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\title{18.757 HW quasi-answers}
%%\author{Professor V. Ka\v{c}\\
%%\,\newline\\
%% Scribe: HwanChul Yoo}

\begin{document}
\maketitle

\section{1st Homework}

\begin{problem} Prove that the restriction of homogeneous
polynomials to sphere is one to one. That is, if $Q$ is a
homogeneous polynomial of degree m and $Q|_{S^{n-1}}=0$, then $Q=0$.
\end{problem}
\begin{proof}
This is true only if the sphere is nonempty; that is, if $n > 0$. So
assume that. For any nonzero vector $v \in \R^n$, ${v \over \|v\|} \in
S^{n-1}$. If $Q$ is a homogeneous polynomial of degree m, and
$Q|_{S^{n-1}} = 0$, then $Q(v)=Q(\|v\|{v \over \|v\|})={\|v\|}^m Q({v
\over \|v\|})=0$. So $Q$ vanishes except perhaps at $0$. Since $Q$ is
continuous,  we have $Q=0$.\\
\end{proof}

\begin{problem} Prove that $P^{m-2} \subset P^m$ for $m \ge 2$.
\end{problem}
\begin{proof} Recall that $P^m$ is the space of restrictions to the
  sphere of homogeneous polynomials of degree $m$. Because of Problem
  1, it makes sense to define $\gi: P^{m-2} \rightarrow P^m$ by
  $\gi(f|{S^{n-1}})=[(x_1^2+ \dots +x_n^2)f]|{S^{n-1}}$.
It is clear that $(x_1^2+ \dots +x_n^2)f|_{S^{n-1}} \in P^m$. Since
multiplication by a nonzero polynomial is one to one,
this map is one to one. Hence $P^{m-2} \subset P^m$.\\
\end{proof}

\begin{problem}Calculate $\dim P_m$ and $\dim P_m/P_{m-2}$
\end{problem}
\begin{proof}By Problem 1, the monomials of degree m in n variables
form a basis of $P_m$. Hence $\dim P_m = {n+m-1 \choose m}={n+m-1
\choose n-1}$, and $\dim P_m/P_{m-2}={n+m-1 \choose n-1}-{n+m-3
\choose n-1}$.\\
\end{proof}

\section{2nd Homework}

\begin{problem}Calculate the eigenspaces of $r^2\Delta$ acting on
$S^m(\R^n)$
\end{problem}
\begin{proof} For $n=1$, $x^k$ are the eigenvectors with
eigenvalue $k(k-1)$. Now we consider $n\ge2$. We can show from a
brute calculation that 
$$r^2\Delta(r^{2k}f)=2k(n+2m-2k-2)r^{2k}f + r^{2k+2}\Delta(f). $$
Therefore $r^{2k}H^{m-2k}$ is an eigenspace of $r^2\Delta$ with
eigenvalue $2k(n+2m-2k-2)$. Observe that we have a decomposition
$S^m(\R^n)=H^m\oplus r^2H^{m-2}\oplus\dots\oplus
r^{2[{m\over2}]}H^{m-2[{m\over2}]}$, hence this is an eigenspace
decomposition of $S^m(\R^n)$. Furthermore the eigenvalues
$2k(n+2m-2k-2)$ are all distinct for $n\ge2$. Therefore
$\{r^{2k}H^{m-2k}|k=0\dots[{m\over2}]\}$ is the complete set of
eigenspaces of $r^2\Delta$ on $S^m(\R^n)$.\\
\end{proof}

\begin{problem} Show that every $n$ variable polynomial can be uniquely written
as a polynomial in $r^2$ with  harmonic polynomial coefficients.
\end{problem}
\begin{proof} First, observe that any polynomial $f$ can be uniquely
written as a sum of homogeneous polynomials $f=f_0+\dots f_d$, $f_i
\in S^i(\R^n)$. Second, we have the decomposition
$S^i(\R^n)=H^i\oplus r^2H^{i-2}\oplus\dots\oplus
r^{2[{i\over2}]}H^{i-2[{i\over2}]}$ for each $i$. The assertion follows.\\
\end{proof}

