TITLE: How do I know what I need to show when using induction? (Peano axioms)
QUESTION [0 upvotes]: As an exercise I wanted to prove that addition was commutative using the Peano axioms.
To quickly restate the definition of addition (where $S(n)$ is the successor function of $n$, which we can show is the same as $S(n) = n+1$):
$$\begin{align}
a + 0 &= a \tag{additive identity} \\
a + S(b) &= S(a + b) \tag{addition of naturals}
\end{align}$$
The axiom of induction then is the statement that:
$$\forall P(P(0) \land (P(k) \implies P(k+1)) \implies \forall n(P(n)))$$
Which means for all predicates, some base case being true and the $k$th case implying the $k+1$th case means that the predicate is true for all $n$.
For the purposes of this question, assume we have just proven that $a + 0 = 0 + a$, that $a + 1 = 1 + a$, and that $(a+b)+c = a+(b+c)$.
Then my question is how we're actually using this induction for proving the general commutative case $a+b=b+a$. Here is what I originally did:
Base case: We have $a + 0 = 0 + a$ as true.
Inductive step: Suppose $a+b = b+a$. Then $a+S(b) = S(a + b) = S(b + a) = b + S(a) = b + (a + 1) = b + (1 + a) = (b + 1) + a = S(b) + a$. We also have $(b+1)+a = (1+b)+a = 1+(b+a) = 1+(a+b) = (1+a)+b = S(a) + b$.
I figure that $P(0)$ here is when $k=0$ and we say $a+k = k+a$ is our base case, same as how we say $a + 0 = 0 + a$ is our base case.
But then what exactly is the $k+1$th case here? I mean I tried to show that $a+b=b+a$ implied all sorts of identities (I basically tried to sledgehammer it with every permutation I could think of) but I don't know exactly what is minimal and sufficient to prove the inductive step.
If the inductive hypothesis is $a+b=b+a$ and this is our $k$th step, i.e. $P(k)$ is true, what is $k$ here? $k=a$? $k=b$? Is it enough to show that $S(a) + b = b + S(a)$? If not, how do I know what all I must prove here?
I'm trying to stay strict within the definitions here so that I don't run the risk of letting intuition and "common sense" cheat myself of the understanding here.
How do I correctly interpret the $P$ statements here? What is $P(0)$? What is $P(k)$? What is $P(k+1)$ and what must I do to show it? Or is my induction axiom not correct? What should I be doing here?

REPLY [0 votes]: In the vein of my previous answer, here's some Agda:
data _≡_ {A : Set} (a : A) : A → Set where
    Refl : a ≡ a

data ℕ : Set where
    Z : ℕ
    S : ℕ → ℕ

_+_ : ℕ → ℕ → ℕ
m + Z = m
m + S n = S (m + n)

cong : {A B : Set} → (f : A → B) → {x y : A} → x ≡ y → f x ≡ f y
cong f Refl = Refl

_trans_ : {A : Set} → {x y z : A} → x ≡ y → y ≡ z → x ≡ z
Refl trans q = q

left-additive-identity : {n : ℕ} → (Z + n) ≡ n
left-additive-identity {Z} = Refl
left-additive-identity {S n} = cong S (left-additive-identity {n})

left-increment : {m n : ℕ} → (S m + n) ≡ S (m + n)
left-increment {m} {Z} = Refl
left-increment {m} {S n} = cong S (left-increment {m} {n})

commutative : {m n : ℕ} → (m + n) ≡ (n + m)
commutative {Z} {n} = left-additive-identity {n}
commutative {S m} {n} = left-increment {m} {n} trans cong S (commutative {m} {n})

This uses the equivalent of three uses of the induction axiom. One each for left-additive-identity which proves that $0+n=n$ ($n+0=n$ comes automatically from the definition of _+_); left-increment which proves $(m+1)+n=(m+n)+1$; and commutativity itself. Refl represents reflexivity of equality. cong represents the congruence property of equality, i.e. that functions take equal inputs to equal outputs. This is only used in the form: if $m=n$, then $m+1=n+1$. trans is the transitivity of equality.


TITLE: Is the first excited state of a superconducting qubit a stationary state?
QUESTION [1 upvotes]: A superconducting qubit is essentially an anharmonic oscillator with uneven spacings of the eigenstates. These states are eigenstates of the overall hamiltonian, which should mean that it is an energy eigenstate. This then seems to imply that like all energy eigenstates it evolves in time as such:
$$
\psi (t) = e^{-\frac{i}{\hbar}E_nt}\psi(0)
$$
Which would still leave $ |\psi(t)|^2$ invariant over time. Given that the superconducting wave function can be equated to this:
$$
 \psi(r) = \sqrt{\rho(r)}e^{-i\theta(r)}
$$
Where $\rho(r)$ is the Cooper pair number density. This evolves in time as such:
$$
\psi(r, t) = \sqrt{\rho(r)}e^{-i\theta(r)}e^{-\frac{i}{\hbar}E_nt}
$$
Which means the probability distribution stays constant over time:
$$
|\psi(r, t)|^2 = \rho(r)
$$
This seems to imply that the first excited state (and the ground state) has no change in number density, i.e. no current flow. This seems very unintuitive given that qubit can be seen as an LC circuit. Any insights?

REPLY [1 votes]: so you are confusing a superconductor whose macroscopic wavefunction describing only the ground state could be described the way you did, with a superconducting qubit which has a totally different wavefunction and is a junction between two blobs of superconducting material.
this introductory article might straighten things out:
http://arxiv.org/abs/1310.1897


TITLE: Help wanted with Chebotarev condition in characteristic 2
QUESTION [13 upvotes]: Having promised a longtime collaborator that I would clear my plate to finish up some joint work of ours, I am swallowing my pride and tossing up the following technical point of function field arithmetic to the experts:
Let $k_0 = \mathbb{F}_2(t)$, and let $k(P)$ be a finite separable extension of $k_0$.  Let $S$ be a finite set of places $v$ of $k_0$ (thus corresponding to monic irreducible polynomials in $\mathbb{F}_2[t]$ together with possibly the place at infinity).  Consider the set of monic irreducible polynomials $w \in \mathbb{F}_2[t]$ with the following properties:
(i) Every place $v$ in $S$ splits completely in the quadratic Artin-Schreier extension of $k_0$ defined
by the polynomial $X^2 + X = \frac{1}{w}$, and
(ii) $w$ splits completely in $k(P)$.
Problem: Show that there are infinitely many such $w$.
The natural strategy is to show that the set of such $w$ satisfies some Chebotarev condition -- presumably it even contains the set of all primes splitting in a certain finite Galois extension of $k_0$ -- and then apply the Chebotarev Density Theorem in this context (e.g. Theorem 9.13A in Rosen's Number Theory in Function Fields).
I have already proved similar statements when $k_0$ is a number field or is $\mathbb{F}_p(t)$ with $p$ an odd prime: for the latter I used the Quadratic Reciprocity Law in such fields.  In the present case I was trying to use Hasse's characteristic 2 Quadratic Reciprocity Law -- for which my only exposure is this nice note of K. Conrad -- but after some hours of fiddling around, I am having trouble making this work: in the above formulation at least, what is given is not exactly a reciprocity law -- i.e., it doesn't directly compare $[w,\ell_v)$ to $[\ell_v,w)$ but only establishes a certain periodicity relation which is, in the classical cases, equivalent to QR.
Many thanks if you can help me out with this!  I really need it in order to complete revisions on a long overdue paper, so a solution will be worth an acknowledgment in the paper at the least.

REPLY [16 votes]: By my comments to the question, we can assume the infinite place is not in $S$ (because its splitting in the indicated quadratic extensions is automatic).  Let the elements of $S$ be $\pi_1,\dots,\pi_r$, where the $\pi_i$'s are distinct (monic) irreducibles in ${\mathbf F}_2[t]$. For each $\pi_i$, the Artin-Schreier map $\wp(x) = x^2 + x$ on the field ${\mathbf F}_2[t]/(\pi_i)$ has image equal to half of that field. Fix a nonzero element of the image, say $g_i \bmod \pi_i$ (why? you'll see shortly). For a monic irreducible $\pi$ in ${\mathbf F}_2[t]$, if $\pi \equiv 1/g_i \bmod \pi_i$ then the congruence $x^2 + x \equiv 1/\pi \bmod \pi_i$ is solvable in ${\mathbf F}_2[t]/(\pi_i)$ by the choice of $g_i$, and then by Hensel's lemma the polynomial $X^2 + X - 1/\pi$ splits in the completion ${\mathbf F}_2(t)$ at $\pi_i$, so $\pi_i$ splits in the quadratic extension of ${\mathbf F}_2(t)$ obtained by adjoining a root of $X^2 + X - 1/\pi$.
So what you would like is to find infinitely many monic $\pi$ such that $\pi \equiv 1/g_i \bmod \pi_i$ for $i = 1,\dots,r$ and $\pi$ splits completely in a predetermined finite separable extension $L/{\mathbf F}_2(t)$. We can assume $L/{\mathbf F}_2(t)$ is Galois by replacing $L$ with its Galois closure over ${\mathbf F}_2(t)$, and then the hypothesis that $\pi$ splits completely in $L$ is the same as saying $\pi$ is unramified in $L$ with trivial Frobenius conjugacy class in the Galois group of $L/{\mathbf F}_2(t)$.
At the same time, the congruence condition $\pi \equiv 1/g_i \bmod \pi_i$ for monic irred. $\pi$ is also a Frobenius constraint using Carlitz extensions. The group $({\mathbf F}_2[t]/(\pi_i))^\times$ is naturally the Galois group of the splitting field over ${\mathbf F}_2(t)$ of the Carlitz polynomial associated to $\pi_i$. Write this splitting field as $K_{\pi_i}$. The natural isomorphism of the  (abelian) Galois group of $K_{\pi_i}/{\mathbf F}_2(t)$ with $({\mathbf F}_2[t]/(\pi_i))^\times$ identifies the Frobenius element attached to $\pi$ with $\pi \bmod \pi_i$ for any monic irred. $\pi$ in ${\mathbf F}_2[t]$ not equal to $\pi_i$. (This is analogous to the way $({\mathbf Z}/(m))^\times$ is identified with the Galois group of a cyclotomic field with the Frobenius at a prime $p$ being $p \bmod m$.) Therefore asking that the monic irred. $\pi$ satisfy $\pi \equiv 1/g_i \bmod \pi_i$ for all $i$ is a bunch of simultaneous Frobenius conditions on $\pi$ in the fields $K_{\pi_1},\dots,K_{\pi_r}$. (Warning: these Frobenius conditions for $\pi$ are usually not conditions for $\pi$ to split completely in $K_{\pi_i}$ because we're working with reciprocals of nonzero elements mod $\pi_i$ that are in the image of the Artin-Schreier map, and the Artin-Schreier map has no good multiplicative properties. Maybe you could get a split completely interpretation using the reciprocal of a nonzero element in the image of the Artin-Schreier map mod $\pi_i$ which is itself in the image of the Artin-Schreier map mod $\pi_i$. A count shows it is very likely that there is such an overlap, but I haven't checked it for certain.)
The Carlitz extensions $K_{\pi_1},\dots,K_{\pi_r}$ are linearly disjoint over $F_2(t)$, so there is no problem finding infinitely many $\pi$ with given Frobenius elements in each of the Galois groups of $K_{\pi_i}/{\mathbf F}_2(t)$. (This is also a consequence of the Chinese remainder theorem and Dirichlet's theorem modulo $\pi_1\cdots\pi_r$.) If the field $L$ is linearly disjoint from all the fields $K_{\pi_i}$ over ${\mathbf F}_2(t)$ then we can express all of Pete's original conditions as a single Frobenius condition in the Galois group of their composite field over ${\mathbf F}_2(t)$ and that has infinitely many solutions by Chebotarev. On the other hand, if $L$ is not linearly disjoint from those Carlitz extensions over ${\mathbf F}_2(t)$ then there could be some subtle compatibility issues to be sure Chebotarev can be applied. I'll stop at this point since I've shown the basic plan for how to interpret everything as a Frobenius condition in a Galois group over ${\mathbf F}_2(t)$.


