TITLE: Are the rank and the corank of the cohomologies of the given Tate module the same?
QUESTION [1 upvotes]: Let $G$ be a Galois group of two kinds:

$G=Gal(K_S/K)$ where $K$ is a number field and $K_S$ is a maximal $S$-ramified extension. ($S$ is a finite set of primes containing primes above $p$ and $\infty$).
$G=Gal(K^{alg}/K)$ where $K$ is a finite extension of $\mathbb{Q}_l$. ($l$ is not necessarily different from $p$.)

Suppose $T$ is a  $\mathbb{Z}_p[G]$-module such that $T\cong \mathbb{Z}_p^r$ for some $r$. Define $W:=\varinjlim_n T/p^nT$.
My question is: $rank_{\mathbb{Z}_p}H^i(G, T) =corank_{\mathbb{Z}_p} H^i(G, W)$ for $i=0, 1, 2$? ($H^i(G, T)$ is a continuous cochain cohomology and  $H^i(G, W)$ is a Galois cohomology.)

REPLY [2 votes]: Your hypothesis $T \cong {\mathbf Z_p}^r$ is not precise enough. For convenience, I suppose that this is an isomorphism of topological $G$-modules, although this assumption is not necessary (1). We have continuous cohomology groups ${H_c}^i(G,T)$, with compact $T$, and Galois cohomology groups $H^i(G, W)$, with dicrete $W \cong T \otimes \mathbf Q_p /\mathbf Z_p$. Usually, the natural map ${H_c}^i(G,T) \to \varprojlim H^i(G,T/p^n)$ is surjective, with a kernel isomorphic to some $\varprojlim^1$ (2). But here, by the choice of your profinite group $G$ (both in the local and global cases), it is known that the groups $H^i(G,T/p^n)$ are finite, and it follows from this finiteness property that $rank_{\mathbf Z_p} {H_c}^i(G,T)= dim_{\mathbf Q_p} ({H_c}^i(G,T) \otimes{\mathbf Q_p})=corank_{\mathbf Z_p} H^i(G,T\otimes \mathbf Q_p /\mathbf Z_p )$ (3).
A convenient reference for all this is the book by Neukirch-Schmidt-Wingberg, "Cohomology of number fields", chap. II, §7, with (1) = the remark just before thm. 2.7.5, (2) = thm. 2.7.5, and (3) = coroll. 2.7.12.


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config: arxiv


TITLE: Finding $\sum\limits_{\substack{p<n\\p\text{ is prime}}}p^{n}\pmod{n}$
QUESTION [2 upvotes]: Suppose $n=2^{a}5^{b}$, where $a\geq 2$. We wish to find
$$S_{n}=\sum_{\substack{p<n\\p\text{ is prime}}}p^{n}\pmod{n}.$$
To do this, we find $S_{n}\pmod{2^{a}}$ and $S_{n}\pmod{5^{b}}$.
Since $\phi(2^{a})=2^{a-1}$, if $p\neq 2$, then $\gcd(p,2^{a})=1$, so
$$p^{n}=p^{2^{a}5^{b}}=\left(p^{2^{a-1}}\right)^{2\cdot5^{b}}\equiv1\pmod{2^{a}}.$$
If $p=2$, then $p^{n}\equiv0\pmod{2^{a}}$, and thus $S_{n}\equiv\pi(n)-1\pmod{2^{a}}$.
Now, we note that $\phi(5^{b})=4\cdot5^{b-1}$, so if $p\neq 5$, then $\gcd(p,5^{b})=1$, and
$$p^{n}=p^{2^{a}5^{b}}=\left(p^{4\cdot5^{b-1}}\right)^{2^{a-2}\cdot5}\equiv1\pmod{5^{b}}$$
since $a\geq 2$. And if $p=5$, we have $p^{n}\equiv0\pmod{5^{b}}$, so again we have $S_{n}\equiv\pi(n)-1\pmod{5^{b}}$, and thus
$$S_{n}\equiv\pi(n)-1\pmod{n}.$$
I believe this "trick" works whenever $n$ satisfies $p\mid n\Rightarrow p-1\mid n$ for all primes $p$ (that is, $S_{n}\equiv\pi(n)-1\pmod{n}$).
What do we know about the behavior of $S_{n}$ for other $n$? Are there ways of computing $S_{n}$ for other $n$ as well (perhaps given things like the factorization of $n$, $\pi(n)$, etc.). How does it behave as $n\rightarrow\infty$? Maybe it has some asymptotic relationship with $\cfrac{n}{\ln n}$?

REPLY [2 votes]: $n=2^a5^b,a\ge 1$.

