TITLE: Stable homotopy groups of complex projective plane
QUESTION [9 upvotes]: We know that there is a cofiber sequence $S^3\xrightarrow{\eta}S^2\to\mathbb{C}\mathbb{P}^2$. It's easy to know that $\pi_3^s(\mathbb{C}\mathbb{P}^2)=0$ so there is a surjection
$$\partial:\pi_7^s(S^2\wedge\mathbb{C}\mathbb{P}^2)\to\pi_7^s(\mathbb{C}\mathbb{P}^2\wedge\mathbb{C}\mathbb{P}^2)$$ by the long exact sequence of $\eta$. The former group $\pi_7^s(S^2\wedge\mathbb{C}\mathbb{P}^2)$ is $\mathbb{Z}/12$, generated by the second hopf element. I wonder whether $\partial$ is an isomorphism?

REPLY [7 votes]: $\newcommand{\Z}{\mathbb Z}\newcommand{\cA}{\mathcal
A}\newcommand{\Sq}{\mathrm{Sq}}\newcommand{\CP}{\mathbb{CP}}\newcommand{\Ext}{\mathrm{Ext}}$It's possible to run
the Adams spectral sequence directly to show that the $2$-torsion subgroup of $\pi_7^s(\CP^2\wedge\CP^2)$ is isomorphic to $\Z/2$, so that
$\partial$ can't be an isomorphism. The calculation is a little longer than Lennart
Meier's answer above, but it is more elementary. In what follows, all
cohomology is with $\Z/2$ coefficients.
As an $\cA$-module, $\tilde H^\ast(\CP^2\wedge\CP^2)$ looks like $\tilde H^\ast(\CP^2)\oplus \Sigma^2\tilde
H^\ast(\CP^2)$, with the lowest- and highest-degree elements joined by a $\Sq^4$. There is a short exact sequence of
$\cA$-modules
\begin{equation}
    0\longrightarrow \Sigma^4\tilde H^\ast(\CP^2)\longrightarrow \tilde H^\ast(\CP^2\wedge\CP^2)\longrightarrow
    \Sigma^2 \tilde H^\ast(\CP^2)\longrightarrow 0.
\end{equation}
Here's a picture of that short exact sequence.

A short exact sequence of $\cA$-modules induces a long exact sequence of Ext groups. Often one computes this long
exact sequence by drawing an Adams chart for the Ext of both the sub and the quotient; the boundary map has
bidegree $(-1, 1)$. Beaudry and Campbell, section 4.6, give some examples of
this technique. To use this, we need to know $\Ext_\cA(\tilde H^\ast(\CP^2), \Z/2)$; you can look it up
in Hood Chatham and Dexter Chua's Adams spectral sequence
calculator (this is for $\Sigma^{-2}\tilde
H^\ast(\CP^2)$, so shift everything to the right by $2$), but it is also possible to compute it by hand in the range
needed using a similar long exact sequence associated to the short exact sequence $0\to\Sigma^4\Z/2\to\tilde
H^\ast(\CP^2)\to\Sigma^2\Z/2\to 0$.
Thus the long exact sequence in Ext looks like this:
We're done once we figure out whether the boundary map in red is nonzero: either way, there are no differentials to
or from the $7$-line, and there are no possible hidden extensions on the $7$-line. We will show the red boundary
map is nonzero by showing the map
$$i^\ast\colon \Ext_\cA^{1,9}(\tilde H^\ast(\CP^2\wedge\CP^2), \Z/2) \longrightarrow
\Ext_\cA^{1,9}(\Sigma^4\tilde H^\ast(\CP^2), \Z/2),$$
which is induced by $i\colon \Sigma^4\tilde H^\ast(\CP^2)\to\tilde H^\ast(\CP^2\wedge\CP^2)$, vanishes;
exactness then forces the red boundary map to be injective, so the $7$-line of the $E_2$-page for $\tilde
H^\ast(\CP^2\wedge\CP^2)$ has only a single $\Z/2$.
Now let's show $i^\ast$ vanishes. $\Ext_\cA^{1,9}(\Sigma^4\tilde H^\ast(\CP^2), \Z/2)\cong\Z/2$ and is generated by
the upside-down question mark extension:

Therefore if $i^\ast\ne 0$, there is some extension $0\to\Sigma^9\Z/2\to M\to \tilde H^*(\CP^2\wedge\CP^2)\to 0$
whose pullback along $i$ is the upside-down question mark extension. The only way for this to work would be the
following diagram:

However, this choice of $M$ is not a valid $\cA$-module: in $M$, $\Sq^5x = \Sq^1\Sq^4x \ne 0$, but $\Sq^5 =
\Sq^4\Sq^1 + \Sq^2\Sq^1\Sq^2$, and $\Sq^4\Sq^1x = \Sq^2\Sq^1\Sq^2x = 0$. Therefore $i^\ast = 0$.


TITLE: Isomorphic Mapping Between $\mathbb{Z}_3 [x] / \langle x^2 + x + 2 \rangle$ and $\mathbb{Z}_3 [x] / \langle x^2 + 1\rangle$
QUESTION [0 upvotes]: The fields $\mathbb{Z}_3 [x] / \langle x^2 + x + 2\rangle$ and $\mathbb{Z}_3 [x] / \langle x^2 + 1\rangle$ are both exactly of the form:
$\{0, 1, 2, x, 1 + x, 2 + x, 2x, 1 + 2x, 2 + 2x\}$.
And since both of these fields have a one-to-one correspondence, we can have an isomorphic mapping $f$ that sends each element to itself, right? So would stating the mapping:
$0 \rightarrow 0$, $1 \rightarrow 1$, .... etc. be sufficient to show an isomorphism? Do I also need to prove the homomorphic property separately? If so, how would I do that?  I know there's a theorem that states two fields of the cardinality $p^n$ are isomorphic, but I need to show a specific mapping. Thanks.

REPLY [1 votes]: You want commutative diagram:
$\require{AMScd}$
\begin{CD}
 \Bbb Z_3[x]@>\varphi>> \Bbb Z_3[x]\\
@VVV @VVV\\
\Bbb Z_3[x]/\langle x^2+x+2\rangle@>\bar\varphi>> \Bbb Z_3[x]/\langle x^2+1\rangle
\end{CD}
where $\varphi$ is a ring homomorphism such that $\varphi(x^2+x+2) \in \langle x^2+1\rangle$ so we could induce $\bar\varphi$ by first homomorphism theorem. All we have to decide is where to send $x$. In general, $\varphi(x)$ can be polynomial of arbitrary degree, but in our case, it is not a loss of generality to assume it is of degree $1$. (Can you see why?)
So, let $\varphi(x) = ax + b$. Then $\varphi(x^2+x+2) = a^2x^2+a(2b+1)x+(b^2+b+2)$ and it must be divisible by $x^2 + 1$, so we conclude that $a^2x^2+a(2b+1)x+(b^2+b+2) = a^2x^2 + a^2$, i.e.
\begin{align}
a(2b+1) &= 0\\
b^2+b+2 &= a^2
\end{align}
which we solve in $\Bbb Z_3$ to get $b = 1$, $a=\pm 1$. This gives us two different $\bar\varphi$, one that maps $x$ to $x+1$ and the other that maps $x$ to $2x+1$, more concretely, $\bar\varphi_1(ax+b) = ax+(a+b)$ and $\bar\varphi_2(ax+b) = 2ax+(a+b)$. One could now use linear algebra to immediately verify that both are bijective (we have $\Bbb Z_3$-linear maps between two-dimensional $\Bbb Z_3$-vector spaces).


TITLE: In how many different ways can this problem be solved?
QUESTION [0 upvotes]: I have a math problem. In many different ways can this problem be solved? Here is the problem:
$$y''-y'-2y=0, \\ y(0)=1 \\ y'(0)=0$$
I have already found $5$ ways:
$(1):$ Characteristic equation (standard)
$(2):$ Laplace Transforms
$(3):$ Series Solution
$(4):$ Numerical approximation
$(5):$ Reduce to autonomous linear system
Need I say more?

REPLY [0 votes]: Differential Operators:
$$(D^2-D-2)y=0$$ $$(D-2)(D+1)y=0$$ and the solution follows.
Random Substitutions:
Let $u=y'-2y$. Note that $$y''-y'-2y=y''-2y'+y'-2y=u'+u=0$$ $$u'=-u$$ $$u=y'-2y=c_1e^{-x}$$ and now use integrating factor to solve the first-order problem.
I'll add more as I think of them.


text:
\begin{document}

\title[Inequivalent Lefschetz fibrations on rational and ruled surfaces]
{Inequivalent Lefschetz fibrations on \\ rational and ruled surfaces}

\author[R. \.{I}. Baykur]{R. \.{I}nan\c{c} Baykur}
\address{Department of Mathematics and Statistics, University of Massachusetts, Amherst, MA 01003-9305, USA}
\email{baykur@math.umass.edu}

\begin{abstract}
In this short note, we give an explicit construction of inequivalent  Lefschetz pencils and fibrations of same genera on blow-ups of all rational and ruled surfaces. This complements our earlier results in \cite{BaykurJSG}, concluding that \emph{every} symplectic $4$--manifold, after sufficiently many blow-ups, admits inequivalent Lefschetz pencils and fibrations, which cannot be obtained from one another even via any sequence of fibered Luttinger surgeries.
\end{abstract}

\maketitle

\setcounter{secnumdepth}{2}
\setcounter{section}{0}

\section{Introduction}

A pair of Lefschetz pencils or fibrations\footnote{We assume that a \emph{pencil}, unlike a Lefschetz fibration, always has base points. We also assume that a Lefschetz pencil or fibration always has critical points, and each singular fiber contains only one critical point and no exceptional spheres (i.e. it is \emph{relatively-minimal}).} on a closed oriented smooth \mbox{$4$--manifold} are said to be \emph{equivalent} (or \emph{isomorphic}) if there is an orientation-preserving self-diffeomorphism of the total space which commutes the two maps. Simon Donaldson's ground-breaking work in \cite{Donaldson} provides arbitrarily large genera \emph{pencils} on any symplectic \mbox{$4$--manifold.} A fundamental problem then is to determine if there are relatively-minimal inequivalent pencils or fibrations of \emph{the same genus and number of base points} on a given symplectic $4$--manifold. For brevity, we refer to them as \emph{inequivalent pencils} and \emph{inequivalent fibrations}, where it should be understood that their genera and number of base points are already the same.

This note is an amendment to our results in \cite{BaykurJSG}, where we proved that any symplectic $4$-manifold, which is not a rational or a ruled surface, after a number of blow-ups, admits arbitrarily many inequivalent Lefschetz pencils and fibrations. Moreover, they are not related via \emph{fibered Luttinger surgeries} \cite{BaykurJSG}. Here we will give similar examples of pencils and fibrations on \emph{rational} and \emph{ruled surfaces}. Recall that rational and ruled surfaces are birationally equivalent to the complex projective plane and $S^2$--bundles over Riemann surfaces, so they are diffeomorphic to blow-ups of $\CP$, $S^2 \x \Sigma_h$, $S^2 \widetilde{\x} \Sigma_h$, which are known to have unique symplectic structures up to symplectomorphisms and deformations.

Our examples in \cite{BaykurJSG} were obtained from Donaldson pencils via \emph{partial doubling sequences}, which involve blow-ups and degree doublings of pencils  introduced in \cite{Smith, AurouxKatzarkov}. These pencils were distinguished by looking at the collection of exceptional spheres, which arise as \emph{multisections} \cite{DonaldsonSmith, BaykurHayano} of different degrees in respective pencils. Thus, blowing-up all the base points, we obtain inequivalent Lefschetz fibrations. However, the same strategy does not work for rational or ruled surfaces, since in this case, one is not guaranteed to have disjoint multisections representing a given collection of  exceptional classes \cite{BaykurJSG}.

We will instead apply partial doublings to hand-picked pencils on rational and ruled surfaces in order to derive inequivalent pencils which have \emph{different numbers of reducible fibers} versus irreducible ones. The point here is that ``doublings are not created equal''; depending on which subcollections of base points we pick, we can arrive at pencils with topologically different singular fibers. Since the distinction comes from the number of reducible fibers, we get inequivalent fibrations after blowing-up all the base points. Our main result is:

\begin{theorem}
Any rational or ruled surface, after sufficiently many blow-ups, admits inequivalent relatively-minimal pencils and fibrations of the same genera and number of base points, which cannot be obtained from one another via any sequence of fibered Luttinger surgeries.
\end{theorem}

\noindent Last part of the theorem comes easy for our examples: performing a fibered Luttinger surgery is equivalent to conjugating a subword of the corresponding monodromy factorization of a pencil by some Dehn twist; see \cite{Auroux2, BaykurJSG}. This can be done whenever the twist curve is fixed, up to isotopy, by the mapping class given by the subword. Clearly, the number of reducible fibers is unchanged under this operation, so the examples with different numbers of reducible fibers we produce are also inequivalent up to fibered Luttinger surgeries.