TITLE: Hyperplane is the zero set of a single variable by a linear change of coordinates?
QUESTION [1 upvotes]: My professor said that a hyperplane in projective space can be assumed to be $V(x_0)$ by a linear change in coordinates. Im not quite sure what he means. A hyperplane in Hartshorne is defined to be the zero set of a linear homogeneous polynomial. So say $H = V(x_0 + x_1).$ Why can we assume $H = V(x_0)?$ Are the two isomorphic under some isomorphism of $\mathbb{P}^n?$ The only isomorphism I can think of is sending $x_0, x_1$ to $x_0$ is the pullback ring homomorphism of the respective coordinate rings but this is not an isomorphism.

REPLY [1 votes]: The map
\begin{align*}
\mathbb{P}^n &\to \mathbb{P}^n\\
[x_0 : x_1 : \cdots : x_n] &\mapsto [x_0 + x_1 : x_1 : \cdots : x_n]
\end{align*}
is an isomorphism of $\mathbb{P}^n$ mapping $V(x_0+x_1)$ to $V(x_0)$.
This is really just an incarnation of changing bases.  More generally, given vector space $V$ of dimension $n+1$ over a field $k$, a choice of basis $e_0, \ldots, e_n$ induces an isomorphism $\mathbb{P}(V) \to \mathbb{P}^n$ such that the following diagram commutes. The homogeneous coordinates on $\mathbb{P}(V)$ are basically just the linear maps $e_0^*, \ldots, e_n^*$ in the dual basis.
$\hspace{3.75cm}$
So, given a hyperplane $H = V(\ell)$ in $\mathbb{P}^n$ where $\ell = a_0 x_0 + \cdots + a_n x_n$ is some linear form, we can complete $v_0 = (a_0, \ldots, a_n)$ to a basis $\{v_0, v_1, \ldots, v_n\}$ for $k^{n+1}$. Letting $e_0, \ldots, e_n$ be the standard basis for $k^{n+1}$, then the change of basis map sending $e_i \to v_i$ induces a map $\mathbb{P}^n \to \mathbb{P}^n$ sending $V(\ell)$ to $V(x_0)$.


TITLE: Projective Banach spaces
QUESTION [4 upvotes]: Injective Banach spaces, with morphisms as contractive linear maps, have been classically studied (and are $C(K)$ spaces with $K$ Stonian).  But what about projectives?
So $P$ will be projective if given Banach spaces $E$ and $F$ and a quotient map (aka metric surjection) $\psi:E\rightarrow F$, given any contractive $\phi:P\rightarrow F$, we can lift this to a contractive $\varphi:P\rightarrow E$ with $\psi\varphi = \phi$.  (If someone can make a nice commutative diagram, go ahead and edit this!)
Claim: The scalar field (say $\mathbb C$, but also works for $\mathbb R$) is not projective.
Proof: Let $f:c_0\rightarrow\mathbb C$ be the contractive functional $f(x) = \sum_{n=1}^\infty 2^{-n} x_n$ for $x=(x_n)\in c_0$.  This induces an isometric isomorphism $c_0 / \ker(f) \cong \mathbb C$.  So if $\mathbb C$ is projective, we can find a contractive $g:\mathbb C\rightarrow c_0$ with $fg=1$.  That is, $g=(g_n)\in c_0$ is a norm-one vector with $f(g) = \sum 2^{-n} g_n =1$, which is impossible.

Question: Are there any projective Banach spaces?

It seems to me that the problem is insisting upon contractive morphisms.  In the proof above, if we just need, for each $\epsilon>0$, to find $g$ with $\|g\|<1+\epsilon$, then this is no problem.  Is anything known in this generalised setting?  (It's easy to see that then $\ell_1(\Gamma)$ is always projective, if I allow myself this wiggle-room).

REPLY [3 votes]: Matthew, you answered your own question with your claim--there are no isometrically projective Banach space (since, for example, a contractively complemented subspace of a projective space is again projective).

REPLY [2 votes]: You essentially answered your first question yourself: the ground field is a (contractive) retract of any nonzero Banach space by Hahn-Banach. If there were a non-zero projective Banach space in your sense then the ground field would be projective as well.
On the other hand: It is a theorem due to Köthe and Pełczyński that every projective Banach space (lifting over surjective maps in the additive category of Banach spaces) is isomorphic to $\ell^1{(S)}$ for some $S$. I don't know how well the norm of the isomorphism is controlled, but as all these spaces already satisfy your relaxed condition, you won't find any others.
The references for the second paragraph are:

A. Pełczyński, Projections in certain Banach spaces, Studia Math. 19 (1960), 209–228. MR0126145.
G. Köthe, Hebbare lokalkonvexe Räume, Math. Ann. 165 (1966), 181–195. MR0196464.


TITLE: What should I know about Sobolev space?
QUESTION [3 upvotes]: I have done some Sobolev spaces with some embedding theorems, trace theorems etc. Sorry that my question is really vague.
If my professor asks me what is great about Sobolev space, what should I answer (details, examples, counterexamples are very much welcome) to make sure that he feels ok this guy knows the concept pretty well.
Any readings, insights, examples, counterexamples are most welcome.
Thanks for your help.

REPLY [0 votes]: n many problems of mathematical physics and variational ca
lculus it is not
sufficient to deal with the classical solutions of differentia
l equations. It is
necessary to introduce the notion of weak derivatives and to
work in the so
called Sobolev spaces
http://www.iadm.uni-stuttgart.de/LstAnaMPhy/Weidl/fa-ws04/Suslina_Sobolevraeume.pdf


text:
\begin{document}
\baselineskip=18pt
\maketitle
\begin{abstract}Given a finite set $\{M_0,\dots,M_{d-1}\}$ of nonnegative $2\times 2$ matrices and a nonnegative column-vector $V$, we associate to each $(\omega_n)\in\{0,\dots,d-1\}^\mathbb N$ the sequence of the column-vectors $\displaystyle{M_{\omega_1}\dots M_{\omega_n}V\over\Vert M_{\omega_1}\dots M_{\omega_n}V\Vert}$. We give the necessary and sufficient condition on the matrices $M_k$ and the vector $V$ for this sequence to converge for all \hbox{$(\omega_n)\in\{0,\dots,d-1\}^\mathbb N$} such that $\forall n,\ M_{\omega_1}\dots M_{\omega_n}V\ne\begin{pmatrix}0\\0\end{pmatrix}$.
\end{abstract}
\vskip20pt
{2000 Mathematics Subject Classification:} {15A48.}\par\par

\vskip20pt

\centerline{\sc Introduction}

\smallskip

Let  ${\mathcal M}=\{M_0,\dots,M_{d-1}\}$ be a finite set of nonnegative $2\times 2$ matrices and $V=\begin{pmatrix}v_1\\v_2\end{pmatrix}$ a nonnegative column-vector. We use the notation $Y_n=Y_n^\omega:=M_{\omega_1}\dots M_{\omega_n}$ and give the necessary and sufficient condition for the pointwise convergence of $\displaystyle{Y_nV\over\Vert Y_nV\Vert}$, \hbox{$(\omega_n)\in\{0,\dots,d-1\}^\mathbb N$} such that $Y_nV\ne\begin{pmatrix}0\\0\end{pmatrix}$ for any $n$, where $\Vert\cdot\Vert$ is the norm-sum. The idea of the proof is that, if the conditions are satisfied, either both columns of $Y_n$ tends to the same limit, or they tend to different limits with different orders of growth, so in case $V$ is positive the limit points of $\displaystyle{Y_nV\over\Vert Y_nV\Vert}$ only depend on the limit of the dominant column. This problem is obviously very different from the one of the convergence of $\displaystyle{Y_n\over\Vert Y_n\Vert}$, or the convegence of the $Y_n$ itselves, see the intoduction of~\cite{OTdXd} for some counterexamples and \cite[Proposition 1.2]{sto} for the infinite products of $2\times2$ stochastic matrices.

The conditions for the pointwise convergence of $\displaystyle{Y_nV\over\Vert Y_nV\Vert}$ also differ from the conditions for its uniform convergence, see \cite{OT2X2}. The uniform convergence can be used for the multifractal analysis of some continuous singular measures called Bernoulli convolutions (see \cite{PSS} for the Bernoulli convolutions and \cite{FO} for their multifractal analysis). We study such measures in \cite{OT2X2}, \cite{OT4X4} and \cite{OT7X7}. The Birkhoff's contraction coefficient \cite[Chapter 3]{S} that we use in \cite{OT4X4} and \cite{OTdXd} but not here, is really not of great help to solve the main difficulties. Moreover the theorem that gives the value of this coefficient is difficult to prove (see \cite[\S3.4]{S}) even in the case of $2\times2$ matrices. In~\cite{OT2X2} we use some other contraction coefficient quite more easy to compute (\cite[Proposition 1.3]{OT2X2}).

