TITLE: Extensions of a simple group by an elementary abelian $p$-group
QUESTION [3 upvotes]: Let $V$ be an elementary abelian $p$-group of size $p^n$. Let $G$ be a finite group with $V\unlhd G$ such that $G/V=H$ is simple (like $\operatorname{PSL}(m,q)$ with $q$ a power of $p$ or any other finite simple group of Lie type in characteristic $p$). Moreover, $V$ is a faithful irreducible $H$-module (equivalently, $V$ is a minimal normal subgroup of $G$ and $C_G(V)=V$).
Ideally, I would like to show that $G$ must be the semidirect product $V\rtimes H$, or $|V|$ must be somewhat large compared to $|H|$. A quick example: when $p=2$ and $H=\operatorname{PSL}(n,2)$, so that $V$ is the natural $n$-dimemsional $\mathbb{F}_2$-module for $H$, according to "U. Dempwolff, On the second cohomology of $\operatorname{GL}(n,2)$", $G$ must be the mentioned semidirect product.
Any suggestions and references are appreciated.

REPLY [7 votes]: I think this can often fail even for the non-trivial representation $V$ of smallest possible dimension.
For a reductive group $G$ over $\mathbf{Z}_p$, the group $G(\mathbf{Z}/p^2 \mathbf{Z})$ will typically be a non-split extension of $G(\mathbf{F}_p)$ by the adjoint representation $V$, and for various $G$ the adjoint representation $V$ will be the smallest non-trivial representation. For example, take $G = E_8$.
An even simpler example is the group $\Gamma = \mathrm{PSL}_2(\mathbf{F}_p)$ for $p > 3$. Then the smallest non-trivial representation over $\mathbf{F}_p$ will be the adjoint representation $V$ of dimension $3$, but $\mathrm{PSL}_2(\mathbf{Z}/p^2 \mathbf{Z})$ gives a non-split extension of $\Gamma$ by $V$, since the lift of any unipotent element in $\Gamma$ will have order $p^2$.


TITLE: existence of a prime $p$ for which $a$ is a primitive root
QUESTION [2 upvotes]: It is known that every prime $p$ has a primitive root modulo $p$. Is every number $a$ which is not a perfect square a primitive root modulo $p$ for some prime $p$? If it is a square, we already have $a^{\frac{p - 1}{2}} \equiv 1 \text{ (mod $p$)}$, so the condition on not being a square is necessary. However, I'm not sure what would suggest that not being a perfect square is sufficient.

REPLY [2 votes]: I believe that the full answer is not yet known. Certainly the (much) more difficult problem of whether every number other than $-1$ and the perfect squares is a primitive root of infinitely many primes is not fully settled. This is Artin's Conjecture.
However, there are many partial results towards Artin's Conjecture. for example, Hooley proved Artin's Conjecture, assuming an (unproved) version of the Generalized Riemann Hypothesis.


text:
\section{Broader Impact}
This research has a direct impact on the theoretical understanding of bilevel optimization employed in various deep-learning applications such as hyperparameter optimization, neural architecture search, adversarial robustness and gradient-based meta-learning methods (MAML), which are used in robotics, language, and vision. We have rigorously demonstrated some of the failure cases for convergence in the bilevel optimization framework, which may benefit both practitioners and theoreticians alike. Our proof techniques may also be extended for future work on the theory of gradient based adaptation. Furthermore, we have shown that our relatively simple modification is memory efficient, which can be scaled for applications, potentially allowing better democratization (i.e. cost reduction) of deep learning and reducing intensive computation usage, energy consumption \cite{energy} and $\text{CO}_2$ emission \cite{co2}.

config: arxiv


TITLE: How does Wiles' proof fail at $n=2$?
QUESTION [46 upvotes]: The content is miles outside what I know about.
So the question is a mixture of idle curiosity and maybe having this answered somewhere on the Internet. It is likely I will not be able to understand the answer.

How exactly does Wiles' proof of Fermat's Last Theorem fail for $n=2$?

