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\title{Homework in Representation theory}
\date{May 14, 2012}
\author{Erik Aas}
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\newcommand{\mcl}{\mathcal} \begin{document} 
\maketitle

{\bf Question 1}

(i) Suppose $W$ is not irreducible. Then it contains a invariant non-trivial subspace $V$ which must have dimension $1$ and so be spanned by some non-zero vector $u = (u_1,u_2,u_3) \in \mbb{C}^3$ such that $u_1+u_2+u_3 = 0$. Since $V$ is invariant under the cyclic shift $(123)$ there must exist a number $k$ such that $u_i = ku_{i+1} = k^2u_{i+2} = k^3u_i$ for each $i$ mod $3$. Since $u\neq 0$, we have $k^3 = 1$. Hence $u = u_1(1,k,k^2)$ ($u_1$ must be non-zero, otherwise $u$ is zero) so $u_1(1+k+k^2) = 0$ which means that $u=(1,\omega,\omega^2)$ where $\omega$ is one of the two third roots of unity different from $1$. Since $V$ is invariant there must also be some $t$ such that $(1,\omega,\omega^2) = t(\omega,1,\omega^2)$. However the vectors $(1,\omega,\omega^2)$ and $(\omega,1,\omega^2)$ are not parallel, a contradiction. We conclude that $W$ is irreducible.


(ii) Since we know there are as many irreps as conjugacy classes, and there are three partitions of $3$ ($3=1+1+1$, $3=2+1$, $3=3$) corresponding to the cycle types in $S_3$, the given list is an exhausting one. The elements in the given list are non-isomorphic since $W$ has different dimension from the other two and the trivial representation is a constant, unlike the alternating representation.
\vspace{2cm}


{\bf Question 2}

I'm not sure I understand this question.
Exactly this is done in a direct manner in Fulton and Harris' book, and the multiplicity of any irrep $V$ is $\dim V$ by character theory.
\vspace{2cm}

{\bf Question 3}

(i)
An arbitrary element of $\bigwedge^2 V$ can be written $\sum_i \wedge_{j=1} ^n (u_{ij}\oplus w_{ij})$ where $u_{ij} \in U$, $w_{ij} \in W$.

Expanding this formally and grouping all terms coming from $U$ to the left, we get a sum of terms of the following form $\wedge_{j=1}^i u\p _j \wedge_{j=1}^{n-p} w\p_j$ for varying $i$. But this is precisely the description of a term of the RHS, giving us an isomorphism.

(ii)
I just compute the dimension of the two sides, without giving a bijective proof. Let $\dim U = r, \dim W = s$, so that $\dim V = rs$. We compute
$\dim(RHS) = {r+1\choose 2}{s\choose 2} + {r\choose 2}{s+1\choose 2} = \frac{1}{4}\left((r+1)rs(s-1) + r(r-1)(s+1)s \right) = \frac{1}{2}\left(rs(rs-1)\right) = \dim(LHS)$.



\vspace{2cm}

{\bf Question 4}

By definition, $\chi_{S^2V}(g)$ is the trace of the mapping $g\otimes g$, which is the sum of its eigenvalues. So we need to examine the eigenvalues of $g \otimes g$.
By finiteness of $G$ and answer 11, $g$ is diagonalizable and has eigenvectors $v_i$ with eigenvalues $\lambda_i$ spanning $V$.

Claim: the vectors $\{v_i \otimes v_j \}_{i\leq j}$ are linearly independent. Suppose not, then there is a relation $\sum_{i\leq j} c_{ij} v_i \otimes v_j = 0$. Since $\{v_j\}$ form a basis, for each $j$ we have $\sum_{i\leq j} c_{ij} v_i = 0$. Again, since $\{v_i\}$ are linearly independent we get $c_{ij} = 0$ for all $i,j$.