\smallskip
There are many other examples of inequivalent Lefschetz pencils and fibrations. A pair of inequivalent \emph{fibrations} on a blow-up of $T^2 \x \Sigma_2$, whose fibers have different divisibility in homology, was discussed by Ivan Smith in \cite{SmithThesis}. Several inequivalent fibrations on homotopy elliptic surfaces, distinguished by their monodromy groups,  were discovered by  Jongil Park and Ki-Heon Yun in \cite{ParkYun, ParkYun2}. As for inequivalent \emph{pencils}, first examples were constructed by Kenta Hayano and the author on \mbox{$4$--manifolds} homeomorphic to blow-ups of the $\K$ surface \cite{BaykurHayano}. The construction in \cite{BaykurHayano} uses pairs of lantern substitutions which modify pencils differently, but land on the same $4$--manifold. Notably, Noriyuki Hamada recently constructed pairs of inequivalent pencils on irrational ruled surfaces $S^2 \widetilde{\times} \Sigma_h$, which \emph{only} differ by the configuration of their base points on the reducible fibers, so they do not yield inequivalent fibrations after blow-ups.

With our aforementioned result from \cite{BaykurJSG}, we arrive at the following conclusion, which demonstrates how diverse the Donaldson-Gompf correspondence between symplectic $4$--manifolds and Lefschetz pencils and fibrations is:

\begin{corollary}
Any closed symplectic $4$--manifold, possibly after blow-ups, admits inequivalent relatively-minimal pencils and fibrations of the same genera and number of base points, which cannot be obtained from one another via any sequence of fibered Luttinger surgeries.
\end{corollary}

\smallskip

\section{Preliminaries}

Here we will quickly review the basic notions and background results on Lefschetz pencils\,/\,fibrations, positive Dehn twist factorizations, fibered Luttinger surgeries and partial conjugations, and doublings of Lefschetz pencils. For further details on these topics, the reader can turn to \cite{GS} and \cite{ADK, AurouxKatzarkov, BaykurJSG}.

Let $X$ be a closed, oriented $4$-manifold, and $B=\{ b_j \}$, $C=\{ p_i \}$ be finite, non-empty sets of points in $X$.  A \emph{Lefschetz pencil} $(X,f)$ is given by a surjective map $f\colon X \setminus B \to S^2$, which is a submersion on $X \setminus C$, such that around each \emph{base point} $b_j$ and each \emph{critical point} $p_i$, it conforms to local complex models $(z_1,z_2) \mapsto z_1/z_2$ and $(z_1, z_2) \mapsto z_1 z_2$, respectively.  A \emph{Lefschetz fibration} is defined similarly for $B = \emptyset$. After blowing-up every base point $b_j$ of a pencil, we obtain a Lefschetz fibration with disjoint exceptional spheres $S_j$ as its sections. We say $(X,f)$ is a \emph{genus--$g$} pencil\,/\,fibration, for $g$ the genus of the regular fiber $F$. We will assume that every  fiber contains at most one critical point, which can always be achieved after a small perturbation. A \emph{singular fiber} has a nodal singularity at the critical point $p_i$, and is obtained by shrinking a simple loop $a_i$ on $F$, called the \emph{vanishing cycle}. We have a \emph{reducible} fiber (with two connected components) if $a_i$ is separating $F$, and \emph{irreducible} if $a_i$ is non-separating.

Let $\Sigma_g^m$ denote a compact, oriented surface of genus $g$ with $m$ boundary components, where $\Sigma_g=\Sigma_g^0$. The \emph{mapping class group} $\Gamma_g^m$ is the group of orientation-preserving self-diffeomorphisms of $\Sigma_g^m$ which restrict to identity along $\partial \Sigma_g^m$, modulo isotopies of the same type. Let $t_a \in \Gamma_g^m$ denote the positive (right-handed) Dehn twist along a simple loop $a$ on $\Sigma_g^m$. A genus--$g$ Lefschetz pencil $(X,f)$ with $m$ base points prescribes a \emph{positive factorization} of the boundary multi-twist (or \emph{monodromy factorization}), given by a relation
\[ t_{\delta_1} \cdot \ldots \cdot t_{\delta_m} = t_{a_1} \cdot \ldots \cdot t_{a_\ell} \]
in $\Gamma_g^m$, where $\delta_1, \ldots, \delta_m$ are curves parallel to distinct boundary components of $\Sigma_g^m$ (which we will also denote the boundary component by),  $a_i$ is a vanishing cycle corresponding to the critical point $p_i$ of $f$, and $\ell=|C|$.  Conversely, given a positive factorization as above, one can construct such a Lefschetz pencil $(X,f)$. For $g \geq 2$, we indeed get a one-to-one correspondence between equivalence classes of  \mbox{Lefschetz pencils} and positive factorizations up to \textit{Hurwitz moves} (trading subwords $t_{a_i}t_{a_{i+1}}$ and $ t_{t_{a_i}(a_{i+1})}  t_{a_i}$) and \textit{global conjugations} (replacing every $t_{a_i}$ in the factorization with $t_{\phi{(a_i)}}$, for the same $\phi \in \Gamma_g^n$). All goes the same for Lefschetz \emph{fibrations} for $m=0$.

By the pioneering works of Donaldson and Gompf  \cite{Donaldson, GS}, every symplectic \mbox{$4$-manifold} admits a  Lefschetz pencil whose fibers are symplectic,  and conversely, the total space of a Lefschetz pencil or a fibration  (recall $C \neq \emptyset$) can always be equipped with a symplectic form for which all regular fibers are symplectic.

As observed in \cite{ADK}, \emph{Luttinger surgery} (where one removes a neighborhood of a Lagrangian torus in a symplectic $4$--manifold and glues it back in differently so as to produce a new symplectic $4$--manifold) can be performed in a way compatible with the pencil structure,  when the restriction of the pencil map to the torus is a circle bundle over a loop on the base, away from the critical values \cite{Auroux2, ADK, BaykurJSG}. This surgery is equivalent to modifying the corresponding positive factorization by replacing a subword $t_{a_i} \cdot \ldots \cdot t_{a_k}$ with a conjugate $t_{\phi(a_i)} \cdot \ldots \cdot t_{\phi(a_k)}$ for some $\phi=t_{b}^{\pm}$, which can be done if and only if the mapping class $t_{a_i} \cdot \ldots \cdot t_{a_k}$ fixes $b$, up to isotopy \cite{Auroux2}.

Another way to produce a new pencil from a given one, and on the same \mbox{$4$--manifold,} is the \emph{doubling} procedure, discussed for holomorphic pencils and Donaldson’s pencils in \cite{Smith}, and for pencils obtained via branched coverings of $\CP$ in \cite{AurouxKatzarkov}. In general, by \cite{BaykurJSG}[Lemma~3.1], one can double  \emph{any} (topological) genus--$g$ pencil $(X,f)$ with $m$ base points, provided $g \geq 2$ and $m \leq 2g-2$. The result is a genus--$g'$ pencil $(X, f')$ with $m'$ base points, where $g'=2g+m-1$ and $m'= 4m$. If the original pencil is compatible with a given symplectic form, then the resulting one is also compatible with a deformation equivalent symplectic form.

Topologically, the doubling procedure can be interpreted roughly as follows: First break up the symplectic pencil $(X,f)$ into two pieces; the regular neighborhood of a smooth fiber (the ``concave piece''), which is a symplectic disk bundle of degree $m$ over a genus--$g$ surface, and its complement (the ``convex piece''). Describe a standard higher genus pencil on the former piece, based on $g$ and $m$ ---the model for which comes from holomorphic pencils. Then extend it over the convex piece, which now becomes a subpiece of the new pencil $(X,f')$. By the enterprise of \cite{AurouxKatzarkov}, the monodromy of $f'$ is explicitly determined by that of $f$ \cite{AurouxKatzarkov}[Theorem~4]. In particular, we observe  the following: any \emph{separating} Dehn twists in the positive factorization of $f'$ comes, in a one-to-one fashion, from  a separating Dehn twist in the positive factorization of $f$, which bounds a subsurface on $F$ that does not contain a base point of $f$. (See Figure~\ref{figure0} for an example.)

\section{Constructions}

It is time to spell out our recipe in full detail. Take a positive factorization for a genus $g \geq 2$ pencil $(X,f)$ with base locus $B$ and regular fiber $F$, whose separating vanishing cycles are \,$a_1, \ldots, a_r$. Let $B'=\{b'_j\}$ and $B''=\{b''_j\}$ be two subsets of the base locus $B$ of $f$ (where the base points correspond to boundary components in the positive factorization), so that $|B'|=|B''| \leq 2g-2$. Set $K'_i, K''_i \in \{0,1\}$ to be the number of components in  $F \setminus a_i$  which do not intersect $B'$ (resp. $B''$).  \linebreak If $K'=\sum_{i=1}^{r'} K'_i$ is unequal to $K''=\sum_{i=1}^{r''} K''_i$, we are game.

We blow-up all the base points of $f$ in $B \setminus B'$ (resp.\,$B \setminus B''$)  and double the resulting pencil on \,\mbox{$\widetilde{X}=X \, \# \,  |B \setminus B'| \, \CPb$} with $m=|B'|$ base points, which naturally come from the inclusion of $X$ minus the blown-up base points into $\widetilde{X}$. In this way, we arrive at a new pencil $(\widetilde{X},f')$ (resp.\,$(\widetilde{X}, f'')$). As we reviewed earlier, each one of the pencils $f'$ and $f''$ on $\widetilde{X}$  has genus $2g+m-1$ and $4m$ base points, whereas the number of reducible fibers they have are $K'$ and $K''$, respectively. (Figure~1 illustrates $2$ distinct partial doublings of sorts.) For each $0 \leq m_0 < m$, we get a pair of inequivalent pencils by blowing-up $m_0$ base points of both pencils, and for $m_0=m$, we get a pair of inequivalent fibrations.

\smallskip
\begin{figure}[ht!]
 \centering
     \includegraphics[width=10cm]{doubling.eps}
     \caption{The lifts of a separating vanishing cycle (in blue) under doubling of a genus--$2$ pencil with $2$ different pairs of base points, marked with $\bf{\times}$ in the figure.}
     \label{figure0}
\end{figure}

It remains to find the main ingredient for our recipe; a pencil $(X,f)$ as prescribed above, on (blow-ups of) rational and ruled surfaces.