REPLY [15 votes]: The other "accepted" answer in my opinion more or less misses the point. So here is another answer.
Consider the elliptic curve
$$E: y^2 = x(x - a^p)(x + b^p).$$
The $p$-torsion points $E[p]$ of $E$ form a $2$-dimensional vector space over $\mathbf{F}_p$ which has an action of the Galois group, and this gives rise to a representation
$$\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_p).$$
The structure of the proof is based on the idea that the shape of $E$ implies that $\overline{\rho}$ has very special ramification properties. Assuming that $E$ and hence $\overline{\rho}$ is
modular, Ribet's argument shows that either $\overline{\rho}$ should come from another modular form of a weight and level which doesn not exist (which is a contradiction, and this is the part which uses modularity) or $\overline{\rho}$ is not absolutely irreducible.
In particular, in order to complete the proof, one has to rule out the possibility that $E[p]$ is not absolutely irreducible. Assuming one can massage the choices of $a$ and $b$ to guarantee that $E$ is semistable, this turns out to be equivalent to showing that $E$ does not have a ratio
nal point of order $p$. But since $E$ already has rational $2$-torsion, this is not possible for
primes $p \ge 5$ by Mazur's theorem (https://en.wikipedia.org/wiki/Torsion_conjecture). Without
Mazur's theorem (or something equivalent or stronger) you can't prove Fermat.
Returning to the case $p = 2$, the key reason the argument fails is that $\overline{\rho}$ can be (and indeed is) reducible, but knowing that $E[2]$ is reducible is not any contradiction to Mazur's theorem. In particular, you can write down $E: y^2 = x(x-9)(x+16)$ and then this is a perfectly good elliptic curve with rational $2$-torsion points.


TITLE: Double checking complex number reduction to x +iy form
QUESTION [0 upvotes]: Express $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n $$
in the form $ x + iy$

$$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n =  \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \cdot \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta} \right)^n $$
$$ \left( \frac{\left( 1 + \sin\theta + i\cos\theta \right)^2}{\left(1 + \sin\theta\right)^2 - \left(i\cos\theta\right)^2} \right)^n  = \left(\frac{\sin^2\theta}{1 + \sin\theta}+i\cos\theta\right)^n$$
then, if necessary, we can further expand via Newton's Binomial Theorem:
$$ \left( x + y \right)^n = {{n}\choose{k}}x^{n-k}y^k$$
Am I missing something?

REPLY [1 votes]: Go through exponential form.
$$\sin \theta+ i \cos\theta = i e^{-i\theta}$$
$$\sin \theta- i \cos\theta = -i e^{+i\theta}$$
Your expression is:
$$\left(\frac{1+ie^{-i\theta}}{1-ie^{i\theta}}\right)^n=$$
$$\left(\frac{(i)e^{-i\theta}(-i)e^{i\theta}(1+ie^{-i\theta})}{1-ie^{i\theta}}\right)^n=$$
$$\left(\frac{i e^{-i\theta}(-ie^{i\theta}+1)}{1-ie^{i\theta}}\right)^n=$$
$$\left(i e^{-i\theta}\right)^n=\left( e^{-i\theta+i\pi/2}\right)^n=e^{in(\pi/2-\theta)}=\cos (n(\pi/2-\theta))+i\sin(n(\pi/2-\theta))$$
You'll notice the most awkward part is that sine and cosine are exchanged compared to the Euler's formula. For that reason, using a new variable $u=\frac{\pi}{2}-\theta$ for most of the calculation makes things much cleaner and show the true beauty of this solution:

$$\sin\theta+i\cos\theta=\cos u +i \sin u = e^{iu}$$
$$\sin\theta-i\cos\theta=\cos u -i \sin u = e^{-iu}$$
Now you don't have the awkward $i$'s there:
$$\left(\frac{1+e^{iu}}{1+e^{-iu}}\right)^n=\left(e^{iu}\right)^n=e^{inu}=\cos nu+i\sin nu$$