\section{Condition for the pointwise convergence of $\displaystyle{Y_nV\over\Vert Y_nV\Vert}$}

\begin{pro}{The sequence $\displaystyle{Y_nV\over\Vert Y_nV\Vert}$ converges for any $\omega\in\{0,\dots,d-1\}^\mathbb N$ such that $\forall n,\ Y_nV\ne\begin{pmatrix}0\\0\end{pmatrix}$, if and only if at least one of the following conditions holds:\hfill\break
(i) $V$ has positive entries and it is an eigenvector of any invertible matrix of the form $\begin{pmatrix}a&0\\0&d\end{pmatrix}$ or $\begin{pmatrix}0&b\\c&0\end{pmatrix}$ that belongs to $\mathcal M$.\hfill\break
(ii) Any invertible matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in{\mathcal M}$ satisfies $a>0$
and, if $b=c=0$, $a\ge d$.\hfill\break
(iii) Any invertible matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in{\mathcal M}$ satisfies $d>0$
and, if $b=c=0$, $d\ge a$.\hfill\break
(iv) $V$ has a null entry and all the invertible matrices $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in{\mathcal M}$ satisfy $ad>0$.
}
\end{pro}
\begin{proof}Let $\omega\in\{0,\dots,d-1\}^\mathbb N$. If there exists N such that $\det M_{\omega_N}=0$, the column-vectors $Y_N V,\ Y_{N+1}V,\ \dots$ are collinear and $\displaystyle{Y_nV\over\Vert Y_nV\Vert}$ is constant for $n\ge N$.
So we look only at the $\omega\in\{0,\dots,d-1\}^\mathbb N$ such that $\forall n,\ \det
M_{\omega_n}\ne0$. In order to use only matrices with positive determinant we set $\Delta:=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and
$$
A_n=A_n^\omega:=\begin{cases}M_{\omega_n}\quad&\hbox{if }\det Y_{n-1}>0\ (\hbox{or }n=1)\hbox{ and }\det M_{\omega_n}>0\\
M_{\omega_n}\Delta\quad&\hbox{if }\det Y_{n-1}>0\ (\hbox{or }n=1)\hbox{ and }\det
M_{\omega_n}<0\\
\Delta M_{\omega_n}\quad&\hbox{if }\det Y_{n-1}<0\hbox{ and }\det M_{\omega_n}<0\\
\Delta M_{\omega_n}\Delta\quad&\hbox{if }\det Y_{n-1}<0\hbox{ and }\det
M_{\omega_n}>0.\end{cases}
$$
We set also
$$
\begin{pmatrix}a_n&b_n\\c_n&d_n\end{pmatrix}:= A_n\quad\hbox{and}\quad\begin{pmatrix}p_n&q_n\\
r_n&s_n\end{pmatrix}:=A_1\dots A_n
=\begin{cases}Y_n&\hbox{if }\det Y_n>0\\
Y_n\Delta&\hbox{if }\det Y_n<0.\end{cases}
$$
The matrices $A_n$ belong to the set
$$
{\mathcal M}^+:=\{M\;;\;\exists i,j,k,\ M=\Delta^iM_k\Delta^j\hbox{ and }\det M>0\}.
$$
Since $\det A_n>0$ we have $a_nd_np_ns_n\ne0$.
If $\{n;\ A_n\ \hbox{not\ diagonal}\}$ is infinite we index this set by an increasing sequence $n_1<n_2<\dots$.
We have $b_{n_1}\ne0$ or $c_{n_1}\ne0$; both cases are equivalent because, using the set of matrices ${\mathcal M}'=\Delta{\mathcal M}\Delta$ and defining similarly $Y'_n$ and $A'_n=\begin{pmatrix}a'_n&b'_n\\c'_n&d'_n\end{pmatrix}$ from this set,
we have
$Y'_n=\Delta Y_n\Delta$, $A'_n=\Delta A_n\Delta$ and
$b'_{n_1}=c_{n_1}$.

So we can suppose $b_{n_1}\ne0$; we deduce $q_n\ne0$ by induction on $n\ge n_1$. The sequences defined for any $n\ge n_1$ by
$$
u_n={r_n\over p_n},\quad
v_n={s_n\over q_n},\quad
w_n={q_n\over p_n},\quad x_n=\begin{cases}v_2/v_1&\hbox{if }\det
Y_n>0\\
v_1/v_2&\hbox{if not}\end{cases}\quad\hbox{and}\quad\lambda_n=(1+w_nx_n)^{-1}
$$
satisfy $0\le u_n<v_n<\infty$, $0<w_n<\infty$ and
$$
\begin{array}{l}\hbox{if the entries of }V\hbox{ are positive, }0<x_n<\infty\hbox{ and }0<\lambda_n<1\\
\hbox{if not, }x_n\in\{0,\infty\}\hbox{ and }\lambda_n\in\{0,1\}\hbox{ according to the sign of }\det Y_n.\end{array}
$$

Since we have assumed that $Y_nV\ne\begin{pmatrix}0\\0\end{pmatrix}$, the ratio $\displaystyle{(Y_nV)_2\over(Y_nV)_1}$ exists in $[0,\infty]$ and we have to prove that it has a finite or infinite limit when $n\to\infty$. If $A_n$ is not eventually diagonal we have for $n\ge n_1$
\begin{equation}\label{1}
{(Y_nV)_2\over(Y_nV)_1}=\lambda_nu_n+(1-\lambda_n)v_n\in I_n:=[u_n,v_n]\hbox{ and }I_n\supseteq I_{n+1}.
\end{equation}
An immediate consequence is the following lemma:

\begin{lem}\label{lem}Suppose $A_n$ is not eventually diagonal, then

(i) the sequences $(u_n)$ and $(v_n)$ converge in $\mathbb R$ and the sequence $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ is bounded;

(ii) $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ converges if $\displaystyle\lim_{n\to\infty}\vert I_n\vert=0$;

(iii) if $V$ has positive entries, $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ converges if $\displaystyle w_n$ has limit $0$ or~$\infty$;

(iv) if $V$ has a null entry, the necessary and sufficient condition for the convergence of $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ is that $\displaystyle\lim_{n\to\infty}\vert I_n\vert=0$ or the sign of $\det Y_n$ is eventually constant.
\end{lem}

We also define for $n>n_1$
$$
\alpha_n=\left(1+{c_n\over a_n}w_{n-1}\right)^{-1},\quad
\beta_n=\left(1+{b_n\over d_n}(w_{n-1})^{-1}\right)^{-1},\quad
\gamma_n=1-{c_n\over a_n}{b_n\over d_n}
$$
that belong to $]0,1]$ and satisfy
\begin{equation}\label{2}
\begin{array}{l}\displaystyle\vert I_n\vert=\alpha_n\beta_n\gamma_n\vert I_{n-1}\vert\\
\displaystyle w_n={d_n\over a_n}{\alpha_n\over\beta_n}w_{n-1}\end{array}
\end{equation}
so $\displaystyle\prod_{n>
n_1}\alpha_n\beta_n\gamma_n=\lim_{n\to\infty}{\vert I_n\vert\over\vert I_{n_1}\vert}$ is positive if and only if $\displaystyle\lim_{n\to\infty}\vert I_n\vert>0$. Using the equivalents of $\log\alpha_n$, $\log\beta_n$ and $\log\gamma_n$,
\begin{equation}\label{complicated}
\displaystyle\lim_{n\to\infty}\vert I_n\vert>0\Leftrightarrow\sum{c_n\over a_n}w_{n-1}<\infty,\quad
\sum{b_n\over d_n}(w_{n-1})^{-1}<\infty\quad\hbox{and}\quad
\sum{c_n\over a_n}{b_n\over d_n}<\infty.
\end{equation}
The set of indexes $\{n;\ A_n\ \hbox{not\ diagonal}\}$ is the union of
$$
L^\omega=\{n;\ c_n\ne0\}
\quad\hbox{and}\quad U^\omega=\{n;\ b_n\ne0\}.
$$
Moreover, since $A_n$ belongs to the finite set ${\mathcal M}^+$ there exists
$K>0$ such that
$$
L^\omega=\left\{n;\ {1\over K}\le{c_n\over a_n}\le K\right\}\quad\hbox{and}\quad U^\omega=\left\{n;\
{1\over K}\le{b_n\over d_n}\le K\right\}.
$$
We deduce a simpler formulation of (\ref{complicated}):
\begin{equation}\label{simple}
\displaystyle\lim_{n\to\infty}\vert I_n\vert>0\Leftrightarrow\sum_{n\in L^\omega}w_{n-1}<\infty,\quad\sum_{n\in U^\omega}(w_{n-1})^{-1}
<\infty\quad\hbox{and}\quad L^\omega\cap U^\omega\ \hbox{is finite}.
\end{equation}
In view of Lemma \ref{lem} we may suppose from now that $\displaystyle\lim_{n\to\infty}\vert I_n\vert>0$. Since $L^\omega\cap U^\omega=\{n\;;\;A_n\hbox{ positive}\}$ is finite, for $n$ large enough the matrix $A_n$ is lower triangular if $n\in L^\omega$, upper triangular if $n\in U^\omega$, diagonal if $n\not\in L^\omega\cup U^\omega$. When $A_n$ is diagonal the second relation in (\ref{2}) becomes $\displaystyle
w_n={d_n\over a_n}w_{n-1}$; consequently any integer $n$ in an interval $]n_i,n_{i+1}[$ with $i$ large enough satisfies
\begin{equation}\label{4}
{w_n\over w_{n_i}}=\prod_{n_i<j\le
n}{d_j\over a_j}.
\end{equation}
Moreover if $L^\omega$ is infinite, (\ref{simple}) implies that $w_{n-1}$ has limit to $0$ when $L^\omega\ni n\to\infty$, and $w_n$ also has limit $0$ because $\displaystyle w_n={d_nw_{n-1}\over a_n+c_nw_{n-1}}$ for any $n\in
L^\omega\setminus U^\omega$. We have a similar property if $U^\omega$ is infinite, so
\begin{equation}\label{5}
\begin{array}{ll}\hbox{if}\ L^\omega\ \hbox{is infinite},&
w_{n-1}\to0\ \hbox{and}\ w_n\to0\hbox{ for }L^\omega\ni n\to\infty;\\
\hbox{if}\ U^\omega\ \hbox{is infinite},&
w_{n-1}\to\infty\ \hbox{and}\ w_n\to\infty\hbox{ for }U^\omega\ni n\to\infty.\end{array}
\end{equation}

\smallskip

{\bf First case:} Suppose that (i) holds. Then the diagonal matrices
of $\mathcal M$
are collinear to the unit matrix. If at least one matrix of ${\mathcal M}$ has the form $M_k=\begin{pmatrix}0&b\\c&0\end{pmatrix}$ with $bc\ne0$, its nonnegative eigenvalue -- namely
$\begin{pmatrix}\sqrt b\\\sqrt c\end{pmatrix}$ -- is collinear to $V=\begin{pmatrix}v_1\\v_2\end{pmatrix}$ hence there exists $\lambda\in{\mathbb R}$ such that $M_k=\lambda\begin{pmatrix}0&v_1^2\\v_2^2&0\end{pmatrix}$.

Notice that if $A_n$ is diagonal from a rank $N$,
the matrix $M_{\omega_n}$ has the form $\begin{pmatrix}a&0\\0&d\end{pmatrix}$ or $\begin{pmatrix}0&b\\c&0\end{pmatrix}$ hence it has $V$ as eigenvector; consequently
$\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ converges because it is $\displaystyle{(Y_NV)_2\over(Y_NV)_1}$ for any $n\ge N$.

Suppose now $A_n$ is non-diagonal for infinitely many $n$. We apply (\ref{4}) on each interval $]n_i,n_{i+1}[$ (if non empty), for $i$ large enough.
Among the integers $n\in]n_i,n_{i+1}[$ we consider the ones for which $\det M_{\omega_n}<0$.
For such $n$ the matrix $A_n$ is alternately
$M_{\omega_n}\Delta$ and $\Delta M_{\omega_n}$, hence alternately proportional to
$\begin{pmatrix}v_1^2&0\\0&v_2^2\end{pmatrix}$ and to $\begin{pmatrix}v_2^2&0\\0&v_1^2\end{pmatrix}$
and, according to (\ref{4}),
\begin{equation}\label{6}
n_i\le n<n_{i+1}\Rightarrow{w_n\over w_{n_i}}\in\Big\{{v_1^2\over
v_2^2},{v_2^2\over v_1^2},1\Big\}.
\end{equation}
In particular this relation holds for $n=n_{i+1}-1$. One deduce -- according to (\ref{5}) -- that there do not exist infinitely many $i$ such that
$n_i\in L^\omega$ and $n_{i+1}\in U^\omega$. Thus $n_i\in L^\omega$ for $i$ large enough (resp. $n_i\in U^\omega$ for $i$
large enough) and, according to (\ref{5}) and (\ref{6}), $\displaystyle\lim_{n\to\infty} w_n=0$ (resp. $\displaystyle\lim_{n\to\infty} w_n=\infty$). In view of Lemma \ref{lem}(iii),
$\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ converges.

