes]: Background
The isoperimetric dimension of a finitely generated group $G$, which we denote by $\dim(G)$, is the largest number $d$ such that any Cayley graph $\Gamma$ of $G$ (constructed with respect to a finite generating set) satisfies a $d$-dimensional isoperimetric inequality, i.e.
\begin{equation}
|\partial A|\geq C|A|^{(d-1)/d}
\end{equation}
for all finite subsets $A\subseteq\Gamma$, where $C$ is some constant (which depends on $\Gamma$ and $d$ but not on $A$). Here $\partial A$, the bounday of $A$, is the set of vertices in $\Gamma\backslash A$ which have a neighbor in $A$.

Suppose now that $G$ is a $d$-generated group, i.e. a quotient of $\mathbb{F}_d$, the free group of rank $d$. Then provided $d>1$, $\dim(G)$ may attain any value in the set $\{0,\ldots,d\}\cup\{\infty\}$, as is evidenced, for instance, by the free Abelian groups $\mathbb{Z}^k$, where $0\leq k\leq d$ (since $\dim(\mathbb{Z}^k)=k$), and the free group $\mathbb{F}_d$ itself (since $\dim(\mathbb{F}_d)=\infty$). My question is:

What are examples of $d$-generated groups $G$ that satisfy $d<\dim(G)<\infty$?

Going a bit further:

Can the isoperimetric dimension of a $d$-generated group attain any value?

REPLY [1 votes]: The simplest Another group answering the question is the Heisenberg group (over $\mathbb{Z}$):
$$
H_3(\mathbb{Z}) = \left\lbrace \left(\begin{smallmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{smallmatrix}\right) \bigg| x, y, z \in \mathbb{Z} \right\rbrace
$$
This group is generated by two elements (the matrices $\left(\begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{smallmatrix}\right)$ and $\left(\begin{smallmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right)$ (since the matrix  $\left(\begin{smallmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right)$ is the commutator of the two other.
On the other hand it has volume growth of the type $V(n) \simeq n^4$ and satisfies a 4-dimensional isoperimetric inequality. So this answers the first question.
As for the second question, note first that your inequality implies a lower bound on the volume growth of the type $n^d$. Indeed, by looking at $B_n$ the ball of radius $n$, one gets that $|\partial B_n| \geq C|B_n|^{1-1/d}$. Since the vertices are of bounded degree, $|B_n| -|B_{n-1}| \geq C'|B_n|^{1-1/d}$. The easiest way to conlcude is to introduce a piecewise affine extension $b(x)$ of the function $n \mapsto |B_n|$. Then your inequality reads $b'(x) \geq C'b(x)^{1-1/d}$ which integrates to a bound of $b(x)^{1/d} \geq C'' x$
On the other hand there is a reverse inequality (see Coulhon, Thierry, and Saloff-Coste, Laurent. "Isopérimétrie pour les groupes et les variétés.." Revista Matemática Iberoamericana 9.2 (1993): 293-314. There is a survey written in English and availiable online by Pittet and Saloff-Coste, see here. Look at section 1 (the function you are interested in is $\tfrac{1}{J}$, but not $J$ or $I$) and section 7 (more precisely Theorem 7.0.10): one has that $V(n) \succeq n^d$ implies $J(t) \preceq t^{1/d}$ (which implies $\tfrac{1}{J(t)} \succeq t^{-1/d}$, i.e. a $d$-dimesional inequality).
So this means the answer to your second question is: the isoperimetric dimension of a group is an integer or $+\infty$ (see Yves' answer for an example of a group with infinite dimension which is amenable [in fact solvable] a group which is amenable [in fact solvable] and has infinite isoperimetric dimension is $\mathbb{Z}^\infty \rtimes \mathbb{Z}$, basically the limit as $k \to \infty$of the groups in Yves' answer).
Indeed, if a group had an isoperimetry of $k +\epsilon$ for some $k \in \mathbb{N}$ and $\epsilon \in ]0,1[$ then, its growth would be at least $k+\epsilon$. Since the growth exponent is an integer, then its growth is actually at least $k+1$. In turns this implies the group has a $(k+1)$-dimensional isoperimetry.


TITLE: Quaternion Rings
QUESTION [12 upvotes]: Let $R$ be a commutative ring. Define the Hamilton quaternions $H(R)$ over $R$ to be the free $R$-module with basis $\{1, i, j, k\}$, that is,
$$H(R)=\{a_0+a_1i+a_2j+a_3k\;\;:\;\;a_l\in R\}.$$
and multiplication is defined by: $i^2=j^2=k^2=ijk=-1$.
Is well-known that over a field $F$ (with char $F\neq 2$) the ring $H(F)$ is a division ring or isomorphic to $M_2(F)$. What can we say  about the Hamilton quaternions over an arbitrary commutative ring $R$? Is still true that $H(R)$ is a division ring or isomorphic to $M_2(R)$? We must impose some conditions to the ring to make it happen?

REPLY [9 votes]: (Your definition of the multiplication looks slightly strange to me.  I would prefer to write $k = ij = -ji$.  This implies your relations; as Arturo shows in his comment below, your relations also imply mine, but to my mind it's sort of mixing two different ways to give an $R$-algebra: you could either give it via generators and relations, or if it's a free $R$-module you can give an explicit basis and the structure constants.)
If $R$ is a field of characteristic different from $2$, then $H(R)$ is by definition the quaternion algebra $\left( \frac{-1,-1}{R} \right)$.  In general, for $a,b \in R^{\times}$, the quaternion algebra
$\left( \frac{a,b}{R} \right) = R\langle i,j \rangle/(i^2 = a, j^2 = b, ji = -ij)$
is indeed always a division algebra or isomorphic to $M_2(R)$: see e.g. $\S 5.1$ of these notes.
(If $R$ is a field of characteristic $2$, then $H(R)$ is a commutative ring, so is not what you want.  There are analogues of quaternion algebras in characteristic $2$ defined slightly differently.)
If $R$ is a domain, then the center of $H(R)$ is $R$.  It follows that if $R$ is not a field, $H(R)$ is not a division ring.  Note also that if $R \subset S$ is an extension of domains, then $H(S) \cong H(R) \otimes_R S$.
In particular $H(\mathbb{Z})$ is not a division ring.  Neither is it isomorphic to $M_2(\mathbb{Z})$: if so, then $H(\mathbb{\mathbb{R}})$ would be isomorphic to $M_2(\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{R} \cong M_2(\mathbb{R})$, but $H(\mathbb{R}) = \left( \frac{-1,-1}{\mathbb{R}}\right)$ is the classical Hamiltonian quaternions, which is well-known to be a division ring.
In general, if $R$ is a domain of characteristic different from $2$ with fraction field $K$, then $H(R)$ is an order in the quaternion algebra $H(K)$.  This is more a definition than anything else, but it gives you a name.  There is a vast literature on quaternion orders: for just a little bit, see e.g. these notes.


TITLE: Sets computable from enough hints
QUESTION [10 upvotes]: Is there a non-computable set $X\subset\omega$ such that, for some $Y\subset\omega$, any infinite subset or cosubset (=subset of the complement) of $Y$ computes $X$?
More generally, call a set $X$ $n$-hintable if there is some $f\in n^\omega$ such that, whenever I am told infinitely many bits of $f$, I can compute all of $X$. Then: What are the $n$-hintable sets?
(I ran into this question while thinking about stable ramsey's theorem: given a stable $k$-coloring of pairs - that is, a $k$-coloring of pairs $c$ such that $\lim_yc(x<y)$ always exists - the map $f: x\mapsto \lim_yc(x<y)$ is an element of $k^\omega$ such that, given infinitely many bits of $f$, we can compute an infinite $c$-homogeneous set. So we could say that the mass problem of solutions to $c$ is $k$-hintable; and now the question arises whether there are individual reals which are hintable.)
I suspect the answer is "every $n$-hintable set is computable," but I can't see how to prove it.
By way of background, a set $X$ is bi-introreducible if any infinite subset or cosubset of $X$ computes $X$. So 2-hintability is a much weaker property than bi-introreducibility. Seetapun (http://projecteuclid.org/download/pdfview_1/euclid.ndjfl/1040136917, theorem 2.19) showed that the bi-introreducible sets are all computable, but his argument doesn't answer my question, as far as I can tell. EDIT: As Liang Yu's answer below shows, Seetapun's arguments do answer my question, at least for $n=2$.

REPLY [8 votes]: There is no 2-hintable set by the same argument from Seetpun.
Suppose that there is such a pair $X$ and $Y$. By Freidberg's Theorem, there is a set $A$ such that $A$ is not above either $X$ or $Y$, but $A'$ computes $X\oplus Y$. By Theorem 2.18 in the paper, there is a function $f\leq_T A:[\omega]^2\to 2$ such that any infinite $f$-homogeneous set is either subset of $Y$ or of the complement of $Y$. Now by Theorem 2.1 in the paper, there is an $f$-homogeneous set $H$ not computing $X$.
I think that, by a simple modification of this argument, one may show that there is no $n$-hintable set for any $n\geq 2$.


TITLE: How can you construct as many intersections as possible with n lines?
QUESTION [9 upvotes]: If you have $n$ lines, it seems to be obvious that you can have at most $\frac{n^2-n}{2}$ intersections:
$n = 1$: Obviously you need two lines to intersect, so the maximum number of intersections is 0
$n=2$: As one line can't intersect with itself, it can only intersect with the old lines. As there is only one line, you can get at most one intersection. So the maximum number of intersections of two lines is 1.
$n=3$: The same argument as before. The new line can only intersect with the two old lines and a given pair of two lines can intersect at most once. So you can get at most two new intersections. With the intersection of the two lines before, you get 3 intersections.
So the pattern is $\sum_{i=1}^{n-1} i = \frac{n^2-n}{2}$
But this only gives an upper border for the maximum number of intersections. I would now like to give a constrution where you can obviously see that you can get that many intersections.
I tried to make such an construction by hand:

but I guess with 10 lines this will get very confusing. Also, I don't know how to make it obvious that this construction can be continues indefinitely.
Do you know how you can construct as many intersections as possible with n lines?

REPLY [0 votes]: I'm not qualified to speak on this subject, but one cool think Ive found is that if you construct a regular polygon with n sides and extend the lines of those sides to infinity, you get that maximum number $\frac{n^2-n}2$ of intersections for odd $n$.
Repeating for even numbers we have $\frac{n^2-2n}2$ intersections, a difference of $\frac{n}2$ intersections.
So, for a polygon $n = \infty$ (a circle), you would get either $\infty$ OR $\frac{\infty}2$ intersections (its meaningless to talk about infinity as even or odd).
Therefore, this construction guaranteed to have as many intersections as possible.
Note:
This construction is basically extending every tangent line of a circle to infinity, which I guess would be a solution to the system in an answer above which involved picking lines such that none are at the same angle and all have different positions.


TITLE: Proving $f\in\mathbb{C}(D)$ if and only if for every closed set $F\subseteq R^m$, $f^{-1}(F)$ is closed (relatively) in $D$.
QUESTION [0 upvotes]: I am confident in my first inclusion but I am quite lost in my second. Can I please have help?
$\def\x{{\mathbf x}}
\def\R{{\mathbb R}}
\def\C{{\mathbb C}}
\def\f{{\mathbf f}}$
Let $\f\colon D\to \R^m$, $D\subseteq\R^n$. Prove $\f\in\C(D)$ if and only if for every closed set $F\subseteq \R^m$, $\f^{-1}(F)$
is closed (relatively) in $D$.
$\textbf{Solution:}$
$(\rightarrow)$ Suppose $\f\in\C(D)$ and $F$ be any arbitrary closed set in $\R^m$. Our claim is that $\f^{-1}(E)$ is relatively closed in $D$. $F$ is a closed set in $\R^m$ if and only if $\R^m \setminus F$ is open set in $\R^n$. As $\f$ is continuous on $D$,
$$\f^{-1}(\R^m\setminus E) \text{ is open (relatively) in } D$$ $$\f^{-1}(\R^m \setminus E) = \f^{-1}(\R^m)\setminus \f^{-1}(E) = D\setminus \f^{-1}(E)$$ because $\f^{-1}(A\setminus B) = \f^{-1}(A)\setminus \f^{-1}(A) \setminus \f^{-1} (B)$ and $\f^{-1}(\R^n) = D$.
Then, $D\setminus \f^{-1}(E)$ is open (relatively) in $D$. Thus, $D\setminus (D\setminus \f^{-1}(E)) = \f^{-1}(E)$ is relatively closed in $D$. This is for all $E$ closed in $\R^n$.
$(\leftarrow)$ Conversely, suppose for every $F$ closed in $\R^m, \f^{-1}(F)$ is relatively closed in $D$. Our claim is that $\f$ is continuous, $\f \in \C(D)$, i.e. for every open set $E \subseteq \R^m, \f^{-1}(E)$ is relatively open in $D$. $E$ is any open set in $\R^m$ which is equivalent to saying $\R^m \setminus E$ is closed in $\R^m$. Then $\f^{-1}(\R^m\setminus E)$ is closed (relatively) set in $D$. So, $\f^{-1}(\R^m\setminus E) = \f^{-1}(\R^m)\setminus \f^{-1}(E) = D\setminus \f^{-1}(E)$ is closed (relatively) set in $D$. So, $\f^{-1}(E)$ is relatively open in $D$ and this is for all $E$ open in $\R^m$. Thus, $\f\in \C(D)$.

