hw06.tex

\documentclass[
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]{homework}


\usepackage{cleveref}
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\hwnum{6}
\duedate{Sun, Apr 11, 2020 10:00 PM}

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\begin{document}

\begin{problem}
Let $G$ by a compact connected matrix Lie group. Given a subset $A$ of $G$, recall that $Z_G(A)\defeq\qty{g\in G : gx = gx \text{ for all } x\in A}$. Also, we write $Z_G(x)$ for $Z_G(\qty{x})$.
\begin{parts}
	\part{Prove that every torus is contained in a maximal torus.}\label{part:1a}
	\part{Let $S$ be a torus, prove $Z_G(S)$ is the union of all maximal tori in $G$ containing $S$.}\label{part:1b}
\end{parts}
\end{problem}

\begin{solution}
	\ref{part:1a}
	Let $A$ be a torus in $G$. If $A$ is maximal, it's contained in itlelf $A\subseteq A$, so we're done. Thus assume $A$ is not maximal. By non-maximality of $A$ there exists a torus $T_1$ containing it. If it's maximal we're done, so assume it's not and hence $A\subsetneq T_1$. Repeat this argument with $T_1$ to obtain $T_2$ and so on. That is we have the following chain of strict inclusions:
	\begin{equation*}
		A\subsetneq T_1 \subsetneq T_2 \subsetneq T_3 \subsetneq \cdots
	\end{equation*}
	We can now pass to the Lie algebra's where we have
	\begin{equation*}
		\mathsf{a} \subseteq \mathsf{t}_1 \subseteq \mathsf{t}_2 \subseteq \mathsf{t}_3 \subseteq \cdots \subseteq \mathsf{g} \eqdef \Lie(G).
	\end{equation*}
	Since $\mathsf{g}$ is a \emph{finite} dimensional vector space, we cannot have an infinite chain of strict inclusions, so there must exist an $n\in \N$ such that $\mathsf{t}_{n + k} = \mathsf{t}_n$ for all $k\in \N$. However on compacted, connected matrix Lie groups the exponential map is surjective and hence $\exp(\mathsf{t}_i) = T_i$ and
	\begin{equation*}
		T_{n + k} = \exp(t_{n + k}) = \exp(\mathsf{t}_n) = T_n
	\end{equation*}
	but we had $T_{n} \subsetneq T_{n + k}$ thus we have a contradiction. Hence $A$ is contained in a maximal torus.

	\ref{part:1b}
	Suppose $T$ is a maximal torus containing $S$. Then by definition we have $T\subseteq Z_G(S)$ and hence $Z_G(S)$ contains all maximal tori containing $S$, and also their unions.

	Now take $g\in Z_G(S)$, or written differently as $S\subseteq Z_G(g)$. Since $S$ is a connected, compact matrix Lie group, so is $Z_G(g)_0$. Take $T\subseteq Z_G(g)_0$ to be a maximal torus that contains $S$. Since the exponential map is surjective in this case there must exist an element $X\in\Lie(G)$ such that $\e^{X} = g\in Z_G(g)_0$. This implies $g$ is in the center of $Z_G(g)_0$, and using the fact that the $Z(G)$ is equal to the intersection of all maximal tori we conclude $g\in T$. Thus we've found a torus that contains both $S$ and $g$.
\end{solution}

\begin{problem}
\begin{parts}
	\part{Let $g\in G$. Prove that $Z_G(g)_0$ is the union of all maximal tori in $G$ containing $g$.}\label{part:2a}
	\part{Specializing to the case $G = \SO{3}$ and let $T$ be the maximal torus corresponding to the Cartain subalgebra given in \#4(b) of Assignment 5. Find $g\in \SO{3}$ such that $Z_G(g)_0 = T$ but that $Z_G(g)$ is disconnected.}\label{part:2b}
\end{parts}
\end{problem}

\begin{solution}
	\ref{part:2a}
	See proof to \#1(b) to see that $Z_G(g)_0$ is connected.