\smallskip

{\bf Second case:} Suppose that (ii) holds
(if (iii) holds the proof is similar).

Suppose first the $M_{\omega_n}$ are diagonal from a rank $N$. From the hypothesis (ii) there exists $\delta_n,\delta'_n$ such that $M_{\omega_N}\dots M_{\omega_n}V=\left(\begin{array}{c}\delta_nv_1\\\delta'_nv_2\end{array}\right)$ and $\delta_n\ge\delta'_n$. Since the $M_{\omega_i}$ belong to a finite set we have $\displaystyle\lim_{n\to\infty}{\delta_n\over\delta'_n}=\infty$, or $\displaystyle{\delta_n\over\delta'_n}$ is eventually constant in case $M_{\omega_n}$ is eventually the unit matrix, or $\delta'_n=0\ne\delta_n$ for $n$ large enough. Denoting by $\left(\begin{array}{cc}p&q\\r&s\end{array}\right)$ the matrix $M_{\omega_1}\dots M_{\omega_{N-1}}$, we have $\displaystyle{(Y_nV)_2\over(Y_nV)_1}=\displaystyle{r\delta_nv_1+s\delta'_nv_2\over p\delta_nv_1+q\delta'_nv_2}$ hence $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ converges in all the~cases.

Suppose now $M_{\omega_n}$ is non-diagonal for infinitely many $n$. There exists from (\ref{5}) an integer $\kappa$ such that
\begin{equation}\label{either}
i\ge\kappa\Rightarrow\left\{\begin{array}{ll}w_{n_i-1}<1\hbox{ and }w_{n_i}<1&\hbox{if }n_i\in L^\omega\\w_{n_i-1}>1\hbox{ and }w_{n_i}>1&\hbox{if }n_i\in U^\omega\end{array}\right.
\end{equation}
and such that the $A_n$ are diagonal for $n\in]n_i,n_{i+1}[$, $i\ge\kappa$. According to (ii), for such values of $n$ the matrix $M_{\omega_n}$ is diagonal and $A_n=M_{\omega_n}$ with $a_n\ge d_n$, or $A_n=\Delta M_{\omega_n}\Delta$ with $a_n\le d_n$.

If there exists $i\ge\kappa$ such that $n_i\in L^\omega$ and $n_{i+1}\in U^\omega$, $\det Y_{n_i}$ is necessarily negative: otherwise $A_n$ should be equal to $M_{\omega_n}$ for $n\in]n_i,n_{i+1}[$, $\displaystyle{d_n\over a_n}\le1$ and, by (\ref{4}), $w_{n_{i+1}-1}\le w_{n_i}<1$ in contradiction with (\ref{either}).

Now $M_{\omega_{n_{i+1}}}$ has positive determinant, otherwise it should have the form $\begin{pmatrix}a&b\\c&0\end{pmatrix}$ and $A_{n_{i+1}}=\Delta M_{\omega_{n_{i+1}}}=\begin{pmatrix}c&0\\a&b\end{pmatrix}$ in contradiction with $n_{i+1}\in U^\omega$.

We have again $\displaystyle{d_n\over a_n}\ge1$ for $n\in]n_{i+1},n_{i+2}[$ and consequently $w_{n_{i+2}-1}\ge w_{n_{i+1}}>1$; so, by induction, $n_j\in U^\omega$ and $\det Y_{n_j}<0$ for any $j\ge i+1$. From ({5}) $w_n$ lies between $w_{n_j}$ and $w_{n_{j+1}-1}$ for any $n\in]n_j,n_{j+1}[$ and $j$ large enough, and from (\ref{5}) its limit is infinite. Distinguishing the cases where $V$ has positive entries or $V$ has a null entry, $\displaystyle{(Y_nV)_2\over(Y_nV)_1}$ converges by Lemma \ref{lem}.

The conclusion is the same if there do not exist $i\ge\kappa$ such that $n_i\in L^\omega$ and $n_{i+1}\in U^\omega$, because in this case $n_i\in L^\omega$ for $i$ large enough, or $n_i\in U^\omega$ for any $i\ge\kappa$.

\smallskip

{\bf Third case:}
Suppose (iv) holds. As we have seen, from (\ref{simple}) $A_n$ is eventually triangular or diagonal, and $M_{\omega_n}$ also is because -- by (iv) -- $\mathcal M$ do not contain invertible matrices of the form $\begin{pmatrix}0&b\\c&d\end{pmatrix}$ or $\begin{pmatrix}a&b\\c&0\end{pmatrix}$. We deduce that the sign of $\det Y_n$ is eventually constant. If $A_n$ is not eventually diagonal the sequence $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ converges by Lemma \ref{lem}(iv) and, if $A_n$~is, the sequence $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ is eventually constant.

\smallskip

{\bf Fourth case:}
Suppose that the set $\mathcal M$ do not satisfy (i), (ii), (iii)
nor (iv), and that at~least one matrix of this set, let $M_k$, has the form $M_k=\begin{pmatrix}0&b\\c&0\end{pmatrix}$ with $bc\ne0$; let us prove that $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ diverges.

$\_$ Suppose first there exists a matrix $M_k$ of this form that do not have $V$
as eigenvector; we chose as counterexample the constant sequence defined by $\omega_n=k$ for any $n$: $Y_{2n}$ is collinear to the unit matrix, hence $Y_{2n}V$ is collinear to $V$ and $Y_{2n+1}V$ to $M_kV$, so $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ diverges.

$\_$ Suppose now that all the matrices of $\mathcal M$ of the form $\begin{pmatrix}0&b\\c&0\end{pmatrix}$ with $bc\ne0$ have $V$
as eigenvector that is, $V$ is collinear to $\begin{pmatrix}\sqrt b\\\sqrt c\end{pmatrix}$ for all such matrix. Since (i) do not hold, at least one matrix $M_h$ of $\mathcal M$ is diagonal with nonnull and distinct diagonal entries. In this case
$M_hM_k$ has the form $\begin{pmatrix}0&b\\c&0\end{pmatrix}$ but do not have $V$
as eigenvector. We recover the previous case; more precisely the counterexample is defined by
$\omega_{2n-1}=h$ and $\omega_{2n}=k$ for any $n\in\mathbb N$.

\smallskip

{\bf Fifth case:}
Suppose that $\mathcal M$ do not satisfy (i), (ii), (iii)
nor (iv),
and that no matrix of the form $\begin{pmatrix}0&b\\c&0\end{pmatrix}$ with $bc\ne0$ belongs to $\mathcal M$.
Since (i) do not hold, at least one matrix of this set has the form $M_k=\begin{pmatrix}a&0\\0&d\end{pmatrix}$ with $ad\ne0$ and $a\ne d$.
We suppose that $a>d$ and we use the negation of (ii) (in case $a<d$ we use similarly the negation of (iii)).
According to the negation of (ii) there exists in $\mathcal M$ at least one matrix of the form $M_h=
\begin{pmatrix}0&\beta\\\gamma&\delta\end{pmatrix}$ with $\beta\gamma\delta\ne0$,
or one of the form $M_\ell=
\begin{pmatrix}\alpha&0\\0&\delta\end{pmatrix}$ with $0<\alpha<\delta$.

$\_$ Consider first the case where $\mathcal M$ contains some matrices $M_k$ and $M_h$ as above.
Let $(n_i)_{i\in\mathbb N}$ be an increasing sequence of positive integers with $n_1=1$, and $\omega$ the sequence defined by
$\omega_n=h$ for $n\in\{n_1,n_2,\dots\}$ and $\omega_n=k$ otherwise.

For $i$ odd, $A_{n_i}$ is lower-triangular and $\forall n\in]n_i,n_{i+1}[,\ A_n=\begin{pmatrix}d&0\\0&a\end{pmatrix}$, $a_n=d$ and $d_n=a$. For $i$ even, $A_{n_i}$ is upper-triangular and $\forall n\in]n_i,n_{i+1}[,\ A_n=\begin{pmatrix}a&0\\0&d\end{pmatrix}$, $a_n=a$ and $d_n=d$.

Using (\ref{4}) for $n=n_{i+1}-1$ and choosing $n_{i+1}-n_i$ large enough one has $w_{n_{i+1}-1}\ge2^i$ if $i$ is odd, $w_{n_{i+1}-1}\le2^{-i}$ if $i$ is even, so the three conditions in (\ref{simple}) are satisfied and the interval $\displaystyle\cap I_n$ is not reduced to one point.
If the entries of $V$ are positive, the first relation in (\ref{1}) and the definition of~$\lambda_n$ imply that
$\displaystyle\liminf_{n\to\infty}\left({(Y_nV)_2\over(Y_nV)_1}\right)$ is the lower bound of this
interval and $\displaystyle\limsup_{n\to\infty}\left({(Y_nV)_2\over(Y_nV)_1}\right)$
its upper bound, so the sequence $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ diverges. If $V$ has a null entry, the divergence of $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ results from Lemma \ref{lem}(iv).

$\_$ In case $\mathcal M$ contains some matrices $M_k$ and $M_\ell$ as above, one defines $\omega$ from a sequence $i_1=1<i_2<i_3<\dots$
by setting, for $j\ge1$ and $i_j\le n<i_{j+1}$,
$$
\omega_n=\left\{\begin{array}{ll}k&\hbox{if }j\hbox{ even}\\\ell&\hbox{if }j\hbox{ odd}.\end{array}\right.
$$
The diagonal matrix $Y_n$ can be easily computed, and $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ obviously diverges if one choose the
$i_{j+1}-i_j$ large enough.

If $V$ has a null entry, since (iv) do not hold $\mathcal M$ contains at least one matrix of the form $M_h=
\begin{pmatrix}0&\beta\\\gamma&\delta\end{pmatrix}$, $\beta\gamma\delta\ne0$ or $M_{h'}=\begin{pmatrix}\alpha&\beta\\\gamma&0\end{pmatrix}$, $\alpha\beta\gamma\ne0$, or $M_{h''}=\begin{pmatrix}0&\beta\\\gamma&0\end{pmatrix}$, $\beta\gamma\ne0$. We already know that $\displaystyle\left({(Y_nV)_2\over(Y_nV)_1}\right)$ diverges if $\mathcal M$ contains $M_k$ and $M_h$. Similarly it diverges if $\mathcal M$ contains $M_\ell$ and $M_{h'}$. If $\mathcal M$ contains $M_k$ and $M_{h''}$ the counterexample is given -- from a sequence $i_1=1<i_2<i_3<\dots$ -- by $\omega_{i_j}=h''$ and $\omega_n=k$ for $n\in]i_j,i_{j+1}[$, $j\in\mathbb N$:

$\displaystyle{(Y_nV)_2\over(Y_nV)_1}$ is alternately $0$ and $\infty$ because $Y_{i_j}$ has the form $\left(\begin{array}{cc}0&q\\r&0\end{array}\right)$ for $j$ odd and $\left(\begin{array}{cc}p&0\\0&s\end{array}\right)$ for $j$ even.
\end{proof}

config: arxiv


TITLE: Suzuki Sporadic Group on Sage
QUESTION [1 upvotes]: I need to call the Suzuki Sporadic group in Sage. I found the following command in the documentation.
G = groups.permutation.SuzukiSporadic()

This gives me a GAPError so I tried similar commands, but they all give me an error.