REPLY [1 votes]: As Reveillark noted: "Your proof seems fine. This is a general fact about arbitrary topological space $f : X \to Y$ is continuous iff $f^{-1}(E)$ is closed in $X$ for every closed $E \subseteq Y$."


TITLE: Proving that $\mathbb{R}$ is a field
QUESTION [2 upvotes]: Assuming that we have the definition of the rational numbers $\mathbb{Q}$ we define $\mathbb{R}$ as the completion of $\mathbb{Q}$ with respect to the usual norm. I want to understand why this gives a field.
By definition of the construction, $\mathbb{R}$ is defined as equivalence classes of Cauchy sequences $\{x_n\}$ with $x_n\in \mathbb{Q}$. That is
$$
\mathbb{R} = \{[\{x_n\}] : \{x_n\}\text{ Cauchy in } \mathbb{Q}\}.
$$
Here $\{x_n\} \sim \{y_n\}$ if $\lim_{n\to \infty} \lvert x_n - y_y\rvert = 0$.
I believe the operations on $\mathbb{R}$ are
$$
[\{x_n\}] + [\{y_n\}] = [\{x_n + y_n\}] \\
[\{x_n\}] [\{y_n\}] = [\{x_ny_n\}]
$$
I believe I have shown that $\mathbb{R}$ is a commutative ring. All I have left to show is that all the (non-zero) elements have multiplicative inverses.
But if I have $[\{x_n\}]\neq 0$, then I can just define $y_n = 1/ x_n$ so that
$$
[\{x_n\}] [\{y_n\}] = [\{x_n y_n\}]  = [\{1\}]
$$
I am under the impression that the inverse part of the proof is hard, so I am guessing that there is something wrong with my approach.
Am I doing the right thing?

REPLY [1 votes]: Let's see what's the meaning of $[\{x_n\}]\ne0$. Since $0$ is the equivalence class of the constant zero sequence, we know that it doesn't hold
$$
\lim_{n\to\infty}|x_n|=0
$$
Thus there exists $\eta>0$ (rational) such that, for every $n$, there is $m>n$ with $|x_m|\ge2\eta$.
On the other hand, $\{x_n\}$ is Cauchy, so there exists $\bar{n}$ such that, for $m,n>\bar{n}$, $|x_n-x_m|<\eta$.
Fix $m>\bar{n}$ such that $|x_m|\ge\eta$; then, for every $n\ge\bar{n}$,
$$
x_m-\eta<x_n<x_m+\eta
$$
If $x_m>0$, then $x_m\ge2\eta$, so $x_n>\eta$; if $x_m<0$, then $x_m<-2\eta$ and so $x_n<-\eta$.
In either case we can state that $x_n\ne0$, for every $n>\bar{n}$.
Define a new sequence $\{y_n\}$ by $y_n=0$ if $0\le n\le\bar{n}$ and $y_n=x_n^{-1}$ if $n>\bar{n}$.
Now the task is to prove that $\{y_n\}$ is Cauchy. Once you have this, you can easily prove that $[\{x_n\}][\{y_n\}]=1$ (the equivalence class of the constant one sequence).
You need to use that $|x_n|\ge\eta$ for every $n>\bar{n}$.


TITLE: For $\mathfrak{g}$ semisimple, is $[\mathfrak{g}, \mathfrak{g}] = \mathfrak{g}$.
QUESTION [0 upvotes]: If $\mathfrak{g}$ is a semisimple Lie algebra, is it true that $$ [\mathfrak{g}, \mathfrak{g}] = \mathfrak{g}.$$
Obviously the derived ideals must stabilize somewhere, but I don't see why $D^{1}(\mathfrak{g}) = \mathfrak{g}$. It seemed my professor was stating this result as a consequence of Weyl's Theorem, but I am not sure why this should follow (or even if this is true).

REPLY [1 votes]: Yes, it is true. A semisimple Lie algebre is a direct sum $\bigoplus_{j=1}^n\mathfrak g_j$ of simple Lie algebras. And $[\mathfrak g_j,\mathfrak g_j]$ is an ideal of $\mathfrak g_j$. Since $\mathfrak g_j$ is simple, this ideal can only be $\mathfrak g_j$ itself. So$$[\mathfrak g,\mathfrak g]=\bigoplus_{j=1}^n[\mathfrak g_j,\mathfrak g_j]=\bigoplus_{j=1}^n\mathfrak g_j=\mathfrak g.$$


text:
\section{Rational elliptic surfaces, del Pezzo surfaces, and phantoms}
\label{sec: del pezzos}

From the classification of admissible subcategories of $\Dbcoh(\P^2)$ given by Theorem~\ref{thm: all admissible subcategories are standard} we easily see that there are no phantom subcategories in $\P^2$. In fact, the full classification is not necessary, and it is not hard to come to the same conclusion right after Lemma~\ref{lem: p2 numerical lemma}. In this section we explore other situations where we can apply the strong structural result given by Proposition~\ref{prop: numerical lemma without numbers} to study phantoms.

To apply the methods of Section~\ref{sec: numerical lemmas} we need a surface with many anticanonical divisors. We concentrate on the case of rational elliptic surfaces. Those are the surfaces whose anticanonical linear system defines a fibration over $\P^1$. First we show that a phantom subcategory in a rational elliptic surface cannot have full support; in fact, in Theorem~\ref{thm: rational elliptic surfaces} we prove that the support of any phantom subcategory is contained in the union of sections and reducible fibers of the anticanonical fibration.

This is a relatively explicit description: all sections are smooth $(-1)$-curves, and Kodaira classified the reducible fibers of (minimal) elliptic surfaces. Since we have managed to prove a classification of admissible subcategories supported on a single smooth $(-1)$-curve in Section~\ref{sec: minus one curves}, it seems reasonable to expect that it may be possible to classify the admissible subcategories supported on this more complicated, but still quite specific, configuration of negative curves. We have done this partially.

Using Proposition~\ref{prop: numerical lemma without numbers} we have proved the classification of admissible subcategories in rational elliptic surfaces which are supported only along the sections of the anticanonical fibration. We show in Theorem~\ref{thm: admissible subcategories on sections} that the only option is to choose several pairwise disjoint sections, pick a line bundle on each one, and span the subcategory by their pushforwards.

Finally, we utilize the classification result above and a relation between del Pezzo surfaces and rational elliptic surfaces to show in Theorem~\ref{thm: no phantoms in del pezzos} that there are no phantoms in del~Pezzo surfaces.

The proofs in this section use the notion of a \emph{point-support} of an admissible subcategory at some point. This notion is introduced in Subsection~\ref{ssec: point-supports}. We also need some additional lemmas about objects set-theoretically supported on curves in surfaces. We study them by pulling them back along curves transverse to the support in Subsection~\ref{ssec: cutting lemmas}. We proceed by studying possible phantom subcategories in rational elliptic surfaces in Subsection~\ref{ssec: rational elliptic surfaces}. The classification of admissible subcategories supported on the union of sections of the elliptic fibration is established in Subsection~\ref{ssec: sections of anticanonical system}. We conclude by showing the non-existence of phantoms in del Pezzo surfaces in Subsection~\ref{ssec: no phantoms in del pezzos}.

\subsection{Point-supports of admissible subcategories}
\label{ssec: point-supports}

\begin{definition}
  \label{def: point-support}
  Let $X$ be an algebraic variety. Let $\mB \subset \Dbcoh(X)$ be an admissible subcategory. Pick a point $p \in X$. The \emph{point-support} of $\mB$ at $p$ is a set-theoretic support $\supp(\mB_R(\O_p))$ of the right projection of the skyscraper sheaf at $p$ to the subcategory $\mB$.
\end{definition}

\begin{remark}
  In this definition we chose to use the right projection functor. I do not know whether the supports of the right projection $\mB_R(\O_p)$ and the left projection $\mB_L(\O_p)$ always coincide.
\end{remark}

\begin{lemma}
  \label{lem: properties of point-supports}
  Let $X$ be an algebraic variety, let $p \in X$ be a point, and let $\mB \subset \Dbcoh(X)$ be an admissible subcategory. Let $S_p$ be the point-support of $\mB$ at $p$.
  \begin{enumerate}
    \item If $S_p$ is not empty, then it is a connected closed subset which contains $p$.
    \item If $S_p$ is empty, then $\O_p \in \mB^\perp$. If $S_p = \{ p \}$, then $\O_p \in \mB$.
    \item Let $\mA = \mB^\perp$ be the orthogonal subcategory. Consider the projection triangle
      \[
        \mB_R(\O_p) \to \O_p \to \mA_L(\O_p)
      \]
      If $S_p$ is neither empty nor $\{ p \}$, then $\supp(\mA_L(\O_p)) = S_p = \supp(\mB_R(\O_p))$.
  \end{enumerate}
\end{lemma}

\begin{proof}
  Let $\mA = \mB^\perp$ be the orthogonal subcategory, and let $A_p = \mA_L(\O_p)$ and $B_p = \mB_R(\O_p)$ denote the projections of the skyscraper sheaf $\O_p$, fitting into the projection triangle
  \[
    B_p \to \O_p \to A_p.
  \]
  Then $S_p = \supp(B_p)$ by definition. Suppose that this set has a nonempty connected component which does not contain $p$. Then by Lemma~\ref{lem: disjoint support forces splitting} the object $B_p$ has a nonzero direct summand~$B^\prime$ which is supported away from $p$. Then the map $B_p \to \O_p$ factors through the projection to $B_p/B^\prime$, and therefore its cone, which is $A_p$, has a direct summand isomorphic to~$B^\prime[1]$. But then $\RHom(B_p, A_p) \neq 0$, which is a contradiction with semiorthogonality. Thus the support of $B_p$ (and by a similar argument the support of $A_p$ as well) is either empty, or a connected subset containing $p$. This proves part (1).

  Suppose now that $S_p = \{ p \}$. The projection triangle implies in this case that $\supp(A_p)$ is contained in the one-point set $\{ p \}$. Assume that both $A_p$ and $B_p$ are nonzero objects supported at a single point $p$. Let $a \in \Z$ be the smallest number such that $\mH^a(A_p) \neq 0$ and let~$b \in \Z$ be the largest number such that $\mH^b(B_p) \neq 0$. Pick nonzero morphisms $\mH^b(B_p) \epiarrow \O_p$ and $\O_p \monoarrow \mH^a(A_p)$, which always exist for coherent sheaves supported at one point. Then the composition
  \[
    B_p \to \mH^b(B_p)[-b] \epiarrow \O_p[-b] \monoarrow \mH^a(A_p)[-b] \to A_p[a-b]
  \]
  with truncation morphisms is a morphism $B_p \to A_p[a-b]$ which by construction is nonzero on cohomology sheaves. This contradicts semiorthogonality. Thus~$A_p$ must be a zero object when~$S_p = \{ p \}$, so part (2) is proved.

  To deal with the last part, note that the long exact sequence of cohomology sheaves proves that $\supp(B_p) \subset \supp(A_p) \cup \{ p \}$ and similarly for $\supp(A_p)$. Since both of those supports are either empty or contain the point $p$, part (3) follows.
\end{proof}

\begin{lemma}
  \label{lem: no semiorthogonality for zero-dimensional intersections}
  Let $Y$ be a smooth variety, $S_1, S_2 \subset Y$ two closed subsets whose set-theoretic intersection $S_1 \cap S_2$ contains an isolated point. Let $F_1, F_2 \in \Dperf(Y)$ be objects whose set-theoretical supports are $S_1, S_2$ respectively. Then $\RHom(F_1, F_2) \neq 0$.
\end{lemma}

\begin{proof}
  We can compute the $\RHom$-space by the dualization:
  \[
    \RHom(F_1, F_2) \caniso \RGamma(Y, F_1^\dual \otimes F_2).
  \]
  The support of the tensor product $F_1^\dual \otimes F_2$ is the intersection $S_1 \cap S_2$. It contains an isolated point, so by Lemma~\ref{lem: disjoint support forces splitting} the object $F_1^\dual \otimes F_2$ has a nonzero direct summand supported only at a single point. Any object with zero-dimensional support has a nonvanishing (hyper)cohomology class given by a nonzero global section of the lowest degree cohomology sheaf, so the lemma is proved.
\end{proof}

\begin{lemma}
  \label{lem: skyscrapers and intersections}
  Let $Y$ be a smooth variety, $p, q \in Y$ distinct points. Let $\mB \subset \Dbcoh(Y)$ be an admissible subcategory. Denote by $S_p$ and $S_q$ the point-supports of $\mB$ at $p$ and $q$ respectively. Then the set-theoretic intersection $S_p \cap S_q$ does not contain any isolated points.
\end{lemma}

\begin{remark}
  This lemma is most useful on surfaces, where all nontrivial point-supports are curves, and curves usually intersect along finitely many points.
\end{remark}

\begin{figure*}[ht]
  \centering
  \def\svgwidth{0.3\columnwidth}
  \import{./}{point-supports.pdf_tex}
  \caption{An impossible situation}
  \label{fig: skyscrapers and intersections}
\end{figure*}

\begin{proof}
  Without loss of generality assume that $S_p$ is larger than just $\{ p \}$. Let $\mA = \mB^\perp$ be the orthogonal subcategory. Let~$A_p = \mA_L(\O_p)$ and~$B_p = \mB_R(\O_p)$ be the projections of the skyscraper sheaf at $p$ to the categories $\mA$ and $\mB$ respectively. By the last part of Lemma~\ref{lem: properties of point-supports} we see that $\supp(A_p) = \supp(B_p) =  S_p$ in this case.