	\ref{part:2b}
	Take $g = \mqty[\dmat{-1, -1, 1}]$.
\end{solution}

\begin{problem}
Dont think Imma do this one.
\end{problem}

\begin{solution}

\end{solution}

\begin{problem}
Let $G$ be a compact matrix Lie group and $V$ and $W$ irreducible complex representations of $G$, equipped with $G$-invariant innter products $(-, -)_V$ and $(-,-)_W$, respectively, which are linear in the first variable and conjugate linear in the second.
\begin{parts}
	\part{Let $\varphi: V \to W$ be an intertwining map. Prove that there exists $\alpha\in\R_{\geq 0}$ such that
		\begin{equation*}
			(\varphi(v), \varphi(v'))_W = \alpha(v, v')_V
		\end{equation*}
		for all $v, v'\in V$.}\label{part:4a}
	\part{Imitate the proof of the orthogonality of characters to prove the following orthogonality relations for matrix coefficients: Given $v_1, v_2\in V$ and $w_1, w_2\in W$, there holds
		\begin{equation*}
			\int_G(g\cdot v_1, v_2)_V\overline{(g\cdot w_1, w_2)_W}\dd{\mu_G} = \frac{(\varphi(v_1), w_1)_W\overline{(\varphi(v_2), w_2)_W}}{\dim V}[V\cong W]
		\end{equation*}
		where $[A]$ is the Iverson bracket and $\varphi: V \to W$ is any intertwining isometry, that is, and intertwining isomoprhism such that the conclusion of part (a) holds with $\alpha = 1$.}\label{part:4b}
\end{parts}
\end{problem}

\begin{solution}
	\ref{part:4a}
	By Schur's lemma $\varphi$ is either the 0 map---in which case $\alpha = 0$---or a scalar multiple of the identity. Thus as long as $\varphi$ is not identically 0, then $V$ and $W$ are isomorphic and by Assignment 3 problem 6 there is only one $G$-invariant inner product up to a positive constant.
	\begin{equation*}
		(\varphi(v), \varphi(v'))_W = (\beta v, \beta v')_W = \abs{\beta}^2(v, v')_W = \underbrace{\abs{\beta}^2}_{\geq 0}\underbrace{\gamma}_{\geq 0}(v, v')_V
	\end{equation*}
	Thus if we take $\alpha \defeq \abs{\beta}^2\gamma$ then the above equation is satisfied.

	\ref{part:4b}
	Let $\Pi$ and $\Sigma$ be the irreps corresponding to $V$ and $W$ respectively. Define the map $L: W \to V$.
	\begin{equation*}
		L\defeq \int_G \Pi(g)\circ \varphi^{-1} \circ \Sigma(g)^\dagger \dd{\mu}\!(g)
	\end{equation*}
	and note that $\Pi(h) \circ L \circ \Sigma(h^{-1}) = L$ by the invariance of the Haar measure and so $\Pi(h) \circ L = L \circ \Sigma(h)$ and hence $L$ is an intertwining map. By Schur's lemma $L = 0$ or $L = \lambda\1$. When $L = 0$ we can take $\varphi^{-1}(w) = (w, w_1)v_1$ and thus
	\begin{align*}
		0 & = (L(w_2), v_2)                                                                   \\
		  & = \int_G(\Pi(x)\circ \varphi^{-1} \circ \Sigma(x)^\dagger w_2, v_2) \dd{\mu}\!(x) \\
		  & = \int_G(\Pi(x)(\Sigma(x^{-1})w_2, w_1)v_1, w_1)\dd{\mu}\!(x)                     \\
		  & = \int_G(\Pi(x) v_1, v_2)(\Sigma(x^{-1})w_2, w_1)\dd{\mu}\!(x)                    \\
		  & = \int_G(\Pi(x) v_1, v_2)\overline{(\Sigma(x)w_1, w_2)}\dd{\mu}\!(x)
	\end{align*}

	Now we have the case where $V\cong W$ and $\varphi$ can be treated as a map $\varphi: V\to V$. As above we have $L = \lambda \1$ and taking the trace of both sides we have $\tr(L) = \lambda \dim V = \tr(\varphi)$. Thus
	\begin{equation*}
		(L(v_2), v_1) = \frac{\tr(\varphi)}{\dim V}\overline{(v_1, v_2)}.
	\end{equation*}
	Similar to above we take $\varphi(v) = (v, w_2)v_1$. so that
	\begin{align*}
		\frac{(v_1, w_2)(\overline{(v_2, w_2)})}{\dim V} & = \frac{\tr(\varphi)}{\dim V}\overline{(v_1, w_2)}                \\
		                                                 & = (L(w_2), v_1)                                                   \\
		                                                 & = \int_G(\Pi(x)\circ\varphi\circ\Pi(x^{-1})w_2, v_2)\dd{\mu}\!(x) \\
		                                                 & = \int_G (\Pi(x) v_1, v_2)(w_2, \Pi(x)w_1)\dd{\mu}\!(x)
	\end{align*}
	Thus, done. I understand this is probably sloppy.
\end{solution}

\begin{problem}
Some character computations.
\begin{parts}
	\part{Let $\chi$ denote the character of the irreducible representation $\mathcal{H}_m(\R^3)$ of $\SO{3}$. Compute $\chi(g)$ for $g\in\SO{3}$ of the form
		\begin{equation*}
			g = \mqty(1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \phantom{-}\cos\theta).
		\end{equation*}}\label{part:5a}
	\part{Recall the irreducible $\SU{2}$-representations $(\Pi_m, V_m(\C^2))$. Use the character computation to prove that, as representations of $\SU{2}$, we have for non-negative integers $m\geq n$ that
		\begin{equation*}
			V_m\otimes V_n\cong V_{m + n}\oplus V_{m + n - 2}\oplus \cdots \oplus V_{m - n + 2} \oplus V_{m - n}
		\end{equation*}}\label{part:5b}
\end{parts}
\end{problem}