\begin{problem}Suppose $n=p+q$, positive integers. Decompose $^nH^m|_{O(p)\times
O(q)}$ into irreducibles.
\end{problem}
\begin{proof} We have the decomposition $S^m(\R^n)|_{O(p)\times O(q)}=\bigoplus_{i=1}^m
S^i(\R^p)\otimes S^{m-i}(\R^q)$. Express both sides as a sum of
$^iH^j$'s using $S^i(\R^n)=H^i\oplus r^2H^{i-2}\oplus\dots\oplus
r^{2[{i\over2}]}H^{i-2[{i\over2}]}$. Since $^pH^i$ and $^qH^j$ are
irreducible representations of $O(p)$ and $O(q)$ respectively,
$^pH^i\otimes ^qH^j$ is an irreducible representation of $O(p)\times
O(q)$. Hence, after cancellation, we have 
$$^nH^m|_{O(p)\times O(q)}= \bigoplus _{0\le i, j \le m \atop i+j
  \equiv m \pmod{2}}{^pH^i\otimes {^qH^j}}$$
\\
\end{proof}

\section{3rd Homework}

\begin{problem} Suppose $\{\lambda_k \in \C | k \in \mathbb{N}\}$ is
a sequence of complex numbers. We may try to define a map $T:
l^2(\C) \rightarrow l^2(\C)$ by the formula $ (a_k) \mapsto
(\lambda_k a_k)$. Then,

(a) $T$ is well-defined if and only if $(\lambda_k)$ is bounded.

(b) $T$ is compact if and only if $\lambda_k \rightarrow 0$.
\end{problem}
\begin{proof}
(a) If $(\lambda_k)$ is not bounded, there is a subsequence
$(\lambda_{k_i})$ s.t. $i\le |\lambda_{k_i}|$. Let $b$ be the vector
with $1\over i$ at each $k_i$th coordinate and $0$ at all other
places. Clearly $b$ is in $l^2(\C)$, but $T(b)$ is not. 

Conversely, if $\lambda_k \le M$ for all $k$, then $\|T(a)\| \le M\|a\|$, hence
$T(a) \in l^2(\C)$.

(b) If $\lambda_k\nrightarrow 0$, then there is an $\epsilon > 0$ and
a subsequence
$(\lambda_{k_i})$ s.t. $|\lambda_{k_i}| > \epsilon$ for some
$\epsilon$. Let $b_i$ be the vector with 1 at the $m_i$th coordinate
and 0 elsewhere. Then $\{b_i\}$'s are all in the unit ball, but the
sequence of vectors $\{T(b_i)\}$ has no convergent subsequence. Hence $T$ is not compact.
Conversely, suppose $\lambda \rightarrow 0$. Define $T_i:
l^{\infty}(\C)\rightarrow l^{\infty}(\C)$ by $\sum a_k e_k \mapsto
\sum_{k\le i}\lambda_k a_k e_k$. Then $T_i$'s are compact operators.
Also, $||T-T_i||=sup_{k>i}|\lambda_k|$, so $T_i \rightarrow T$ in
the operator norm topology. Hence $T$ is compact.\\
\end{proof}