Since the given set of vectors has size ${n+1\choose 2}$ which also is the dimension of $S^2 V$, they form an eigenbasis of $S^2V$. Moreover the eigenvalue of $v_i\otimes v_j$ is $\lambda_i\lambda_j$ since $g(v_i\otimes v_j) = gv_i \otimes gv_j = \lambda_i\lambda_j v_i\otimes v_j$.
So we need to express the sum $\sum_{i \leq j} \lambda_i \lambda_j$ in terms of $\left(\sum_i \lambda_i\right)^2$ and $\sum_i \lambda_i ^2$, the latter being the sum of the eigenvalues of $g^2$. This is easy, we have 
$$
\sum_{i\leq j} \lambda_i\lambda_j = \frac{1}{2}\left(\left(\sum_i\lambda_i\right)^2+\sum_i \lambda_i^2 \right),
$$
which is what we wanted to prove.

\vspace{2cm}

{\bf Question 5}

I follow the solution given in Fulton-Harris, proving some assertions more carefully.

Let $\chi$ be the character of $\varrho$ and $\varphi$ the character of any (irreducible) representation. We need to prove that the sequence $a_i:=(\varphi,\varrho^i)$, $i\geq 0$ is not identically zero, as $a_i$ is the multiplicity of representation corresponding to $\varphi$ in $V^{\otimes i}$.

The generating function of this sequence is $\sum_{i=0} ^\infty a_it^i = \sum_{i=0}^\infty \frac{1}{\# G} \sum_C \#C \overline{\varphi(C)} \chi(C)^n t^n = \frac{1}{\# G} \sum_C \frac{\#C \overline{\varphi(C)}}{1-\chi(C)t}$, and we need to prove that it is not identically zero.
We do this by showing that the coefficient in front of $\frac{1}{1-t\dim V}$ is not zero.
This follows from two claims: (i) the only term in the sum contributing to this coefficient is the one where $C$ contains the identity element (ii) that term is non-zero.

Proof of claim (i): $\chi(C) = \tr \varrho g$, the sum of the eigenvalues of $\varrho g$, where $g\in C$.
Since $g^{\#G}=1$, the eigenvalues $\lambda_1,\dots,\lambda_n$, where $n=\dim V$, of $\varrho g$ are roots of unity. Consequently $|\lambda_1+\dots+\lambda_n| \leq |\lambda_1|+\dots+|\lambda_n| = \dim V$ with equality iff $\lambda_i = 1$ for all $i$, by the triangle inequality. Hence $\varrho g$ is the identity, since $\varrho g - Id$ has all eigenvalues equal to $0$ and is semisimple, hence zero.
Since $\varrho$ is injective, $g = 1$.
Proof of claim (ii): $\overline{\varphi(C)} = '\dim' \varphi > 0$

\vspace{2cm}

{\bf Question 6}

I follow the solution given in Fulton-Harris, proving some assertions more carefully.

For any conjugacy class $C$ of $G$, define the $G$-linear map $\varphi_C = \sum _{g\in C} g$.
By Schur's lemma, there is a complex number $\lambda_C$ such that $\varphi_C$ is the identity times $\lambda_C$. Hence $\tr \varphi = \lambda_C \dim V$. Also, $\tr \varphi = \sum_{g\in C} \tr g = \# C \chi_V(C)$ by definition of $\chi_V$.

We now show that $\sum_{g\in C} g$, as $C$ varies of conjugacy classes of $G$, generate the center of $\mbb{Z}[G]$.
That $z \in Z(\mbb{Z}[G])$ means that $z = \sum_{g\in G} c_g g$ and that $zg = gz$ for each $g \in G$.
Hence $c_{ghg^{-1}}$ is equal to $c_h$ for any $g,h\in G$, which means that $c_g$ only depends on the conjugacy class of $g$, and so is a $\mbb{Z}$-linear combination of the elements $\sum_{g\in C} g$.
So $Z(\mbb{Z}[G])$ is a finitely generated abelian group. In such a group, for any element $x$, the chain $\{ \{1,x,\dots,x^r\} \}_r$ stabilizes. This means some power of $x$ is equal to a linear combination of lower powers of $x$, or $x$ is an algebraic integer (root of monic polynomial with integer coefficients). Thus the $\lambda_C$ are all algebraic integers.