  First, suppose that $S_q = \{ q \}$ and $S_p$ contains $q$. Then by Lemma~\ref{lem: properties of point-supports} this implies that~$\O_q \in \mB$. But then the graded space $\RHom(\O_q, A_p)$ is zero by semiorthogonality, which implies that $\supp(A_p) = S_p$ does not contain $q$, a contradiction.

  Thus we may assume that $S_q \neq \{ q \}$. Then by the same lemma $\supp(\mA_L(\O_q)) = S_q$. Consider the space $\RHom(B_p, \mA_L(\O_q))$. If the intersection $S_p \cap S_q$ contains an isolated point, then by Lemma~\ref{lem: no semiorthogonality for zero-dimensional intersections} this space is not zero, but this again is impossible by semiorthogonality of $\mA$ and $\mB$.
\end{proof}

\subsection{Cutting lemmas}
\label{ssec: cutting lemmas}

This subsection contains a few observations about objects in the derived categories of surfaces which are set-theoretically supported on curves.

\begin{definition}
  Let $S$ be a smooth surface, $F \in \Dbcoh(S)$ an object. Assume that the set-theoretic support of $F$ is a reduced curve $C \subset S$. A \emph{slice} of $F$ at a point $p \in C$ is the derived pullback $j^*F$ to a curve $j\colon D \to S$ which is smooth at the point $p$ and does not intersect $C$ anywhere else.
\end{definition}

Note that an alternative way to state the definition would be to let $D$ intersect $C$ at some other points, but replace the derived pullback by the largest direct summand supported at the point $p$. This is equivalent to replacing $D$ with an open neighborhood of $p$ in $D$.

\begin{lemma}
  \label{lem: equicutting lemma}
  Let $U$ be a smooth surface, let $i\colon C \monoarrow U$ be a curve, and let $p \in C$ be a point. Let $F \in \Dperf(U)$ be an object whose set-theoretic support is $C$. Suppose that there exists a slice of $F$ at $p$ which is a torsion object of length one. Then $C$ equipped with a reduced scheme structure is smooth at the point $p$, and, after possibly replacing $U$ by a Zariski-neighborhood of $p$, the object $F$ is isomorphic to $i_*\O_C[a]$ for some shift $a \in \Z$.
\end{lemma}

\begin{proof}
  Let $j\colon D \to U$ be a smooth curve such that the derived pullback $j^*F \in \Dbcoh(D)$ is a torsion object of length one, i.e., it is isomorphic to a shift of a skyscraper sheaf $\O_p[a]$ for some $a \in \Z$. By Lemma~\ref{lem: derived restriction to a divisor} the isomorphism $j^*F \iso \O_p[a]$ implies that in a neighborhood of the point $p$ the object $F$ has only one nonzero cohomology sheaf, $F \iso \mathcal{F}[a]$ for some coherent sheaf $\mathcal{F} \in \mathrm{Coh}(U)$. By shrinking $U$ we may assume that $U \iso \Spec(A)$ for some ring $A$, the coherent sheaf $\mathcal{F}$ corresponds to a module $M$ over the ring $A$, and the smooth curve $D \subset U$ is defined by an equation $\{ d = 0 \}$ for an element $d \in A$. The assumption that $M/dM$ is isomorphic to a skyscraper sheaf at the point $p$ implies that $M/\mathfrak{m}_pM$ is also a skyscraper sheaf, so by Nakayama's lemma $M$ is locally isomorphic to a cyclic module, i.e., there exists an isomorphism $M \iso A/I$ for some ideal $I \subset A$.

  Since $(A/I)/d = (A/d)/I$ has length one, and $A/(d)$ is a discrete valuation ring, this means that the image of $I$ in the quotient ring $A/(d)$ is generated by one regular element $\tilde{f} \in A/(d)$ such that $\tilde{f}$ generates the maximal ideal of $A/(d)$. Pick a preimage $f \in A$ of $\tilde{f}$ in the ideal $I$. We will show that $f$ generates $I$. Consider the short exact sequence
  \[
    0 \to I/fA \to A/fA \to A/I \to 0.
  \]
  The derived pullback to the smooth curve $j\colon D \to U$ produces a long exact sequence of modules over the quotient ring $A/d$. Consider the following fragment:
  \[
    L_1j^*(A/I) \to L_0j^*(I/fA) \to L_0j^*(A/fA) \to L_0j^*(A/I) \to 0
  \]
  Since $L_1j^*(A/I) \iso L_1j^*M = 0$ by the assumption of the theorem, this is in fact a short exact sequence. Note that $L_0j^*(A/fA)$ by the definition is isomorphic to the quotient $A/(d, f)$. Since $f$ is equal to $\tilde{f}$ modulo $(d)$, this quotient is isomorphic to $(A/d)/\tilde{f}$, which by the choice of $\tilde{f}$ is isomorphic to $(A/d)/I$. Thus the last two terms of the short exact sequence are both torsion sheaves of length one. Therefore $L_0j^*(I/fA)$ can only be zero. In particular, $I/fA$ is not supported at the point $p \in U$.

  Thus the inclusion $(f) \subset I$ is an isomorphism at the point $p$, so after shrinking $U$ we can assume that $I = (f)$, so the module $M \iso A/fA$ is the structure sheaf of the curve $\{ f = 0 \}$. Note additionally that since $\tilde{f} \in A/d$ has valuation 1, the curve $C = \{ f = 0 \}$ is smooth at the point $p$. This finishes the proof of the lemma.
\end{proof}

This local description may be improved to a global one if we consider the slices at all points of the curve instead of a single point.

\begin{lemma}
  \label{lem: global equicutting lemma}
  Let $S$ be a smooth surface, and let $i\colon C \monoarrow S$ be a connected curve. Let $F \in \Dperf(S)$ be an object whose set-theoretic support is $C$. Suppose that at each point $p \in C$ there exists a slice of $F$ which is a torsion object of length one. Then the curve $C$ is smooth, and the object $F$ is isomorphic to a pushforward $i_*(L)[a]$ for some line bundle $L \in \Pic(C)$ and a shift $a \in \Z$.
\end{lemma}

\begin{remark}
  If $C$ is not connected, the pushforwards of line bundles from different connected components may have different shifts, but otherwise the conclusion is the same.
\end{remark}

\begin{proof}
  By Lemma~\ref{lem: equicutting lemma}, applied at all points of $C = \supp(B)$, the object $B$ is a shift of some coherent sheaf $\mathcal{F} \in \mathrm{Coh}(Y)$. Moreover, the scheme-theoretic support of $\mathcal{F}$ is equal to the reduced scheme structure on the curve $C$. Therefore $\mathcal{F}$ is a pushforward of a coherent sheaf $\mathcal{F}^\prime \in \mathrm{Coh}(C)$. Locally the sheaf $\mathcal{F}^\prime$ is isomorphic to the structure sheaf of $C$, thus $\mathcal{F}^\prime$ is in fact a line bundle on $C$.
\end{proof}

The minimal possible length of any nonzero slice is equal to one, and lemmas above show what happens in this minimal case. In particular, we see that the support curve is necessarily smooth at the point of the slice. When the curve is singular, we will see below that the length of the slice is always greater or equal to the multiplicity of the singular point. Moreover, the minimal possible length only happens in some special situations. The rest of this subsection is dedicated to the study of that minimal case.

\begin{lemma}
  \label{lem: thin slices are minimal}
  Let $U$ be a smooth surface, and let $F \in \Dbcoh(U)$ be an object whose set-theoretic support is a reduced curve $C \subset U$. Let $j\colon D \monoarrow U$ be a smooth curve which intersects $C$ at a single point $p \in C$. Denote by $d$ the multiplicity of the curve $C$ at $p$.
  \begin{enumerate}
    \item For the length of the torsion object $j^*F \in \Dbcoh(D)$ we have an inequality $\mylength(j^*F) \geq d$.
    \item If the tangent vector to the curve  $D$ at $p$ lies in the tangent cone of the curve $C$ at $p$, then $\mylength(j^*F) > d$.
    \item If $F \iso \mathcal{F}[0]$ is a single coherent sheaf, then the bounds above hold for $\mylength(L_0 j^*\mathcal{F})$.
  \end{enumerate}
\end{lemma}

\begin{proof}
  Since $D \subset U$ is a Cartier divisor, by Lemma~\ref{lem: derived restriction to a divisor} we know that
  \[
    \mylength(j^*F) = \sum_{n \in \Z} \mylength(j^*\mH^n(F)).
  \]
  Thus we may replace the object $F$ with the coherent sheaf $\bigoplus_{n \in \Z} \mH^n(F)$ without changing the lengths of the slices. Thus it is enough to prove the bounds for the length of the nonderived pullback $L_0 j^* \mathcal{F}$ of a coherent sheaf $\mathcal{F}$ on $U$.

  In the proof we use the notion of a (zeroth) Fitting ideal of a coherent sheaf. Recall the definition: given a finitely generated module $M$ over a Noetherian ring $A$, pick an arbitrary free presentation:
  \[
    A^{k} \xrightarrow{Q} A^n \to M \to 0.
  \]
  The Fitting ideal $\Fit(M)$ is defined to be the ideal of $A$ generated by the $(n \times n)$-minors of the matrix $Q$. This construction globalizes to coherent sheaves. The Fitting ideal is contained in the annihilator ideal, and the formation of Fitting ideals is compatible with arbitrary base change (see, e.g., \cite[Tag~07Z6]{stacks-project}).

  Consider the coherent sheaf $L_0 j^*\mathcal{F}$ on a curve $D$. It is supported at a single point $p \in D$. Let $\mathfrak{m} \subset \O_D$ be the maximal ideal sheaf of the point $p$. Using Lemma~\ref{lem: smooth torsion sheaves on curves} it is easy to compute that for any coherent sheaf on a smooth curve $D$ supported at the point $p$ the Fitting ideal is equal to $\mathfrak{m}^{\mylength}$, where $\mylength$ is the length of the torsion sheaf. Thus in order to bound the length of $L_0 j^*\mathcal{F}$ it is enough to understand the Fitting ideal of this sheaf.

  By passing to an \'etale neighborhood of the point $p \in U$ we may assume that $U$ is the affine plane $\A^2 = \Spec \k[x, y]$ and $p$ is the origin. Let $f \in \k[x, y]$ be the reduced equation of the curve $C$. Since the set-theoretic support of $\mathcal{F}$ is $C$, we know that the annihilator ideal of the sheaf $\mathcal{F}$ is contained in the ideal $(f)$. The Fitting ideal $\Fit(\mathcal{F})$ is contained in the annihilator ideal, so $\Fit(\mathcal{F}) \subset (f)$.

  Since Fitting ideals are compatible with base change, we know that $\Fit(L_0 j^* \mathcal{F})$ is contained in the restriction of the ideal $(f)$ to the curve $D$. The pullback of $(f)$ is contained in $\mathfrak{m}^d$, where $d$ is the multiplicity of $C$ at $p$, which is the lowest degree of a monomial occuring in $f$ with nonzero coefficient. Thus $\mylength(L_0 j^*\mathcal{F}) \geq d$. Moreover, if the tangent vector to $D$ lies in the tangent cone of $C$ at the point $p$, by definition this means that the degree-$d$ part of the polynomial $f$ restricts to zero in the quotient $\mathfrak{m}^{d}/\mathfrak{m}^{d+1}$. Thus in this case the pullback of $(f)$ to the curve $D$ is contained in $\mathfrak{m}^{d+1}$, and then $\mylength(L_0 j^*\mathcal{F}) > d$, as claimed.
\end{proof}

Motivated by the bound above, we introduce the following definition.

\begin{definition}
  \label{def: thin objects}
  Let $U$ be a smooth surface, and let $F \in \Dbcoh(U)$ be an object whose set-theoretic support is a reduced curve $C \subset U$. We say that the object $F$ is \emph{thin at the point $p \in C$} if there exists a slice of $F$ at the point $p$ which is a torsion object of the length equal to the multiplicity of the curve $C$ at $p$.
\end{definition}

\begin{lemma}
  \label{lem: thin coherent sheaves}
  Let $U$ be a smooth surface, let $C \subset S$ be a reduced curve, and let $p \in C$ be a point. Let $\mathcal{F}$ be a coherent sheaf on $U$ whose set-theoretic support is $C \subset S$. Suppose that $\mathcal{F}$ is thin at the point $p \in C$. Then, after possibly replacing $U$ by a Zariski neighborhood of $p \in U$, the sheaf $\mathcal{F}$ is a pushforward of a torsion-free rank one sheaf $\mathcal{F}^\prime$ on $C$.
\end{lemma}

\begin{proof}
  The foundational case is when the multiplicity of the curve $C$ at the point $p$ is equal to one. Then by definition $\mathcal{F}$ is thin at $p$ if and only if there exists a slice of $\mathcal{F}$ of length one. This case is proved in Lemma~\ref{lem: equicutting lemma}. Otherwise, let $d > 1$ be the multiplicity of the curve $C$ at the point $p$.

  By shrinking $U$ we may assume that $U$ is affine. Let $f \in \mathrm{H}^0(\O_U)$ be the equation for the reduced scheme structure on the curve $C$. Since the characteristic of the base field is zero, by further shrinking $U$ we may assume that all points in $C \setminus \{ p \}$ are smooth in the curve $C$. Let $j\colon D \monoarrow S$ be a smooth curve passing through $p$ such that the derived pullback $j^*\mathcal{F}$ is a torsion object of the length $d$.