\smallskip

\subsection{Inequivalent pencils and fibrations on rational surfaces} \

There is the following positive factorization of the boundary multi-twist in $\Gamma_2^3$, which was found by Sinem Onaran in \cite{Onaran}:
\begin{equation*}
\begin{array}{ccl}
t_{\delta_1} t_{\delta_2} t_{\delta_3} &=&  t_{a_3} t_{b_1} (t_{a_1}t_{a_2}t_{a_3}t_{b_1})^2 t_{a_3}  t_{\psi(b_2)} t_{a_3} t_{b_2} t_{a_4} t_{a_5} t_{a_3} t_{b_2} t_{a'_4} t_{a_6} \, .
\end{array}
\end{equation*}
Here the twist curves $a_i, b_j$ and $a'_4$ are as in Figure~\ref{figure1}, and $\psi = t_{a_5}^{-1} t_{a_4}^{-1}$. \linebreak (See \cite{Onaran}[Figure~6]. This is a lift of a well-known positive factorization of identity in $\Gamma_2$ \cite{BirmanHilden}, which corresponds to a hyperelliptic Lefschetz fibration on $\CP \# 13 \CPb$.)

\begin{figure}[ht!]
 \centering
     \includegraphics[width=12.5cm]{figure1b.eps}
     \caption{Twist curves $a_i, b_j$ and $a'_4$ on the surface $\Sigma_{2}^3$. The boundary components are $\delta_1, \delta_2, \delta_3$. The new curves $x,y$ are given in the second copy of $\Sigma_2^3$, where the lantern sphere sits in the middle. The separating curve $x$ is in blue.}
     \label{figure1}
\end{figure}

We rewrite the above expression as
\begin{equation*}
\begin{array}{ccl}
t_{\delta_1} t_{\delta_2} t_{\delta_3} &=&
t_{a_3} t_{b_1} t_{a_1}t_{a_2}t_{a_3} t_{b_1} t_{a_1} t_{a_2}  t_{a_4} t_{a_5}
\psi (t_{a_3}t_{b_1} t_{a_3} t_{\psi(b_2)} t_{a_3} t_{b_2}) \psi^{-1}
t_{a_3} t_{b_2} t_{a'_4} t_{a_6}  \\ &=&
t_{a_3} t_{b_1} t_{a_1}t_{a_2}t_{a_3} t_{b_1} \underline{t_{a_1} t_{a_2}  t_{a_4} t_{a_5} }
t_{\psi(a_3)} t_{\psi(b_1)} t_{\psi(a_3)} t_{\psi^2(b_2)} t_{\psi(a_3)} t_{\psi(b_2)}
t_{a_3} t_{b_2} t_{a'_4} t_{a_6}
\end{array}
\end{equation*}
and then apply a lantern substitution along the underlined subword to obtain a new positive factorization:
\begin{equation*}
\begin{array}{ccl}
t_{\delta_1} t_{\delta_2} t_{\delta_3}  &=&
t_{a_3} t_{b_1} t_{a_1}t_{a_2}t_{a_3} t_{b_1} \underline{t_{a_3} t_{x}  t_{y}}
t_{\psi(a_3)} t_{\psi(b_1)} t_{\psi(a_3)} t_{\psi^2(b_2)} t_{\psi(a_3)} t_{\psi(b_2)}
t_{a_3} t_{b_2} t_{a'_4} t_{a_6} \, ,
\end{array}
\end{equation*}
which involves new twist curves $x, y$ given in Figure~\ref{figure1}. Note that $x$ is the only separating twist curve here.  Importantly, the ``lantern sphere'' we utilized did not contain the base points. The final positive factorization we obtained above describes a genus--$2$ Lefschetz pencil $(X,f)$ with $3$ base points.

The Seiberg-Witten adjunction inequality implies that in any closed symplectic $4$--manifold, except for rational or ruled surfaces, we have \,$2g(F)-2=-\eu(F) \geq [F]^2$ for any connected symplectic surface $F$ \cite{Ta, LiLiu}. However, $(X,f)$, equipped with the Gompf-Thurston symplectic form, has a symplectic fiber $F$ of genus $g(F)=2$ and self-intersection $[F]^2=3$ (the number of base points), violating the inequality. So we conclude that $X$ is a rational or ruled surface. We can then determine its exact diffeomorphism type through a simple calculation of its algebraic topological invariants. The first integral homology group $H_1(X)$, which is isomorphic to the quotient of $H_1(F)$ by the subgroup normally generated by  homology classes of vanishing cycles, is easily seen to be trivial: the vanishing cycles $a_1, b_1, a_3$ and $b_2$ already kill the whole group. Since the Euler characteristic of $X$ is given by \,$\eu(X)=4-4g+|C|-|B|$, where $C$ and $B$ are the critical locus and the base locus, we calculate $\eu(X)=12$. It follows that $X \cong \CP \# 9 \CPb$.

Now let $b_j$ be the base point of $(X,f)$ corresponding to the boundary components $\delta_j$ in the factorization, for $j=1,2,3$. For $B'=\{b_2, b_3\}$ and $B''=\{b_1, b_2\}$, our recipe generates a pair of inequivalent genus--$5$ pencils \mbox{$f'$ and $f''$} on the rational ruled surface \mbox{$\widetilde{X} \cong \CP \# 10 \,\CPb \cong S^2\x S^2 \#\,9 \,\CPb$} with $8$ base points, where $f'$ has $1$ reducible fiber but $f''$ has none. (As illustrated in Figure~\ref{figure0}.)

\smallskip

\subsection{Inequivalent pencils and fibrations on irrational ruled surfaces} \

We next consider the following positive factorization of the boundary multi-twist in $\Gamma_{2h}^3$, for each $h \geq 1$, due to Noriyuki Hamada \cite{Hamada}:
\begin{equation*}
\begin{array}{ccl}
t_{\delta_1} t_{\delta_2} t_{\delta_3} &=& t_{B_{0,1}} t_{B_{1,1}} \cdots t_{B_{h,1}} t_{C_1} t_{B_{0,2}} t_{B_{1,2}} \cdots t_{B_{h,2}} t_{C_2} \, ,
\end{array}
\end{equation*}
where the curves $B_{i,k}, C_k$ for $i=0,1, \ldots, h$ and $k=1,2$ are as in Figure~\ref{figure2}.
(This is the positive factorization ``$W_{IIA}$ for even genus'' in \cite{Hamada}[Figure~23] with only $3$ boundary components we picked and relabeled here. It is a lift of a well-known positive factorization of identity in $\Gamma_2$ \cite{Matsumoto, Korkmaz, Cadavid}, which corresponds to a hyperelliptic Lefschetz fibration on $S^2 \x \Sigma_h \# 4 \CPb$.) Corresponding to this positive factorization is a genus--$2h$ pencil $(X,f)$ with $3$ base points, where \mbox{$X\cong S^2 \x \Sigma_h \# \, \CPb$.} Note that $f$ has $2$ separating vanishing cycles.

\begin{figure}[ht!]
 \centering
     \includegraphics[width=13cm]{figure2b.eps}
     \caption{From left to right: twist curves $B_{0,1}, B_{1,1}, \ldots, B_{h,1}, C_1$, and $B_{0,2}, B_{1,2}, \ldots, B_{h,2}, C_2$ on (identical copies of) the surface $\Sigma_{2h}^3$. The boundary components are $\delta_1, \delta_2, \delta_3$. The separating curves $C_1$ and $C_2$ are given in blue.}
     \label{figure2}
\end{figure}

Once again, let $b_j$ be the base point of $(X,f)$ corresponding to the boundary component $\delta_j$ in the factorization, for $j=1,2,3$. We see that for $B'=\{b_2, b_3\}$ and $B''=\{b_1, b_2\}$, our recipe generates a pair of inequivalent genus--$(4h+1)$ pencils \mbox{$f'$ and $f''$} on the irrational ruled surface $\widetilde{X} \cong S^2 \x \Sigma_h \# \, 2\, \CPb \cong S^2 \widetilde{\x} \Sigma_h \# \, 2\, \CPb$ with $8$ base points, where $f'$ has $2$ reducible fibers but $f''$ has $1$.

\smallskip

\subsection{Generalizations.} \

We end with a few remarks on our recipe for constructing inequivalent pencils and fibrations in general:

1. From each pair of inequivalent pencils $f'$ and $f''$ we obtained, we can easily derive infinitely many more pairs of inequivalent pencils and fibrations of arbitrarily high genera. This is  achieved by taking further doubles of  $f'$ and $f''$; the base points of the new pencil will lie only on one component of each reducible fiber in $f'$ or $f''$ (see the monodromy in \cite{AurouxKatzarkov}[Theorem~4]), so the number of reducible fibers will remain different.

 2. One can produce more than just a pair of inequivalent pencils and fibrations, whenever there is  an input pencil $(X,f)$ with several subcollections of base points $B', B'', \ldots, B^{(n)}$ of equal rank, which have different numbers $K', K'', \ldots, K^{(n)}$ of reducible fiber components they miss. For instance, a further lift of Hamada's positive factorization to $\Gamma_{2h}^4$ in \cite{Hamada} can be used to generate yet another pencil $f'''$ with no reducible fibers.

 3. Our recipe can be improved in several ways. For example, we only compared the \emph{number }of reducible fibers, yet one can as well compare the \emph{topological types} of reducible fibers (determined by the genera of the subsurfaces they bound). It seems plausible that one can produce \emph{arbitrarily many} inequivalent fibrations and pencils by employing doubling \emph{sequences} as in \cite{BaykurJSG} for pencils on blow-ups of the same manifold, which contain  different topological types of reducible fibers.

\vspace{0.1in}
\noindent \textit{Acknowledgments.} These examples of inequivalent pencils and fibrations  came to life during my visit to Seoul, Korea in 2015, and were first announced during my lectures at the  highly stimulating \emph{2015 Seoul National University Topology Winter School}. This work is supported by the NSF Grant DMS-$1510395$.

\vspace{0.25in}

config: arxiv


text:
\section{Problem statement and preliminary results}

\subsection{Notation}

By $\|v\|$, we denote the Euclidean norm of a vector $v \in \mathbb{R}^n$ and by $\|\bA\|$ the spectral norm of a matrix $\bA \in \mathbb{R}^{n \times n}$. The spectrum of a matrix will be denoted by $\sigma(\bA)$ and for the spectral radius we will write $\rho(\bA)$. We will also often use the notation $\|v\|_{\bA}^2 = v^T\bA v$
for the semi-norm defined by a positive semi-definite matrix $\bA$, which is a full norm whenever $\bA$ is positive definite. If $\bA_1,\bA_2$ are two symmetric matrices of the same dimensions, then we write $\bA_1 \succ \bA_2$ ($\succeq$) if $\bA_1 - \bA_2$ is positive (semi-) definite and $\bA_1 \prec \bA_2$ ($\preceq$) if $\bA_1 - \bA_2$ is negative (semi-) definite. With $\bA^\dagger$, we will denote the Moore-Penrose pseudo inverse of a matrix, while $\bA^T$ will denote its transpose. The orthogonal projection matrix onto the kernel of a matrix $\bA$ will be denoted by $\bPi_{\ker \bA}$. In large matrix equations, we will sometimes write $\bA^T\bB(\star)$. In that case, $(\star)$ is to be understood as a copy of the matrix $\bA$.

\subsection{Problem statement}

Consider the gradient based algorithm defined by
\begin{align}
\label{eq:algo}
x_{k+1} &= \bA x_k + \bB \nabla f(\bC x_k),
\end{align}
where $x_k \in \bbR^{n}$ and the matrices $\bA \in \bbR^{n\times n},\bB \in \bbR^{n \times d},\bC \in \bbR^{d\times n}$ are the algorithm parameters to be designed.
The objective function $f \in C^1(\bbR^d)$ is assumed to satisfy the following generalized sector condition for all $z_1,z_2 \in \bbR^d$:

\begin{align}

		\frac{1}{2}\| z_1 - z_2\|_{\bM}^2 &\leq f(z_2) - f(z_1) + (\nabla f(z_1))^T(z_1-z_2) \nonumber\\
		&\leq \frac{1}{2}\| z_1 - z_2 \|_{\bL}^2,

	\label{eq:4SCLcharacterisation}
\end{align}
where $\bM \preceq \bL \in \bbR^{d\times d}$ are given symmetric matrices.
In the following, $S(\bM,\bL)$ denotes the set of all $C^1$ functions that satisfy \eqref{eq:4SCLcharacterisation}.
Note that $S(m \bI_d, l \bI_d)$, $m<l$, is a set of strongly convex functions,
as typically found in the literature.
In the case $f \in C^2(\bbR^d)$, \eqref{eq:4SCLcharacterisation} is equivalent to $\bM \preceq H_f(z) \preceq \bL$ for all $z \in \bbR^d$, where $H_f$ denotes the Hessian of $f$.

The algorithm design problem addressed in this paper is formally stated as:

\begin{problem}
	\label{problem1}
    For given $n \ge d$, $\bM\preceq \bL$, and convergence rate $\rho \in [0,1[$,  we aim to \emph{design} matrices $( \bA, \bB,  \bC) \in  \bbR^{n\times n} \times \bbR^{n\times d} \times \bbR^{d\times n} $ such that for any $f \in S(\bM,\bL)$ there exist $x_f^* \in \bbR^n$ and $c_f \in \bbR_{\geq 0}$ such that
    \begin{align*}
        \nabla f(z_f^*) = 0 \mathrm{~for~} z_f^* := \bC x_f^*
    \end{align*}
    and the iterates $x_k$ of \eqref{eq:algo} satisfy
    \begin{align*}
        \| x_f^* -  x_k\| &\leq c_f \rho^k \| x_f^*- x_0\|,
    \end{align*}
    for any $x_0 \in \bbR^{n}$, $k \in \bbN_0$.

\end{problem}

 In our setting, \emph{design} (synthesis) refers to computing the algorithm parameters $(\bA,\bB,\bC)$ by solving a  \emph{convex} optimization problem, i.e. a semi-definite program.