REPLY [2 votes]: You need to install the optional gap_packages Sage package. If you look at the documentation for groups.permutation.SuzukiSporadic(), you should see:
sage: groups.permutation.SuzukiSporadic?
Init signature: groups.permutation.SuzukiSporadic()
Docstring:
   Suzuki Sporadic Group

   EXAMPLES:

      sage: G = groups.permutation.SuzukiSporadic(); G # optional - gap_packages internet
      Sporadic Suzuki group acting on 1782 points

The part after the command "optional -gap_packages ..." is the clue.
How to actually install that package may depend on how you've installed Sage. Running sage -i gap_packages or make gap_packages from the top-level Sage directory should do it. But this is not a math question anymore (if it was at all), so please direct inquiries about installation to the sage-support Google group or to ask.sagemath.org.


TITLE: cubic root of negative numbers
QUESTION [32 upvotes]: Excuse my lack of knowledge and expertise in math,but to me it would came naturally that the cubic root of $-8$ would be $-2$ since $(-2)^3 = -8$.
But when I checked Wolfram Alpha for $\sqrt[3]{-8}$, real it tells me it doesn't exist.
I came to trust Wolfram Alpha so I thought I'd ask you guys, to explain the sense of that to me.

REPLY [3 votes]: In the years since the question was asked and answered, Wolfram introduced the $\operatorname{Surd}(x,n)$ function (Mathematica 9 circa $2012$, then Alpha) to designate the real single-valued $n^{th}$ root of $x$.
For example $\sqrt[3]{-8}$ and $\sqrt[5]{-243}$ result directly in $-2$ and $-3$ respectively:

The $\operatorname{Cbrt}$ function discussed in Mark McClure's answer - which had changed behavior around the same time to return the real cube root by default - appears to be identical to $\operatorname{Surd}(\,\cdot\,,3)$:


TITLE: Cube Tumbleweeding on a Plane
QUESTION [3 upvotes]: Am not a mathematician so please forgive any ambiguity in construction of the question below.
Given:

a cube, with faces $a_1$,... $a_6$, with $a_1$ and $a_6$ forming opposite sides of the cube, and face $a_1$ resting on the $x$-$y$ plane,
and with a transformation $r(p, q)$ defined as a $90$ degree rotation around a side between adjacent faces $a_p$ and $a_q$ such that at least one of the cube's faces is always touching the $x$-$y$ plane:

Does there exist a series of transformations, $r(p_1, q_1)$, $r(p_2, q_2)$, ..., $r(p_n, q_n)$ such that:

every face $(a_1, ..., a_6)$ shall touch the $x$-$y$ plane an equal number of times $(k)$
at the end of which the center of the cube returns to its original location?

If so, what is the minimum possible value of $k$?

REPLY [1 votes]: Suppose you roll the cube 6 times such that each face is turned to the bottom once. Imagine also, that it leaves a trail of squares behind. Those squares could then be rolled up into a cube, so they form a net of the cube. There is however no cube net that is a loop, so it is not possible to return to the starting location already.
The next smallest number of rolls that could work is 12, where every face is visited twice. For simplicity we might as well visit all faces once with the first 6 rolls, and then roll back the cube in reverse to revisit the same squares and visit all 6 faces again.
On way to roll is in the compass directions NENENE and then back WSWSWS.
If the faces of the cube are denoted with D=Down (=$a_1$), U=Up (=$a_6$), L=Left, R=Right, F=Front, B=Back, where F is initially on the south side, then the cube starts on D and rolls onto the faces BRUFLD. When tracing back the moves, it visits them in reverse, rolling onto LFURB and then D again.
You can do the reverse trip as SWSWSW instead. Then the faces visited in the second half would be FLUBR and D.
In both cases the cube returns in the same orientation.


TITLE: Are differential rings monoids in a monoidal category?
QUESTION [9 upvotes]: $\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}$A number of algebraic structures can be defined as monoids in some appropriate monoidal category:

A monoid                          is a monoid in $(\mathsf{Sets},\times,\pt)$;
A semiring                        is a monoid in $(\mathsf{CMon},\otimes_{\N},\N)$;
A ring                            is a monoid in $(\mathsf{Ab},\otimes_\Z,\Z)$;
An $R$-algebra                    is a monoid in $(\mathsf{Mod}_R,\otimes_R,R)$;
A graded $R$-algebra              is a monoid in $(\mathsf{Gr}_\Z\mathsf{Mod}_R,\otimes_R,R)$;
A differential graded $R$-algebra is a monoid in $(\mathsf{Ch}_\bullet(\mathsf{Mod}_R),\otimes_R,R)$.

Is this also the case for differential rings?

REPLY [7 votes]: $\newcommand{\defeq}{\overset{\mathrm{def}}{=}}\newcommand{\id}{\mathrm{id}}\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\d}{\mathrm{d}}\newcommand{\dAb}{\mathsf{End}(\mathsf{Ab})}$DGAs are monoids in chain complexes. To get differential rings as monoids in some monoidal category, it suffices to remove the grading and the $\d^{2}=0$ condition.
In detail, consider the category $\mathsf{End}(\mathsf{Ab})\defeq\mathsf{Fun}(\mathbf{B}\N,\mathsf{Ab})$ whose

Objects are pairs $(A,\d)$ with $A$ an abelian group and $d\colon A\to A$ a morphism of abelian groups.
Morphisms $(A,\d_A)\to(B,\d_B)$ are morphisms of abelian groups preserving the derivation, i.e. such that the diagram
$$
\require{AMScd}
\begin{CD}
A @>\d_A>> A\\
@V f V V @VV f V\\
B @>>\d_B> B
\end{CD}
$$
commutes.

We can then put a monoidal structure $\otimes_\Z$ on $\dAb$ by defining
$$(A,\d_A)\otimes(B,\d_B)=(A\otimes_\Z B,\d_A\otimes_\Z1_B+1_A\otimes_\Z\d_B),$$
where the unit is given by the pair $(\Z,\d_\Z)$ with $\d_\Z\overset{\mathrm{def}}{=} 0$. Note that a morphism in $\dAb$ from $(\Z,\d_\Z)$ to $(A,\d_A)$ is just a "constant" element of $A$, i.e. an element with $\d_A a = 0$.
A monoid in $(\dAb,\otimes_\Z,(\Z,\d_\Z))$ will then be a triple $((A,\d),\mu,\eta)$ with

$(A,\d)$ an object of $\dAb$; this accounts for the underlying additive abelian group of a differential ring and the derivation $\d$, which is $\Z$-linear;
$\mu\colon(A,\d_A)\otimes_\Z(A,\d_A)\to(A,\d_A)$ a morphism of $\dAb$; this accounts for the multipication and the Leibniz rule: asking for the diagram
$$
\require{AMScd}
\begin{CD}
A\otimes_\Z A @>\d_A\otimes_\Z1_A+1_A\otimes_\Z\d_A>> A\otimes_\Z A\\
@V \mu V V @VV \mu V\\
A @>>\d_A> A
\end{CD}
$$
to commute is equivalent to asking
$$\d(ab)=\d(a)b+a\d(b)$$
to hold for all $a,b\in A$;
$\eta\colon(\Z,\d_\Z)\to(A,\d_A)$ a morphism of $\dAb$, determining an element $1_A$ of $A$;

such that the usual associativity and unitality diagrams commute, which makes $(A,\mu,\eta)$ into a ring, and together with $\d$, this makes the quadruple $((A,\d),\mu,\eta)$ into a differential ring.


TITLE: $f$ is T periodic and $f(x) + f'(x) \ge 0 \Rightarrow f(x) \ge 0$
QUESTION [3 upvotes]: Let $f: \Bbb R \to \Bbb R$ be a function such that $f'(x)$ exists and is continuous over $\Bbb R$. Moreover, let there be a $T > 0$ such that $f(x + T) = f(x)$ for all $x \in \Bbb R$ and let $f(x) + f'(x)\ge 0$ for all $x \in \Bbb R$.
Show that $f(x) \ge 0$ for all $x \in \Bbb R$.

My attempt:
$f(x) \ge 0 \iff f(x) \ge f'(x) - f'(x) \iff f(x) + f'(x) \ge f'(x)$.
Thus, it is enoguh to show that $0 \ge f'(x)$.
$\iff 0 \ge \lim_{h\to0}\frac{f(x + h) - f(x)}{h}$
I do not know how to proceed from here. I know that $f'$ also has a periodicity of T but I do not know how to use that here.
Am I on the right track? How can I use the periodicity of $f$ to solve the problem?

REPLY [5 votes]: We don't even need that $f'(x)$ is continuous! It is enough to show that $f(x) \ge 0$ on $[0,T]$, because of periodicity. Since $[0,T]$ is compact, $f$ has a minimum on $[0,T]$. Let $x_0 \in [0,T]$ be such that $f(x_0)$ is minimal. If $x_0 =0$ or $x_0 =T$, then $f(x_0)$ is also minimal in $[-T,2T]$, because of periodicity.
A necessary condition for a minimum is that $f'(x_0) =0$. (And it is also valid in the case $x_0 =0$ or $x_0 =T$, because of the above remark.) Thus we get $f(x_0) = f(x_0)+f'(x_0) \ge 0$. This shows that $f(x) \ge 0$ everywhere.


TITLE: I want to find a closed form of the sum $\sum_{i=1}^n x^{i^2}$
QUESTION [0 upvotes]: Is there any closed form of the following sum?
$\sum_{i=1}^n x^{i^2}$, where $x$ is a variable.
Calculating the sum for a first few $n$ does not give any pattern.

REPLY [1 votes]: EDIT: the OP actually wants a finite sum. I have no idea what could be done with that in the way of simplification.
ORIGINAL:
This is standard, see the definition in formula (1.3) and the relationship to the Dedekind eta function. See also THETA
==============

================


TITLE: Problem with solving quadratic equation
QUESTION [2 upvotes]: I've been solving various quadratic equation for some time now, and I found 2 equations that I am unable to solve:
$x^2-y^2=5 $
$\frac{1}{x}+\frac{1}{y}=-\frac{1}{6}$
This is one of the equations. The problem I have is if I exchange $y$ with and expression with $x$ I will always get $xy$, and I just can't get rid of $xy$ and I can't express $xy$ by anything else.

REPLY [4 votes]: I understand that you want to solve these equations simultaneously?
First take the equation
$$\frac{1}{x}+\frac{1}{y}=−\frac{1}{6}$$
We'll rearrange this to find $y$:
$$\frac{1}{y}=−\frac{1}{6}-\frac{1}{x}$$
Now put the RHS into one fraction:
$$\frac{1}{y}=-\frac{6+x}{6x}$$
Then reciprocate both sides:
$$y=-\frac{6x}{6+x}$$
Now, you can substitute this into the first equation (or, another method that you prefer).
Be careful though - this may result in a quartic equation.
Consider writing the first equation as $(x-y)(x+y) = 5$ before substituting.
Hopefully it should be clear what to do from here.