  We first show that $\mathcal{F}$ has no point-torsion at $p$. We know that the length
  \[
    \mylength(j^*\mathcal{F}) := \mylength(L_0 j^*\mathcal{F}) + \mylength(L_1 j^* \mathcal{F})
  \]
  is equal to $d$. By Lemma~\ref{lem: thin slices are minimal} the summand $\mylength(L_0 j^*\mathcal{F})$ is greater or equal to $d$. Thus $L_1 j^* \mathcal{F}$ has length zero, so it is a zero object. Consider the subsheaf $\mathcal{T} \subset \mathcal{F}$ spanned by sections supported only at the point $p$. Consider the short exact sequence
  \[
    0 \to \mathcal{T} \to \mathcal{F} \to \mathcal{F}/\mathcal{T} \to 0.
  \]
  Since $j$ is an inclusion of a Cartier divisor, $L_2 j^*(-)$ vanishes at every argument. Thus the long exact sequence of derived pullbacks along $j\colon D \monoarrow S$ shows that $L_1 j^* \mathcal{F}$ has a subsheaf isomorphic to $L_1 j^* \mathcal{T}$. If $\mathcal{T}$ is a nonzero sheaf, then by Lemma~\ref{lem: lines detect point-torsion} the sheaf $L_1 j^*\mathcal{T}$ is also nonzero, but this leads to a contradiction with the fact that $\mathcal{F}$ is thin at $p$. Thus $\mathcal{T} = 0$, i.e., the sheaf $\mathcal{F}$ has no point-torsion.

  Let $g \in \mathrm{H}^0(\O_U)$ be the equation of the smooth curve $D \subset U$. Consider a family of inclusions $j_t\colon D_t \monoarrow U$, where the curve $D_t$ is given by the equation $\{ g = t \}$. By Lemma~\ref{lem: thin slices are minimal} the tangent vector of $D$ at the point $p$ does not lie in the tangent cone of $C$, and thus for a general value of $t$ the curve $D_t$ intersects $C$ transversely in exactly $d$ distinct points (see, e.g., \cite[\S 5A]{mumford-algeom}; we use the assumption of characteristic zero here). Thus, after possibly shrinking $U$, by semicontinuity we may assume that at each point of $C \setminus \{ p \}$ the sheaf $\mathcal{F}$ has a slice which is a torsion object of length one (see Figure~\ref{fig: semicontinuity and slices}).

  \begin{figure*}[ht]
    \centering
    \def\svgwidth{0.3\columnwidth}
    \import{./}{slices.pdf_tex}
    \caption{Semicontinuity and slices}
    \label{fig: semicontinuity and slices}
  \end{figure*}

  In particular, by Lemma~\ref{lem: global equicutting lemma} this implies that on $U \setminus \{ p \}$ the sheaf $\mathcal{F}$ is a direct sum of pushforwards of line bundles from the irreducible components of $C \setminus \{ p \}$. Assume that on the unpunctured surface $U$ the sheaf $\mathcal{F}$ is not a pushforward from the curve $C$. Since $U$ is affine, this is equivalent to the fact that the equation $f \in \mathrm{H}^0(\O_U)$ does not annihilate $\mathcal{F}$. Then there exists a section $s \in \mathrm{H}^0(\mathcal{F})$ such that $f \cdot s$ is not zero. The equation $f$ annihilates any section on the open set $U \setminus \{ p \}$. Therefore $f \cdot s$ is a section of $\mathcal{F}$ which is supported only at a single point $p$. But we proved that $\mathcal{F}$ has no zero-dimensional torsion, a contradiction. Thus the scheme-theoretic support of the sheaf $\mathcal{F}$ is equal to $C$.
\end{proof}

\begin{lemma}
  \label{lem: thin objects}
  Let $S$ be a smooth surface, and let $F \in \Dperf(S)$ be an object whose set-theoretic support is a reduced curve $C \subset S$. Suppose that $F$ is thin at every point of $C$. Then $F$ is a formal complex, and each cohomology sheaf $\mH^n(F)$ is isomorphic to a pushforward of a torsion-free rank one sheaf from some subcurve $C_n \subset C$.
\end{lemma}

\begin{proof}
  Let $p \in C$ be a point, and let $j\colon D \monoarrow S$ be a smooth curve passing through $p$ such that $j^*F$ is a torsion object of length equal to the multiplicity of $C$ at $p$. By Lemma~\ref{lem: derived restriction to a divisor} we know that $\mylength(j^*F) = \sum_n \mylength(j^*\mH^n(F))$. Let $C_n := \supp(\mH^n(F)) \subset C$ be the set-theoretic support of the $n$'th cohomology sheaf. Denote by $I \subset \Z$ the subset of those indices $n \in \Z$ such that $C_n$ contains $p$ and $p$ is not an isolated point in $C_n$. For $n \in I$, let $m_n$ be the multiplicity of $C_n$ at the point $p$. By Lemma~\ref{lem: thin slices are minimal} we have $\mylength(j^*\mH^n(F)) \geq m_n$ for each $n \in I$. Let $m$ be the multiplicity of the curve $C$ at $p$. Since $C = \cup_{n \in I} C_n$ near the point $p$ set-theoretically, we have $\sum_{n \in I} m_n \geq m$. Taking all this information into account, we get a chain of inequalities:
  \[
    \mylength(j^*F) = \sum_{n \in \Z} \mylength(j^* \mH^n(F)) \geq
    \sum_{n \in I} \mylength(j^* \mH^n(F)) \geq \sum_{n \in I} m_n \geq m.
  \]
  The assumption that $F$ is thin at $p$ implies that each inequality is in fact an equality. Note that this holds for any point $p \in C$. Thus we conclude that:
  \begin{enumerate}
    \item For any $n \in \Z$ such that $\mH^n(F) \neq 0$, the subset $C_n = \supp(\mH^n(F))$ is a curve, and the sheaf $\mH^n(F)$ is thin at any point of its support $C_n$.
    \item For any two distinct $n, n^\prime \in \Z$ the intersection $C_n \cap C_{n^\prime}$ is a zero-dimensional set.
  \end{enumerate}

  Consider a nonzero cohomology sheaf $\mH^n(F)$. It is thin at every point of the curve $C_n$, so by Lemma~\ref{lem: thin coherent sheaves} the sheaf $\mH^n(F)$ is isomorphic to a pushforward of a torsion-free rank one sheaf from $C_n$. It only remains to show that $F$ is a formal complex.

  By Lemma~\ref{lem: complexes on a surface are given by adjacent glueings} the glueing data for $F$ consists of classes in $\Ext^2(\mH^n(F), \mH^{n-1}(F))$ for each~$n \in \Z$. Since the supports $C_n \cap C_{n - 1}$ intersect along a zero-dimensional set, the $\Ext$-group may be computed locally at each intersection point, i.e.
  \[
    \Ext^2(\mH^n(F), \mH^{n-1}(F)) = \mathrm{H}^0(\mathcal{E}xt^2(\mH^n(F), \mH^{n-1}(F)).
  \]
  Let $p \in S$ be any point. Since $\mH^n(F)$ is a pushforward of a torsion-free sheaf from a curve via an inclusion $C_n \monoarrow S$, it has no point-torsion at $p$. By definition this means that the depth of the coherent sheaf $\mH^n(F)$ at $p \in S$ is not zero. Since $S$ is a smooth surface, Auslander--Buchsbaum formula implies that the projective dimension of $\mH^n(F)$ over the local ring $\O_{S, p}$ is at most one. Since this true for every point $p \in S$, the local $\Ext$-sheaf~$\mathcal{E}xt^2(\mH^n(F), -)$ vanishes for any second argument. Therefore $\Ext^2(\mH^n(F), \mH^{n-1}(F)) = 0$. This shows that complex $F$ splits into a direct sum of its cohomology sheaves, and the lemma is proved.
\end{proof}

\subsection{Rational elliptic surfaces}
\label{ssec: rational elliptic surfaces}

We are interested in the following class of surfaces.

\begin{definition}
  \label{def: rational elliptic surface}
  A smooth proper surface $S$ is a \emph{rational elliptic surface} if the anticanonical linear system $| {-K_S} |$ defines a regular morphism $\pi\colon S \to \P^1$.
\end{definition}

There are other, sometimes incompatible, definitions of rational elliptic surfaces in the literature. We follow the terminology of \cite{heckman-looijenga}. Note that in this convention the elliptic fibration $\pi\colon S \to \P^1$ is relatively minimal, i.e., there are no smooth $(-1)$-curves in the fibers. See the reference for more information about this class of surfaces.

Since the anticanonical divisors of a rational elliptic surface $S$ are nothing but the fibers of the morphism $\pi\colon S \to \P^1$, the methods of Section~\ref{sec: numerical lemmas} may be productively used to study admissible subcategories in rational elliptic surfaces. In this subsection we study phantom subcategories using Proposition~\ref{prop: numerical lemma without numbers}, the notion of a point-support introduced in Subsection~\ref{ssec: point-supports}, and the observations about objects set-theoretically supported on curves from Subsection~\ref{ssec: cutting lemmas}. The main result is Theorem~\ref{thm: rational elliptic surfaces}.

We start by a couple of general observations about phantom subcategories and objects with zero-dimensional support.

\begin{lemma}
  \label{lem: zero-dimensional is not a phantom}
  Let $X$ be a smooth projective variety, and let $\mB \subset \Dbcoh(X)$ be an admissible subcategory. If $\mB$ contains an object with zero-dimensional support, then $\mB$ is not a phantom subcategory.
\end{lemma}

\begin{proof}
  Let $G \in \mB$ be an object with zero-dimensional support. Let $i \in \Z$ be the smallest integer such that $\mH^i(G) \neq 0$. Then $\mH^i(G)$ is a coherent sheaf with zero-dimensional support, and in particular there exists an embedding $\O_q \monoarrow \mH^i(G)$ of a skyscraper sheaf $\O_q$ at some point $q \in X$ into that coherent sheaf. Then the composition $\O_q \monoarrow \mH^i(G) \to G$ induces a nonzero map on cohomology sheaves. Thus by Lemma~\ref{lem: maps from torsion cannot be nonzero on cohomology} we see that any object of $\mB^\perp$ is supported on the complement $X \setminus \{ q \}$. In particular, the rank of any object of $\mB^\perp$ at the generic point of $X$ is zero. This shows that $K_0(\mB^\perp)$ cannot be equal to $K_0(X)$. Since there exists a direct sum decomposition $K_0(X) = K_0(\mB^\perp) \oplus K_0(\mB)$, we see that $K_0(\mB) \neq 0$.
\end{proof}

\begin{lemma}
  \label{lem: splitting from single sheaf}
  Let $X$ be a smooth projective variety, and let $\mB \subset \Dbcoh(X)$ be an admissible subcategory. Let $B \in \mB$ be an object. Suppose that $\mH^n(B)$ for some $n \in \Z$ is the only cohomology sheaf of $B$ which is supported at a point $p \in X$. If $p \in \supp(\mH^n(B))$ is an isolated point, then $\mB$ is not a phantom subcategory.
\end{lemma}

\begin{proof}
  Assume that $p \in \supp(\mH^n(B))$ is an isolated point. Since all other cohomology sheaves of $B$ are supported away from $p$, this point is still isolated in the union
  \[
    \supp(B) := \cup_{i \in \Z} \supp(\mH^i(B)).
  \]
  Then by Lemma~\ref{lem: disjoint support forces splitting} the object $B \in \mB$ has a direct summand $T$ with $\supp(T) = \{ p \}$. Since admissible subcategories are closed under direct summands, we have $T \in \mB$. Then by Lemma~\ref{lem: zero-dimensional is not a phantom} the category $\mB$ is not a phantom.
\end{proof}

We continue with more specific properties of phantom subcategories in rational elliptic surfaces. As in Section~\ref{sec: projective plane}, we use the projections of skyscraper sheaves to study admissible subcategories. We fix the notation used in the rest of this subsection below.

\begin{setting}
  \label{set: rational elliptic surfaces}
  Let $\pi\colon S \to \P^1$ be a rational elliptic surface. Let $\mB \subset \Dbcoh(S)$ be an admissible subcategory. Pick a point $p \in S$ lying on the fiber $F \subset S$ of the projection map $\pi\colon S \to \P^1$ such that
  \begin{enumerate}
  \item $F$ is irreducible;
  \item $p \in F$ is a smooth point of that curve.
  \end{enumerate}
  Let $B := \mB_R(\O_p)$ be the (right) projection of the skyscraper sheaf at the point $p$ to the subcategory $\mB$. Denote by $S_p := \supp(B) \subset S$ the set-theoretic support of $B$.
\end{setting}

\begin{lemma}
  \label{lem: fiber and point-support}
  Let $p, \mB, B, F$ be as in Setting~\textup{\ref{set: rational elliptic surfaces}}. If $\mB$ is a phantom subcategory, then the restriction $B|_F$ is isomorphic to either
  \begin{itemize}
  \item a zero object, in which case $B = 0$ and $\O_p \in \mB^\perp$; or
  \item to a direct sum $\O_p[0] \oplus \O_q[a]$ for some smooth point $q \in F$, possibly coinciding with the point $p$, and some odd shift $a \in \Z$.
  \end{itemize}
\end{lemma}

\begin{proof}
  Consider the restriction $B|_F \in \Dperf(F)$. The fiber $F$ is an anticanonical divisor of the surface $S$. Note that $F$ is a reduced irreducible projective curve of arithmetic genus one. Since $p \in F$ is a smooth point, the skyscraper sheaf $\O_p$ is a perfect object on $F$. Then we can apply Proposition~\ref{prop: numerical lemma without numbers} to see that the object $B|_F$ is either
  \begin{enumerate}
  \item a cone between two skyscrapers on smooth points; or
  \item a cone of a map from the skyscraper $\O_p$ to a simple vector bundle on $F$.
  \end{enumerate}

  In the case (2) the object $B|_F$ has a nonzero rank at the generic point of $F$. Then the class of $B|_F$ in $K_0(F)$ is nonzero, and hence the class of $B \in \mB$ is $K_0(S)$ is also nonzero. This shows that $\mB$ is not a phantom subcategory, a contradiction.