 Our goal is solving Problem \ref{problem1}. The following Problem \ref{problem2}  is similar to Problem \ref{problem1} with the slight modification that all the functions $f$ under consideration have their critical points in $z_f^* = 0$. This is favourable for the application of tools from robust control theory, which are often formulated for fixed-points in zero.

\begin{problem}
	\label{problem2}
	For given $n \geq d$, symmetric matrices $\widetilde{\bL} \succeq 0$ and $\bM$, and $\rho \in [0,1[$, design matrices
	$(\widetilde \bA, \widetilde \bB, \widetilde \bC) \in  \bbR^{n\times n} \times \bbR^{n\times d} \times \bbR^{d\times n} $ satisfying the constraint
	\begin{align}
	\widetilde \bC(\widetilde{\bA} - \bI_n)^{-1} \widetilde \bB \bM = \bI_d \label{eq:Constraint}
	\end{align}
	such that for any $f \in S_0(0,\widetilde{\bL}) := \{ f\in S(0,\widetilde{\bL}): \nabla f(0) = 0 \}$ there exists $c_f \in \bbR_{\geq 0}$ such that the iterates of \eqref{eq:algo} satisfy
	\begin{align*}
	\| x_k\| &\leq c_f \rho^k \| x_0\|
	\end{align*}
	for any $x_0 \in \bbR^{n}$ and $k \in \bbN_0$.
\end{problem}

The subsequent theorem states that the two problems are equivalent.

\begin{theorem}
	\label{thm:ConvTrInv}
	Let symmetric matrices $\bM \preceq \bL$ be given, set $\widetilde{\bL}:= \bL - \bM$ and fix $\rho \in [0,1[$.
	Then the matrices $(\bA,\bB,\bC)$ solve Problem \ref{problem1} if and only if the matrices $(\widetilde{\bA},\widetilde \bB,\widetilde \bC)$ solve Problem \ref{problem2}, where $\widetilde{\bA} = \bA + \bB\bM\bC$, $\widetilde{\bB} = \bB$, $\widetilde{\bC} = \bC$.
\end{theorem}

This theorem justifies that we can solve Problem \ref{problem2} instead of Problem \ref{problem1}.

\subsection{Properties of the class $S(\bM,\bL)$}

This subsection serves the purpose of introducing some important properties of $S(\bM,\bL)$.
The first result gives some equivalent characterizations for when $f \in S(0,\bL)$ holds true. Note, that these conditions can be applied to any class $S(\bM,\bL)$ by using the fact $f \in S(\bM,\bL) \Leftrightarrow (z\mapsto f(z) - \frac{1}{2}z^T\bM z) \in S(0,\bL-\bM)$.

\begin{lemma}[Characterizations for $f \in S(0,\bL)$]
	\label{lem:LipschitzConv}
	Let $\bL \succeq 0$ and $f \in C^1(\bbR^d)$. All conditions below, holding for all $z_1,z_2 \in \bbR^n$, are equivalent to $f\in S(0,\bL)$:
	\begin{enumerate}
	    \setlength\itemsep{1em}
		\item $0 \leq f(z_2)-f(z_1) - (\nabla f(z_1))^T(z_2-z_1) \leq \frac{1}{2}\|z_1-z_2\|_{\bL}^2$,
		\item $0\leq(\nabla f(z_1) - \nabla f(z_2))^T(z_1-z_2)\leq \|z_1-z_2\|_{\bL}^2$,
		\item $\frac{1}{2}\|\nabla f(z_1) - \nabla f(z_2)\|_{\bL^\dagger}^2
		\leq f(z_2) - f(z_1) + (\nabla f(z_1))^T(z_1 - z_2)$ and $\bPi_{\ker \bL} (\nabla f(z_1) - \nabla f(z_2)) = 0$,
		\item $\|\nabla f(z_1) - \nabla f(z_2)\|_{\bL^\dagger}^2
		\leq (\nabla f(z_1) - \nabla f(z_2))^T (z_1-z_2)$ and $\bPi_{\ker \bL} (\nabla f(z_1) - \nabla f(z_2)) = 0$.
	\end{enumerate}
\end{lemma}

Not all possible variations of matrices $\bM\preceq \bL$ should be considered for optimization. For example, if there exists a singular matrix $\bQ$ such that $\bM \preceq \bQ \preceq \bL$, then the function $f$ defined by $f(z) = \frac{1}{2}z^T\bQ z + v^T z$, where $v$ is not in the range of $\bQ$, would be an element of $S(\bM,\bL)$ without any critical point. Therefore this set $S(\bM,\bL)$ would not make sense as a set of objective functions, since we cannot solve Problem \ref{problem1} for it. The following Lemma characterizes when such cases can be avoided.

\begin{lemma}[Well-posed pairs $\bM, \bL$]
	\label{lem:4.1EigCond}
	Let $\bM,\bL \in \bbR^{d\times d}$ be symmetric matrices with $\bM \preceq \bL$. Then the following five statements are equivalent:
	\begin{enumerate}
		\item The matrices $\bM$ and $\bL$ have the same numbers of positive and negative, and no zero eigenvalues.
		\item Any symmetric matrix $\bQ \in \bbR^{n\times n}$ with $\bM \preceq \bQ \preceq \bL$ is non-singular.
		\item $\bL + \bM$ is non-singular and the spectral radius of $(\bL + \bM)^{-1}(\bL - \bM)$ is smaller than one.
		\item $\bM$ is non-singular and $\bM^{-1} \bL$ has only positive eigenvalues.
		\item $\bM$ and $\bL$ are non-singular and congruent, i.e. there exists a non-singular matrix $\bT \in \bbR^{d\times d}$ with $\bM = \bT^T\bL\bT$.
	\end{enumerate}
\end{lemma}

\begin{remark}
    In Lemma \ref{lem:4.1EigCond}, statement 1) serves the purpose of giving the reader a good intuition for the property under consideration. Statement 2) and 3) will be useful in later proofs. Note that in particular 2) prevents the counter-example we constructed in the motivation of this lemma. Statement 4) offers the most efficiently verifiable test of the considered property, by the fact that the verification whether a matrix has positive eigenvalues can be done by solving a Lyapunov equation.
\end{remark}

Because of the importance of this property we define a new notation for matrices $\bM,\bL$ fulfilling one and thus all conditions in Lemma \ref{lem:4.1EigCond}.

\begin{definition}[Loewner-congruence ordering on symmetric matrices]
	For symmetric matrices $\bM,\bL\in \bbR^{d\times d}$, we introduce the partial ordering
	\begin{align*}
	\bL \succeq_c \bM : \Leftrightarrow \begin{cases}
	\bL - \bM \text{ is positive semi-definite}\\
	\bL \text{ and } \bM \text{ are congruent}
	\end{cases}.
	\end{align*}
\end{definition}

Under the Loewner-congruence ordering, a critical point exists, is unique,
and a simple gradient method converges to the critical point, as stated in
the following results.

\begin{proposition}[A simple gradient method]
	\label{prop:4.1GradContraction}
	Let $\bL \succeq_c \bM$ be non-singular. Then for any convergence rate $\rho > \rho \left( (\bL + \bM)^{-1}(\bL - \bM) \right)$ there exists $r\in \bbR_{>0}$ such that
	\begin{align}
	z \mapsto z - 2(\bM + \bL)^{-1}\nabla f(z) \label{eq:StructuredGD}
	\end{align}
	is a contraction for all $f \in S(\bM,\bL)$ with contraction constant $\rho$ on the Banach space $(\bbR^d,\|\cdot\|_{\bP})$, where $\bP = (\bL + \bM)((\bL - \bM)^{\dagger} + r\bPi_{\ker (\bL-\bM)})(\bL + \bM)$.
\end{proposition}

\begin{remark}
    \label{rem:StructuredGD}
    For $\bM \preceq_c \bL$, the optimizer defined by $(\bA,\bB,\bC)$ with $\bA = \bC = \bI_d$ and $\bB = -2(\bL+\bM)^{-1}$ realizes the contraction in Proposition \ref{prop:4.1GradContraction}. As a consequence of the Banach fixed-point theorem, it converges faster than any convergence rate $\rho > \rho_\mathrm{grad}:= \rho \left( (\bL + \bM)^{-1}(\bL - \bM) \right)$ and converges monotonically in the norm $\|\cdot\|_{\bP}$ to the unique critical point.
    Finally, notice that in the case $\bL-\bM$ is singular, the infimal convergence rate may not be attained, since $r$ can go towards infinity if $\rho$ goes towards $\rho_\mathrm{grad}$. However, if $\bL-\bM$ is non-singular, then $r$ disappears from the equation and the constructed gradient method converges at the rate $\rho_\mathrm{grad}$.

\end{remark}

\begin{theorem}[Existence and uniqueness of critical points]
	\label{thm:ExMinCoCa}
	Let $\bM, \bL \in \bbR^{d\times d}$ be given symmetric matrices. Then the following three statements are equivalent:
	\begin{enumerate}
		\item The matrices $\bM$, $\bL$ are non-singular and satisfy $\bM \preceq_c \bL$.
		\item $S(\bM,\bL)$ is not empty and for all $f \in S(\bM,\bL)$ there exists at least one $z_f^* \in \bbR^d$ with $\nabla f(z_f^*) = 0$.
		\item $S(\bM,\bL)$ is not empty and for all $f \in S(\bM,\bL)$ there exists at most one $z_f^* \in \bbR^d$ with $\nabla f(z_f^*) = 0$.

	\end{enumerate}
\end{theorem}

\begin{remark}
	Theorem \ref{thm:ExMinCoCa} shows that if we aim to design algorithms that are convergent for the whole class $S(\bM,\bL)$, we must necessarily require $\bM \preceq_c \bL$, because otherwise there would be elements of $S(\bM,\bL)$ without critical points. Hence the introduced partial ordering plays a key role in our results.
	Note that it is no coincidence that in Theorem \ref{thm:ExMinCoCa} the existence of critical points for all functions in $S(\bM,\bL)$ and the uniqueness of critical points are two separate, equivalent statements. Similar to solutions of linear equation systems, here a solution for the equation $\nabla f(z) = 0$ exists for all $f \in S(\bM,\bL)$ if and only if the solution is unique for all $f \in S(\bM,\bL)$.
\end{remark}


TITLE: Examples of rings satisfying some of the following properties
QUESTION [7 upvotes]: I am trying to find at least one example of rings for every combination of the following properties
Any ideas for the remaining cases?
Thank you

REPLY [1 votes]: I see three rows of ???? that you are apparently asking about.
For the first one occurring, you could use the ideal $(x)/(x^3)$ in $F_2[x]/(x^3)$. It is properly inside a local ring, so it doesn't have any idempotents aside from $0$, so no identity is there. It's not a domain since $x^3=0$.
For the second one, the integer quaternions (the subring of the quaternions generated by $i,j,k$ and the integers) (or also the Hurwitz quaternions) will do. They're a domain with identity that isn't commutative.
For the third one, you can take any nontrivial ideal of the integer quaternions.


TITLE: Calculate the variance of $X \text{~} Pascal(m,p)$
QUESTION [0 upvotes]: Calculate the variance of $X \text{~} Pascal(m,p)$.
I know that $\operatorname{Var}(X) = E[X^2] - [EX]^2$, and that $X$ can be thought of as $m$ independent geometric random variables.  This leads to $[EX]^2 = (\frac{m}{p})^2$.  But I'm confused for $E[X^2]$.
By the multinomial theorem:
$E[X^2] = mE[X_i^2]$ for an arbitrary $i$ in $\{1, ..., m\}$  $+ \binom{m}{2}4E[X_jX_k]$ for arbitrary $j$ and $k$ in $\{1, ..., m\}$.
But can I replace $X_jX_k$ with any $X_q^2$ and evaluate normally?

REPLY [1 votes]: To answer your last question, note that the $X_j$ and $X_k$ are independent so $E[X_j X_k] = E[X_j] E[X_k]$.
You can continue with your approach (computing $E[X^2]$) and you will get the right answer. However, it may be slightly easier to work with the variance directly. Since the $X_i$ are independent, we have $\operatorname{Var}(X) = \operatorname{Var}(X_1)+\cdots+\operatorname{Var}(X_m)$, and you can use the variance of your geometric distribution. Note that the above variance formula is a consequence of your "multinomial theorem."