TITLE: Non-existence of space with measure with following property
QUESTION [0 upvotes]: I need to show that there is no such space $(\Omega , \Sigma , \mu)$ for which Every self-adjoint operator unitary equivalent to multiplication operator in $L_2 (\Omega , \mu)$. I think I just need some hints on attempting this problem. Any ideas?

REPLY [1 votes]: Consider a self-adjoint operator $A$ with empty point spectrum.  Show that if $A$ is unitarily equivalent to a multiplication operator on $L^2(\Omega,\mu)$, then $(\Omega, \mu)$ contains no atoms.
Consider a rank-one orthogonal projection operator $P$.  Show that if $P$ is unitarily equivalent to a multiplication operator on $L^2(\Omega,\mu)$, then $(\Omega, \mu)$ contains an atom.


TITLE: Mochizuki's "phenomena in number theory" outside the scope of Langlands
QUESTION [28 upvotes]: (Crossposted from math.stackexchange by suggestion)
On page 12 of Shinichi Mochizuki's "On the Verification of Inter-universal Teichmuller Theory: A Progress Repor", he writes

"The representation-theoretic approach exemplified by the Langlands
  program does indeed constitute one major current of research in modern
  number theory. On the other hand, my understanding is that the idea
  that every essential phenomenon in number theory may in fact be
  incorporated into, or somehow regarded as a special case of, this
  representation-theoretic approach is simply not consistent with the
  actual content of various important phenomena in number theory."

Does anyone know what these 'various important phenomena' are that he's referring to?

REPLY [30 votes]: As noted by Lucia, large parts of number theory are completely "beyond the scope of the Langlands program": most of analytic number theory, obviously, is, but also many important, active and beautiful subfields of algebraic number theory -- for a list of examples, see for instance the list of publications of Bjorn Poonen, which cover a large scope of subjects in algebraic number theory, but touch the Langlands program only tangentially.
To make sense of the proposed quote of Mochizuki, it is therefore necessary to give a much more restrictive interpretation of "number theory",
which I imagine would be in this context the set of number-theoretic questions that can be solved or at least attacked by understanding the Galois groups of number fields, together with all its attached tractors of decomposition groups, inertia subgroups, and Frobenius elements. This set contains all the reciprocity laws, proved or conjectured, and many diophantine equations and problems, from Fermat's last theorem to Mordell's conjecture and its generalizations, through Birch and Swinnerton-Dyer.
My guess is that Mochizuki means that even in this relatively restricted sense, many phenomenons of number theory lie beyond the scope of Langlands. In saying so, he is following an idea that Grothendieck tried to disseminate in the 80's ("avec force", as Deligne said), which I will try to summarize as follows. The Langlands program tries to understand Galois groups by looking at their representations, that is, their action over vector spaces over fields or slightly more generally over modules over commutative rings. At any rate, these objects form an abelian category (even Tannakian), a linear object. Grothendieck's theory of Motives
should be attached to this general program, because the motives also
are  "linear objects", and as Langlands himself famously argued in his famous 1979 paper "Automorphic Representations, Shimura Varieties, and Motives. Ein Märchen".
But, Grothendieck argues, there are many other kind of objects on which we may let Galois group act in order to study them in a different direction than with representations: sets, for instance, as in the version of Galois theory developed in SGA 1 (1960), or non-abelian (profinite) groups, as the étale fundamental groups of various variety over $\mathbb Q$: to this aspect belong all the theory of Grothendieck-Teichmüller, the Dessins d'enfants, the anabelian geometry. That is of course a subject in which Mochizuki has himself proved some remarkable results, and which he claims to have enormously developed up to the point of deducing the ABC conjecture.
An example of a number theoretic (in the restricted sense mentioned above) problem that arguably lies beyond the Langlands program but that
the practicians of anabelian geometry plans to study successfully is Mordell's conjecture (proved by Faltings, it is true, with methods not so far from the Langlands program, at least in that they study "linear objects", namely abelian varieties) and all its generalizations (still open). For a very interesting, if speculative, discussion of these questions, see "Galois Theory and Diophantine Geometry" by Minhyong Kim.

REPLY [9 votes]: The basic idea is that sometimes arithmetic objects aren't "nice enough" for the tools that support the Langlands program (representation theory, noncommutative harmonic analysis, etc) to work.
They work fine, pretty much by definition, as long as the arithemtic object is essentially linear ($GL_n(\mathbb{A})$, cohomologies, Galois groups...).
But when you consider the kind of highly non-linear arithmetic objects (étale fundamental groups, moduli spaces...) that anabelian geometry studies, those theories just aren't meant to apply.
I would recomend section 1.3.3 of "Survey of the Hodge-Arakelov Theory of Elliptic Curves I" (pages 17 and 18), called "The Meaning of Nonlinearity", for a concrete (yet philosophical) example of this kind of reasoning in Mochizuki's own words.
For a pre-anabelian example of the perceived limitations of the Langlands tools and techniques, notice that after all these years, no natural proof (from the point of view of the theory of automorphic representations) exists of the Ramanujan conjecture for classical modular forms.


TITLE: Auxiliary trigonometric identities [Identidades Trigonométricas Auxiliares]
QUESTION [1 upvotes]: I need to show two auxiliary trigonometric identities:
1) $\sec^2x = \tan ^2x + 1 (\cos x \neq 0)$
2) $\csc^2x = \cot^2x +1 (\sin x \neq 0)$
How could I do it?

[Original Portuguese]
Identidades Trigonométricas Auxiliares
Sendo preciso mostrar a demonstração das duas fórmulas das IDENTIDADES TRIGONOMÉTRICAS AUXILIARES:
1) sec²x = tg²x + 1 (cos x ≠ 0)
2) cossec²x = cotg²x = 1 (sen x ≠ 0)
Como seria possível fazê-lo???

REPLY [1 votes]: Starting from $\sin^2 x + \cos^2 x = 1$, you get, dividing by $\cos^2 x$ (if this is $\ne 0$)
\[ \frac{\sin^2x}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x} = \frac 1{\cos^2 x} \]
which gives
\[  \tan^2 x + 1 = \mathrm{sec}^2 x \]
Dividing the first identity by $\sin^2 x$ we get
\[ \frac{\sin^2x}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x} = \frac 1{\sin^2 x} \]
which gives
\[  1 + \cot^2 x = \mathrm{csc}^2 x \]


TITLE: Dimension of the intersection of $2$ subspaces: $\dim(W \cap X) \leq \min \{\dim(W), \dim(X)\}$
QUESTION [1 upvotes]: Let $V$ be an $n$-dimensional vector space, $W$ a subspace of $V$, and $X$ another subspace of $V$.
Prove that
$$
\dim(W \cap X) \leq \min \{\dim(W), \dim(X)\}.
$$
I'm not sure the best way to prove it. Any idea? Thanks!

REPLY [1 votes]: $W \cap X \subseteq W$ and $W \cap X \subseteq X$. So, any basis for $W$ spans $W \cap X$. Hence, the dimension of $W \cap X$ is at most the dimension of $W$.
The same applies for $X$, and the result follows immediately.


TITLE: Computing $\sum\limits_{k=-\infty}^{+\infty}\frac{\left(-1\right)^{k}}{\alpha k +\beta}$
QUESTION [2 upvotes]: I wonder how we can prove that
$$
\sum_{k=-\infty}^{+\infty}\frac{\left(-1\right)^{k}}{\alpha k +\beta}=\frac{\pi}{\alpha \sin\left(\displaystyle \frac{\beta}{\alpha}\pi\right)}
$$
without writing it as an integral. I've really no idea on how to proceed, any hints?

REPLY [0 votes]: Assume we may write some function $f$ as a product over its roots, i.e. we may write
$$f(x)=\prod_{f(w)=0}(x-w).$$
Then on one hand we see that
$$\frac1{f(x)}=\prod_{f(w)=0}\frac1{x-w}$$
and we can do some fraction decomposition and assume that we may write
$$\prod_{f(w)=0}\frac1{x-w}=\sum_{f(w)=0}\frac{b(w)}{x-w}$$
for some coefficients $b(w)$. We can multiply through by $f(x)$ and see that this becomes
$$1=\sum_{f(w)=0}b(w)\prod_{{f(a)=0}\atop{a\ne w}}(x-a)$$
so we plug in $x=\phi$ for some $f(\phi)=0$ and see that every term on the RHS vanishes except for the case $w=\phi$ and we're left with
$$b(\phi)=\prod_{{f(a)=0}\atop{a\ne \phi}}\frac1{\phi-a}.$$
On the other hand, we have
$$\ln f(x)=\sum_{f(w)=0}\ln(x-w)$$
which implies that
$$\frac{f'(x)}{f(x)}=\sum_{f(w)=0}\frac1{x-w}$$
which is
$$f'(x)=\sum_{f(w)=0}\prod_{{f(a)=0}\atop{a\ne w}}(x-a).$$
Then assuming that $f(w)=0\Rightarrow f'(w)\ne 0$ we have, for any $f(\phi)=0$
$$f'(\phi)=\prod_{{f(a)=0}\atop{a\ne\phi}}(\phi-a)$$
and consequently
$$\frac1{f(x)}=\sum_{f(w)=0}\frac1{(x-w)f'(w)}.\tag 1$$
We take this a step further by seeing that the roots of $f(x)=\sin\pi x$ are the integers. Indeed, it was Euler who wrote that
$$\sin\pi x=\pi x\prod_{n\ge1}\left(1-\frac{x^2}{n^2}\right)=\pi x\prod_{n\ne0}\left(1+\frac{x}{n}\right).$$
We apply $(1)$ to $f(x)=\sin\pi x$ and get
$$\frac1{\sin\pi x}=\sum_{k\in\Bbb Z}\frac{1}{(x+k)\pi\cos\pi k}=\frac1\pi\sum_{k\in\Bbb Z}\frac{(-1)^k}{x+k}.$$
Plugging in $x=\beta/\alpha$ gives the result in question.


TITLE: How do I apply this maximum principle?
QUESTION [2 upvotes]: I have the maximum principle: $$\text{If } \psi\geq 0 \text{ on }\Gamma. \text{Then }L\psi\geq0\text{ implies }  \psi\geq 0 \text{ in } \bar{D},$$ where $D=(0,1)\times (0,T], \Gamma$ is parabolic boundary. The operator is defined as $$Lu:=-u_{xx}+bu+du_t=f,$$ where $b,d,f$ are functions and $b\geq\beta>0,d\geq\delta>0$. I would like to show that $$||u||\leq(1+\alpha T)\max\{||Lu||,||u||_\Gamma\},$$ where $\alpha=\max\{0,(1-b)/d\}\leq1/\delta$. It is supposed to be an immediate consequence of the Maximum principle but I am stuck.
I have tried to apply the Maximum principle on $$\phi^{\pm}=\max\{||Lu||,||u||_\Gamma\}\pm u$$ but I don't seem to go anywhere.

