2 Character theory 2.5 Universal projections 2.7 The character table of S₅

2.6 Linear algebra constructions

We construct new representations from old in various ways, using linear algebra. We continue to work over $k=\mathbb{C}$ throughout, though many of these constructions work over any field.

2.6.1 The dual representation

Recall that we earlier defined, for any $G$ -representations $V$ and $W$ , a $G$ -representation on the space $\operatorname{Hom}(V,W)$ of linear maps from $V$ to $W$ . It has character $\overline{\chi}_{V}\chi_{W}$ .

Definition 2.37.

If $V$ is any vector space, then the dual space of $V$ is

V^{*}=\operatorname{Hom}(V,\mathbb{C}).

We have $\dim V^{*}=\dim V$ . To see this, given a basis $v_{1},\dots,v_{n}$ of $V$ , we have the dual basis $v_{1}^{*},\dots,v_{n}^{*}$ of $V^{*}$ given by

v_{i}^{*}(v_{j})=\delta_{ij}=\begin{cases}1&\text{if $i=j$}\\ 0&\text{otherwise.}\end{cases}

There is a bilinear map $V^{*}\times V\rightarrow\mathbb{C},(\phi,v)\mapsto\phi(v)$ . The choice of basis identifies $V$ with $\mathbb{C}^{n}$ (as column vectors) and then the dual basis realizes $V^{*}$ as $1\times n$ matrices, with the above pairing being the usual matrix/dot product.

If $V$ has a $G$ -representation $\rho$ , then we take $\mathbb{C}$ to have the trivial representation and get an action $\rho^{*}$ of $G$ on $V^{*}$ defined by $\rho^{*}(g)(\phi)=\phi(\rho(g)^{-1}v)$ . From the formula for the character of $\operatorname{Hom}(V,W)$ , we see

\chi_{V^{*}}=\overline{\chi}_{V}.

If the matrix of $\rho(g)$ with respect to some basis is $A$ , then the matrix of $\rho(g)$ with respect to the dual basis is $(A^{T})^{-1}$ , the inverse of the transpose of $A$ .

2.6.2 Tensor products

Let $V, W$ be two vector spaces. Then the tensor product

V\otimes W

is the $\mathbb{C}$ -vector space generated by the symbols $v\otimes w$ for $v\in V$ and $w\in W$ , with the “bilinear” relations

	$\displaystyle\lambda(v\otimes w)$	$\displaystyle=(\lambda v)\otimes w=v\otimes(\lambda w);$
	$\displaystyle(v+v^{\prime})\otimes w$	$\displaystyle=v\otimes w+v^{\prime}\otimes w;$
	$\displaystyle v\otimes(w+w^{\prime})$	$\displaystyle=v\otimes w+v\otimes w^{\prime}.$

Remark 2.38.

Rigorously, we are taking the (infinite dimensional) vector space with a basis element $v\otimes w$ for every $v\in V$ and $w\in W$ and then forming its quotient by the (infinite dimensional) subspace generated by vectors of the form $(v+v^{\prime})\otimes w-v\otimes w-v^{\prime}\otimes w$ and by similar expressions corresponding to the other relations; this quotient is then finite dimensional, as the proposition below shows.

Proposition 2.39.

Let $e_{1},\ldots,e_{n}$ be a basis of $V$ and $f_{1},\ldots,f_{m}$ be a basis of $W$ . Then

\{e_{i}\otimes f_{j}:i=1,\ldots,n,j=1,\ldots,m\}

is a basis of $V\otimes W$ .

Proof.

Omitted.∎

In particular, the dimension of the tensor product is the product of the dimensions of the vector spaces:

\dim(V\otimes W)=\dim V\dim W.

Contrast the direct sum, which has dimension the sum of the dimensions of the vector spaces.

Exercise 2.40.

It is not true that every vector in $V\otimes W$ is of the form $v\otimes w$ . For example, if $V=W$ is two-dimensional with basis $e, f$ then $e\otimes e+f\otimes f\in V\otimes V$ cannot be written in this form.

Then next remark is non-examinable.

Remark 2.41.

Tensor products can be difficult to get used to. Perhaps the most important principle for understanding them is the following:

A linear map $V\otimes W\rightarrow U$ is the same as a bilinear map $V\times W\rightarrow U$ .