For $p\nmid n$ then $ \phi(2^a)| n$ thus $p^n \equiv 1\bmod 2^a$ and $\phi(5^b)| 2n$ thus $p^{2n} \equiv 1\bmod 5^b$, since $\Bbb{Z/5^b Z}^\times$ is cyclic it means $p^n \equiv \pm 1 \bmod 5^b$.
The sign depends on if $a\ge 2$ and if $p$ is a square $\bmod 5^b$ or not ie. if $p$ is a square $\bmod 5$ or not. That is to say with $$f(l,n)=(l^n \bmod n)= l^n - n\lfloor l^n/n \rfloor$$ then $$ \gcd(l,n)= \gcd(l+5k,n)=1 \implies f(l,n)=f(l+5k,n)$$
$7$ is not a square $\bmod 5$
thus you are looking at $$\sum_{p<n}f(p,n)=f(2,n)+f(3,n)+f(5,n)+\frac{1+f(7,n)}{2}\sum_{5<p<n}1+\frac{1-f(7,n)}{2}\sum_{5<p<n} (\frac{p}{5})$$
By the PNT in arithmetic progression we get as $n\to \infty$ $$\sum_{p<n}f(p,n)\sim \frac{1+f(7,n)}{2}\frac{n}{\log n}$$
If $a\ge 2$ then $f(7,n)=1$, otherwise it is an integer $\in [\sqrt{n},n]$.


TITLE: Shorter notation for sine function
QUESTION [1 upvotes]: TL;DR: Are there shorter, established notations for the sine function and other trigonometric functions than "$ \sin(x) $" etc.?
In a some homeworks and tests, there are exercises (e.g. differential equations or matrices) which contains sine and cosine, which I have to write very often - obviously, this becomes quite annoying.
My usual "workaround" is, that I define $ \sin(x) = s $ and $ \cos(x) = c $ to safe me from writing the rest and loose time unessecarily. As far as I can see, this is totally correct and can't be marked as wrong.
Yet this reduces the readability (and probably annoys the corrector). So are there any shorter notations for the trigonometric functions, which are  established or at least sometimes used?

REPLY [4 votes]: In general, there is no such other notation. Your relabelling approach is certainly fine (except maybe if the work is being marked by someone unwilling to look through your custom symbol definitions). But I'd concentrate on whether your proofs are more verbose than necessary for other reasons. Writing in terms of $e^{ix}$ would shorten many proofs, especially if it's appropriate in that context to write e.g. $z=e^{ix}$.
In contexts where multiple angles are indexed viz. $\theta_i$, it's common to use notation such as $c_i:=\cos\theta_i$ etc. See e.g. the tabulated results in terms of proper Euler angles & Tait–Bryan angles here. Credit goes to @JohnDouma for this observation.

REPLY [1 votes]: I would from my experience assume that there is nothing shorter for the sine or the cosine etc other than the usual ones. Think of the following, sin is a three-letter word the best we could shorten it to is one or two-letter word, now naming sin as si and cosine as co would just create a headache to anyone correcting (to me at least it would) and what is the benefit? Even if it was 100 letters saved I assure you it is not worth it.
P.S I think such a question would be more appropriate to a StackExchange more towards the linguistic aspect.


TITLE: Remainder term of Taylor polynomial in matrix calculus
QUESTION [3 upvotes]: This is my first time asking a question here and it might be rather specific. Hopefully there is still someone who can answer.
Assume for a function $f:\mathbb{R} \rightarrow \mathbb{R}$ that is at least one time differentiable that we have the Taylor approximation $ f(x) = f(a) + f'(x)(x-a) + r_a(X) $ with $  r_a(x) = O(|x-a|)         $.
The question now becomes: Are there results concerning the error term when instead of scalars we plug in $n \times n$ symmetric, positive definite matrices X and A (assume that all the operations are well defined, e.g. inverses exist if neccessary)
\begin{align}
  f(X) = f(A) + f'(A)(X-A) + r_A(X),
\end{align}
such that for example $\|r_A(X)\| = O(\|X - A\|)$?
Thanks in advance for any help!

REPLY [0 votes]: First one correction. The approximation is not only $O$ but also $o$ for a one time differentiable map:
$$r_a(x) = o(|x-a|)$$
Then coming back to your question, if $V, W$ are two Banach normed spaces, you can define the Fréchet derivative for a map $f : V \to W$. And you'll have also $$f(X) = f(A) + f'(A)(X-A) + o(\Vert X-A \Vert_V)$$
You can apply this to the Banach space of $n$-dimensional real matrices. And plug in positive definite matrices in the formula. Just take car that the set of positive definite matrices itself is not a vector space.


TITLE: Differentiate $\sin^{-1} \sqrt {1-x^2} $ wrt $\cos^{-1} $ if $x\in (-1,0)$
QUESTION [0 upvotes]: Differentiate $\sin^{-1} \sqrt {1-x^2} $ wrt $\cos^{-1} $ if $x\in (-1,0)$
Let $x=\cos y$
Then $f(x)= \sin^{-1} \sqrt{1-\cos^2y}=\sin^{-1} |\sin y| =\pm y$
And $g(x)=\cos^{-1} \cos y =y$
$$\frac{f’(x)}{g’(x)} = \pm 1$$
Since $\sin y$ can be both positive or negative if $\cos y \in (-1,0)$
Which is the right answer?

REPLY [1 votes]: For real $x,\sqrt{1-x^2}\ge0$
Let $\sin^{-1}\sqrt{1-x^2}=y,\implies y\ge0, x^2=\cos^2y$ and $0\le y\le\dfrac\pi2$
As $x<0,\cos y=-x$
$y=\cos^{-1}(-x)=\pi-\cos^{-1}x$
using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?