\begin{problem} let $G \subset \mathbb{H}^{\times}$ be the Lie group
of unit quaternions, which is diffeomorphic to $S^3$. The
quaternionic product defines left and right actions of $G$ on
$\mathbb{H}\simeq \R^4$, defining a map $\Phi: G\times G \rightarrow
O(4)$. Recall that $L^2(S^3)=L^2(G)= \bigoplus_{m\ge0} {^4H^m}$ as
$O(4)$-modules, hence as $(G\times G)$-modules. Moreover, Peter-Weyl
tells us that $L^2(G)= \bigoplus_{\pi\in\hat{G}} V_{\pi}\otimes
V_{\pi}^*$ as $G\times G$-modules. Reconcile these two orthonormal
decompositions.
\end{problem}
\begin{proof} The given two actions of $G\times G$ on $L^2(S^3)$
coincide. It can be shown that $\Phi^*(^4H^m)$'s are distinct
irreducible $G\times G$ representations, so by Schur's lemma, the
given two decompositions are the same (up to some permutation). More
precisely, observe that $\Phi$ is onto the identity component of
$O(4)$ which is $SO(4)$(one can compute its kernel
$\{(1,1),(-1,-1)\}$). Also $^4H^m$ is irreducible as $SO(4)$
representation, so $\Phi^*(^4H^m)$ is irreducible as $G\times G$
representation.

[Additional hint for proving that $^4H^m$ remains irreducible on
  restriction to $SO(4)$: we know that $^4H^m$ restricted to $SO(3)$
  is the sum of the representations $^3H^k$ restricted to $SO(3)$, for
  $0 \le k \le m$.  Because $SO(3)$ acts transitively on $S^2$, the
  trivial representation of $SO(3)$ can appear in functions on $S^2$
  only in the constant functions.  That is, $^3H^k$ contains the
  trivial representation of $SO(3)$ only if $k=0$.  It follows that
  $^4H^m|_{SO(3)}$ contains at most one copy of the trivial
  representation of $SO(3)$.  Now copy the proof given in class that
  $^4H^m$ is irreducible for $O(4)$.]

We can reconcile these two decompositions more directly. Since
$\Phi^*(^4H^m)$ is an irreducible $G\times G$ representation,
$\Phi^*(^4H^m)=V\otimes W$ for some irreducible $G$ representations
$V$ and $W$. Let $\sigma : G\times G \rightarrow G \times G$ be the
interchanging map of two factors, and $\tau: O(4)\rightarrow O(4)$
be the inner automorphism defined by
$\tau(r)x=\overline{r(\overline{x})}$ for $r\in O(4)$, $x\in
\mathbb{H}$, and $\overline{x}$ the conjugation of $x$ in
$\mathbb{H}$. Then we have $\Phi \circ \sigma = \tau \circ \Phi$,
hence $\sigma^* \Phi^* (^4H^m) = \Phi^* \tau^* (^4H^m) \cong
\Phi^*(^4H^m)$ as $\tau$ is an inner automorphism. Therefore
$V\otimes W \cong \Phi^*(^4H^m) \cong \sigma^* \Phi^*(^4H^m) \cong W
\otimes V$ as $G \times G$ representations. Moreover, $^4H^m$ is
self dual as $O(4)$ representation, so $V \otimes W \cong
\Phi^*(^4H^m) \cong \Phi^*({^4H^m}^*) \cong \Phi^*(^4H^m)^* \cong (V
\otimes W)^* \cong V^* \otimes W^*$. So we have $V \cong V^*$,
$W\cong W^*$. As a result, we finally get $\Phi^*(^4H^m) \cong V
\otimes V^*$ for some irreducible $G$ representation $V$.
\end{proof}

\begin{problem} Find a two-dimensional (complex) representation of a Lie group
that is not a direct sum of irreducible representations.
\end{problem}
\begin{proof} Let $G=\R$ act on $V=\C^2$ by
$\pi(r)\dot(x,y)=(x+ry,y)$. Equivalently, $G$ is the subgroup of
$GL(2,\C)$ consist of the matrices of the form $\left(
                                                     \begin{array}{ccc}
                                                     1 & r \\
                                                     0 & 1 \\
                                                     \end{array}
                                                     \right)$.
Then, the only proper $G$-invariant subspace is the one spanned by (1,0).
Hence $V$ is neither irreducible nor direct sum of irreducible
representations.\\
\end{proof}