REPLY [3 votes]: The boundary $\Gamma$ is
$$
\Gamma=\{(t,x)\in \{0\}\times (0,1)\}\cup \{(t,x)\in (0,T)\times \{0\}\}\cup
\{(t,x)\in (0,T)\times \{1\}\}.
$$
You consider the problem
$$
-u_{xx}+bu+du_t=f \mbox{ in }(0,1)\times(0,T), \quad u=\phi \mbox{ on } \Gamma.
$$
Consider $v=u+M$. It satisfies
$$
-v_{xx}+bv+dv_t=f+bM \mbox{ in }(0,1)\times(0,T), \quad u=\phi+M \mbox{ on } \Gamma.
$$
Choose $M$ so that $f+bM\geq0$ and $\phi+M\geq0$ and you obtain from the maximum principle you stated that $v\geq0$, in other words
$$
u\geq-M=-\max(\inf(-\phi),\inf(-f/b))\geq -\max(\|\phi\|_\infty,\|\frac{f}{b}\|_\infty).
$$
Symmetrically, consider now the problem satisfied by $-u$, (which means just replacing $f$ and $\phi$ by $-f$ and $-\phi$). By exactly the same argument,
$$
-u\geq -N = - \max(\inf(\phi),\inf(f/b)) \geq -\max(\|\phi\|_\infty,\|\frac{f}{b}\|_\infty)..
$$
so altogether
$$
\|u\|_\infty\leq \max(\|\phi\|_\infty,\|\frac{f}{b}\|_\infty)\leq \max\left(1,\frac{1}{\beta}\right)\max(\|\phi\|_\infty,\|f\|_\infty).
$$
This is not the constant you wanted but your constant is puzzling me. You seem to say
$$
b\leq\beta,d\geq\delta\implies\max(0,\frac{b-1}{d})\leq\delta^{-1},
$$
this is not true in general, take for example $\delta<1$, and $b=\beta=1+\frac{1}{\delta^N}$, you obtain $$\max(0,\frac{b-1}{d})=\delta^{-N-1}>\delta^{-1}.$$
If you want to put some $\delta$ in the formula, you can, before you start, consider the problem satisfied by $\tilde{u}=u\exp(-\lambda t)$. Then, $f$ is changed to $\exp(-\lambda t)f$, $\phi$ is changed to $\exp(-\lambda t)\phi$, and $b$ is changed to $b+\lambda d$. The same argument then applies to $\tilde u$
Then you have instead, as long as $\beta+\lambda \delta >0$,
$$
\|u\|_\infty\leq\exp(\lambda T)\|\tilde{u}\|_\infty\\ \leq \max\left(1,\frac{1}{\beta+\lambda \delta}\right)\exp(\lambda T)\max(\|\exp(-\lambda t)\phi\|_\infty,\|\exp(-\lambda t)f\|_\infty),
$$
so if $f$ and $\phi$ do not depend on time, you obtain, for $\lambda>0$,
$$
\|u\|_\infty\leq\max\left(1,\frac{1}{\beta+\lambda \delta}\right)\exp(\lambda T)\max(\|\phi\|_\infty,\|f\|_\infty),
$$.
(and a worse bound than the direct one if $-\beta/\delta<\lambda<0$).


TITLE: Formulating Linear Programming for Production Planning Problem
QUESTION [1 upvotes]: I'm going through some past papers for an exam which has the production planning problem and formulating linear programming problems in it but I've come across a question that is an odd mix of both.
It gives a limit to weekly production, the demands for the next four weeks, the costs of backlogging and inventory storage (plus their limits) and the cost of production. The objective function you're seeking is for the optimal production plan (so I assume maximising profit) and it specifically states that you should not solve the problem.
It gives all the information you'd need to solve this as a production planning problem but I've never come across a combination of the two. How do I formulate this?
So far all I've worked out is that the decision variables are the four amounts produced which must be above zero and below the weekly production limit.
Any help greatly appreciated.
EDIT: I have typed out the whole question as best I can below
A factory can make at most $10$ units a week and there is demand forecast to be $8, 14, 11, 7$ for the next four weeks. $7$ units can be held in inventory at a cost of £10 a unit/week, orders for up to $3$ units can be backlogged at a cost of £15 a unit/week. There is currently no inventory and no orders backlogged and we want none at the end of the four weeks.
The cost of production is
$0$ if $n=0$
$100$ if $0<n\leq6$
$130$ if $6<n\leq8$
$150$ if $9<n\leq10$

REPLY [1 votes]: In addition to production volumes, the amount of inventory carried and the amount of demand backlogged in each week are decisions, and you will need constraints (sometimes referred to as "flow balance" constraints) to make sure the amounts match up correctly.
You also need to be concerned whether the production costs are per-unit or for a batch, and if per-unit whether they are incremental or cumulative. Looking at what you wrote and postulating a production run of 9.2 units, I can picture three possibilities for production cost:

9.2 * 150;
150 (for a batch of anywhere between 9 and 10); or
100 * 6 + 130 * 2 + 150 * 1.2 (incremental).

I believe that two of those cases will require integer variables.


TITLE: What is the genus of an unramified extension of a function field?
QUESTION [2 upvotes]: Suppose L/K is a degree $n$m separable, geometric extensions of function fields where both function fields have constants $\mathbb{F}_q$. Suppose also that $L$ is unramified over $K$. Let $g_L$ be the genus of $L$ and let $g_K$ be the genus of $K$.
Riemann-Hurwitz tells us that
$$2g_L - 2 = n(2g_K -2) + \text{deg}_L\left(\text{Diff}(L/K)\right)$$
where $\text{Diff}(L/K)$ is the different divisor. So $(\text{Diff}(L/K) = \sum_{\frak{p}}c_{\frak p}\frak p$ where all but finitely many of the $c_{\frak p}$s are zero. Further, since $L/K$ is an unramified extensions here, $c_{\frak p}=0$ for all primes $\frak p$.
Then by Riemann-Hurtiz we have then that $g_L=\frac{n}{2}(2g_K-2)+1$.
Two questions:

Is this computation correct?
If so, suppose $g_K=0$. If $n>1$ we have $g_L < 0$. Does this mean that genus $0$ function fields don't have unramified extensions?

REPLY [1 votes]: Yes, this is true.

This is analogous to the result from algebraic number theory that any algebraic extension of $\Bbb{Q}$ is ramified above some prime $p$.
The genus zero rational function field $K=\Bbb{F}_q(x)$ has no unramified extensions with $\Bbb{F}_q$ as its field of constants. In many cases this can be seen directly. For example, when $q$ is odd, a quadratic extension $L/K$ is of the form $L=K(y)$ with $y^2=f(x)$, $f(x)\in K\setminus \Bbb{F}_q$ square-free. It is then easy to see that the places associated to factors of the numerator or the denominator of $f(x)$ ramify.
Another consequence is that an unramified extension of the function field of an elliptic curve (so $g=1$) also has genus one. IIRC this has applications to the study of morphisms between elliptic curves.


TITLE: Convergence of $\prod_{n=1}^{\infty} (1 - z^n)$ when $|z| = 1$
QUESTION [0 upvotes]: Clearly, $\prod_{n=1}^{\infty} (1 - z^n)$ converges to $0$ when $1 - z^{n_0} = 0$ for some $n_0$.
Suppose $z$ is a complex number with $|z| = 1$ and $1 - z^{n} \ne 0$ for all $n$.
I think $\prod_{n=1}^{\infty} (1 - z^n)$ diverges, but it is not clear how to prove this.
I know that $\prod_{n=1}^{\infty} (1 - z^n)$ converges to a non-zero number iff $\sum_{n=1}^{\infty} -z^n$ converges absolutely. Hence, if $\prod_{n=1}^{\infty} (1 - z^n)$ converges, it must converge to $0$. It is not clear how to proceed from here.
I would appreciate any hints/pointers for proving (or disproving) that $\prod_{n=1}^{\infty} (1 - z^n)$ is divergent.

REPLY [3 votes]: If the product converges to a  non-zero number then  $1-z^{n } \to 1$ so $z^{n} \to 0$ as $ n \to \infty$. But $|z^{n}|=1$ so the product does not converge (to a non-zero number) for any $z$ on the unit circle.
[Convergence of an infinite product to $0$ is considered as divergence].


TITLE: How to obtain this partial fraction decomposition?
QUESTION [2 upvotes]: I am studying Laplace transforms right now and got stuck at this step that involves a weird partial fraction decomposition. It looks like the instructor skipped a bunch of steps and assigned numerators to a bunch of the fractions without assigning them dummy variables. Any idea how he got to this step so quickly?
I tried doing this the standard way (i.e. breaking each polynomial in the denominator into its own fraction, assigning variables for each numerator, and trying to solve), but it got messy very quickly.
laplace transform up to the part where I got stuck
$$
X=\frac{2}{s^4-1}\left(e^{-s}\cdot\frac{1}{s}-e^{-2s}\cdot\frac{1}{s}\right)=\frac{2}{\left(s^4-1\right)s}\left(e^{-s}-e^{-2s}\right)=2\cdot\frac{1}{\left(s^2+1\right)\left(s+1\right)\left(s-1\right)s}\left(e^{-s}-e^{-2s}\right)=2\left(\frac{As+B}{s^2+1}+\frac{\left(\frac{1}{4}\right)}{\left(s+1\right)}+\frac{\left(\frac{1}{4}\right)}{s-1}+\frac{\left(-1\right)}{s}\right)\left(e^{-s}-e^{-2s}\right)
$$

REPLY [2 votes]: Let's start from the general form of the decomposition of
$${1\over s(s-1)(s+1)(s^2+1)}={a\over s}+{b\over s-1}+{c\over s+1}+{ds+e\over s^2+1}$$
Now multiply both sides by $s$ then set $s=0$ you get $a=-1$
Multiply both sides by $s-1$ then set $s=1$ you get $b={1\over 4}$
Multiply both sides by $s+1$ then set $s=-1$ you get $c={1\over 4}$
Finally multiply by $s^2+1$ then set $s=i$ you get
$${1\over i(i^2-1)}=di+e={i\over 2}$$
And this leads to $d=1/2$ and $e=0$


TITLE: Confusion over naming Line and Ray
QUESTION [0 upvotes]: <-A-----------B--------------C->

A line is passing through points A, B and C.
As far I know, here :
Line: AC
Segment: AB, BC and AC
Ray: BA and BC
My confusion is about:
In case of segment we are considering point A and C as end point.
Now can I take, in case of Ray, A as an end point, and write,
Ray: AB or AC
Can I write Line AC as AB ???
Thank you.

REPLY [1 votes]: The line can be defined equivalently through any 2 distinct points on it. So $AC, AB, BC, CA, CB, BA$ all refer to the same line. Segments are uniquely identified by 2 points, segment $AB$ is the same as $BA$ but it's different than $AC, BC$. Finally for a ray the order in which you give the points matters, so $AB$ is the opposite ray to $BA$.
More formally you can define a line parametrically as the affine linear combination of two points $A,B \in \mathbb{R}^n$:
$$(1-\lambda)A + \lambda B, \lambda \in \mathbb{R}$$
You can define a line segment as the convex linear combination of two points:
$$(1-\lambda)A + \lambda B, \lambda \in [0,1]$$
Finally you can define a ray as an affine linear combination where $\lambda$ is additionally restricted to be non-negative:
$$(1-\lambda)A + \lambda B, \lambda \in [0,+\infty)$$


TITLE: Exercise 18b in Schutz's First course in GR
QUESTION [1 upvotes]: The question is as follows:

Show that a timelike vector and a non-zero null vector cannot be orthogonal.