TITLE: What is this sampling method called?
QUESTION [0 upvotes]: I was trying to implemented normal distribution sample and I came upon this answer (java implementation). It was very simple to understand. Essentially, we start with a list of probabilities (decreasing with sum = 1) and then create a increasing list out of it such that value of last element in the list is 1. Then I read about normal distribution and their probability distribution function (PDF). So, I thought I can modify the code so that it uses PDF of normal distribution to generate list of probabilities. The following is my Sage implementation (run the code):
import random

sigma = 8/sqrt(2 * pi)  # standard deviation
variance = sigma^2      # variance
mu = 0                  # mean (or center of distribution)

# probability density of the normal distribution
def pdf(x, sigma, variance, mu):
    return N((1 / sqrt(2 * variance * pi)) * e^(-(x - mu)^2 / (2 * variance)))

# note that sum of probabilities that this function generates is equal to 1/2
# because we are ignoring probabilities of negative numbers
def create_positive_domain_probability():
    probability_list = [0] * ceil(variance)

    for x in range(0, variance):
        probability_list[x] = pdf(x, sigma, variance, mu)

    return probability_list

# normalize probability list such that sum of probability would equal to 1
def normalize_probabilities(probability):
    s = sum(probability)
    return map((lambda x: x / s), probability)

# use output of this function for sampling
def make_increasing_distribution(probability_list):
    size = len(probability_list)
    distribution = [0] * size
    sum = 0

    for i in range(size):
        sum = sum + probability_list[i]
        distribution[i] = sum

    return distribution

# sample from (increasing) distribution
def sample():
    rand = random.uniform(0, 1)
    index = 0

    while(rand > distribution[index]):
        index = index + 1

    return index

probability_list = create_positive_domain_probability()
probability_list = normalize_probabilities(probability_list)
print "1. bar chart of probability list"
bar_chart(probability_list).show() # bar chart of probability list

distribution = make_increasing_distribution(probability_list)
print "2. bar chart of increasing distribution list"
bar_chart(distribution).show()     # bar chart of increasing distribution list

arr = [0] * len(probability_list)
for i in range(10000):    # count of samples
    s = sample()
    arr[s] = arr[s] + 1

print "3. bar chart (histogram) of samples generated"
bar_chart(arr).show()     # bar chart (histogram) of samples generated

My question is:

what is this method called? is it truly rejection sampling?
what is the increasing sequence that is generated from list of probabilities called?

REPLY [1 votes]: Your program is definitely not doing rejection sampling, because it never rejects any output of random.uniform.
As noted in a comment, the array distribution is a cumulative distribution
function, abbreviated CDF.
Assuming it works, the program performs inversion sampling on this CDF.
You might want to consider simply setting the last value of distribution
to $1$ rather than counting on the sum to add up exactly.
If the sum does not add up exactly (due to roundoff error or some other effect), and happens to be slightly less than $1,$ there is a very, very small chance that random.uniform will return a number greater than the last entry in distribution. It's not clear what happens then; it appears sample() will attempt the read values beyond the end of the declared length of the array.
You're not really using a normal distribution here.
For one thing, all your sampled values are integers, whereas the normal
distribution is continuous.
You're not even getting a histogram of the normal distribution, since you truncate it just below its mean and somewhere further above.
You're just getting some unnamed discrete distribution that is very heavily biased toward low numbers.
Notice what happens if you simply double the value of sigma at the start
of the program. The random values range from $0$ through $40$ inclusive
($41$ values) instead of $0$ through $10$ ($11$ values)
and about $90\%$ of the samples are in the first one-quarter of the range
($0$ to $10$). Values in the second half of the range (above $20$)
have almost no chance of occurring.
I wonder if that is what you wanted.


TITLE: Third-order autonomous ODE - if $y' = p$, $y'' = p'p$, what is $y''' = $?
QUESTION [0 upvotes]: This question has been in my mind for quite some time.
Suppose we have an autonomous ODE which we can solve by substituting:
$$y' = p(y)$$. From there on, we know that $$y'' = p'_y \cdot y' $$
which is $$y'' = p'p$$
But what would the third derivative be? I tried a calculation and I got that
$$y''' = p''p^2 + p'p^2$$ but I'm not sure if I'm right or wrong!

REPLY [1 votes]: We have
$$y'''=(y'')'=\left(p'p\right)'=p\left(p''y'\right)+p'\left(p'y'\right)=p''p^2+p'^2p$$


TITLE: Prove that if $AC^T = |A|I \implies \det C = (\det A)^{n-1}$
QUESTION [2 upvotes]: Prove that if $AC^T = |A|I \implies \det C = (\det A)^{n-1}$
Ran into trouble with a proof for linear algebra. $C$ is the cofactor matrix of $A \in \mathbb{R}^{n\times n}$, and I'm not sure how to even approach this problem. Any tips for starting? Not the entire proof, please.

REPLY [2 votes]: Hint:$$\det(AC^T)=\det(A)\det(C^T) =\det(A)\det(C)$$
What is $\det(\alpha I)$, for any scalar $\alpha$?
If $\det(A)=0$, note that $AC^T=CA^T =0_{n\times n}$. What does this say about the nullspace of $C$?


TITLE: effect of vertical collision on kinetic friction and subsequent change in horizontal velocity
QUESTION [1 upvotes]: Suppose somehow a block of mass $m$ is moving on ground, and the coefficient of kinetic friction between the block and the block is $\mu_k$. If I drop a tennis ball(of same mass) on it from a height, say $h$ and it bounces back, coefficient of restitution being $e$ for the collision, I need to work out what effect would this have on its velocity, which was say at the time of collision equal to $v(t_0)=v$. The speed may or may not be changing due to everything else, but the change the collision would cause is what I want to find.
My view is that, at the time of collision, when the ball and the block are in contact, at that time ($t_0$), the normal reaction to the block will become $2mg$ instead of $mg$, which would change the kinetic friction at that time.
We say there is no change in momentum as impulse is due to opposite and equal magnitude forces. We say nothing about time. But here that time of contact is needed so that change in friction somehow changes the velocity.
Otherwise there is no extra change and so horizontal velocity shouldn't be affected by the collision. Does the velocity change due to collision?
Thanks in advance.[NO ROTATIONAL EFFECTS]

REPLY [1 votes]: 1st: you must take care that friction is a impulsive force and momentum never remains conserved when $f$ changes.
2nd:the normal reaction will become $mg+I/\Delta t$ where $I$ is the impulse imparted  on the ball ie:$mv(1+e);v=\sqrt{2gh}\ and\ \Delta t .$is the small time of impact.So, $$N=m(g+v(1+e)/\Delta t)$$
As,the momentum and energy both are not conserved , it is difficult to comment on the further motion of plank.

REPLY [1 votes]: The kinetic energy the ball has when it hits the block is $mgh$. We have the relation:
$$
\frac{mv^2}{2}=mgh
$$
This can be rewritten:
$$
m^2v^2=2m^2gh
$$
Which means the downward momentum of the ball is:
$$
p=mv=m\sqrt{2gh}
$$
The block will excert a force on the ball to cancel this momemntum and give it a momentum of $e\cdot p$ in the opposite direction:
$$
F\Delta t=m\sqrt{2gh}(1+e)
$$
This will require a corresponding increase in the normal force on the block and thereby in the friction force:
$$
\Delta F_f=\mu\Delta N=\mu F
$$
The change in moementum of the block will be:
$$
\Delta p_B=\Delta F_f\Delta t=\mu F\Delta t=\mu m\sqrt{2gh}(1+e)
$$
The change (decrease) in speed of the block will thus be $\mu\sqrt{2gh}(1+e)$.
I have not included the extra force due to the weight of the ball or due to giving it a horizontal velocity component or spin, but these are likely to be small.


TITLE: Intuition of normal bundle to a manifold
QUESTION [1 upvotes]: So i recently learned about fibre bundles and tangent bundles in particular. While the definition of tangent bundles seems quite intuitive, i really struggle to understand any definition of the normalbundles to a submanifold. I looked everywhere and not only did the definitions all somehow seem to be different, i just havent managed to understand either one of them to be honest.
Can someone please help me in understanding the quotientbundle definition of the normal bundle and maybe provide some geometric intuition?
Thank you very much!

REPLY [2 votes]: I don't think there is really that much good intuition in the non-Riemannian setting.
Imagine the circle $S^1$ sitting in $\mathbb{R}^2$. At every point, $z=e^{i\theta}\in S^1$, you have the tangent plane of $z$ in $\mathbb{R}^2$ spanned by your two favourite cardinal directions. The tangent space of $z$ in $S^1$ is just one-dimensional, though, and these two spaces are naturally comparable. Indeed, if you draw the situation, it's clear that the line $z+tiz$ should realise the tangent space. Now, $iz\mathbb{R}$ has an orthogonal complement, namely $z\mathbb{R}$. In this way, the tanget plane of $z$ in $\mathbb{R}^2$ naturally decomposes into the direct sum of the tangents of $z$ in $S^1$ and their orthogonal complement. It's this orthogonal complement that we refer to as the normal bundle.
Formally, you let $z=e^{i\theta}$ be a local chart, and compute the differential as a smooth map $(a,b)\to \mathbb{R}^2$ with $b-a<2\pi$. Then you'll find, of course that the differential is given by $\theta\mapsto ie^{i\theta}$, which is exactly the above identification of the tanget space of $z$.
You could do the same analysis in any dimension: If $M\subseteq \mathbb{R}^n$ is an embedded manifold, you can simply identify the tangent space of $M$ at any point, and the normal bundle is then the bundle of orthogonal complements.


text:
\begin{definition}[Definition:Degree of Polynomial/Polynomial Form]
Let $f = a_1 \mathbf X^{k_1} + \cdots + a_r \mathbf X^{k_r}$ be a [[Definition:Polynomial Form|polynomial]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\family {X_j: j \in J}$ for some [[Definition:Multiindex|multiindices]] $k_1, \ldots, k_r$.
Let $f$ '''not''' be the [[Definition:Null Polynomial/Polynomial Form|null polynomial]].
Let $k = \family {k_j}_{j \mathop \in J}$ be a [[Definition:Multiindex|multiindex]].
Let $\ds \size k = \sum_{j \mathop \in J} k_j \ge 0$ be the [[Definition:Degree of Element of Free Commutative Monoid|degree of the monomial]] $\mathbf X^k$.
The '''degree of $f$''' is the [[Definition:Supremum of Set|supremum]]:
:$\ds \map \deg f = \max \set {\size {k_r}: i = 1, \ldots, r}$
\end{definition}

config: wiki


TITLE: Controling mixed derivatives
QUESTION [2 upvotes]: This is a cross-post from Math.SE since the question got nothing (but upvotes) even after offering a decent bounty. If it is too trivial or in other ways not suited for this site, please let me know and I'll delete it.
In Muscalu, Schlag - Classical and Multilinear Harmonic Analysis (Cambridge Universitv Press 2013), page 299 there is a rather odd estimate for wich I cannot find any justification:
Functions used:
$$\def\supp{\mathop{\rm supp}}\begin{align*}
    \psi & \in C_c^\infty(\mathbb R) \\
    \supp \psi & \subset [-2,2] \\
    \psi|_{[-1,1]} & \equiv 1 \\
    \chi & \in C_c^\infty(\mathbb R) \\
    \supp \chi & \subset [-1,1] \\
    \chi(0) & = 1 \\
    \psi(\mathbb R) = \chi(\mathbb R) & = [0,1]\\
    z & \in\mathbb C\\
    \tau & \in\mathbb R
\end{align*}$$
The claim is that
$$\begin{align*}
\int_0^\infty \left| \frac{\mathrm d^N}{\mathrm dt^N} (t^z (1-\psi(t\tau)) \chi(t)) \right| \mathrm dt & \le C_N \int_0^\infty \left| \prod_{k=0}^{N-1} (z-k) t^{z - N} (1-\psi(t\tau)) \chi(t) \right| \\
& \qquad \qquad + \left| t^{z} (-\psi^{(N)}(t\tau) \tau^N) \chi(t) \right| \\
& \qquad \qquad + \left| t^{z} (1-\psi(t\tau)) \chi^{(N)}(t) \right| \mathrm dt \\
& = C_N \int_0^\infty \left| \prod_{k=0}^{N-1} (z-k) \right| t^{\Re z - N} (1-\psi(t\tau))\chi(t) \\
& \qquad \qquad + t^{\Re z} |\psi^{(N)}(t\tau)| \tau^N \chi(t) \\
& \qquad \qquad + t^{\Re z} (1-\psi(t\tau)) |\chi^{(N)}(t)| \mathrm dt
\end{align*}$$
So basically that we can control
$$\int |\partial^N (uvw)| \le C_N \int |\partial^N u vw| + |u \partial^N vw| + |uv\partial^N w|$$
Wich is certainly not true in general (chose $u=v=w=x$ and $N=3$, for example)
So how can we justify that estimate in this special case?