The dictionary as follows: given a linear map $\phi:V\otimes W\rightarrow U$ , we define a bilinear map $V\times W\rightarrow U$ by sending $(v,w)$ to $\phi(v\otimes w)$ . The bilinearity is then a consequence of the relations that hold in the tensor product. Conversely, given a bilinear map $\psi:V\times W\rightarrow U$ , define a linear map $V\otimes W\rightarrow U$ by sending $v\otimes w$ to $\psi(v,w)$ . We have to check that this is well-defined, i.e. that the bilinear relations are respected, and this is equivalent to the bilinearity of $\psi$ .

This all seems very abstract, but it is useful: defining bilinear maps is ‘easy’! In fact, tensor products ‘in real life’ often arise in situations where you have a number that depends on the choice of two vectors (in $V$ , say); then this dependency can be expressed as a linear map $V\otimes V\rightarrow\mathbb{C}$ .

All that said, for this course it will not be important to have such a deep theoretical understanding of tensor products as long as you are able to do calculations with them and are willing to take Proposition 2.39 on trust.

Naturally, we can consider more factors $V_{1}\otimes V_{2}\otimes\cdots\otimes V_{r}$ , with linearity in each slot.

Now, if $V$ and $W$ are both representations of $G$ then $V\otimes W$ becomes a representation via

g(v\otimes w)=gv\otimes gw.

We also write $\rho_{V}\otimes\rho_{W}$ for this representation. If we have bases $e_{1},\ldots,e_{n}$ of $V$ and $f_{1},\ldots,f_{m}$ of $W$ , with respect to which the matrices of $g$ acting on $V$ and $W$ are $A$ and $B$ , and we order the resulting basis of $V\otimes W$ as

e_{1}\otimes f_{1},e_{1}\otimes f_{2},\ldots,e_{1}\otimes f_{m},e_{2}\otimes f% _{1},\ldots,

then the matrix of $g$ on $V\otimes W$ is $A\otimes B$ where this is the block matrix

\begin{pmatrix}a_{11}B&a_{12}B&\ldots\\ a_{21}B&a_{22}B&\ldots\\ \vdots&\vdots&\ddots.\end{pmatrix}.

Proposition 2.42.

If $\rho_{V}$ and $\rho_{W}$ have characters $\chi_{V}$ and $\chi_{W}$ , then

\chi_{V\otimes W}=\chi_{V}\chi_{W}.

Proof.

Let $g\in G$ and choose bases $e_{1},\ldots,e_{n}$ and $f_{1},\ldots,f_{m}$ such that $ge_{i}=\lambda_{i}e_{i}$ and $gf_{j}=\mu_{j}f_{j}$ . Then

g(e_{i}\otimes f_{j})=\lambda_{i}\mu_{j}e_{i}\otimes f_{j},

so with respect to the basis $\{e_{i}\otimes f_{j}\}$ of $V\otimes W$ , $g$ acts diagonally with entries $\lambda_{i}\mu_{j}$ . So

\chi_{V\otimes W}(g)=\sum\lambda_{i}\mu_{j}=\left(\sum\lambda_{i}\right)\left(% \sum\mu_{j}\right)=\chi_{V}(g)\chi_{W}(g).\qed

The tensor product generalises the ’twisting’ construction earlier. If $V$ is any vector space then $V\otimes\mathbb{C}$ is isomorphic to $V$ via the map $v\otimes\lambda\mapsto\lambda v$ . If $(\chi,\mathbb{C})$ is a $1$ -dimensional representation and $(\rho,V)$ is any representation of $G$ , then $\rho\otimes\chi$ is a representation acting on $V\otimes\mathbb{C}$ . We have

(\rho\otimes\chi)(g)(v\otimes\lambda)=(\rho(g)v)\otimes(\chi(g)\lambda)\mapsto% \chi(g)\rho(g)\lambda v

via the above isomorphism so that

\rho\otimes\chi\cong\chi\rho.

Note that if $\rho$ is irreducible, so is $\chi\otimes\rho$ . Furthermore, $\chi\otimes\rho$ might or might not be isomorphic to $\rho$ .

Lemma 2.43.

Let $V$ and $W$ be two finite-dimensional representations of a group $G$ . Then

V^{*}\otimes W\cong\operatorname{Hom}(V,W)

as $G$ -modules.