  Suppose we are in the case (1). Then by Corollary~\ref{cor: autoequivalence triangle} the object $B|_F$ fits into a distinguished triangle
  \[
    B|_F \to \O_p[0] \to \O_q[n]
  \]
  where $q \in F$ is some smooth point and $n \in \Z$ is some shift. If $n$ is an odd integer, then the class of $B|_F$ in $K_0(F)$ is equal to the class of $\O_p \oplus \O_q$, a nonzero element, and this gives a contradiction with the assumption that $\mB$ is a phantom subcategory. Thus we see that $n$ is necessarily an even integer.

  If the morphism $\O_p \to \O_q[n]$ is zero, then $B|_F \iso \O_p \oplus \O_q[n-1]$, as claimed. For even~$n$, the morphism of skyscrapers can be nonzero only when $q = p$, $n = 0$, and the map is an isomorphism $\O_p[0] \isoarrow \O_p[0]$. In this case the cone $B|_F$ is a zero object. Then the intersection of the support $\supp(B)$ with $F$ is empty. In particular, $\supp(B)$ does not contain the point~$p$. Since $B$ is the projection of the skyscraper sheaf, by Lemma~\ref{lem: properties of point-supports} this implies that $B = 0$ and hence the skyscraper sheaf $\O_p$ lies in the orthogonal subcategory $\mB^\perp$, as claimed.
\end{proof}

\begin{lemma}
  \label{lem: phantoms like minus one curves}
  Let $p, \mB, B, F$ be as in Setting~\textup{\ref{set: rational elliptic surfaces}}. If $\mB$ is a phantom subcategory and $B \neq 0$, then the point $p$ lies on some section of the map $\pi\colon S \to \P^1$.
\end{lemma}
\begin{proof}
  For any torsion object $G \in \Dperf(F)$ denote by $\mylength_p(G)$ the sum of lengths of cohomology sheaves of $G$ at the point $p \in F$. By Lemma~\ref{lem: derived restriction to a divisor} we have an equality
  \begin{equation}
    \label{eqn: additivity of lengths}
    \mylength_p(B|_F) = \sum_{i \in \Z} \mylength_p (\mH^i(B)|_F).
  \end{equation}

  From Lemma~\ref{lem: fiber and point-support} we know that if $B \neq 0$, then $B|_F$ is isomorphic to a direct sum of two skyscrapers $\O_p[0] \oplus \O_q[a]$ where $a$ is an odd integer.

  Consider first the case where the point $q$ is distinct from the point $p$. Then $\mylength_p(B|_F) = 1$. From the formula (\ref{eqn: additivity of lengths}) we see that there exists exactly one cohomology sheaf $\mH^n(B)$ which is nonzero at the point $p$, and moreover the length $\mylength_p(\mH^n(B)|_F)$ is equal to one. By Lemma~\ref{lem: splitting from single sheaf} we see that $p \in \supp(\mH^n(B))$ is not an isolated point, so there exists an irreducible curve $C \subset \supp(\mH^n(B))$ passing through $p$.

  Suppose that $C$ also intersects the fiber $F$ at some point $r \in F$ other than $p$. Then by Lemma~\ref{lem: derived restriction to a divisor} this implies that $\mH^n(B|_F)$ is not zero at both points $p$ and $r$. But this contradicts the fact that $B|_F \iso \O_p[0] \oplus \O_q[a]$ with odd $a$, and each cohomology sheaf is supported only at a single point. Therefore $C$ intersects $F$ only at the point $p$.

  Moreover, since $\mylength_p(\mH^n(B)|_F) = 1$, by Lemma~\ref{lem: thin slices are minimal} we see that $C$ intersects $F$ at $p$ transversely, i.e., with multiplicity one. Since $F$ is a fiber of the fibration $\pi\colon S \to \P^1$, this implies that the composition $C \monoarrow S \xrightarrow{\pi} \P^1$ is a morphism of curves of degree one into a smooth curve. Any such map is necessarily an isomorphism. Thus $C \iso \P^1 \subset S$ is a section of the map $\pi$, and it contains the point $p$ by construction, so the lemma is proved in the case where $q \neq p$.

  Consider now the case where $B|_F \iso \O_p[0] \oplus \O_p[a]$ is supported only at the point $p$. Here the length $\mylength_p(B|_F)$ is two. As above, from the formula (\ref{eqn: additivity of lengths}) we see that there are at most two cohomology sheaves of $B$ which are nonzero at $p$. If there are two distinct ones, $\mH^i(B)$ and $\mH^j(B)$ with $i \neq j$, then we have
  \[
    \mylength_p(\mH^i(B)|_F) = \mylength_p(\mH^j(B)|_F) = 1.
  \]
  In this case we can repeat the argument above using either of those two cohomology sheaves.

  Thus it only remains to deal with the case where only one cohomology sheaf, $\mH^n(B)$, is nonzero at $p$ and has $\mylength_p(\mH^n(B)) = 2$. Again, using Lemma~\ref{lem: splitting from single sheaf} we can pick an irreducible curve $C \subset \supp(\mH^n(B))$ containing the point $p$. It intersects $F$ only at the point $p$ since $\supp(B|_F) = \{ p \}$. If $C$ intersects $F$ at $p$ with multiplicity one, then it is a section of the fibration $\pi\colon S \to \P^1$, and we are done by the same argument as above. If the intersection multiplicity is $m > 1$, then either $C$ is singular at $p$, or it intersects $F$ non-transversely. In both those cases from the last part of Lemma~\ref{lem: thin slices are minimal} we see that the length of the nonderived restriction of $\mH^n(B)$ along the inclusion $j\colon F \monoarrow S$ is at least~$2$. From the short exact sequences in Lemma~\ref{lem: derived restriction to a divisor} we see that
  \[
    \mylength_p(\mH^n(B|_F)) \geq \mylength_p(L_0 j^* \mH^n(B)) \geq 2.
  \]
  But this is a contradiction with the fact that $B|_F \iso \O_p[0] \oplus \O_p[a]$ with odd $a \in \Z$, and each individual cohomology sheaf has length at most one. Therefore, the intersection of $C$ with $F$ is always transverse, and the lemma is proved.
\end{proof}

\statemaintheoremRES

\begin{remark}
  Note that any section is a $(-1)$-curve by the adjunction formula. Moreover, each smooth $(-1)$-curve is a section, and by Kodaira's classification of singular fibers each reducible fiber is a union of several $(-2)$-curves. In other words, the conclusion is that any phantom subcategory is supported on the union of $(-1)$-curves and $(-2)$-curves.
\end{remark}

\begin{proof}
  Let $p \in S$ be a smooth point of an irreducible fiber $F \subset S$, and suppose that $p$ does not lie on any section of $\pi\colon S \to \P^1$. Then by Lemmas~\ref{lem: fiber and point-support} and~\ref{lem: phantoms like minus one curves} we see that the projection of the skyscraper sheaf $\O_p$ to $\mB$ is zero, so $\O_p$ lies in the subcategory $\mB^\perp$. By semiorthogonality this implies that any object of $\mB$ is supported on the complement $S \setminus \{ p \}$. Since this holds for any smooth point of an irreducible fiber, we see that the support of $\mB$ is contained in the union of
  \begin{itemize}
  \item sections of the fibration $\pi\colon S \to \P^1$;
  \item reducible fibers of the fibration $\pi\colon S \to \P^1$; and
  \item singular points of irreducible fibers of $\pi$.
  \end{itemize}

  Note that there are finitely many irreducible singular fibers of the morphism $\pi$, and each has only a single singular point. So singular points of irreducible fibers form a discrete subset of the surface $S$. By Lemma~\ref{lem: support of a category} the support of $\mB$ does not contain any isolated points. Thus the subset $\supp(\mB)$ is in fact a union of curves, as claimed in the statement.
\end{proof}

\subsection{Admissible subcategories supported on sections}
\label{ssec: sections of anticanonical system}

Theorem~\ref{thm: rational elliptic surfaces} shows that any phantom subcategory in a rational elliptic surface is supported on the union of sections and reducible fibers. Recall that each section is a $(-1)$-curve in a rational elliptic surface, and in Section~\ref{sec: minus one curves} we have classified admissible subcategories supported on a single $(-1)$-curve. In this subsection we generalize the classification of admissible subcategories to arbitrary configurations of sections.

The main result of this subsection is Theorem~\ref{thm: admissible subcategories on sections}. We show that the only option is to choose several disjoint sections, pick a line bundle on each one, and span the subcategory by their pushforwards. As for the classification of admissible subcategories on a projective plane in Section~\ref{sec: projective plane}, the key tool is Proposition~\ref{prop: numerical lemma without numbers}. The most difficult part is to deal with configurations of two intersecting sections. We use various lemmas from Subsection~\ref{ssec: cutting lemmas} to handle this situation.

\begin{setting}
  \label{set: sections}
  Let $\pi\colon S \to \P^1$ be a rational elliptic surface. Let $\mB \subset \Dbcoh(S)$ be an admissible subcategory such that $\supp(\mB)$ is contained in the union of sections of the map $\pi$. Denote the support by $C := \supp(\mB)$. Note that by Lemma~\ref{lem: support of a category} the subset $C$ is a curve, i.e., a union of finitely many sections of the fibration $\pi$.

  Define a subset $C^\circ \subset C$ by saying that a point $c \in S$ lies in the complement $C \setminus C^\circ$ if the fiber $F_c$ of the fibration $\pi\colon S \to \P^1$ containing $c$ is singular or if it contains an intersection point of two distinct sections in $C$. Note that all points in $C^\circ$ are smooth in the curve $C$.

  \begin{figure*}[ht]
    \centering
    \def\svgwidth{0.3\columnwidth}
    \import{./}{openpart.pdf_tex}
    \caption{The subset $C^\circ \subset C$ is the complement to the punctured points.}
  \end{figure*}

  We also fix a point $p \in C^\circ$ and denote by $S_p \subset S$ the point-support of the admissible subcategory $\mB \subset \Dbcoh(S)$ at the point $p$. Denote by $B \in \mB$ the (right) projection $\mB_R(\O_p)$ of the skyscraper sheaf at the point $p$ to the subcategory $\mB$.
\end{setting}

\begin{lemma}
  \label{lem: projection from support is nonzero}
  Let $S, \mB, p, B$ be as in Setting~\textup{\ref{set: sections}}. Then $B \neq 0$.
\end{lemma}
\begin{proof}
  Assume that $B$, the projection of the skyscraper sheaf $\O_p$ to $\mB$, is a zero object. Then the skyscraper sheaf $\O_p$ lies in the orthogonal subcategory $\mB^\perp$. By Lemma~\ref{lem: maps from torsion cannot be nonzero on cohomology} this implies that $\supp(\mB)$ does not contain the point $p$, but this is a contradiction, since by assumption the point $p$ lies in $C^\circ \subset C = \supp(\mB)$.
\end{proof}

\begin{lemma}
  \label{lem: at most two irreducible components}
  Let $S, \mB, C, C^\circ, p, S_p$ be as in Setting~\textup{\ref{set: sections}}. Then $S_p$ is a curve with at most two irreducible components.
\end{lemma}

\begin{proof}
  By Lemma~\ref{lem: projection from support is nonzero} the projection $B$ of the skyscraper sheaf $\O_p$ to $\mB$ is nonzero and we have $S_p = \supp(B)$. Since $\supp(\mB)$ is not the whole surface $S$, the subset $S_p$ cannot be equal to the point $\{ p \}$, as in that case we could deform $B$ by moving the point $p$ along the surface, and admissible subcategories are closed under small deformations by Proposition~\ref{prop: admissible subcategories are open}. Then by Lemma~\ref{lem: properties of point-supports} we know that $\supp(B) = S_p \subset \supp(\mB)$ is a connected union of several sections of the map $\pi\colon S \to \P^1$.