REPLY [1 votes]: As stated, the inequality indeed fails. But the final estimate (11.27) in the book is essentially correct. From the product rule the derivative
$$ (\frac{d}{dt})^n (t^z (1 - \psi(t\tau)) \chi(t) $$
contains three types of terms:

A term where all derivatives fall on the weight $t^z$. This gives the first term in your expansion.
Terms, excluding the one above, where no derivatives hit $\psi$.
Terms where some number of derivatives hit $\psi$.

The first term you estimate exactly as the authors do, which gives you
$$ \leq \int_{1/\tau}^\infty \prod (z-k) t^{\Re z - N}~\mathrm{d}t = \prod(z-k)\cdot \frac{1}{\Re z - N + 1} \cdot \tau^{N -1 -\Re z}  $$
assuming, as they did, that $\Re z - N < -1$.
For the second types of terms, note that since $\chi$ is support on $[-1,1]$ this means that $|t^z| < 1$ on its support. So every single term of type two is estimated by
$$ \leq \sup_{k \in \{1, \ldots, N\}} \sup_{t \in [0,1]} | \chi^{(k)}(t) | $$
in the context of the estimates we will take this to just be a constant depending on $N$. So all terms of type two together is estimated by
$$ \leq C_N$$
We are left with terms of type three. Since a derivative hits $\psi$, the terms of type three are supported in the region
$$ t \in [1/\tau, 2/\tau ] $$
which has width $1/\tau$. In this region $t \approx 1/\tau$. So a typical term can be controlled by
$$ \leq C\cdot \frac{1}{\tau} \cdot \prod(z-k) \cdot \frac{1}{\tau}^{\Re z - N_1} \cdot \tau^{1 + N_2} \cdot |\psi^{(1+N_2)}|_\infty \cdot |\chi^{(N_3)}|_\infty $$
where $1 + N_1 + N_2 + N_3 = N$. We can simplify a little bit putting the $L^\infty$ norms of derivative of $\psi$ and $\chi$ into a constant
$$ \leq C_N \prod_{k = 0}^{N_1 - 1}(z - k) \cdot \tau^{N_1 + N_2 - \Re z} $$
since we are interested in the $\tau \gg 1$ case
$$ \leq C_N \prod_{k = 0}^{N_1 - 1}(z-k)  \cdot \tau^{N - 1 - \Re z} $$
So up to some factors of products of the form $z(z-1)\ldots (z-N_1)$ the estimate (11.27) is correct. Since I haven't read the rest of the chapters to see why the authors chose to keep track of the $z$ dependence, I don't know how much this changes the exposition.


TITLE: Show that this matrix product is bounded
QUESTION [0 upvotes]: Suppose  we have a symmetric real matrix $\pmb M$ satisfying:
$$\underset{\pmb\alpha\in\mathbb{R}^p:||\alpha||=1}{\min}\;\pmb \alpha\pmb M\pmb \alpha^{\top}\geqslant k>0.$$
Then, I am trying to show that the norm:
$$\underset{\pmb z\in\mathbb{R}^p}{\max}\;\left|\left|\left(\pmb M+\pmb z^{}\pmb z^{\top}\right)^{-1}\pmb z^{\top}\right|\right|^2\leqslant\; ?$$
is bounded. I can see how intuitively that would be true, but I fail to build a connection between (the inverse of) $k$ and an upper bound for the second expression.

REPLY [1 votes]: Using the Sherman–Morrison–Woodbury formula we arrive at
$$(M + zz^\top)^{-1} z = \frac{M^{-1} z}{1+z^\top M^{-1} z}.$$
From here it is easy to conclude by using
$$z^\top M^{-1} M^{-1}z \le \|M^{-1}\| z^\top M^{-1} z.$$


TITLE: Two-Torsion of Elliptic Curves and Their Twists
QUESTION [2 upvotes]: Let $E/\mathbb{Q}$ be an elliptic curve such that $E(\mathbb{Q})[2]=\{\mathcal{O}\}$. Does there exist a quadratic twist $E^{(d)}$ of $E$ with a $\mathbb{Q}$-rational two torsion point? I'm particularly interested in the case where $E$ has complex multiplication.

REPLY [2 votes]: If $E : y^2 = x^3 + ax + b$, then the two torsion points are exactly of the form $(r, 0)$ where $r$ is a root of $x^3 + ax + b$. So your question becomes, if $x^3 + ax + b$ has no rational roots, is there an integer $d$ so that $x^3 + ad^2 x + bd^3$ has rational roots? But the roots of the right hand side are just $dr$, and so it also has no $\mathbb{Q}$-rational two torsion.


TITLE: Incorrect comparison test on series
QUESTION [0 upvotes]: Test whether the series $\sum_{n=1}^\infty\frac{1+\sin(n)}{2^n}$ is converges or diverges. I used the limit comparison method in order to test its convergence. I chose $a_n = \frac{1+\sin(n)}{2^n}$ and $b_n = \frac{1}{2^n}$.
$$
\lim_{n \rightarrow \infty}\frac{\frac{1+\sin(n)}{2^n}}{\frac{1}{2^n}} = \lim_{n \rightarrow \infty}({1+\sin(n)}) = \infty
$$
However, WolframAlpha suggests that the series is in fact convergent. If this is the case, then is the way I set up the limit comparison test incorrect or is there another test that proves this series is convergent? Any help would be appreciated!

REPLY [2 votes]: First of all, $\lim_{n \to \infty} 1 + \sin(n)$ is not equal to $\infty$. It doesn't exist, since $\sin(n)$ oscillates between $-1$ and $1$ as $n$ tends to infinity.
Try instead to compare $a_n =\sum_{i=1}^n \frac{1+\sin(n)}{2^n}$ with $b_n = \sum_{i=1}^n\frac{1+1}{2^n} = \sum_{i=1}^n\frac{2}{2^n}$. Note that since $\sin(n) \leq 1$ for all $n$, we have $a_n \leq b_n$ for all $n$. Thus,
$$\lim_{n \to \infty} \sum_{i=1}^n a_n \leq \lim_{n \to \infty} \sum_{i=1}^nb_n$$
Since $\lim_{n\to \infty} \sum_{i=1}^n b_n = \lim_{n \to \infty}  \sum_{i=1}^n\frac{2}{2^n} = 2$, $\lim_{n \to \infty} \sum_{i=1}^n a_n \leq 2$ must converge.
Edit: As pointed out by zwim, you also have to make sure that $\sum a_n$ doesn't tend to $-\infty$. To do this, note that since $-1 \leq \sin(n)$ for all $n$, we have
$$ 0 = \sum_{i=1}^n \frac{1-1}{2} \leq \sum_{i=1}^n \frac{1+\sin(n)}{2} = \sum_{i=1}^n a_n .$$
Since $\sum_{i=1}^n a_n$ is bounded from below and from above, it must converge.


TITLE: Under what conditions do all singletons belong to sigma algebra?
QUESTION [0 upvotes]: Let's suppose that (A, F) is a measurable space (A underlying set, F the sigma algebra), and that F arises from some topology T.
I would like a theorem of the form:
All singletons belong to F if and only if (A, F, T) has properties P1,P2,...
where P1,P2,... may be measure theoretic or topological conditions on the triple (A, T, F).
Thanks a lot for your help.

REPLY [0 votes]: A necessary condition is that the underlying topological space is $T_0$. This is sufficient if the space has a countable basis.
A sufficient condition is that the underlying space is $T_1$.
If the $\sigma$-algebra is generated by an arbitrary family $\mathcal{F}$, a necessary and sufficient condition for $\{x\}$ to be measurable is that there is a countable subset $\mathcal{C}\subseteq\mathcal{F}$ such that for all $y\neq x$ there is $F\in\mathcal{C}$ such that $x\in F$ and $y\notin F$ or $y\in F$ and $x\notin F$.


TITLE: How to prove the existence and uniqueness of the solution of a second order linear ODE?
QUESTION [7 upvotes]: Considering homogeneous, linear, second order differential equation
$$y'' + p(x)y' + q(x)y = r(x), a<x<b$$
$p, q, r$ are given. We already know $y(a) = A, y(b) = B$ and $q(x) < o$. How to prove that this equation has unique solution in $[a,b]$ if it has a solution?

REPLY [5 votes]: Suppose $y_1(x)$ and $y_2(x)$ both satisfy
$y'' + p(x) y' + q(x)y = r(x) \tag 1$
with
$y_1(a) = y_2(a) = A, \; y_1(b) = y_2(b) = B; \tag 2$
then we have
$y_1'' + p(x) y_1' + q(x)y_1 = r(x) \tag 3$
and
$y_2'' + p(x) y_2' + q(x)y_2 = r(x), \tag 4$
and, subtracting,
$(y_1 - y_2)'' + p(x) (y_1 - y_2)' + q(x)(y_1 - y_2) = 0, \tag 5$
and
$(y_1 - y_2)(a) = y_1(a) - y_2(a) = 0 = y_1(b) - y_2(b) = (y_1 - y_2)(b); \tag 6$
setting
$z(x) = y_1(x) - y_2(x), \tag 7$
we have
$z'' + p(x) z' + q(x) z = 0, \; z(a) = z(b) = 0; \tag 8$
now if $z'(a) = 0$, then by uniqueness of solutions we must have $z(x) = 0$, $x \in [a, b]$, since $z(a) = 0$, and thus $y_1(x) = y_2(x)$, and we are done; so we assume $z'(a) \ne 0$; in fact, since (8) is linear, we may take $z'(a) > 0$, and thus, by the continuity of $z'(x)$, there is some $\delta > 0$ such that
$z'(x) >  0, \; x \in [a, a + \delta); \tag 9$
we then have
$z(x) = z(x) - z(a) = \displaystyle \int_0^x z'(s) \; ds > 0, x \in (a, a + \delta); \tag{10}$
since $z(x) > 0$ somewhere on $[a, b]$, it must have a positive absolute maximum $x_M \in (a, b)$; at such a point
$z'(x_M) = 0, \tag{11}$
and thus by (8),
$z''(x_M) = -q(x_M)z(x_M) > 0 \tag{12}$
by virtue of the facts that $z(x_M) > 0$, $q(x_M) < 0$; but (12) is impossible at a maximum of $z(x)$, where we must have $z''(x_M) \le 0$; this contradiction forces $z'(a) = 0$ and thus also
$z(x) = 0, \; x \in [a, b], \tag{13}$
and hence
$y_1(x) - y_2(x) = z(x) = 0, \; x \in [a, b]; \tag{14}$
we conclude the equation (1) with boundary conditions (2) has at most one solution on $[a, b]$.


text:
\section{Numerical Validation}\label{sec:numeric}

To support our theory we aim to conclude by showing the results of the following numerical experiments. Basically we run the boxed-constrained basis pursuit \eqref{Pbin} and the box-constrained least squares \eqref{LSBin} for biased circulant matrices $\Phi(b)$ and Toeplitz matrices $T(b)$ for either Gaussian or Rademacher input vectors. For all experiments we choose the ambient dimension to be $N=500$. For each $s,M\in \left\{5i:i\in [100]\right\}$ we choose a random binary vector $x_0\in\R^N$ with $s$-non-zero elements. These $s$-non-zero positions of $x_0$ are chosen uniformly randomly from $1$ to $500$ without repetition (Matlab method \textit{randperm}). Then we constructed the different structured measurement matrices and run boxed constrained linear least squares (Matlab method \textit{lsqlin}) as well as the linear program \textit{linprog} with box constraints to obtain reconstructions $x_{\text{LS}}$ and $x_{\text{PBin}}$ of $x_0$. Finally we computed the $\ell_2$-error $\|x_{\text{LS}}-x_0\|_2$ and $\|x_{\text{PBin}}-x_0\|_2$ and repeated the computation for each combination of $s$ and $M$ $100$-times.

The measurement matrices were constructed as follows. For the Rademacher Toeplitz matrix we choose  $c=[c_1,\dots, c_N]\in \R^N$ and $r=[r_1,\dots,r_{N-1}]\in \R^{N-1}$ such that $c_i,r_j$ are either $0$ or $1$ with equal probability for $i\in[N]$, $j\in [N-1]$. Then we define $A=\text{toeplitz}(c,[c_1\; r])$, thus $c$ is the first column of $A$ and $[c_1\; r]$ the first row. Finally we choose a random subset  $\Theta\subset[N]$ of size $\abs{\Theta}=M$ by randomly permutating $[N]$ and choosing the first $M$ numbers (matlab method \textit{randperm}$(N,M)$). The measurement matrix $\Phi$ then consist of the columns $A$ which correspond to the subset $\Theta$.