Proof.

The best way to prove this is to show that the map $\phi\otimes w\mapsto T$ , where $T(v)=\phi(v)w$ , is a $G$ -isomorphism. This is straightforward but a bit technical.

Another proof which works in our situation is simply to observe that both sides have character $\overline{\chi}_{V}\chi_{W}$ . ∎

2.6.3 Symmetric and alternating powers

The symmetric square $\operatorname{Sym}^{2}(V)$ of $V$ is the vector space spanned by symbols $vv^{\prime}$ subject to the bilinear relations above and, additionally,

vv^{\prime}=v^{\prime}v

for all $v,v^{\prime}\in V$ .

Remark 2.44.

Formally, $\operatorname{Sym}^{2}(V)$ is the quotient of $V\otimes V$ by the subspace spanned by all elements of the form $v\otimes v^{\prime}-v^{\prime}\otimes v$ , and then $vv^{\prime}=v^{\prime}v$ is the image of $v\otimes v^{\prime}$ in $\operatorname{Sym}^{2}V$ .

Proposition 2.45.

Given a basis $e_{1},\dots,e_{n}$ of $V$ , the $e_{i}e_{j}$ with $i\leq j$ are a basis of $\operatorname{Sym}^{2}(V)$ . ∎

Hence

\dim\operatorname{Sym}^{2}(V)=\frac{n(n+1)}{2}.

The alternating square $\sideset{}{{}^{2}}{\bigwedge}(V)$ of $V$ is spanned by elements of the form $v\wedge v^{\prime}$ subject to the bilinear relations above and, additionally,

v\wedge v^{\prime}=-v^{\prime}\wedge v

for all $v,v^{\prime}\in V$ .

Remark 2.46.

Formally, it is the quotient of $V\otimes V$ by the subspace spanned by all elements of the form

v\otimes v^{\prime}+v^{\prime}\otimes v,

and then $v\wedge v^{\prime}=-v^{\prime}\wedge v$ is the image of $v\otimes v^{\prime}$ in $\Lambda^{2}(V)$ .

If $V$ and $W$ are representations of $G$ , we define actions of $G$ on these spaces as for the tensor product.

We can define linear maps $\operatorname{Sym}^{2}V\rightarrow V\otimes V$ and $\sideset{}{{}^{2}}{\bigwedge}(V)\rightarrow V\otimes V$ sending $vv^{\prime}\mapsto v\otimes v^{\prime}+v^{\prime}\otimes v$ and $v\wedge v^{\prime}\mapsto v\otimes v^{\prime}-v^{\prime}\otimes v$ . Any $v\otimes w\in V\otimes V$ can be written

v\otimes w=\frac{1}{2}\left((v\otimes w+w\otimes v)+(v\otimes w-w\otimes v)\right)

and this shows that

V\otimes V\cong\operatorname{Sym}^{2}(V)\oplus\sideset{}{{}^{2}}{\bigwedge}(V).

In fact this decomposition holds as $G$ -representations.

Remark 2.47.

The space $V\otimes V$ has an involution²² 2 Map whose square is the identity.

\sigma:v\otimes w\mapsto w\otimes v.

As $\sigma^{2}=I$ , its eigenvalues are $\pm 1$ . The decomposition above is the the eigenspace decomposition for $\sigma$ : $\operatorname{Sym}^{2}V$ is the $(+1)$ -eigenspace, $\Lambda^{2}(V)$ the $(-1)$ -eigenspace.

Proposition 2.48.

If $(\rho,V)$ has character $\chi$ , then

\chi_{\operatorname{Sym}^{2}V}(g)=\frac{1}{2}\left(\chi(g)^{2}+\chi(g^{2}))\right)

and

\chi_{\Lambda^{2}V}(g)=\frac{1}{2}\left(\chi(g)^{2}-\chi(g^{2}))\right).

Proof.

If $\rho(g)$ has eigenvalues $\lambda_{1},\ldots\lambda_{n}$ , then diagonalise it as usual to get an eigenvector basis $v_{1},\ldots,v_{n}$ . Using the basis $v_{i}\wedge v_{j}$ of $\Lambda^{2}(V)$ you find that

\chi_{\Lambda^{2}V}(g)=\sum_{i<j}\lambda_{i}\lambda_{j}.