Since $\# G= \sum_C \# C \overline{\chi_V(C)}\chi_V(C) = \dim V \sum_C \lambda_C \overline{\chi_V(C)}$ (equivalent to $(\chi_V, \chi_V)=1$). We see that $\#G / \dim V$ too is an algebraic integer. However $\#G$ and $\dim V$ are both positive integers hence $\#G / \dim V$ is a positive rational algebraic integer, or simply put, a positive integer.


\vspace{2cm}

{\bf Question 7}
Since $g^2 = 1$, the eigenvalues of $g$ square to $1$ and hence are $1$ or $-1$. Let $n_+$ be the number of $+1$s and $n_-$ the number of $-1$s. We have $n_++n_- = \dim V = \chi_V(1)$ and consequently $\chi_V(g) = n_+ - n_- = \chi_V(1) - 2n_-$, and $n_-$ clearly is a nonnegative integer.
\vspace{2cm}


{\bf Question 8}

The character table will be indexed by conjugacy classes and irreps.
We first find the conjugacy classes.
They are the conjugacy classes of $S_4$ split up further according to parity (which results in some empty classes, which we omit).

The nonempty classes are represented by $1$ (identity), $(123)$, $(132)$, and $(12)(34)$. This gives us the number of elements in each conjugacy class. So the top of the character table will look as follows

$$
\begin{array}{c|cccc}
     & 1 & 4 & 4 & 3 \\
A_4  & 1 & (123) & (132) & (12)(34) \\
\hline
\end{array}
$$

We should fill in four rows corresponding to the four ($=\#$conjugacy classes) irreps of $A_4$.
As always, we have the trivial representation $\begin{array}{c|cccc} U & 1 & 1 & 1 & 1 \end{array}$.

Since $A_4$ is a subgroup of $S_4$ we could examine the representation given by permuting coordinates in $\mbb{C}^4$, call it $V_1$.
The value of the character of a permutation now becomes the number of fixed points it has, its 'row' is $\begin{array}{c|cccc} V_1 & 4 & 1 & 1 & 0 \end{array}$.
We know that this representation certainly is not irreducible as it is the sum of the trivial representation and some representation $V_2$.
This representation $V_2$ will have row $\begin{array}{c|cccc} V_2 & 3 & 0 & 0 & -1 \end{array}$.
Since the inner product of this row with itself is $1$, we find that $V_2$ is an irrep.

Since $A_4$ has a cyclic (hence abelian) subgroup of order $3$ (but none of order $4$) it is natural to consider it. The irreps. in for this group will be one-dimensional so we use representation and character interchangeably in what follows.

The character $\chi$ of any non-trivial representation $V_3$ of this group assigns $\omega$ or $\omega^2$ to the cycle $(123)$, where $\omega$ is a third root of unity.
Moreover, since $(123)$ and $(132)$ are inverses, we have $\chi((132)) = \overline{\chi((123))}$.
Finally, since $(123)\cdot (12)(34) = (134)$, the latter belonging to the conjugacy class of $(123)$, we have $\chi((123)) \chi((12)(34)) = \chi((123))$, so $\chi((12)(34)) = 1$.
The two choices for $\chi((123))$ gives us the last two rows.

$$
\begin{array}{c|cccc}
     & 1 & 4 & 4 & 3 \\
A_4  & 1 & (123) & (132) & (12)(34) \\
\hline
U             & 1 & 1      & 1        & 1  \\
V_2           & 3 & 0      & 0        & -1 \\
V_{3\omega}   & 1 & \omega & \omega^2 & 1  \\
V_{3\omega^2} & 1 & \omega^2& \omega  & 1  \\
\end{array}.
$$

Since this represents a square matrix with orthonormal rows, we have found all the irreps.


