let $U$ be an $n$-by-$n$ unitary matrix and let $\Lambda = \text{diag}(e^{-i \beta m})$ where $m=-j,-j+1,\cdots,j$ and $j=(n-1)/2$. I'm trying to show that the matrix elements of $U\Lambda U^{-1}$ are polynomials in $\cos(\beta/2),\sin\beta(/2)$
I'm doing things in a physics-standard way and enumerating the matrix elements of $U$ according to $m_1,m_2\in\{-j,-j+1,\cdots,j\}$ rather than $i,j\in\{1,2,\cdots,n\}$
@TedShifrin On the bright side I've gained a lot of math confidence from doing Spivak (and other math) semi-independently. Some stuff I've figured out, I never would have been able to before - or I would have been too uncomfortable to try.
Let {R} be a noetherian local ring with maximal ideal {\mathfrak{m}} and residue field {k}, {S} be a local finitely generated {R}-algebra with {\mathfrak{m}S \subset \mathfrak{n}} for {\mathfrak{n}} the maximal ideal of {S}, and {M} a finitely generated {S}-module.
(okay it's important that $\mathfrak{m}$ is maximal)
so you can show by induction that $R/\mathfrak{m}^t$ is of finite length, the argument goes like this: for $t=1$, $R/\mathfrak{m}$ is simple as $\mathfrak{m}$ is maximal, thus the length is $1$
assume that you already know that $R/\mathfrak{m}^t$ has finite length, then consider the exact sequence $0 \to \mathfrak{m}/\mathfrak{m}^{t+1} \to R/\mathfrak{m}^{t+1} \to R/\mathfrak{m} \to 0$
Now, if $n$ is odd, then $k-(n+1)/2$ is always even, so I end up with a linear combination of even powers of $e^{-i\beta/2}=\cos(\beta/2)+i\sin(\beta/2)$
Can we insert a limit inside a trig function? For example, can we do?: $$\displaystyle\lim_{n\to\infty}{\sin\frac{2n+1}{-2n^2+7}}=\sin\lim_{n\to\infty}{\frac{2/n+1/n^2}{-2+7/n^2}}=\sin(0^-)=0^-?$$
note that $\mathfrak{m}^t$ annihilates $\mathfrak{m}/\mathfrak{m}^{t+1}$, so it's a module over $R/\mathfrak{m}^t$, but it is also finitely generated as $R$ is Noetherian (that's where we need that part), but $R/\mathfrak{m}$ is an Artinian ring as it has finite-length over itself, thus $\mathfrak{m}/\mathfrak{m}^{t+1}$ has finite length
using the exact sequence $0 \to \mathfrak{m}/\mathfrak{m}^{t+1} \to R/\mathfrak{m}^{t+1} \to R/\mathfrak{m} \to 0$ and the fact that length is additive wrt to exact sequences, you get that $R/\mathfrak{m}^{t+1}$ has finite-length
I'm using the following facts about length: - f.g. modules over an Artinian ring have finite length - If $0 \to M' \to M \to M'' \to 0$ is exact, then $M$ has finite length iff $M'$ and $M''$ have finite-length
The polynomial 1-x^2+y^2 is not homogeneous. However, if we restrict ourselves to x,y such that x^2+y^2=1, then we have 1-x^2+y^2=2y^2 which is homogeneous
@MatheinBoulomenos I know that if module is finite length then tor_1 function is 0 but don't get the point that how tor zero implies sequence is exact?
@ninjahatori you have the long exact sequence and there's a zero at the right point, so you get a short exact sequence as a part of the long exact sequence
I guess I should start by thinking about the following. Suppose I've got a polynomial $f(x,y)=ax^2+by^2+c$. For what a,b,c can I write that as $f(x,y)=H(x,y)+\lambda(x^2+y^2-1)$ where $H(x,y)$ is homogeneous
More generally: Suppose $f(x,y)$ is some even polynomial of degree $2n$. Can I always find a polynomial $g(x,y)$ such that $f(x,y)-(x^2+y^2-1)g(x,y)$ is homogeneous?
Hints are usually counterproductive, if you could have came up with it yourself you would get less practice and if you could not come up with it yourself you waste time trying to figure out the hint
it must come from {\mathfrak{m}^t \cap I \otimes M} for all {t}. Thus it belongs to {\mathfrak{m}^t(I \otimes M)} for all {t} only this part? I don't know how to write it ?
"You waste your time trying to figure out the hint"
It's entirely possible that you don't necessarily know what avenue you should approach something from, and a hint can, if nothing else, tell you what to look for
@TedShifrin linear isometry $F$ is linear transformation and isometry of the plane at the same time, i. e. $\|X-Y\| = \|F(X) - F(Y)\|$ and we're talkin about $X,Y \in \mathbb{R}^2$
@CaptainAmerica: You should not number steps in proofs. Just write sentences. So what you forgot to say is that $ (b+d)-(a+c)=(b-a)+(d-c)\in P$. That's the most important statement.
Right, @chandx. Now suppose $X$ is an eigenvector with eigenvalue $\lambda$. And finish.
daym, if $X$ is eigenvector, then $F(X)=\lambda X$ so $\|F(X)\| = \| \lambda X \|$ and $\|F(X)\| = | \lambda | \|X\|$, but since $\|F(X)\| = \|X\|$, then $|\lambda | = 1$, thus $\lambda = \pm 1$
I actually do know someone who took the discrete math class I graded (and was a friend of mine) who had carpal tunnel. He was the exception to the requirement of of handing in psets in LaTeX, he started off writing them but even that was painful so he'd write it either on a chalkboard or whiteboard and take a picture
So, I'm trying to use sentences for this proof...suddenly realizing my grammar/english skills are a little subpar. It's going to read like a high schooler was trying to be fake proper, so be prepared.
I think it goes something like this. I start with an even polynomial in x,y and write it as $f(x,y)=f_0(x,y)+f_2(x,y)+\cdots +f_{2n}(x,y)$ where each term is homogeneous of order $2k$
what I think I want to do is subtract off multiples of $x^2+y^2-1$ such that I get rid of the constant terms, then the terms of homogeneous order 2, then terms of homogeneous order 4, etc
Given the inequalities $a<b$ and $c<d$, $a, b, c$ and $d$ are numbers such that $a<b \iff b-a \in P$ and $c<d \iff c-d \in P$. With this, we can create the statement $(b-a)+(d-c) = (b+d) - (a+c) \in P$. Therefore, $a+c < b+d$.
@CaptainAmerica. You don't need to be so verbose when the words get you nothing. Given $a<b$, we know that $b-a\in P$. Similarly, $c<d\iff d-c\in P$. Since the sum of elements of $P$ is again in $P$, we conclude that ...