This is

\frac{1}{2}\left(\left(\sum\lambda_{i}\right)^{2}-\left(\sum\lambda_{i}^{2}% \right)\right)

as required.

The proof for the symmetric square is similar, or use the decomposition of $V\otimes V$ . ∎

Remark 2.49.

One can also define spaces $\operatorname{Sym}^{k}(V)$ and $\Lambda^{k}(V)$ for any $k$ , the latter vanishing if $k>\dim V$ . The former is spanned by expressions $v_{1}v_{2}\ldots v_{k}$ where ‘the order doesn’t matter’, while the latter is spanned by expressions $v_{1}\wedge v_{2}\wedge\ldots\wedge v_{k}$ where ‘switching two vectors introduces a minus sign’.

A particular special case is $k=\dim(V)$ . In this case, $\Lambda^{k}(V)$ is exactly one dimensional (it is easy to see it is spanned by $e_{1}\wedge\ldots\wedge e_{k}$ for $e_{1},\ldots,e_{k}$ any basis of $V$ , showing that the dimension is at most one, and there is an injective map to $V^{\otimes k}$ given by

v_{1}\wedge\ldots\wedge v_{k}\mapsto\sum_{\sigma\in S_{n}}\epsilon(\sigma)e_{1% }\otimes\ldots\otimes e_{k}

which shows that the dimension is at least one).

If $T:V\rightarrow V$ is any linear map, then we get a linear map $\Lambda^{k}(T):\Lambda^{k}(V)\rightarrow\Lambda^{k}(V)$ by setting $T(v_{1}\wedge\ldots\wedge v_{k})=T(v_{1})\wedge\ldots\wedge T(v_{k})$ . Since it is a map from a one-dimensional vector space to itself, $\Lambda^{k}(T)$ is just multiplication by some scalar. This scalar is exactly the determinant of $T$ !

2.6.4 Matrices in dimension 2

Suppose that $(\rho,V)$ is a representation of $G$ and that $\dim V=2$ , with $e_{1},e_{2}$ being a basis of $V$ . Let $g\in G$ , and let

\rho(g)=\begin{pmatrix}a&b\\ c&d\end{pmatrix}

be the matrix of $\rho(g)$ in this basis. We compute the matrices of $\Lambda^{2}\rho(g)$ and $\operatorname{Sym}^{2}\rho(g)$ .

The space $\Lambda^{2}V$ is one-dimensional, with basis vector $e_{1}\wedge e_{2}$ . Then

	$\displaystyle g\cdot(e_{1}\wedge e_{2})$	$\displaystyle=(ge_{1})\wedge(ge_{2})$
		$\displaystyle=(ae_{1}+ce_{2})\wedge(be_{1}+de_{2})$
		$\displaystyle=(ad-bc)e_{1}\wedge e_{2}$
		$\displaystyle=\det(\rho(g))e_{1}\wedge e_{2}$

using that $e_{1}\wedge e_{1}=e_{2}\wedge e_{2}=0$ and $e_{1}\wedge e_{2}=-e_{2}\wedge e_{1}$ . We see that

\operatorname{Sym}^{2}\rho\cong\det\rho.

The space $\operatorname{Sym}^{2}V$ is three-dimensional with basis $e_{1}^{2},e_{1}e_{2},e_{2}^{2}$ , and

	$\displaystyle g(e_{1}^{2})$	$\displaystyle=(ae_{1}+ce_{2})^{2}$
		$\displaystyle=a^{2}e_{1}^{2}+2ace_{1}e_{2}+c^{2}e_{2}^{2}$
	$\displaystyle g(e_{1}e_{2})$	$\displaystyle=abe_{1}^{2}+(ad+bc)e_{1}e_{2}+cde_{2}^{2}$
	$\displaystyle g(e_{2}^{2})$	$\displaystyle=b^{2}e_{1}^{2}+2bde_{1}e_{2}+d^{2}e_{2}^{2}$

whence the matrix of $\operatorname{Sym}^{2}\rho(g)$ (in this basis) is

\begin{pmatrix}a^{2}&ab&b^{2}\\ 2ac&ad+bc&2bd\\ c^{2}&cd&d^{2}\end{pmatrix}.