\vspace{2cm}

{\bf Question 9}
A one-dimensional representation is the same thing a morphism from $G$ to $\mbb{C}^*$, the latter being commutative. Since every such morphism factors through $G/G\p$, we may instead count the number of representations of $G/G\p$. Since $G/G\p$ is commutative, all the irreps. have dimension $1$. Since the sum of their squares should be $\#G/G\p = [G:G\p]$ on the one hand and $\#(irreps) \cdot 1$ on the other hand, the number of such representations is $[G:G\p]$.

\vspace{2cm}
{\bf Question 10}

So $G$ is the dihedral group of order $2p$.

(i) This follows from answer 9 since $G\p$ in this case are(as is well known) the rotations, which have index $2$ in $G$.

(ii) This implies and is consistent with the definition 
$$
\rho_m(a) = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}\right),
$$

$$
\rho_m(b) = \left(\begin{array}{cc} \varepsilon^m & 0 \\ 0 & \varepsilon^m \\ \end{array}\right)
$$

To show that $\rho_m$ is a representation it is enough to calculate the products $\rho_m(a)^2$, $\rho_m(b)^p$, $\rho_m(a)\rho_m(b)\rho_m(a)\rho_m(b)$ and verify that they are equal to $\left(\begin{array}{cc} 1 & 0 \\ 0 & 1\\ \end{array}\right)$. I did that.

We now check that they are irreducible. If $\rho_m$ is not irreducible there is some invariant $1$-dimensional subspace spanned by some $u=(u_1,u_2)\in\mbb{C}^2$. Since $\rho_m(a)u$ is parallel to $u$ we have WLOG $u=(1,1)$ or $(1,-1)$.
If $u=(1,1)$ then $\rho_m(b)u$ being parallel to $u$ is the same thing as $\varepsilon^2 = 1$ which contradicts the choice of $p$. If $u=(1,-1)$ we similarly obtain $\varepsilon^2 = 1$. Thus $\rho_m$ is irreducible.

We now show the irreps are pariwise nonisomorphic. Suppose $\rho_m$ and $\rho_{m\p}$ are isomorphic. We need to show that $m=m\p$.

By hypothesis, we have the following equalities for the characters: $\chi_{\rho_m}(b) = \varepsilon + \varepsilon = \vartheta + \vartheta = \chi_{\rho_{m\p}}(b)$, where $\varepsilon = \exp(2\pi i m/p)$ and $\vartheta = \exp(2\pi i m\p / p)$.
Since $z + \frac{1}{z} = w + \frac{1}{w}$ (being equivalent to $(z-w)(zw-1)=0$ if $z,w\neq 0$) iff $z = w$ or $z = \frac{1}{w}$ we have $\varepsilon = \vartheta$ or $\varepsilon = \vartheta^{-1}$.

Suppose $\varepsilon = \vartheta^{-1}$.
Then $m = -m\p$ but this contradicts the choice of $m$ and $m\p$ being between $1$ and $(p-1)/2$.
If $\varepsilon = \vartheta$ then $m=m\p$ and we are done.
\vspace{2cm}

{\bf Question 11}
Suppose $A$ is not diagonalizable. Then there is a block $J$ ($J = \left(\begin{array}{cccc} 1 & 1 & 0 & \dots \\ 0 & 1 & 1 & \dots \\ 0 &0& 1&  \dots \\ \vdots \end{array}\right)$) in the Jordan form of $A$ of size at least $2$.
If $B$ is any matrix of size at least $2\times 2$, let $TL_{2,2}(B)$ be the top left $2\times 2$ submatrix of $B$. It is easy to see that $TL_{2,2}(J^k) = TL_{2,2}(J) ^k$ so it suffices to show that no positive power of $\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)$ is the identity. The $k$:th power of that matrix is $\left(\begin{array}{cc} 1 & k \\ 0 & 1 \end{array}\right)$, so we are done.
\vspace{2cm}

{\bf Question 12} I use the construction of $V := Ind_G ^H U$ given in Fulton-Harris, 3.3.