TITLE: Hint: vector space $V=\{f:S\rightarrow F\}$ is a finite dimensional vector space over F if and only if $S$ is finite.
QUESTION [0 upvotes]: No answers, please! Just a hint.
Let $S$ be a set and let $F$ be a field. Prove that the vector space $V=\{f:S\rightarrow F\}$ of all maps from $S$ to $F$ is a finite-dimensional vector space over $F$ if and only if $S$ is finite.
EDIT: Thank you all for the hints. This is my proof. How is it?
Suppose $S$ is finite. Let's define $\delta_s$ as follows:
$$\delta_s(x)=\begin{cases}
1&x=s\\
0&x\neq s
\end{cases}$$
We can express any function $f$ in terms of $\{\delta_s \mid s \in S\}$: $f = \sum_{s \in S} a_i \delta_s$, so $\{\delta_s \mid s \in S\}$ is a finite spanning set. Thus, $V$ is a finite-dimensional vector space. Conversely, suppose to the contrary that $S$ is infinite. Then  $\{\delta_s \mid s \in S\}$ is an infinite spanning set. However, this contradicts the fact that $V$ is a finite-dimensional vector space, and so $S$ must be finite.

REPLY [1 votes]: Hint: think about the functions $(\delta_s:s\in S)$ defined by $\delta_s(x)=1$ if $x=s$ and $\delta_s(x)=0$ if $x\ne s$.

REPLY [0 votes]: Hints:
Define $f+g$ and $af$ for $a\in F$ and $f,g\in V$.
Find a basis of $V$.


TITLE: Transform a positive integer to find its next greatest factor
QUESTION [1 upvotes]: Suppose I am trying to find factors of a particular positive integer num. Suppose I also have a function findGreatestFactor(num) that finds the greatest factor of a positive integer.
Is there a way to transform my integer num such that it would still have all of the same factors as the original num but findGreatestFactor(num) would find the next greatest factor of the original num?
Example:
let num = 100

findGreatestFactor(num) gives you 50 (the greatest factor of 100)

let num2 = some transformation of num

findGreatestFactor(num2) gives you 25 (the second greatest factor of 100)

let num3 = same transformation of num2

findGreatestFactor(num3) gives you 20 (the third greatest factor of 100)

REPLY [4 votes]: If you consider the prime factorization $N = p_1 ^{q_1} \times p_2 ^{q_2} \times \ldots \times {p_n}^q_n$, where the $p_i$ are arranged in increasing order, then the smallest proper factor is $p_1$, and the next smallest factor is either $p_1 ^2 $ or $p_2$.
Hence. Let $FGF$ be your findgrestestfactor. $Num/FGF(Num)$ will give you $p_1$. $Take FGF(Num)$ and divide out by $p_1$ repeatedly until we no longer get an integer. Say that the last integer is $N_1$. Then taking $\frac {N_1} {FGF(N_1)}$ will give you $p_2$. If $p_1^2 \not \mid N$, then $p_2$ is the second smallest factor. Otherwise, compare $p_1 ^2 $ and $p_2$. Whichever is smaller, then $N/x$, will give you the second smallest factor.

REPLY [2 votes]: In your example, num2 would have 25 as greatest factor and 20 as second greatest factor. Then num2 must be a multiple of 100, and the greatest factor of num2 would be at least 50.
So this is impossible.


text:
{\bf Problem.} Write $0.\overline{43}$ as a simplified fraction.
{\bf Level.} Level 4
{\bf Type.} Prealgebra
{\bf Solution.} Let $a=0.\overline{43}$. Then $100a=43.\overline{43}$. Subtracting the left-hand sides $100a$ and $a$, and subtracting the right-hand sides $43.\overline{43}$ and $0.\overline{43}$ yields \begin{align*} 100a- a &= 43.\overline{43}- 0.\overline{43}\\ \Rightarrow 99a &= 43\\ \Rightarrow a &= \boxed{\frac{43}{99}}. \end{align*}

set_name: MATH


TITLE: The Hasse-Weil L-function and some equations
QUESTION [2 upvotes]: Let $f$ be an analytic function verfifying
$f(s)=\epsilon f(2-s)$
where $\epsilon=\pm 1$. The expression of Hasse-Weil L-function $f$ is
$$f(s)=N^{s/2}(2\pi)^{-s}\Gamma(s)\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$$
where $N$ is an integer and $\Gamma(s)$ is the gamma function.
Let $r$ be an integer. I have a set of equations of the form
$$f^{(k)}\left(1-2\prod_{j=1}^{k}s_{j}\right)=f^{(k)}\left(1-2\prod_{j=1}^{k}t_{j}\right)$$
for all $k=1,...,r$. Here $f^{(k)}$ is the $k$-th derivative of $f$.
Can I deduce that
$$(t_1,t_2,\ldots,t_{r})=(s_1,s_2,\ldots,s_{r})$$
under some conditions on the derivatives of $f$?
The injectivity is not possible for this case.