  Let $F \subset S$ be the fiber of the fibration morphism $\pi\colon S \to \P^1$ containing the point $p$. By definition each section of the map $\pi$ intersects $F$ at a single point transversely. Since by assumption $p \in C^\circ$, we know that the intersection points of $F$ with sections in $C$ are all distinct. Therefore the number of irreducible components of $S_p$ equals the cardinality of the set-theoretic intersection $S_p \cap F$, which is the same as $\supp(B|_F)$. Since $p \in C^\circ$ by assumption, the curve $F$ is a reducible irreducible anticanonical divisor of the surface $S$, so by Proposition~\ref{prop: numerical lemma without numbers} we know that $B|_F$ is a cone of a map between two skyscrapers. In particular, the curve~$S_p$ has at most two irreducible components.
\end{proof}

The next several lemmas prove by contradiction that $S_p$ cannot have exactly two irreducible components. The overall plan is to deform the curve $S_p$ into a curve  that does not lie in the closed subset $\supp(\mB) \subset S$, and prove that we may also deform the object $B \in \Dbcoh(S)$, set-theoretically supported on $C_1 \cup C_2$, along with the curve. This would be a contradiction with the fact that admissible subcategories are closed under small deformations. It is impossible to control the deformations of an object in the derived category if the only thing we know about it is its set-theoretic support, so in order to realize our plan we study other properties of the object $B$ in the following lemmas.

\begin{lemma}
  \label{lem: unions of minus one curves}
  Let $S$ be a smooth projective surface. Let $C_1$ and $C_2$ be two distinct smooth $(-1)$-curves on $S$. Let $m = C_1 \cdot C_2$ be the intersection number, counted with multiplicity. Then the dimension $\dim H^0(C_1 \cup C_2, N_{C_1 \cup C_2 / S})$ of the space of global sections of the normal bundle to the union $C_1 \cup C_2$ is at least $m$.
\end{lemma}

\begin{proof}
  If $m = 0$, there is nothing to prove, so we assume that $C_1$ and $C_2$ are not disjoint. By the adjunction formula the normal bundle $N_{C_1 \cup C_2 / S}$ to the divisor $C_1 \cup C_2 \subset S$, restricted to a subcurve $C_i$ for $i = 1, 2$, has degree $C_i \cdot (C_1 + C_2) = m-1$. Since both components of $C_1 \cup C_2$ are isomorphic to $\P^1$, on each component the dimension of the space $H^0(C_i, (N_{C_1 \cup C_2 / S})|_{C_i})$ is equal to $m$. Since the intersection multiplicity $C_1 \cdot C_2$ is $m$, there are $m$ linear conditions on those sections in order for them to glue. Thus there are at least $2m - m = m > 0$ global sections of the normal bundle $N_{C_1 \cup C_2 / S}$.
\end{proof}

Recall the notion of a thin object from Definition~\ref{def: thin objects}.

\begin{lemma}
  \label{lem: two components imply thinness}
  Let $S, \mB, C, C^\circ, p, S_p$, and $B$ be as in Setting~\textup{\ref{set: sections}}. If $S_p = C_1 \cup C_2$ with irreducible components $C_1$ and $C_2$, then $S_p$ is a nodal curve and $B$ is a thin object at each point of its support $S_p$.
\end{lemma}

\begin{proof}
  Consider the fiber $j\colon F \to S$ of the fibration $\pi\colon S \to \P^1$ which contains the point $p$. By the assumption that $p \in C^\circ$ we know that $F$ is smooth and the intersection $F \cap S_p$ is $\{ p, q \}$ for some $q \neq p$. Since $F$ is an anticanonical divisor of the surface $S$, from Proposition~\ref{prop: numerical lemma without numbers} we get that the only option for the object $j^* B$ to be supported on two smooth points of the curve $F$ is the direct sum $\O_p[0] \oplus \O_q[a]$ for some integer $a \in \Z$.

  Since both components $C_1$ and $C_2$ of $S_p$ are sections of the map $\pi\colon S \to \P^1$, this implies that the intersection of each $C_i$ with an arbitrary fiber $F^\prime \subset S$ has multiplicity one, in particular the intersection $S_p \cap F^\prime$ consists of smooth points of the fiber $F^\prime$ by Lemma~\ref{lem: thin slices are minimal}.

  Now we can apply Lemma~\ref{lem: arbitrary anticanonical divisors} to see that the restriction of $B$ to any other fiber $F^\prime$ is a torsion object of length at most two. Thus, if the fiber $F^\prime$ intersects $S_p$ in two distinct points, each intersection point has length exactly one, and if both curves $C_1$ and $C_2$ intersect $F^\prime$ at the same point, then this is a torsion object of length two.

  Since each intersection point of $C_1$ and $C_2$ has multiplicity at least $2$, Lemma~\ref{lem: thin slices are minimal} applied to the intersections $S_p \cap F^\prime$ for all fibers $F^\prime$ implies that
  \begin{enumerate}
  \item each singular point of $C_1 \cup C_2$ has multiplicity exactly $2$, i.e., it is a node;
  \item the object $B$ is thin at each point of its support $C_1 \cup C_2$.
  \end{enumerate}
  This finishes the proof of the lemma.
\end{proof}

\begin{lemma}
  \label{lem: two components exts}
  Let $S, \mB, C, C^\circ, p, S_p$, and $B$ be as in Setting~\textup{\ref{set: sections}}. If $S_p = C_1 \cup C_2$ with irreducible components $C_1$ and $C_2$, then
  \(
    \REnd(B) \iso \k[0] \oplus \k[-1].
  \)
\end{lemma}
\begin{proof}
  By Lemma~\ref{lem: two components imply thinness} we know that the object $B$ is thin everywhere, in particular at the point $p \in S_p$. Since by assumption $p \in C^\circ$, this is a smooth point of $S_p$. Thus by Lemma~\ref{lem: equicutting lemma} in a Zariski neighborhood of the point $p \in S$ the object $B$ is isomorphic to the structure sheaf of the smooth curve passing through $p$ transversely. Using this local description, we compute
  \[
    \RHom(B, \O_p) \iso \k[0] \oplus \k[-1].
  \]

  Since the object $B$ is the (right) projection of the skyscraper sheaf $\O_p$, by the universal property of the projections (Corollary~\ref{cor: universal property of projection triangles}) we see that, as claimed in the statement,
  \begin{equation}
    \label{eqn: two components endomorphisms}
    \RHom(B, B) \iso \RHom(B, \O_p) \iso \k[0] \oplus \k[-1]. \qedhere
  \end{equation}
\end{proof}

\begin{lemma}
  \label{lem: two components implies torsion-free sheaf}
  Let $S, \mB, C, C^\circ, p, S_p$, and $B$ be as in Setting~\textup{\ref{set: sections}}. If $S_p = C_1 \cup C_2$ with irreducible components $C_1$ and $C_2$, then $B$ is isomorphic to a pushforward $i_*\mathcal{F}$ of a simple rank one torsion-free sheaf $\mathcal{F} \in \Coh(S_p)$ along the inclusion $i\colon S_p \monoarrow S$.
\end{lemma}

\begin{proof}
  Lemma~\ref{lem: two components imply thinness} shows that $B$ is thin at each point of its support. We described the general structure of thin objects in Lemma~\ref{lem: thin objects}. This lemma shows that $B$ is a formal complex and each cohomology sheaf is a rank one torsion-free sheaf on (a subcurve of) $S_p$. However, by Lemma~\ref{lem: two components exts} we know that the object $B$ is simple, in particular indecomposable. Thus there exists only one cohomology sheaf and hence $B \iso i_*\mathcal{F}$ for some torsion-free rank one sheaf on $S$. Since $i\colon S_p \monoarrow S$ is a closed embedding, we have
  \[
    \mathrm{R}^0\Hom_{S_p}(\mathcal{F}, \mathcal{F}) \iso \mathrm{R}^0\Hom_S(i_*\mathcal{F}, i_*\mathcal{F}) = \mathrm{R}^0\Hom_S(B, B) \iso \k.
  \]
  Thus the torsion-free rank one sheaf $\mathcal{F}$ on $S_p$ is simple.
\end{proof}

\begin{lemma}
  \label{lem: two components imply line bundle}
  Let $S, \mB, C, C^\circ, p, S_p$, and $B$ be as in Setting~\textup{\ref{set: sections}}. If $S_p = C_1 \cup C_2$ with irreducible components $C_1$ and $C_2$, then the category $\mB$ contains a pushforward $i_*L[0]$ of some line bundle $L \in \Pic(S_p)$ along the inclusion $i\colon S_p \to S$. Moreover, we can assume that
  \[
    \dim \Ext^1(i_*L, i_*L) \leq 1.
  \]
\end{lemma}
\begin{proof}
  By Lemma~\ref{lem: two components implies torsion-free sheaf} we see that $B \iso i_*\mathcal{F}$ for a simple rank one torsion-free sheaf on the curve $S_p$. Since $B \in \mB$ and admissible subcategories are closed under small deformations by Proposition~\ref{prop: admissible subcategories are open}, it is enough to deform the sheaf $\mathcal{F}$ on $S_p$ into a line bundle, and the inequality for the dimension follows from Lemma~\ref{lem: two components exts} and semicontinuity.

  Since $S_p \subset S$ is a connected reduced curve with planar singularities, by \cite[Th.~2.3]{melo-rapagnetta-viviani} the moduli space of simple rank one torsion-free sheaves on $S_p$ has a dense open subset consisting of line bundles. Moreover, by \cite[Fact~2.2]{melo-rapagnetta-viviani} there exists a universal sheaf on the moduli space of simple rank one torsion-free sheaves. Then we can find a family of sheaves on $S_p$ containing $\mathcal{F}$ such that the generic member is a line bundle. Thus the lemma is proved.
\end{proof}

\begin{proposition}
  \label{prop: never two components}
  Let $S, \mB, C, C^\circ, p, S_p$, and $B$ be as in Setting~\textup{\ref{set: sections}}. Then the curve $S_p$ is irreducible, i.e., $S_p$ is a section of the fibration $\pi\colon S \to \P^1$.
\end{proposition}

\begin{proof}
  By Lemma~\ref{lem: at most two irreducible components} we know that $S_p$ has at most two irreducible components. Suppose that $S_p = C_1 \cup C_2$ with two irreducible components $C_1$ and $C_2$. The chain of lemmas above, culminating in Lemma~\ref{lem: two components imply line bundle}, shows that in this case the category $\mB$ contains a pushforward~$i_*L$ of a line bundle $L \in \Pic(S_p)$ along the inclusion $i\colon S_p \to S$. Below we construct a deformation of an object~$i_*L$ into an object whose support does not lie inside the union of the sections of the fibration $\pi\colon S \to \P^1$, and this leads to a contradiction with the assumption on the subset $\supp(\mB) \subset S$ in Setting~\ref{set: sections}.

  Let $N$ be the normal bundle to the curve $S_p = C_1 \cup C_2$ in the surface $S$. Since $i$ is the inclusion of the Cartier divisor $S_p$, by \cite[Cor.~11.4]{HuybFM}\footnote{In the reference the triangle is constructed only for smooth divisors, but the same proof works for perfect objects on an arbitrary divisor.} the object $i^*i_*L$ fits into a distinguished triangle :
  \[
    L \otimes N^\dual[1] \to i^*i_*L \to L.
  \]
  in $\Dperf(S_p)$. An application of the functor $\RHom_{S_p}(-, L)$ produces a long exact sequence of vector spaces. Consider the following fragment:
  \[
    \Ext^{-1}(L \otimes N^\dual, L) \to \Ext^1(L, L) \to \Ext^1(j_*L, j_*L) \to \Hom(L \otimes N^\dual, L) \to \Ext^2(L, L).
  \]
  Since $L$ is a coherent sheaf, there are no negative $\Ext$'s from $L \otimes N^\dual$ to $L$. Since $L$ is a line bundle, we have $\Ext^\bullet(L, L) \caniso H^\bullet(\O_{S_p})$. In particular, $\Ext^2(L, L) = 0$. Thus we obtain a short exact sequence
  \[
    0 \to H^1(\O_{S_p}) \to \Ext^1(i_*L, i_*L) \to H^0(N) \to 0.
  \]

  By Lemma~\ref{lem: unions of minus one curves} the dimension of the space $H^0(N)$ is greater or equal to the intersection multiplicity $C_1 \cdot C_2$. However, by Lemma~\ref{lem: two components imply line bundle} we have the bound $\dim \Ext^1(i_*L, i_*L) \leq 1$. Since the curves $C_1$ and $C_2$ are not disjoint, we see that the only option is $C_1 \cdot C_2 = 1$. This may only happen if there is a single intersection point, and the intersection is transverse.

  Since there is only one intersection point and each irreducible component is a rational curve, we see that $\Pic(S_p) \iso \Z \oplus \Z$, and a line bundle is uniquely determined by the degrees of its restrictions to the components $C_1, C_2 \subset S_p$. By Lemma~\ref{lem: unions of minus one curves} we know that $H^0(N) \neq 0$. Since $S_p$ is a curve in a smooth surface, there are no obstructions to deformations of $S_p$. We can deform the divisor of $L$ together with the curve:

  \begin{figure*}[ht]
    \centering
    \def\svgwidth{0.3\columnwidth}
    \import{./}{divisors.pdf_tex}
    \caption{Deforming the divisor in a family of curves.}
  \end{figure*}

  By Proposition~\ref{prop: admissible subcategories are open} admissible subcategories are closed under small deformations. Since $\mB$ contains $i_* L$, it also contains the pushforward of a line bundle from a small deformation of the curve $S_p$. This deformation is clearly not supported in $\supp(\mB)$, since that subset is by assumption a union of finitely many rigid curves. This is a contradiction, and therefore the situation where $S_p$ has two irreducible components is impossible.
\end{proof}

\begin{remark}
  After establishing that $C_1 \cdot C_2 = 1$ in Proposition~\ref{prop: never two components} above, one may use the classification of rank one torsion-free sheaves on nodal curves \cite[Prop.~10.1]{oda-seshadri} to prove that in fact the object $B$ itself is a pushforward of a line bundle from $S_p = C_1 \cup C_2$, but we do not need this.
\end{remark}

\statemaintheoremSections

\begin{proof}
  Let $E$ be a section of $\pi$ contained in the subset $C := \supp(\mB)$. Let $C^{\circ}$ be the open subset of $C$ defined as in Setting~\ref{set: sections}. Since by definition $C \setminus C^\circ$ is a finite set of points, the intersection $E \cap C^\circ$ is nonempty. By Lemma~\ref{lem: projection from support is nonzero} the projection of any point $p \in E \cap C^\circ$ is nonzero, and from Proposition~\ref{prop: never two components} we know that the support of the projection $B := \mB_R(\O_p)$ of the skyscraper sheaf can only be the curve $E$. Thus the point-support $S_p$ at the point $p$ is equal to $E$.