Let $g \in C$, and $u_\sigma$ a vector spanning $U^\sigma$ for each left coset $\sigma$ of $H$ in $G$.
Furthermore let $g_\sigma\in G$ be a representative of the left coset $\sigma$ for each $\sigma$.
By construction of $V$, $gg_\sigma = g_\tau$, where $\tau$ is the left coset to which $gg_\sigma$ belongs. 
Writing $g$ in the basis $\{u_\sigma\}$ we find that the trace of $g$ is the same as the number of $\sigma$ such that $gg_\sigma = g_\sigma h$ for some $h\in H$, that is, the number of $\sigma$ such that $g_\sigma ^{-1} g g_\sigma\in H$.

Since this does not depend on $g$ we have 
$$
\chi_V(C) =
\frac{1}{\# C} \sum_{g\in C} \sum_{\sigma} I(g_\sigma ^{-1} gg_\sigma \in H) = 
\frac{1}{\# C} \sum_{\sigma} \sum_{g\in C} I(g_\sigma ^{-1} gg_\sigma \in H).
$$

In order to show this is equal to $\frac{[G:H]}{\#C} \#(C\cap H)$ it is sufficient(noting that the number of $\sigma$:s is $[G:H]$ by definition) to show that $g_\sigma^{-1} gg_\sigma\in H$ iff $g \in H$.

Clearly $g_\sigma^{-1} g g_\sigma \in H$ if $g\in H$ and since the action of $g \pmod{H}$ permutes the cosets $\pmod{H}$, the reverse implication follows.
\vspace{2cm}

{\bf Question 13}

-

{\bf Question 14}

We should show that $\frac{d!}{l_1!\dots l_k!} \prod_{i<j} (l_i-l_j) = \frac{d!}{\prod_{ij} h_{ij}}$, where $h_{ij}$ are the hook lengths of the partition $\lambda$, $l_i = \lambda_i+k-i$ and $d=|\lambda|$. Note that this is the same as showing that $\prod_{i<j} (l_i - l_j) \prod_{ij} h_{ij} = l_1!\dots l_k!$.

We use induction on $|\lambda|$, observing that the statement clearly is true if $\lambda=\emptyset$.

Denote by $\lambda\p$ the result of removing the first column from $\lambda$.
Define $l\p, h\p_{ij}$ et.c. in the obvious way.
For $i\leq r$ we have $l\p _i = l_i - 1 - (k-r)$ where $r$ is the length of the second column of $\lambda$ (or the length of the first column of $\lambda\p$).

We know by induction that $\prod_{1\leq i < j \leq r} (l\p_i - l\p_j) \prod_{(i,j)\in\lambda\p} h\p_{ij} = l\p_1!\dots l\p_r!$.
So to show that
$\prod_{1\leq i < j \leq k} (l_i-l_j) \prod_{(i,j)\in \lambda} h_{ij} = l_1!\dots l_k!$
it is enough to show that
$\prod_{1\leq i\leq r < j \leq k} (l_i-l_j) \prod_{r \leq i < j \leq k} (l_i-l_j) \prod_{i=1} ^k l_i =
\prod_{1\leq i \leq r} \frac{l_i!}{(l_i-1-(k-r))!} \prod_{i=r+1}^k l_i!$,
where we have used the following facts:
$l_i-l_j = l\p_i-l\p_j$ whenever both sides are defined,
$h_{i1} = l_i$ for all $i$,
the numbers $h\p_{ij}$ and $h_{ij}$ equal whenever both are defined (that is, whenever $j\neq 1$).
By the choice of $\lambda\p$, we have $l_i = k-i+1$ for $r < i \leq k$.
From this it follows that $\prod_{1\leq i \leq r < j \leq k} (l_i-l_j) = \prod_{i=r+1} ^k l_i!$ and we are left with proving that 
$\prod_{1\leq i\leq r < j \leq k} (l_i-l_j) \prod_{i=1} ^k l_i =
\prod_{1\leq i \leq r} \frac{l_i!}{(l_i-1-(k-r))!}$.
This identity is clearly true since both sides may be written as $\prod_{i=1} ^r \prod_{j=0}^{k-r} (l_i-j)$.


%kvar: (3) 
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