REPLY [1 votes]: No, that is impossible. The $k$-th derivative of a L function has necessarily infinitely many zeros. So you can choose $s_j$ and $t_j$ inductively such that the products give distinct zeros of $f^j$. Moreover, if one of $s_j$ is zero you can't say anything clever either, but I assume that you simply have forgotten that condition in your question.


TITLE: Proving the continuity of the tangent function from first principles
QUESTION [1 upvotes]: Okay, so I'm finally actually stuck on a question. I don't really have an idea of how to go forward from where I'm stuck. The question is:
Prove that $\lim_{x \to x_0} \tan(x) = \tan(x_0)$, if $x \neq (n+\frac{1}{2})\pi for n \in \mathbb{Z}$.

Proof Attempt:
Let $\epsilon>0$ be given. Then, we need a $\delta>0$ so that:
$$0<|x-x_0| < \delta \implies |\tan(x)-\tan(x_0)| < \epsilon$$
So, we play around with $|\tan(x)-\tan(x_0)|$:
$$|\tan(x)-\tan(x_0)| = |\frac{\sin(x-x_0)}{\cos(x)\cos(x_0)}| \leq \frac{|x-x_0|}{\cos(x)\cos(x_0)} < \frac{\delta}{|\cos(x)\cos(x_0)|}$$
I don't really know where to go from there? Like, I was thinking of letting $\delta = \epsilon \cdot |\cos(x)| \cdot |\cos(x_0)|$ but that would mean that my $\delta$ would depend on $x$. I'm not too sure if that's allowed.
I was also thinking of using, maybe, the continuity of $\cos(x)$ by introducing a separate $\delta$ and taking the minimum of that $\delta$ and another random number but I'm not quite sure how that would help.
I would appreciate if someone could give me a good hint on how to continue. I don't mind a full proof as well but I'm probably not going to look at your proof till I've completely had it with this problem.
Also, I know very well that I can prove this by showing that $\sin(x)$ and $\cos(x)$ are continuous, before showing that the ratio of two continuous functions is, itself, continuous. But I don't want to do it that way, if it's possible.

REPLY [2 votes]: You have the right idea where you stated

I was also thinking of using, maybe, the continuity of $\cos(x)$ by introducing a separate $\delta$ and taking the minimum of that $\delta$ and another random number but I'm not quite sure how that would help.

However, you don't need a "random number". Instead, you can use that if the second $\delta$ is small enough, $\cos(x)$ is non-zero over it's entire range. For example, you have, for all integers $n$, that
$$x_0 \neq \left(n + \frac{1}{2}\right)\pi \tag{1}\label{eq1A}$$
This means there exists an integer $n_0$ such that
$$\left(n_0 + \frac{1}{2}\right)\pi \lt x_0 \lt \left((n_0 + 1) + \frac{1}{2}\right)\pi \tag{2}\label{eq2A}$$
Choose the second delta, call it $\delta_2$, so it's a fraction less than $1$ of the minimum of the difference between $x_0$ and $\left(n_0 + \frac{1}{2}\right)\pi$, and that of the difference between $\left((n_0 + 1) + \frac{1}{2}\right)\pi$ and $x_0$. Let's say, rather arbitrarily, this fraction is $\frac{1}{2}$. In this region, then, $\cos(x)$ is not $0$ for any $x \in (x_0 - \delta_2, x_0 + \delta_2)$.
Since you stated you would appreciate just a good hint, I believe this qualifies. Nonetheless, since you also don't mind a full proof, I've provided the rest of the solution below, but put them in spoilers below.
Using this concept, you get

$$\delta_2 = \left(\frac{1}{2}\right)\min\left(x_0 - \left(n_0 + \frac{1}{2}\right)\pi,\left((n_0 + 1) + \frac{1}{2}\right)\pi - x_0\right) \tag{3}\label{eq3A}$$

Next, you can reason as follows:

Since $\cos(x) \neq 0$ for any $|x - x_0| \lt \delta_2$, this means the absolute value of $\cos(x)$ in that region has a positive minimum value, say $m$.