\begin{solution}
	\ref{part:5a}

	\ref{part:5b}
	Let $g = \mqty[\e^{\iu\theta} & 0 \\ 0 & \e^{-\iu\theta}]$ be an element of the maximal torus of $\SU{2}$.
	\begin{align*}
		\chi_{V_m\otimes V_n}(g) & = \chi_{m}(g)\chi_n(g)                                                                 \\
		                         & = \qty(\sum_{k = 0}^m\e^{\iu(m - 2k)\theta})\qty(\sum_{j = 0}^n\e^{\iu(n - 2j)\theta}) \\
		                         & = \sum_{k, j = 0, 0}^{m, n}\e^{\iu(m + n - 2k - 2j)\theta}                             \\
		                         & = \sum_{l = 0}^n\sum_{j = p}^{m + n - l}\e^{\iu(m + n - 2k)\theta}                     \\
		                         & = \sum_{l = 0}^n\sum_{j = 0}^{m + n - 2l}\e^{\iu(m + n - 2l - 2k)\theta}               \\
		                         & = \sum_{l = 0}^n\chi_{m + n - 2l}(g)
	\end{align*}
	This shows the representations are equal, and since every element in $\SU{2}$ can be written as $x = yty^{-1}$ with $t$ in the torus, and characters are class functions this must be true on the whole of $\SU{2}$.
\end{solution}

\begin{problem}
Let $G$ be a compact matrix Lie group.
\begin{parts}
	\part{Let $(\Pi, V)$ be a complex representation of $G$ and $\chi$ it's character. Prove that $\abs{\chi(g)} \leq \dim V$, with equality holding if and only if $\Pi(g)$ is multiplication by a scalar. Here $g\in G$ is an arbitrary element.}\label{part:6a}
	\part{Prove that $g$ belongs to $Z(G)$, the center of $G$, if and only if $\abs{\chi_V(g)} = \dim V$ for every irreducible complex representation $V$ of $G$. Here $\chi_V$ denotes the character of $V$.}\label{part:6b}
\end{parts}
\end{problem}

\begin{solution}
	\ref{part:6a}
	The compactness of $G$ implies $(\Pi, V)$ is unitary, and hence $\Pi(g)$ is a normal matrix, with eigenvalues $\e^{\iu\theta_i}$ where $i$ ranges from $1$ to $k \leq \dim V$. Now since the trace is equal to the sum of the eigenvalues we have
	\begin{equation*}
		\abs{\chi(g)} = \abs{\sum_{i=1}^k\e^{\iu\theta_i}} \leq \sum_{i=1}^k \abs{\e^{\iu \theta_i}} = \sum_{i=1}^k 1 = k \leq \dim V.
	\end{equation*}
	In the case when $\Pi(g)$ has full rank ($k = \dim V$) then it's not hard to see that $\abs{\sum_{i = 1}^{\dim V}\e^{\iu \theta_i}} = \dim V$ implies that all of the $\theta_i$ are equal (up to $2\pi$). We can then rewrite all the eigenvalues as $\e^{\iu\alpha + \iu\tilde{\theta}_i} = \e^{\iu\alpha}\e^{\iu\tilde{\theta}_i}$. Thus $\Pi(g) = \e^{\iu\alpha}\1_V$.

	If $\Pi(g) = \beta\1_V$, then since the representation is unitary $\Pi(g)\Pi(g)^\dagger = \1_V$ which implies $\beta = \e^{\iu\varphi}$. Thus all of the eigenvalues are $\e^{\iu\varphi}$ and since the identity map is full rank $\abs{\chi(g)} = \dim V$.

	\ref{part:6b}
	Suppose $g\in Z(G)$. Then $\Pi(g)$ is an intertwining map and by Schur's lemma $\Pi(g) = \alpha\1$ which implies $\abs{\chi_V(g)} = \dim V$ as above.

	Now take $\abs{\chi_V(g)} = \dim V$. As we've shown above $\Pi(g)$ must be a multiple of the identity and hence
	\begin{equation*}
		\Pi(gx) = \Pi(g)\Pi(x) = \alpha\1_V\Pi(x) = \Pi(x)\alpha\1_V = \Pi(x)\Pi(g) = \Pi(xg)
	\end{equation*}
	Since $G$ is a matrix Lie group $\Pi$ is a faithful representation or isometrically similar to one, so thus we can conclude $gx = xg$.
\end{solution}

\end{document}