  Suppose $E, E^\prime$ are two distinct sections of $\pi$ contained in $\supp(\mB)$. By the argument above we can find points $p, p^\prime$ in $E \cap C^\circ$ and $E^\prime \cap C^\circ$, respectively, such that $S_p = E$ and $S_{p^\prime} = E^\prime$. If the curves $E$ and $E^\prime$ are not disjoint, then they intersect along a zero-dimensional subset. But the point-supports of distinct points $p, p^\prime$ cannot have zero-dimensional intersections by Lemma~\ref{lem: skyscrapers and intersections}. Thus $E \cap E^\prime = \emptyset$.

  We conclude that $\supp(\mB)$ is a disjoint union of several sections of the morphism $\pi$. The objects supported at disjoint curves are completely orthogonal to each other, so $\mB$ splits into an orthogonal sum of subcategories, where for each section $E \subset \supp(\mB)$ we have a nonzero subcategory $\mB_E$ supported on $E$. Since each section in a rational elliptic surface is a $(-1)$-curve by the adjunction formula, the possible options for subcategories $\mB_E$ are classified in Proposition~\ref{prop: new local classification on blow-ups}. This finishes the proof of the theorem.
\end{proof}

\subsection{Non-existence of phantoms in del Pezzo surfaces}
\label{ssec: no phantoms in del pezzos}

In the previous two subsections we have studied rational elliptic surfaces. In this subsection we apply the results of the preceding subsections to del Pezzo surfaces. The upshot is that there are no phantom subcategories in del Pezzo surfaces.

\begin{lemma}
  \label{lem: del pezzo and rational elliptic surfaces}
  Let $Y$ be a del Pezzo surface. Then there exists a blow-up $f\colon Y^\prime \to Y$ of several distinct points such that $Y^\prime$ is a rational elliptic surface whose fibration map $\pi\colon Y^\prime \to \P^1$ has no reducible fibers.
\end{lemma}

\begin{proof}
  By blowing up some general points we may assume that $Y$ is a del Pezzo surface of degree one. Then the anticanonical system has a single base point, and its blow-up is a rational elliptic surface $Y^\prime$. The fibers of the fibration map $Y^\prime \to \P^1$ are isomorphic to anticanonical divisors in $Y$. Suppose that some anticanonical divisor of $Y$ is reducible. Then we can write $(-K_Y)$ as a sum $C_1 + C_2$ with two effective curves $C_1, C_2$. Then
  \[
    (-K_Y) \cdot (-K_Y) = (-K_Y \cdot C_1) + (-K_Y \cdot C_2)
  \]
  Since $Y$ is a del Pezzo surface of degree one, the left hand side is equal to one. However, since the line bundle $(-K_Y)$ is ample, the intersection with each effective divisor is positive, so the right hand side is at least two. This is a contradiction. Therefore each anticanonical divisor in a del Pezzo surface of degree one is irreducible, and the map $\pi\colon Y^\prime \to \P^1$ has only irreducible fibers.
\end{proof}

\statemaintheoremDelPezzo

\begin{proof}
  Let $Y$ be a del Pezzo surface, and assume that $\mB \subset \Dbcoh(Y)$ is a phantom subcategory. By Lemma~\ref{lem: del pezzo and rational elliptic surfaces} there exists a blow-up $f\colon Y^\prime \to Y$ of several distinct points which is a rational elliptic surface with no reducible fibers. Consider the category $f^*\mB \subset \Dbcoh(Y^\prime)$ consisting of the pullbacks of objects in $\mB$. By Orlov's theorem on the semiorthogonal decomposition of the blow-up~\cite{orlov93} the category $f^*\mB$ is equivalent to $\mB$, and it is an admissible subcategory of~$\Dbcoh(Y^\prime)$. The pullback $f^*$ induces an isomorphism of Grothendieck groups $K_0(\mB) \to K_0(f^*\mB)$, hence $f^*\mB$ is a phantom subcategory of $\Dbcoh(Y^\prime)$.

  By Theorem~\ref{thm: rational elliptic surfaces} we see that the support of $f^*\mB$ is contained in the union of sections of the anticanonical fibration $\pi\colon Y^\prime \to \P^1$ and reducible fibers of this map. By construction $Y^\prime$ has no reducible fibers, so $f^*\mB$ is supported on the union of sections. Admissible subcategories like that are completely classified in Theorem~\ref{thm: admissible subcategories on sections}, and in particular there are no phantom subcategories of this kind. This is contradiction. Therefore there are no phantoms in the derived category of $Y$.
\end{proof}

It would be interesting to improve this result to the full classification of admissible subcategories, like we have managed to do for the projective plane in Section~\ref{sec: projective plane}. Note that exceptional objects and exceptional collections on del Pezzo surfaces have been studied in the paper \cite{kuleshov-orlov}, but the classification there is not as explicit as the one for the projective plane in \cite{gorodentsev-rudakov}. Another obstacle is that, unlike the case of the projective plane, a semiorthogonal decomposition on a del Pezzo surface is not uniquely determined by the restriction of a projection of a skyscraper sheaf at a single point to a single anticanonical divisor.

config: arxiv


TITLE: How complicated is the theory of $\mathbb Q_p$?
QUESTION [9 upvotes]: Let $p=7$, say.  How complex is the theory of the field $(\mathbb Q_p,0,1,+,\cdot)$ ?  One extreme would be that the theory is like $\mathbb R$: decidable

... admits quantifier elimination, is o-minimal.  Do I need to include $\;<\;$ for $\mathbb R$ for these?  Or is it enough that $<$ is definable in $(\mathbb R, 0,1,+,\cdot)$ ?

The other extreme would be that $\mathbb Z$ is definable in $\mathbb Q_p$, so the theory is as complex as the theory of $\mathbb Z$.

I thought of this when reading a recent question, asking whether $\mathbb C$ and $\mathbb C_p$ admit a Borel-measurable isomorphism.

REPLY [9 votes]: There is a strong analogy between the model theory of $\mathbb{R}$ and the model theory of $\mathbb{Q}_p$. Here's a brief history of some of the main results (I've necessarily left a lot out, so I encourage others to edit and add to the list):

In the '60s, the theory of $\mathbb{Q}_p$ was shown to be decidable, independently by Ershov (On the elementary theory of maximal normed fields, 1965) and Ax and Kochen (Diophantine problems over local fields. III. Decidable fields, 1966).
In the '70s, Macintyre proved that the structure $(\mathbb{Q}_p;0,1,+,-,\times,V,(P_n)_{n\in \mathbb{N}})$ has quantifier elimination, where $V$ is a unary predicate naming the valuation ring, and each $P_n$ is a unary predicate naming the set of $n^{\text{th}}$ powers. Like $\mathbb{R}$ (and like every non-algebraically-closed field, by another result of Macintyre's - see below for more on this), $\mathbb{Q}_p$ does not have quantifier elimination in the language of rings. (On definable sets of $p$-adic fields, 1976).
In the '80s, Denef proved a cell decomposition theorem for $\mathbb{Q}_p$, analogous to Tarski's cell decomposition for real closed fields. (See The rationality  of the Poincaré  series associated  to  the  p-adic  points  on  a  variety, 1984, and p-adic semi-algebraic sets and cell decomposition, 1986).
In the '90s, Haskell and Macpherson defined a $p$-adic analogue of o-minimality, called "P-minimality" (A version of o-minimality for the p-adics, 1997). They did not obtain a cell decomposition theorem for general P-minimal structures, and since then there have been a number of papers about cell decomposition in the P-minimal setting, usually under additional hypothesis, such as definable Skolem functions.
It follows easily from cell decomposition that the theory of $\mathbb{Q}_p$ is NIP, and this was probably folklore for some time. The dp-rank is a measure of complexity for NIP theories. In the 21st century, Dolich, Goodrick, and Lippel showed that the theory of $\mathbb{Q}_p$ is dp-minimal, meaning that it has dp-rank 1 (dp-minimality: basic facts and examples, 2011). The theory of real closed fields is also dp-minimal, as is any o-minimal theory.

Regarding your side question: most model-theoretic notions are impervious to the exact choice of language, as long as the languages considered are interdefinable. This includes notions like decidability, dp-minimality, and NIP. Characterizations of definable sets, like o-minimality, are mostly language-independent, except that e.g. the definition of o-minimality requires the symbol $<$ to be in the language. On the other hand, quantifier elimination is highly sensitive to the choice of language.
So for $\mathbb{R}$, decidability (and dp-minimality, and NIP) holds in the  language of rings, quantifier elimination does not hold in the language of rings, and it doesn't even make sense to ask about o-minimality in the language of rings.

P.S. Above I alluded to a theorem of Macintyre on fields admitting quantifier elimination. The original reference is On $\omega_1$-categorical theories of fields, 1971, where Macintyre proves that the following are equivalent for infinite fields (in the language of rings):

$K$ is algebraically closed
$\text{Th}(K)$ is $\aleph_1$-categorical
$K$ is totally transcendental ($\omega$-stable)
$\text{Th}(K)$ has quantifier elimination

A simpler argument for the equivalence of (1) and (4) was given by Macintyre, McKenna, and van den Dries in Elimination of quantifiers in algebraic structures, 1983. The main results are the following:

If a field admits quantifier elimination in the language of rings, it is algebraically closed.
If an ordered field admits quantifier elimination in the language of ordered rings, it is real closed.
If a non-trivially valued field admits quantifier elimination in the language of of rings together with a binary relation symbol for $v(x)\leq v(y)$,  it is an algebraically closed valued field.
If a $p$-field (see the paper for the definition) admits quantifier elimination in the Macintyre language described above (with $V$ and $(P_n)_{n\in \mathbb{N}}$), it is $p$-adically closed (elementarily equivalent to $\mathbb{Q}_p$).


TITLE: Metal rod in an electric field
QUESTION [1 upvotes]: A metallic rod is placed in a uniform electric field.
Why is an electric field induced in the opposite direction?

REPLY [1 votes]: Let us start from the very basic intution ......a charge placed in a field experiences some force .the direction of the forces matches with the field if the charge is positive. Now we know that a conductor is a cluster of free electrons hence when placed in a field the charged particles experience a force that drift them in a direction opposite to the field hence developing an increased concentration of electrons on that side and a deficiency on the other causing  development of a potential difference between the two sides of the conductor giving rise to an electric field which grows in magnitude until net force experienced by every charge particle is zero


TITLE: Different degrees of freedom with RGB, CMYK and Frequency
QUESTION [4 upvotes]: Ok, so I understand the eye has 3 different types of receptors and I've seen the process of converting from RGB to CMYK. However if in physics I can specify a color using a single number (its frequency) why do computers need 3 (RGB) or 4 (CMYK) numbers to specify the same color? Is there some kind of disagreement in the amount of degrees of freedom involved?

REPLY [1 votes]: The problem is "color" is an ambiguous term, depending on the context it can refer to

An individual wavelength
A spectrum of intensity verses wavelength
The response of the human eye to a spectrum

A laser produces light that is very close to being a single wavelength, plain LEDs are fairly narrow too*. Most other light sources produce a much broader spectrum. Most colors we see in day to day life are not individual wavelengths, they are the result of taking a light source with a very broad spectrum and filtering it down to a spectrum that is not quite so broad (but still much broader than that from a Laser or plain LED) and often has multiple peaks.
The human eye has three types of cone, so we perceive colour tri-chromatically. There are many many different combinations of wavelengths that trigger the same response.
By carefully choosing three primary colors of light (notionally "red","green" and "blue") we can "mix" most of the colors that people see in everyday life. It's not perfect, whatever combination of primary colors we choose, there will be some colors that the human eye can see, but we can't mix. Most of the time though it's "good enough".
In some crude sense, CMY mixing is the opposite of RGB mixing. We start with white light, and then cyan ink subtracts red, magenta ink subtracts green and yellow ink subtracts blue.
Unfortunately while with RGB, we could pick light sources with relatively narrow spectra, with CMY mixing we are stuck with the light we have. This means that a CMY red, green or blue has a much broader spectrum and hence a less-saturated color than a RGB red, green or blue.
Another problem with CMY mixing is that, even if you layer all three of the inks some light still leaks through, this leads to dark colors (particularly black) being poorly represented in CMY. To fix this a separate Black (called Key) is added to the printing process, giving CMYK.
* Note: Some LEDs (particularly those used for general illumination) are coated with a phosphor coating to give them a broad spectrum.


TITLE: Is deciding whether all satisfying assignments are NAE assignments coNP-complete?
QUESTION [3 upvotes]: Let the language $L$ consist of the $k$-CNF formulas $\phi$ with the property that any satisfying assignment $x$ of $\phi$ is a Not-All-Equal (NAE) assignment, i.e. every clause of $\phi$ has at least one false literal under $x$. Equivalently, $L$ includes those $k$-CNF formulas $\phi$ for which $\phi(x) \implies \phi(\neg x)$.
Clearly $L$ is in coNP, since a certificate for a "no" instance is given by a satisfying assignment which is not a NAE assignment. Is $L$ also coNP-complete for $k > 2$?
As a simple example consider the $2$-CNF formula $(a \vee b) \wedge (\neg a \vee \neg b)$ whose satisfying solutions have exactly one of $a$ or $b$ true, but not both.