This gives

$$0 \lt m \le |\cos(x)| \implies \frac{1}{|\cos(x)|} \le \frac{1}{m} \; \forall \; x \in (x_0 - \delta_2, x_0 + \delta_2) \tag{4}\label{eq4A}$$

Thus, continuing your line of inequalities, you then have for $x \in (x_0 - \delta_2, x_0 + \delta_2)$ that

$$|\tan(x)-\tan(x_0)| \lt \frac{\delta}{|\cos(x)\cos(x_0)|} \le \frac{\delta}{m|\cos(x_0)|} \tag{5}\label{eq5A}$$

You would then next use the following statement:

Since your final $\delta$ needs to be at least as small as $\delta_2$ because, for any value of $\delta_2$ which is used, any smaller value will also still fulfill all of the conditions.

You would then have

$$\delta = \min(\delta_2, \epsilon \cdot m \cdot |\cos(x_0)|) \tag{6}\label{eq6A}$$

Of course, there are simpler ways to prove the continuity of $\tan(x)$, such as using that $\sin(x)$ and $\cos(x)$ are continuous, which you've stated yourself, but this shows how you can also do it directly as well.


text:
\begin{definition}[Definition:Commutative/Set]
Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $X \subseteq S$ be a [[Definition:Subset|subset]] of $S$ such that:
:$\forall a, b \in X: a \circ b = b \circ a$
That is, every [[Definition:Element|element]] of $X$ [[Definition:Commute|commutes]] with every other [[Definition:Element|element]].
Then $X$ is a '''commuting set of elements''' of $S$.
\end{definition}

config: wiki


TITLE: How does radiation transfer of heat fit in with the Clausius statement of second law of thermodynamics?
QUESTION [2 upvotes]: The Clausius statement of the second law of thermodynamics says that heat flows from a hotter body to a colder body. Heat can flow in many different mechanisms. In the mechanism of radiation for transferring heat, the body emits radiation though there may not be a temperature difference between it and outside. A simple example: consider a body in a complete vacuum, the vacuum has no defined temperature (acc. to this stack) but it still should emit radiations.
And this radiation that is emitted may travel through space and strike another body which may have hotter temperature than the body emitted it and then cause it to heat up. So, this seems like a violation of the second law.

A possible resolution: The lightwave radiated out by the body will redirect itself (somehow?) to strike only bodies colder than it... but this almost seems ridiculous to think about.

REPLY [1 votes]: Clausius's differential inequality $dS\ge \frac{\delta Q}{T}$ can also be written as an inequality between rates as follows
$$\frac{dS}{dt} = \dot S \ge \oint_{\partial \mathcal B} \frac{\dot q}{T} dA \tag{1}\label{1}.$$
In $\eqref{1}$ $\mathcal B$ is the body of the system receiving heat through its boundary $\partial \mathcal B$ at a rate $\dot q$ and the temperature of the surface element $dA$ is $T=T(dA)$. As written this inequality has only "surface heat sources" but it can be generalized to include "volume heat sources"; Truesdell calls it the Clausius-Duhem inequality[1] :
$$\frac{dS}{dt} = \dot S \ge \oint_{\partial \mathcal B} \frac{\dot q}{T} dA + \int_{\mathcal B} \frac{\dot s}{T} dm\tag{2}\label{2}.$$
In $\eqref{2}$ the quantity $\dot s$ represents the heat supply per unit mass $dm$ and per unit time (it is a rate) at temperature $T=T(dm)$. When the process including the heat transfer is reversible one has equality in $\eqref{2}$. This is a very natural generalization of Clausius's inequality and it also includes radiation that is absorbed "bodily". Just as with $\dot q$ the sign of $\dot s$ tells you in what direction does "heat", i.e., energy and entropy may flow; more specifically when $\dot s$ is the radiated heat supply between two bodies then depending on their relative temperatures one body may be the source while the other the sink, or vice versa. Of course, if they have the same temperature then there is no net flow between them, for whatever one absorbs it will also radiate it out.
[1] Truesdell: Rational Thermodynamics, page 117