 The Gaussian Toeplitz matrix is constructed analogously but with $c\in \R^N$ and $r\in \R^{N-1}$ shifted Gaussian random vectors, i.e, $c=g_c+\one$ and $r=g_r+\one$, where $g_c\in \R^N$ and $g_r\in\R^{N-1}$ are Gaussian vectors.

To construct the partial circulant matrices we choose $c\in \R^N$ as in the Toeplitz case either as shifted Rademacher or Gaussian vector. Then we flipped $c$ and shifted it circularly by one position to obtain the row vector $r\in \R^N$ ($r=$\textit{circshift(fliplr}($c$),$1,2$)). Then we define $A=\text{toeplitz}(c,r)$ and the measurement matrix $\Phi$ as in the Toeplitz case.

\begin{figure}[H]
	\begin{center}
		\includegraphics[trim=40 200 50 200,clip, width=0.45\textwidth]{BoxedPartToepBernPbinNeu.pdf}
		\includegraphics[trim=40 200 50 200,clip, width=0.45\textwidth]{BoxedPartToepBernLSNeu.pdf}
		\caption{Reconstruction from biased Rademacher Toeplitz measurements via \eqref{Pbin} (left) and \eqref{LSBin} (right). The experiment yielding this figure is explained above.}
		\label{ToeplitzBernoulli}
	\end{center}
\end{figure}
\begin{figure}[H]
	\begin{center}
		\includegraphics[trim=40 200 50 200,clip, width=0.45\textwidth]{BoxedPartToepGausPbinNeu.pdf}
		\includegraphics[trim=40 200 50 200,clip, width=0.45\textwidth]{BoxedPartToepGausLSNeu.pdf}
		\caption{Reconstruction from biased Gaussian Toeplitz measurements via \eqref{Pbin} (left) and \eqref{LSBin} (right). The experiment yielding this figure is explained above.}
		\label{ToeplitzGaussian}
	\end{center}
\end{figure}
\begin{figure}[H]
	\begin{center}
		\includegraphics[trim=40 200 50 200,clip, width=0.45\textwidth]{BoxedPartCircBernPbinNeu.pdf}
		\includegraphics[trim=40 200 50 200,clip, width=0.45\textwidth]{BoxedPartCircBernLSNeu.pdf}
		\caption{Reconstruction from biased Rademacher circulant measurements via \eqref{Pbin} (left) and \eqref{LSBin} (right). The experiment yielding this figure is explained above.}
		\label{CirculantBernoulli}
	\end{center}
\end{figure}
\begin{figure}[H]
	\begin{center}
		\includegraphics[trim=40 200 50 200,clip, width=0.45\textwidth]{BoxedPartCircGausPbinNeu.pdf}
		\includegraphics[trim=40 200 50 200,clip, width=0.45\textwidth]{BoxedPartCircGausLSNeu.pdf}
		\caption{Reconstruction from biased Gaussian circulant measurements via \eqref{Pbin} (left) and \eqref{LSBin} (right). The experiment yielding this figure is explained above.}
		\label{CirculantGaussian}
	\end{center}
\end{figure}


TITLE: Correct vector calculus divergence identity
QUESTION [0 upvotes]: I came across this vector calculus identity (in this wikipedia article, but also in other sources):
$$
\vec{\nabla} \cdot (\vec{B}\times \vec{C}) = \vec{C}\cdot(\vec{\nabla}\times \vec{B})-\vec{B}\cdot(\vec{\nabla}\times \vec{C})
$$
where $\vec{\nabla}=(\partial/\partial x,\partial/\partial y,\partial/\partial z)$. However, based on equation (2.5) of the Feynman Lectures on Physics, vol. II, which states that
$$\vec{A} \cdot (\vec{B}\times \vec{C}) = (\vec{A}\times \vec{B})\cdot \vec{C},$$
the above identity should only contain the first term on the right-hand side.
I'm getting confused. Does anyone know which one is correct and what's the difference between the two?

REPLY [2 votes]: The divergence operator looks like a vector but it's still an operator and contains derivatives. Most importantly you have to use the product rule, since the derivative acts on both $\mathbf{A}$ and $\mathbf{B}$. In general for these types of problems working with index notation and Einstein Summation is the best choice. For example
\begin{align*}
\nabla\cdot(\mathbf{A}\times\mathbf{B}) = \partial_i \varepsilon_{ijk} a_j b_k &= \varepsilon_{ijk} \left(\frac{\partial a_j}{\partial x_i} b_k + a_j \frac{\partial b_k}{\partial x_i}\right) \\
&= \varepsilon_{kij} b_k \partial_i a_j - \varepsilon_{jik} a_j \partial_i b_k \\
&= \mathbf{B}\cdot\left(\nabla\times\mathbf{A}\right) - \mathbf{A}\cdot\left(\nabla\times\mathbf{B}\right)
\end{align*}


TITLE: Sequence equation. Looking for a value at n = infinity. Basic stuff.
QUESTION [0 upvotes]: I don't know how to do sigma equations, but my question is:
1/1 + 1/2 + 1/3 + 1/4 + 1/5.. .. + 1/infinity = ?
can this be calculated? is the answer infinity or does it stop at a value?
I tried to create the sequence in c# (very simple program).
        double x = 0;
        double n = 1;
        while (true)
        {
            x = x + 1 / n;
            Console.WriteLine(x);
            i++;
        }

cmd is capable of calculating 10k sequences per second. I can see that the value is slowing down. At stage 10.000.000 the value is about 16,7.
How much is the value at n=infinity?

REPLY [2 votes]: Let's call $S_n = \sum_{i=1}^{n} \frac{1}{i}$. $S_{2n} = \sum_{i=1}^{2n} \frac{1}{i}$.
If $S_n$ converges, then $S_{2n}$ would converge to the same limit.
However, $S_{2n}-S_{n} = \sum_{i=n+1}^{2n} \frac{1}{i} \geq \sum_{i=n+1}^{2n} \frac{1}{2n} = n*\frac{1}{2n} = \frac{1}{2}$
As a consequence, $S_{n}$ does not converge and $\lim_{n \rightarrow \infty} S_n = \infty$.


TITLE: Solve this inequality: $\sin x<1$
QUESTION [0 upvotes]: it may be a simple question, but as simple as it is  I want to know how to solve it:
$$\sin {x}<1$$
What is the solution of this inequality? Thank you!

REPLY [4 votes]: The solution to
$$\sin x = 1$$
is $\left\{\ldots,\frac{-11\pi}{2},\frac{-7\pi}{2},\frac{-3\pi}{2},\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},\ldots\right\}$ (why?).
So the solution set to
$$\sin x < 1$$
should be
$$\mathbb{R}\setminus\left\{\ldots,\frac{-11\pi}{2},\frac{-7\pi}{2},\frac{-3\pi}{2},\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},\ldots\right\}$$
since, for real $x$, we have $\sin x < 1$ if just $\sin x\ne 1$.


TITLE: Prove that if for all $g \in G$ that $gNg^{-1} \subseteq N$, then we must have for all $g\in G$ that $gNg^{-1}=N$
QUESTION [4 upvotes]: I feel kind of shaky about this proof so I'd really appreciate some input. I know this is equivalent to some other forms (which I'll prove afterward so please don't spoil it). Here's the statement again:

Prove that if for all $g \in G$ that $gNg^{-1} \subseteq N$, then we must have for all $g\in G$ that $gNg^{-1}=N$.

Attempt: Assume $gNg^{-1}\subsetneq N$. Then $\exists gng^{-1}, gn'g^{-1} \in gNg^{-1}$ with $n\neq n'$ such that
$$gng^{-1}=gn'g^{-1}$$
but that implies $n=n'$, thus no such elements exist and $gNg^{-1}=N$.
Is this correct? I know I should've instead tried to show if $n \in N$ then $n\in gNg^{-1}$ but this seemed to accomplish the same.

REPLY [6 votes]: Let $g \in G$. We know that $gNg^{-1} \subseteq N$. We want to show that $$gNg^{-1} \supseteq N.$$
Let's just multiply this containment on the left by $g^{-1}$ and on the right by $g$ to get that we want to show $$N \supseteq g^{-1}Ng.$$
So, we just need to show that $g^{-1}Ng \subseteq N$. But we already know that for all $x \in G$ that $xNx^{-1} \subseteq N$. Hence let $x = \ldots$? (Once you know what to pick for $x$, you will have finished a proof.)

The way your attempt goes, I would instead say this: Assume $gNg^{-1} \subsetneq N$.  Thus there exists $n_0 \in N$ such that for all $n \in N$, $gng^{-1} \neq n_0$. But this can't actually be true because all we need to do is consider $n = g^{-1}n_0g$.

As can be seen in the comments below, I interpreted the 'for all' quantifier a little differently than what the OP originally wrote. Indeed, the statement as it was written does not hold for all infinite groups. Here is a counterexample:
Let $G = \mathbb{Z}[1/2] \rtimes \mathbb{Z}$. Let $(1,0)$ be a normal generator of $\mathbb{Z}[1/2]$, and let $(0,1)$ generate the $\mathbb{Z}$ on the right side of the semidirect product. Let $N = \langle (1,0) \rangle$. Notice that $N$ is not a normal subgroup of $G$. Indeed,  if we conjugate $N$ by $(0,1)$ we get $\langle (2,0) \rangle$, which is a proper subset of $N$. But if we conjugate $N$ by $(0,-1)$, then we get $\langle (1/2, 0) \rangle$, which is not a subset of $N$.

REPLY [2 votes]: The statement is true for all groups, not just finite ones.  What you've written here is not well explained, it looks like you're showing that conjugation is injective and so it's a bijection.  But this style proof will only work when $N$ is finite.
Instead, what happens if you conjugate $N$ by $g^{-1}$ instead of $g$?


TITLE: Ideal as kernel of a homomorphism in Gaußian integers
QUESTION [1 upvotes]: Consider the ring $\mathbb{Z}[i]$ of Gaußian integers. The principal ideal $(1+i)$ is maximal ideal in this ring. Since ideals are kernels of some homomorphisms, I would like to see a homomorphism from $\mathbb{Z}[i]$ to some ring whose kernel is $(1+i)$.
Of course, it is kernel of the natural homomorphism

$$f:\;\mathbb{Z}[i]\rightarrow \mathbb{Z}[i]\big/(1+i).$$

However, I tried to define some nice map $f$ whose kernel will be $(1+i)$. I tried this one: $f\colon \mathbb{Z}[i]\rightarrow \mathbb{Z}$, $m+ni\mapsto m-n$, but this $f$ is not a homomorphism. Can you help me to modify (or to define another homomorphism) whose kernel is the ideal $(1+i)$.

REPLY [0 votes]: You can also think of this in the following way:
Look at the composition
$\mathbb Z[i] \to \mathbb Z[i]/(1+i) \cong \mathbb Z[X]/(X^2+1,X+1) = \mathbb Z[X]/(2,X+1) \cong (\mathbb Z/2\mathbb Z[X])/(X+1) \cong \mathbb Z/2\mathbb Z$
The first isomorphism is given by $a+bi \mapsto a+bX$ and the last isomorphism is given by $X \mapsto -1$. Hence the composition is given by $a+bi \mapsto a-b \mod 2$.
The same calculation works for every maximal ideal $(c+di) \subset \mathbb Z[i]$ with $c \neq 0 \neq d$ (this imples $c^2+d^2=p$ for some prime $p$ and we have the equality of ideals $(X^2+1,dX+c)=(p,dX+c)$ in $\mathbb Z[X]$). We get an explicit isomorphism
$\mathbb Z[i]/(c+di) \to \mathbb Z/p\mathbb Z, a+bi \mapsto a-bcd^{-1} \mod p$


TITLE: Is my proof correct? (group theory, Lagrange)
QUESTION [5 upvotes]: $G$ is a finite group.
Prove, that $\exists k\in G:k\neq 1_G, k^2=1_G \iff |G|$ is even.
My proof
1) $\Rightarrow$:
$\exists k\in G: k^2 = 1_G\Rightarrow \exists G'\leq G$ s.t. $ G'=\{1_G,k\}$, then $|G'|=2$
Using Lagrange Theorem:
$|G'|$ divides $|G|\Rightarrow |G|$ is even.
2) $\Leftarrow$:
$|G|=2n\Rightarrow |G|\geq 2$ and $|G|=2n\Rightarrow\exists G'\leq G$ with $|G'|=2\Rightarrow\{1_G=1_{G'}, k\}= G'$ and $k=k^{-1}\Rightarrow |k|=2$
$\blacksquare$
Could you please review my proof? I'm currently learning how to write proofs using as few words as possible, so could you also check if the syntax is correct?