REPLY [6 votes]: Consider the following reduction from coNP-complete problem $k$-UNSAT to your language $L$ (where $k$-UNSAT is the language of all unsatisfiable $k$-CNF formulas):
On input a $k$-CNF formula $\psi$, create new variables $x_1, \ldots, x_k$ and output formula $\phi = \psi \land (x_1 \lor x_2 \lor \cdots \lor x_k)$.
If $\psi$ is unsatisfiable then $\phi$ is also unsatisfiable. In that case, it is vacuously true that every satisfying assignment $x$ of $\phi$ is a NAE assignment. (Equivalently, $\phi(x) \implies \phi(\neg x)$ is always true since $\phi(x)$ is always false.) This shows that $\phi \in L$. Thus, we see that if $\psi \in k$-UNSAT (that is, $\psi$ is unsatisfiable) then $\phi \in L$.
On the other hand, if $\psi$ is satisfiable then $\phi$ is also satisfiable by keeping the old assignment for the variables in $\psi$ and setting $x_1 = x_2 = \ldots = x_k = 1$. Notice that this is a satisfying assignment of $\phi$ that is not NAE. In other words, $\phi \not\in L$. Thus, we see that if $\psi \not\in k$-UNSAT (that is, $\psi$ is satisfiable) then $\phi \not\in L$.
The above shows that $\psi \in k$-UNSAT if and only if $\phi \in L$ and therefore that the reduction is correct. Since $k$-UNSAT is coNP-complete, we can conclude that $L$ is coNP-hard.


TITLE: Electric dipole potential (Taylor expansion)
QUESTION [3 upvotes]: In the x-y plane, I have a charge of $-e$ at $\mathbf{r} = x \mathbf{i} + y \mathbf{j}$, and another of $+e$ at some point a distance of $\mathbf{s} = s\mathbf{i}$ from $\bf{r}$, such that the resulting potential is:
$$4\pi \epsilon_0 F= -\dfrac{e}{\bf{|r|}}+\dfrac{e}{|\bf{r}-\bf{a}|} = -\dfrac{e}{\sqrt{x^2+y^2}}+\dfrac{e}{\sqrt{(x-s)^2 + y^2}}$$
If we assume that $s\to 0$, my lecturer has said that "we may expand the second term in a Taylor series for small $s$", such that:
$$4\pi \epsilon_0 F\approx -\dfrac{e}{r} + \dfrac{e}{\sqrt{x^2 + y^2}} \dfrac{1}{\sqrt{1-\frac{2xs}{r}+\frac{s^2}{r}}}$$
Edit: The lecturer then goes on to write that the above:
$$=\dfrac{e}{r}\left[-1 + \left(1+ \dfrac{xs}{r^2}\right) \right]$$
However, I do not understand how the transition to this line has come about.  Could anyone kindly explain?

REPLY [2 votes]: There's no need to use an approximation sign there yet; if you fill in the dots, you get an equality. With $r=\sqrt{x^2+y^2}$, we have
$$
\begin{align}
4\pi\epsilon_0F
&=
-\frac er+\frac e{\sqrt{(x-s)^2+y^2}}
\\
&=
-\frac er+\frac e{\sqrt{x^2+y^2-2xs+s^2}}
\\
&=
-\frac er+\frac e{\sqrt{x^2+y^2}}\frac1{\sqrt{1+(-2xs+s^2)/(x^2+y^2)}}
\\
&=
-\frac er+\frac er\frac1{\sqrt{1-\frac{2xs}{r^2}+\frac{s^2}{r^2}}}\;.
\end{align}
$$


TITLE: Prove that $\int_0^1\int_x^1 \frac{f(y)}ydy\,dx=\int_0^1f(x)\,dx$ if $f$ is Lebesgue integrable
QUESTION [2 upvotes]: Suppose f is a Lebesgue integrable function on [0,1] and define a new function by $$g(x)= \int_x^1 \frac{f(y)}y\text dy$$ for all x in [0,1]. Prove that
$$\int_0^1{g(x)}\text dx=\int_0^1{f(x)}\text dx$$
My approach: I thought of Fubini or Tonelli's theorem to use. But looks like that didn't work well. How do I approach this?

REPLY [1 votes]: differentiating:
$$
g'(x) = -\frac{f(x)}{x}
$$
so
$$
f(x) = - xg'(x)
$$
integrating, then using parts for RHS
$$
\int_0^1 f(x)\text dx = -\int_0^1 g'(x) x \text dx = [-xg(x)]_0^1 + \int_0^1 g(x) \text dx
$$
and
$$
[-xg(x)]_0^1 =0
$$
since $g(1)=0$


TITLE: How to extract Euler angles from a a point in a plane?
QUESTION [4 upvotes]: Given a certain coordinate frame, I can compute a new one by applying a set of rotations in a given order (what I call Euler $Z-Y-X$). So I yaw, then pitch then roll.
Now imagine that I want to do exactly the opposite: given two coordinate frames (same origin to simplify), how do I find out roll, pitch and yaw angles that were used to transform from one to the other?

REPLY [1 votes]: The details will vary according to what you mean exactly by yaw, pitch, roll. But here is a general way of transforming the axes. It may not be the best theoretical way to do it, but it works fine computationally.
Let rotations about the $X$-axis by an angle $\theta$ be denoted by $R_{X,\theta}$, and similarly for the other axes. A frame is then obtained by $$T_{\theta,\phi,\psi}:=R_{X,\phi}R_{Y,\theta}R_{Z,\psi}$$
If I understand your question correctly, you want the three angles that would give $T_1T_2^{-1}$ given two frames $T_1$, $T_2$.

Calculate the matrix $S:=T_1T_2^{-1}$
Calculate $\phi=\arctan_2(S_{3,3},-S_{2,3})$, $\theta=\arctan_2(\sqrt{S_{1,1}^2+S_{1,2}^2},S_{1,3})$, $\psi=\arctan_2(S_{1,1},-S_{1,2})$.

Then $\theta$, $\phi$, and $\psi$ are the required angles. Here $\arctan_2(x,y)$ is the modified arctan function that gives the angle in the correct quadrant.
Note that there may be different values of $\phi$, $\theta$, $\psi$ that give the same transformation $S$.


TITLE: Every open set $U \subset \Bbb R$ is a part of $Q(\Bbb R)^\uparrow$
QUESTION [0 upvotes]: Let $Q(\Bbb R)$ be the set of every half-open interval $[a, b)$ $\subset$ $\Bbb R$, and let $Q(\Bbb R)^\uparrow$ be the set of every $A \in \mathscr P(\Bbb R)$ with $A_k \uparrow A$, $A_k \in Q(\Bbb R)$, which means that $A_1 \subset A_2 \subset \ ... \ $ and $\bigcup_{k = 1}^\infty A_k = A$. Plus, take a look at the following theorem:
Every open set $U \subset \Bbb R$ is a disjoint union of countable half-open cubes with rational endpoints.
Now, my book states the following:
From the theorem above, it follows that every open set is a part of $Q(\Bbb R)^\uparrow$.
I don't see why this is the case. The theorem works under the premise that the cubes are disjoint to each other, but when we have an open set that is a part of $Q(\Bbb R)^\uparrow$, it is a union of sets that obviously aren't disjoint to each other. So how does this fit together?

REPLY [1 votes]: I'm assuming each $A_k$ is supposed to be a union of elements of $Q(\mathbb{R})$ (If $A_k$ is constrained to be a member of $Q(\mathbb{R})$, then any disconnected open set is a counterexample to the theorem - any increasing union of connected sets is going to remain connected, and half-open intervals are connected). If we can write $U$ as the disjoint union $U=\bigcup_{k=1}^\infty [a_k,b_k)$, then we can also write
$$
[a_1,b_1) \subset [a_1,b_1) \cup [a_2,b_2) \subset \ldots \subset \bigcup_{k=1}^m [a_k,b_k) \subset \ldots
$$
which increases to $U$.


TITLE: $f(x)=\sum_{n=1}^{\infty} {4^{1-n}} {h(4^{n-1}x)}$ is continuous and nowhere monotonic,$h(x)=\vert x \vert$ with period 1,$\vert x \vert \leq 1/2$
QUESTION [2 upvotes]: Given $h: \mathbb{R} \to \mathbb{R}$ be a periodic function with period 1 defined for $\vert x \vert \leq 1/2$ by $h(x)=\vert x \vert$ .
To show $f: \mathbb{R} \to \mathbb{R}$ be such that
$f(x)=\sum_{n=1}^{\infty} \frac{h(4^{n-1}x)}{4^{n-1}}$, for all $x \in \mathbb{R}$ is continuous and nowhere monotonic.
We have, $h(x+1)=h(x)$ for  $\vert x \vert \leq 1/2$.
Consider $f_n(x)=\frac{h(4^{n-1}x)}{4^{n-1}}$ and let $x \leq y$. Then $f_n(x)-f_n(y)=\frac{h(4^{n-1}x)}{4^{n-1}}-\frac{h(4^{m-1}x)}{4^{m-1}}= \vert x \vert -\vert y \vert$.
I am not sure how to use the periodicity to prove continuity and nowhere monotonic.

REPLY [2 votes]: Here is a fairly detailed proof.
Proof that f is continuous:
This is the easier half of the proof. First note that $ h(x)\leq1$ for all x. Thus, we have $$ \Big\lvert\frac{h(4^{n-1}x)}{4^{n-1}}\Big\rvert\ \leq \frac{1}{4^{n-1}}$$.
Since $\sum_{n=1}^{\infty}\frac{1}{4^{n-1}}<\infty$, the Weirstrass M-test implies $\sum_{n=1}^{N}\frac{h(4^{n-1}x)}{4^{n-1}}$ converges uniformly to f as N goes to $\infty.$ Thus, f is the uniform limit of continuous functions, so f is continuous.
Proof that f is not monotone on any interval:
This is the more delicate part. First, we fix some notations. Let
$$f_{1}(x)=h(x)$$ and $$f_{n}(x)=\frac{f_{1}(4^{n-1}x)}{4^{n-1}},$$ so that
$$ f(x) = \sum_{n=1}^{\infty}f_{n}(x)=\sum_{n=1}^{\infty}\frac{f_{1}(4^{n-1}x)}{4^{n-1}}.$$ Note that $f_{n}$ is bounded above by $\frac{1}{2}4^{-n+1}$ and is periodic with period $4^{-n+1}$.
Now let $$A= \left\{x\in\mathbb{R}:x=k 4^{-m}, k\in\mathbb{Z},m=0,1,2,.. \right\}$$ and note that $A$ is dense in $\mathbb{R}$ by the Archimedean property of $\mathbb{R}$ for instance. Now note that for $a\in A$, say $a=k 4^{-m}$, we have $$f_{n}(a)= 0$$ for all $n>m$, since $f_{n}$ vanishes at integers. Thus for such $a$, we have $$f(a)= \sum_{n=1}^{m}f_{n}(a).$$ Now let $m$ be a positive integer and let $$x_{m}=4^{-2m-1}.$$ Then for similar reasons as above, we have $$f_{n}(a+x_{m})=0$$ for all $n>2m+1.$ It then follows that $$f(a+x_{m})-f(a)=\sum_{n=1}^{m}f_{n}(a+x_{m})-f_{n}(a) + \sum_{n=m+1}^{2m+1}f_{n}(a+x_{m}) \geq -mx_{m}+(m+1)x_{m} = x_{m}>0.$$
Similarly, $$ f(a-x_{m})-f(a) \geq -mx_{m}+(m+1)x_{m}=x_{m}>0.$$ Thus by the density of $A$, it follows $f$ is not monotone on any interval.


text:
\begin{definition}[Definition:Separated Sets]
Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
Let $A, B \subseteq S$.
=== [[Definition:Separated Sets/Definition 1|Definition 1]] ===
{{:Definition:Separated Sets/Definition 1}}
=== [[Definition:Separated Sets/Definition 2|Definition 2]] ===
{{:Definition:Separated Sets/Definition 2}}
$A$ and $B$ are said to be '''separated sets (of $T$)'''.
\end{definition}

config: wiki


TITLE: How to find $A = M^{A}_{B}$in linear transformation $F = \mathbb{P_{2}} \rightarrow \mathbb{R^{2}} $?
QUESTION [0 upvotes]: How to find $A = M^{A}_{B}$ in linear transformation $F = \mathbb{P_{2}} \rightarrow \mathbb{R^{2}} $, where $ F(p(t)) = \begin{pmatrix} p(0) \\ P(1) \end{pmatrix},$
$ A = \{1,t,t^{2}\},$
$B=\left \{ \begin{pmatrix} 1\\ 0\end{pmatrix},\begin{pmatrix} 0\\ 1\end{pmatrix} \right \}$?

REPLY [1 votes]: Anoher way, $$F(a+bt+ct^2)=\begin{pmatrix}{a}\\{a+b+c}\end{pmatrix}=a\begin{pmatrix}{1}\\{1}\end{pmatrix}+b\begin{pmatrix}{0}\\{1}\end{pmatrix}+c\begin{pmatrix}{0}\\{1}\end{pmatrix}$$ $$=a\begin{pmatrix}{1}\\{0}\end{pmatrix}+(a+b+c)\begin{pmatrix}{0}\\{1}\end{pmatrix}.$$  Then, in coordinates in the basis $A$ and $B$, $$F\begin{pmatrix}{a}\\{b}\end{pmatrix}=\begin{pmatrix}{a}\\{a+b+c}\end{pmatrix}=\underbrace{\begin{pmatrix}{1}&{0}&{0}\\{1}&{1}&{1}\end{pmatrix}}_{M_B^ A}\begin{pmatrix}{a}\\{b}\\ c\end{pmatrix}.$$