\section{4th Homework}

\begin{problem} Suppose $G$ is a compact group. Prove that
$\{\chi_\pi : \pi \in \hat{G}\}$ is an orthonormal basis for
$L^2_{\textrm{class}}(G)$.
\end{problem}

\begin{proof} From Schur orthogonality, we know that $\{\chi_\pi : \pi \in
\hat{G}\}$ is an orthonormal set. From Peter-Weyl, we know that they
span a dense subset of $L^2_{\textrm{class}}(G)$. Hence the
assertion follows.\\
\end{proof}

\begin{problem} Suppose that $G$ is a finite group. Then,

(a) $\#$ conjugacy classes of $G$ = $\#$ irreducible
representations;

(b) $\Sigma_{\pi\in\hat{G}} (\dim \pi)^2 = |G|$;
\end{problem}

(c) $\#$ 1-dimensional representations of $G$ = $|G/[G,G]|$.

\begin{proof} (a) Since $G$ is finite, $L^2_{\textrm{class}}(G)$ is the
space of all class functions, hence the dimension of it is just the
number of conjugacy classes of $G$. On the other hand, by the
previous problem, its dimension is the number of irreducible
representations of $G$.

(b)  Since $G$ is finite, $L^2(G)$ is the same as the space of all
complex functions on $G$, hence the dimension of it is just $|G|$.
On the other hand, by Peter-Weyl, we have $L^2(G) =
\bigoplus_{\pi\in\hat{G}} V_\pi \otimes V_\pi^*$. Hence the
dimension is $\Sigma_{\pi\in\hat{G}} (\dim \pi)^2$

(c) Giving a 1-dimensional representation is equivalent to giving a
group homomorphism $G \rightarrow GL(\C)$. Since $GL(\C)$ is
abelian, this is equivalent to giving a homomorphism $G/[G,G]
\rightarrow GL(\C)$, which is again equivalent to giving a
1-dimensional representation of $G/[G,G]$. All irreducible
representations of $G/[G,G]$ are
1-dimensional, so the assertion follows.\\
\end{proof}

\begin{problem} List all irreducible representations of $G=S_3$, and
$G=\{\pm1,\pm i,\pm j,\pm k\}\subset\mathbb{H}.$
\end{problem}

\begin{proof}

1) $G=S_3$: The conjugacy classes of $G$ is characterized by
permutation type, hence they are $\{id\}, \{(12),(13),(23)\},
\{(123),(132)\}$. And $G/[G,G]=\{\pm1\}$. By the previous problem,
there are 3 irreducible representations of $G$, and 2 of them are
1-dimensional. We have trivial representation and sign
representation which are nonequivalent 1-dimensional
representations. Consider the defining representation of $G=S_3$ on
$\C^3=\C\{\mathbf{1,2,3}\}$. We have 1 copy of the trivial
representation $\C\{\mathbf{1+2+3}\}$ in it. Define a $G$-invariant
inner product: $<\mathbf{i,j}> = \delta_{i,j}$. Now the complement
of the trivial representation if also a $G$-submodule. While $S_3$
is not abelian, the defining representation is faithful, so it
cannot be decomposed into 1-dimensional representations. Therefore,
the complement of the trivial representation is the 2-dimensional
irreducible representation.

2) $G=\{\pm1,\pm i,\pm j,\pm k\}\subset\mathbb{H}$: Similarly to the
above problem, we find the conjugacy classes of $G$, which are
$\{1\},\{-1\},\{\pm i\},\{\pm j\},\{\pm k\}$. And $|G/[G,G]|=4$.
Hence there are 5 irreducible representations and 4 of them are
1-dimensional. The 1-dimensional representations are given by the
trivial representation and $\pi(\pm1)=-\pi(\pm i)=-\pi(\pm j)=\pi(\pm
k)=1$ and similarly 2 more. We can view quaternion $\mathbb{H}$ as
$\{a+jb | a,b\in \C\}$, so $\C$ vector space (by right
multiplication). This naturally gives a faithful 2-dimensional $G$
representation (by left multiplication), which is irreducible because
of the similar reasoning as in 1).
\end{proof}


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