REPLY [6 votes]: For the reverse implication I think you were trying to express this:-
If for all the elements $x\neq e$ in $G$, $x\neq x^{-1}$ then including identity total number of elements in $G$ is odd, which is not the case, so there must exist a non-identity element in $G$ which is its own inverse.


TITLE: Two dice are rolled. Find the probability that at least one die is 6 if it is known that at least one die had an odd number.
QUESTION [0 upvotes]: So I have a question:

Two dice are rolled. Find the probability that at least one die is 6 if it is known that at least one die had an odd number.

Is this the right way to calculate conditional probability?
If I understood correctly, then I need to find this
\begin{align}
P(A \mid B) &= \frac{AB}{B}
\end{align}
There is an event A -> at least one die is 6. First, I will find probability, that both dice don't have 6.
\begin{align}
P(\neg A) &= \frac{5}{6} \cdot \frac{5}{6} = \frac{25}{36} \\\\
\end{align}
So \begin{align}
P(A) &= 1 - \frac{25}{36} = \frac{11}{36} \\\\
\end{align}
Next, we have event B -> came odd number
\begin{align}
P(B) &= \frac{3}{6} = \frac{1}{2} \\\\
\end{align}
In the end we have:
\begin{align}
P(A|B) &= \frac{\frac{11}{36} \cdot \frac{1}{2}}{\frac{1}{2}} = \frac{11}{36} = 0.305 \\\\
\end{align}
Does this look fine?

REPLY [3 votes]: That's not correct. Also the question is not stated clearly. It says at least one of the dice is $6$. But if at least one of the dice is odd then either exactly one die is $6$ or none of them is $6$.
Event $A$ - one of the dice is $6$
Event $B$ - at least one of the dice is odd
Probability that at least one of the dice is odd
$ \displaystyle P(B) = 1 - \frac 12 \cdot \frac 12 = \frac 34 ~$, we take complement of the probability that none of the dice is odd.
$ \displaystyle P(A \cap B) = 2 \cdot \frac 12 \cdot \frac 16~$, as one of them is $6$ and the other is odd.
Now, $ \displaystyle P(A \mid B) = \frac{P(A \cap B)}{P(B)}$


TITLE: Boundary Maslov index of holomorphic disks in Calabi-Yau manifolds
QUESTION [5 upvotes]: Let $u \colon \Sigma^2 \to M^{2n}$ be a holomorphic disk (so $\Sigma = \{z \in \mathbb{C} \colon |z| \leq 1\}$) in a compact Calabi-Yau manifold $M$ of real dimension $2n$ with boundary on a Lagrangian submanifold $L^n \subset M^{2n}$.  Note that the normal bundle $N\Sigma \to \Sigma$ is a complex vector bundle of complex rank $n-1$.  At points of $\partial \Sigma$, let's orthogonally decompose $TL|_{\partial \Sigma} = T(\partial \Sigma)|_{\partial \Sigma} \oplus F$, so that $F \subset N\Sigma|_{\partial \Sigma}$ is a totally real subbundle of real rank $n-1$.
I am interested in the boundary Maslov index $\mu(N\Sigma, F) \in \mathbb{Z}$.  My question is: Among all possible pairs $(u, L)$ of holomorphic curves with Lagrangian boundary, can $\mu(N\Sigma, F)$ attain every possible integer value, or are there restrictions (coming from, say, the Calabi-Yau assumption on $M$)?
I am not a symplectic geometer, so I don't know what to expect.  In particular, I don't know whether there are standard examples of holomorphic curves with Lagrangian boundary for which $\mu(N\Sigma, F)$ is easily computable.

REPLY [2 votes]: I assume you're asking about embedded or at least immersed discs, in
order to make sense of the normal bundle. If so, let's fix an
immersion $\iota\colon\Sigma\to M$ and observe that the pullback
bundle-pair $(\iota^*TM,\iota^*TL)$ splits as $(T\Sigma\oplus
N\Sigma,TS^1\oplus F)$ for some totally real subbundle
$F\subset\iota^*TL$. The total Maslov number $\mu$ of the disc
splits as $2+n$ where $n$ is the Maslov index of the normal
component (which is what you're interested in). So we might as well
ask about the ordinary Maslov number and then subtract 2.
First thing to notice is that the Maslov number needs to be even if
$L$ is orientable (this is because the orientation allows you to
lift the classifying map for the bundle pair to the double cover which
classifies orientable bundle pairs).
Now it's easy to get any positive even number as the Maslov index of a
disc. To see this, note that any symplectic manifold (in particular a
compact Calabi-Yau) contains a Darboux ball (symplectomorphic to an
open ball in $\mathbb{C}^n$, and inside a Darboux ball you can find a
Lagrangian product torus which bounds discs of Maslov index 2, 4, 6,
etc. This gives you normal Maslov indices 0, 2, 4, etc. I imagine it's
possible to cook up examples with any negative Maslov index.
If you're interested in things like this, you might be interested in
the following papers by Globevnik and Oh:
Josip Globevnik. Perturbation by analytic discs along maximal real submanifolds of C N . Math. Z.,
217(2):287–316, 1994.
Yong-Geun Oh. Riemann-Hilbert problem and application to the perturbation theory of analytic
discs. Kyungpook Math. J., 35(1):39–75, 1995
They study holomorphic discs in complex manifolds with boundary on a
totally real submanifold, and show that there is a splitting theorem
like the Birkhoff-Grothendieck splitting for holomorphic vector
bundles over $\mathbb{CP}^1$. Namely, every holomorphic bundle pair
(holomorphic bundle over the disc together with a totally real
subbundle around the boundary) splits as a direct sum of line
bundles.


TITLE: What is a puiseux series and what is wolfram-alpha doing with this antiderivative?
QUESTION [0 upvotes]: I asked wolfram alpha to compute the antiderivative of the function $x^x$. It gave me some really large confusing polynomial-esque thing called a puiseux series. However, from what I can gather on the internet, the puiseux series is used to approximate points on a function. So what is the puiseux series and what does it have to do with integration? Also could you please explain it so that it would be understandable by someone who's highest mathematics taken was calculus BC.

REPLY [1 votes]: A Puiseux series is formal Laurent power series in $T^{\frac1n}$  for some $n$ (the $n$ may vary  with the series). The set of Puiseux series is a field, denoted $k{\ll} T{\gg}$, and it is the algebraic closure of the field of formal power series $k[[T]]$.


TITLE: Definition of characters in the context of twists of elliptic curves
QUESTION [3 upvotes]: In the Wikipedia article about Twists of curves, I encountered the term character several times and I do not know what the definition of it is (and there is no reference to another Wikipedia article explaining this definition). To cite the relevant passages of the article:

It is possible to "twist" elliptic curves with j-invariant equal to 1728 by quartic characters; twisting a curve E by a quartic twist, one obtains precisely four curves: one is isomorphic to E, one is its quadratic twist, and only the other two are really new. Also in this case, twisted curves are isomorphic over the field extension given by the twist degree.

Analogously to the quartic twist case, an elliptic curve over {\displaystyle K}K with j-invariant equal to zero can be twisted by cubic characters. The curves obtained are isomorphic to the starting curve over the field extension given by the twist degree.

Could you please explain me the definition of the character in this context?

REPLY [2 votes]: Let $E$ be an elliptic curve over a (perfect) field $K$ and let $G_K = \mathrm{Gal}(\overline K/K)$. As explained in Chapter X.2 of Silverman's Arithmetic of Elliptic Curves, the twists of $E$ are in bijective correspondence with elements of $H^1(G_K, \mathrm{Aut}(E))$.
In general, $\mathrm{End}(E) = \mathbb Z$, so $\mathrm{Aut}(E) = \{\pm1\}$. Since $G_K$ acts trivially on $\{\pm1\}$, in this case,
$$H^1(G_K, \mathrm{Aut}(E)) = \mathrm{Hom}(G_K, \{\pm1\})$$
i.e. the twists of $E$ are in bijective correspondence with quadratic Galois characters $G_K\to\{\pm 1\}$. Alternatively, by Kummer theory, $H^1(G_K,\{\pm 1\})\cong K^\times/K^{\times 2}$, which explains why we usually talk about twisting by an element $d\in K^\times$. The character $\chi\colon G_K\to\{\pm 1\}$ corresponding to $d$ is exactly the lift of the canonical character $\mathrm{Gal}(K(\sqrt d)/K)\to \{\pm 1\}$.
If $\mathrm{Aut}(E)$ is strictly larger than $\{\pm1\}$, there will be extra types of twists. When $K$ is a number field, since $\mathrm{End}(E)$ is at worst an order in an imaginary quadratic field, the only way in which $E$ can gain extra automorphisms is if $\mathrm{End}(E) = \mathbb Z[i]$ or $\mathbb Z[\zeta_3]$. This happens exactly for the elliptic curves of $j$-invariant $1728$ and $0$.
In the case of a $j$-invariant $1728$ curve (e.g. take $E\colon y^2 = x^3 - x$), we have $\mathrm{End}(E) = \mathbb Z[i]$, so $\mathrm{Aut}(E) = \{\pm 1, \pm i\} = \mu_4$. Kummer theory gives an isomorphism
$$H^1(G_K, \mu_4)\cong K^\times/K^{\times 4}$$
and the twist corresponding to $d\in  K^\times/K^{\times 4}$ is $E_d\colon y^2 = x^3 - dx$. Note that, over $K(\sqrt[4]d)$, the map $(x, y)\mapsto (\sqrt dx,\sqrt[4]{d^3}y)$ identifies $E$ and $E_d$. In particular, if $d$ is a square, then $E_d$ is just a quadratic twist.
If $K\supset \mathbb Q(i)$, then $G_K$ acts trivially on $\mu_4$, and again $H^1(G_K, \mu_4)\cong\mathrm{Hom}(G_K, \mu_4)$ so that we get a bijection between twists and quartic Galois characters $G_K\to\mu_4$. The character corresponding to $d\in K^{\times}$ is exactly the map $\sigma\mapsto \frac{\sigma(\sqrt[4]d)}{\sqrt[4]d}\in\mu_4$.
In the case of $j$-invariant $0$ (e.g. $E\colon y^2 = x^3 + 1$), $\mathrm{End}(E) = \mathbb Z[\zeta_3] = \mathbb Z[\zeta_6]$ and you get sextic twists and twists by sextic characters $G_K\to\mu_6$ (if $K\supset \mathbb Q(\zeta_3)$) in the same way. I don't know why the Wikipedia article focuses on just the cubic twists in this case.


TITLE: Find $p(x)\in \mathbb{K}(\alpha)[x]$ of degree 2 such that $p(\beta)=0$
QUESTION [3 upvotes]: I am trying to do this exercise:
Consider the field extension $F/K$. Let $f(x)\in K[x]$ be a monic and irreducible polynomial of degree 3. If $\alpha, \beta, \gamma$ are three different roots of $f$ in $F$, prove that $\beta$ is a root of a polynomial of degree 2 with coefficients in $\mathbb{K}(\alpha)$.
I really do not know how to approach this problem. Some things I have thought about are:

For $u\in\{\alpha, \beta, \gamma\}$, we have that Irr(u,K)=f(x).
|K(u):K|=3, for $u\in\{\alpha, \beta, \gamma\}$, and $\{1,u,u^2\} $is a K-base of $K(u)$.
If $p(x)\in {K}(\alpha)[x]$ of degree 2 is such that $p(\beta)=0$, then $Irr(\beta,K(\alpha))$ divides $g(x)$.
Since $f(x)\in K[x]\subset K(\alpha)[x] $ and $f(\beta)=0$,$Irr(\beta,K(\alpha))$ divides $f(x)$.
Since $f(x)$ is irreducible and of degree 3, none of the roots belong to $K$.

None of these things have taken me anywhere.
Any hints for solving this problem would be very helpful. Thanks in advance.

REPLY [2 votes]: $\beta $ is a root of $q(x)=(x-\beta)(x-\gamma)$. And $q$ is an element of $K(\alpha)[x]$ as it is the result of the long division of $f$ by $x-\alpha$ which are both elements of $K(\alpha)[x]$.