So we have a timelike vector $\vec{A}$, s.t $\vec{A}^2<0$; and a non-zero null vector, $\vec{B}$ s.t $\vec{B}^2=0$.
So we get that $\vec{A}\cdot \vec{B} = \frac{(\vec{A}+\vec{B})^2-\vec{A}^2}{2}$.
I need to show that $\vec{A}\cdot \vec{B} \ne 0$, if it were otherwise then:
$$(\vec{A}+\vec{B})^2 = \vec{A}^2$$
How to continues from here?

REPLY [0 votes]: It is tempting to generalize OP's question to the following proposition.

Proposition. Given an $n$-dimensional vector space $V$ over a field $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$ with (possibly degenerate) indefinite sesquilinear form $\beta:V\times V \to \mathbb{F}$ of signature $(n_+,n_0,n_-)$, with  $n_+ + n_0 + n_-=n$, where it is assumed that $$n_-~=~1.$$ Then
  $$ \forall x,y~\in~ V :\quad \beta(x,x)~\leq~0\quad \wedge\quad \beta(y,y)~\leq~0\quad \wedge\quad \beta(x,y)~=~0$$
  $$\quad \Rightarrow\quad  \beta(x,x)~=~0~=~\beta(y,y)\quad\vee\quad x~\in~V^{\perp}\quad\vee\quad y~\in~V^{\perp}. $$

In the non-degenerate case, the proposition states that two non-zero, mutually orthogonal, non-spacelike vectors are both lightlike.
Sketched proof:

Pick an arbitrary (but fixed) positive definite, sesquilinear form $\langle \cdot, \cdot \rangle: V\times V \to \mathbb{F}$.
Find the unique Hermitian linear operator $B:V\to V$ such that
$$\forall x,y~\in~ V:\quad\beta( x, y )~=~\langle x, By \rangle.$$
Decompose the vector space $$V~=~V_+\oplus V_0 \oplus V_-$$ in eigenspaces for $B$, which are mutually orthogonal wrt. $\langle \cdot, \cdot \rangle$. Here $V_0=V^{\perp}$ and
$$n_+ ~=~\dim V_+,\quad n_0 ~=~\dim V_0,\quad n_- ~=~\dim V_-~=~1.$$
Note that, in a hopefully obvious notation,
$$\beta(x,y)~=~\beta(x_+,y_+)+\beta(x_-,y_-),$$
etc.
Assume that $x,y\notin V^{\perp}$. From assumptions, we get that
$$ 0~\leq~\beta( x_+, x_+)~\leq~ -\beta( x_-, x_-)~>~0,$$
$$ 0~\leq~\beta( y_+, y_+)~\leq~ -\beta( y_-, y_-)~>~0, $$
$$ 0~\leq~\beta( x_+, y_+)~=~ -\beta( x_-, y_-). $$
From Cauchy-Schwarz inequality for a positive definite sesquilinear form, we get that
$$|\beta( x_+, y_+ )|^2~\leq~\beta( x_+, x_+)\beta( y_+, y_+)$$
and
$$ \beta( x_-, x_-)\beta( y_-, y_-) ~=~|\beta( x_-, y_- )|^2.$$
The latter since $ x_- \parallel y_- $.
Combine the above to conclude that
$$ \beta( x_+, x_+)~=~ -\beta( x_-, x_-)$$
and
$$ \beta( y_+, y_+)~=~ -\beta( y_-, y_-). $$


TITLE: What is the best reference for Spectral theory?
QUESTION [12 upvotes]: I'm studying  Bernard Aupetit: A Primer on Spectral Theory
but the textbook we are using is a little bit heavy going for me. Is there a best book to learn about these things?
Thank you.

REPLY [7 votes]: As nobody has mentioned Reed and Simon yet, I will do so:

Reed, Michael and Simon, Barry.  Methods of Modern Mathematical Physics, Academic Press, 1980.

Of course "spectral theory" means different things to different people, depending on what they plan on doing with it.  As the title suggests, Reed and Simon is in principle aimed at mathematical physicists (quantum mechanics, etc) but it is an honest mathematics textbook (all theorems are proved, etc).  The first volume begins with the basics of functional analysis and ends with the spectral theorem, and volumes 2-4 proceed from there.  I think it's a good treatment for any "working analyst"; it's well motivated and down to earth.  Applications to topics such as PDE are made more explicit than in texts such as Conway.  There are also a ton of really good exercises.
This book is a little unusual (in what I think is a good way) in that it includes, and shows the benefits of, Halmos's "multiplication operator" version of the spectral theorem, mentioned already by Jon Bannon.
Unfortunately the current edition is very expensive, so you may want to try to borrow it from a library or colleague at first.


text:
\begin{definition}[Definition:P-Product Metric]
Let $M_{1'} = \struct {A_{1'}, d_{1'} }$ and $M_{2'} = \struct {A_{2'}, d_{2'} }$ be [[Definition:Metric Space|metric spaces]].
Let $A_{1'} \times A_{2'}$ be the [[Definition:Cartesian Product|cartesian product]] of $A_{1'}$ and $A_{2'}$.
Let $p \in \R_{\ge 1}$.
The '''$p$-product metric''' on $A_{1'} \times A_{2'}$ is defined as:
:$\map {d_p} {x, y} := \paren {\paren {\map {d_{1'} } {x_1, y_1} }^p + \paren {\map {d_{2'} } {x_2, y_2} }^p}^{1/p}$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in A_{1'} \times A_{2'}$.
The [[Definition:Metric Space|metric space]] $\mathcal M_p := \struct {A_{1'} \times A_{2'}, d_p}$ is the '''$p$-product (space)''' of $M_{1'}$ and $M_{2'}$.
\end{definition}

config: wiki


TITLE: What does "the map $i$ is transverse to the vector field $X$ everywhere" mean?
QUESTION [0 upvotes]: Geometric Theory of Foliations, Page 28. The Lines 7 and 8 from below.
What does "the map $i$ is transverse to the vector field $X$ everywhere" mean?

REPLY [1 votes]: It means that $i(B^{m-1})$ is nowhere tangent to $X$. You can infer this from the next two sentences as well since $\Phi^{-1}|_{V}$ is a diffeomorphism.


TITLE: How to find the closest bounded rational approximation to a rational number?
QUESTION [2 upvotes]: Say I have a rational number $a/b$ and I want to find its closest rational approximation $x/y$ where
$$x_- \leq x \leq x_+$$
$$y_- \leq y \leq y_+$$
for some constants $x_\pm$, $y_\pm$.
How can I compute the integers $x$ and $y$ efficiently, given the rest?

REPLY [1 votes]: Go through the following PDFs:
http://www.ams.org/bookstore/pspdf/mbk-48-prev.pdf &
https://www.dropbox.com/s/7ahnld1wtvqiurz/Continued%20fraction%20expansion.pdf?dl=0
This will clarify your issue.


TITLE: Irreducible Representations from Cartan Matrix
QUESTION [5 upvotes]: I need some help understanding how you can construct the irreducible representations of an algebra from knowing its roots or its Cartan matrix, which I am being told you most certainly can. Unfortunately, despite my sincerest attempts, I have been unable to understand all the instructions for how to do this that I have hitherto come across or been sent. The closest I have come to something that I understand is chapter 9.2 of Howard Georgi's Lie Algebras in Particle Physics, but I am still stumbling in the dark here, and so an example of the procedure would be very, very nice.
Let's say we're looking at the algebra $SU(3)$ which has roots $$\pm (1,0) ,\quad  \pm (\frac{1}{2}, \frac{\sqrt{3}}{2}) , \quad  \pm (\frac{1}{2},- \frac{\sqrt{3}}{2})$$ and Cartan matrix
\begin{equation*}
K_{ij} = \begin{pmatrix}
2 & -1 \\ -1 & 2
\end{pmatrix}
\end{equation*}
How do I go about constructing the matrices for the generators of this algebra in the representation with Dynkin coefficients (1,1), that is, $\mathbf{8}$?

REPLY [3 votes]: This task is more complex than the task to solve a quadratic equation, for example, and one must master a significant portion of a textbook – such as Georgi's textbook – and perhaps something beyond it to have everything he needs.
For the 8-dimensional representation of $SU(3)$, things simplify because it's the "adjoint rep" of $SU(3)$ – the vector space that formally coincides with the Lie algebra itself. And the action of the generator $G_i$ on the basis vector $V_j=G_j$ of the adjoint representation is given by
$$ G_i (V_j) = [G_i,V_j]= \sum_k f_{ij}{}^k G_k $$
This implies that the structure constants $f$ directly encode the matrix elements of the generator $G_i$ with respect to the adjoint representation – $j$ and $k$ label the row and the column, respectively.
The structure constants $f$ determining the commutators may be extracted from all the roots. The whole mathematical structure is beautiful but the decomposition of the generators under the Cartan subalgebra has several pieces, and therefore an even greater number of different types of "pairs of pieces" that appear as the commutators.
Some ($r$, rank) of the generators $G_i$ are identified with the Cartan generators $u_a$. The rest of the generators $G_j$ are uniquely associated with all the roots.
If you only have the Cartan matrix, you effectively have the inner products of the simple roots only. You first need to get all the roots, and those are connected with the $d-r$ (dimension minus rank) root vectors $r_j$.
The commutators of two Cartan generators vanish,
$$[h_i,h_j]=0$$
The commutator of a Cartan generator with a non-Cartan generator is given by
$$[h_i,G_{r(j)}] = r_i G_{r(j)}$$
because we organized the non-Cartan generators as simultaneous eigenstates under all the Cartan generators. Finally, the commutator
$$[G_{r(i)},G_{r(j)}]$$
is zero if $r_i=r_j$. It is a natural linear combination of the $h_i$ generators if the root vectors obey $r_i=-r_j$. If $r_i+r_j$ is a vector that isn't a root vector, the commutator has to vanish. And if $r_i+r_j$ is a root vector but $r_i\neq \pm r_j$, then the commutator is proportional to $G_{r(i)+r(j)}$ corresponding to this "sum" root vector. The coefficient (mostly sign) in front of this commutator is subtle.
Once you have all these commutators, you have effectively restored all the structure constants $f$, and therefore all the matrix entries with respect to the adjoint representation.
To find matrix elements for a general representation is much more complex. You must first figure out what the representations are. Typically, you want to start with the fundamental (and/or antifundamental) rep, and all others may be obtained as terms composing a direct sum decomposition of tensor products of many copies of the fundamental (and/or antifundamental, if it is different) representation.
All the representations may be obtained from the weight lattice which is a subset of the root lattice and is similar. In fact, the weight lattice is the dual (in the "form" vector space sense) of the root lattice under the natural inner product.
In practice, physicists don't ever do the procedures in this order because that's not how Nature asks us to solve problems. We learn how to deal with the groups we need – which, at some moment, includes all the compact simple Lie groups as the "core" (special unitary, orthogonal, symplectic, and five exceptional), and we learn the reps of these Lie groups – the obvious fundamental ones, the adjoint, and the pattern how to get the more complicated ones.
I am afraid that it doesn't make any sense to fill in any "gaps" if you would need to elaborate upon something because in this way, one would gradually be forced to write another textbook on Lie groups and representation theory as this answer, and I don't think that such a work would be appropriately rewarded – even this work wasn't. ;-)