# Abstract Nonsense

## The Tensor Product and the Tensor Product of Representations

Point of post: In this post we discuss the notion of the tensor product and the tensor product of two representations.

Motivation

I have previously discussed the notion of the tensor product, but will be considering a slightly different (although isomorphic) construction. In general any vector space $\mathscr{X}$ with a blinear map $t:\mathscr{V}\times\mathscr{W}\to \mathscr{X}$ such that if $\{x_1,\cdots,x_k\}$ and $\{y_1,\cdots,y_m\}$ are bases for $\mathscr{V}$ and $\mathscr{W}$ then $t(x_i,y_j),\text{ }i\in[k]\text{ }j\in[m]$ can be considered a tensor product in some sense, since this generalized construction defines a unique, up to isomorphism structure. That’s the view we’ll take here with a special focus on the particular case of treating $\mathscr{V}\otimes\mathscr{W}$ as $\text{AntiBil}\left(\mathscr{V},\mathscr{W}\right)$ which is the space of all anti-bilinear forms. We then define the tensor product of two representations.

Bilinear Maps and the Tensor Product

We’ve previously discussed the notion of bilinear forms as being maps $B:\mathscr{V}\boxplus\mathscr{W}\to F$ which are linear in each entry (where $\mathscr{V},\mathscr{W}$ are some $F$-spaces) in the sense that for each fixed $v\in \mathscr{V}$ the map $B_v:\mathscr{W}\to F:w\mapsto B(v,w)$ is linear as is the map $B_w:\mathscr{V}\to F:v\mapsto B(v,w)$ for each fixed $w\in\mathscr{W}$. We can easily extend this notion of “bilinearity” to more general cases. In particular, if $\mathscr{V},\mathscr{W},\mathscr{U}$ are $F$-spaces then a map $B:\mathscr{V}\boxplus\mathscr{W}\to\mathscr{U}$ is called a bilinear map (or just bilinear) if the maps $B_v:\mathscr{W}\to\mathscr{U}:w\mapsto B(v,w)$ and $B_w:\mathscr{V}\to\mathscr{U}:v\mapsto B(v,w)$ are linear for each fixed $v\in\mathscr{V}$ and $w\in\mathscr{W}$.

Let $\mathscr{V,W,U}$ be $F$-spaces and $B:\mathscr{V}\boxplus\mathscr{W}\to\mathscr{U}$ a distinguished bilinear map such that if $\{v_1,\cdots,v_n\}$ and $\{w_1,\cdots,w_m\}$ are bases for $\mathscr{V}$ and $\mathscr{W}$ then $\left\{B(v_i,w_j):i\in[n]\text{ and }j\in[m]\right\}$ is a basis for $\mathscr{U}$. We then call $\mathscr{U}$ a tensor product of $\mathscr{V}$ and $\mathscr{W}$ under $B$ and denote this $\mathscr{U}=\mathscr{V}\underset{B}{\otimes}\mathscr{W}$. It’s clear that for any tensor product of $\mathscr{V}$ and $\mathscr{W}$ the dimension of the tensor product is $\dim(\mathscr{V})\dim\left(\mathscr{W}\right)$ and thus all tensor products of $\mathscr{V}$ and $\mathscr{W}$ are isomorphic.

Anti-Bilinear Forms As A Tensor Product

Let $\mathscr{V}$ and $\mathscr{W}$ be pre-Hilbert spaces with inner products $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$.  We define the space of anti-bilinear forms, denoted $\text{AntiBil}\left(\mathscr{V},\mathscr{W}\right)$, to be the set of all maps $A:\mathscr{V}\boxplus\mathscr{W}\to\mathbb{C}$ such that $A(\cdot,\cdot)$ is antilinear in each entry. This is clearly a vector space with pointwise addition and scalar multiplication of anti-bilinear forms. Our current goal is to show that we can think of $\text{AntiBil}\left(\mathscr{V},\mathscr{W}\right)$ as a tensor product with distinguished bilinear map

$\otimes:\mathscr{V}\boxplus\mathscr{W}\to\text{AntiBil}\left(\mathscr{V},\mathscr{W}\right):(v,w)\mapsto v\otimes w$

where

$\left(v\otimes w\right)(x,y)=\langle v,x\rangle_1\langle w,y\rangle_2$

Of course this is equivalent to showing that $\otimes$ is bilinear map which carries bases to bases. But before we do this we make  a quick observation in the form of the following theorem

Theorem: Let $\mathscr{V}$ and $\mathscr{W}$ be pre-Hilbert spaces (really it’s true even for spaces without specified inner products, but this will suffice) then $\text{AntiBil}$ is $\dim\left(\mathscr{V}\right)\dim\left(\mathscr{W}\right)$ dimensional.

Proof: It’s easily verified, using the same techniques as in the theory of the space of all bilinear forms that if $\{v_1,\cdots,v_n\}$ is a basis for $\mathscr{V}$ and $\{w_1,\cdots,w_m\}$ a basis for $\mathscr{W}$ then

$\displaystyle \left\{A_{i,j}:i\in[n]\text{ and }j\in[m]\right\}$

is a basis for $\text{AntiBil}\left(\mathscr{V},\mathscr{W}\right)$ where $A_{i,j}$ is the unique anti-bilinear form such that $A_{i,j}(v_k,w_\ell)=\delta_{i,k}\delta_{j,\ell}$. $\blacksquare$

With this in mind we go on to prove that $\text{AntiBil}\left(\mathscr{V},\mathscr{W}\right)$ can be viewed as a tensor product. Since the fact that the map $\otimes$ is evidently bilinear it suffices to prove that $\otimes$ carries bases to bases. But, this is true. Indeed:

Theorem: Let $\otimes$ be defined as above, then if $\left\{v_1,\cdots,v_n\right\}$ is a basis for $\mathscr{V}$ and $\{w_1,\cdots,w_m\}$ a basis for $\mathscr{W}$ then $\mathscr{B}=\left\{v_i\otimes w_j:i\in[n]\text{ and }j\in[m]\right\}$  is a basis for $\text{AntiBil}\left(\mathscr{V},\mathscr{W}\right)$.

Proof: By our previous theorem it suffices to show that $\mathscr{B}$ is linearly independent, since $\#\left(\mathscr{B}\right)=\dim\left(\mathscr{V}\right)\dim\left(\mathscr{W}\right)$. Indeed, suppose that

$\displaystyle \sum_{\substack{1\leqslant i\leqslant n\\ 1\leqslant j\leqslant m}}\gamma_{i,j}\left(v_i\otimes w_j\right)=\bold{0}$

Then, fix $i_0\in[n]$ and $j_0\in[m]$. Note that since $\dim\left(\text{span}\{v_i:i\ne i_0\}\right)^{\perp}=1$ there exists some non-zero $x\in\left(\text{span}\{v_i:i\ne i_0\}\right)^{\perp}$. We see then that $\langle x,v_{i_0}\rangle_1\ne 0$ otherwise $x$ would be perpendicular to a basis and thus zero, contrary to construction. Similarly, we may choose $y\in\left(\text{span}\{w_j:j\ne j_0\}\right)^{\perp}$ such that $\langle y,w_{j_0}\rangle_2\ne 0$. It follows then that

$\displaystyle \gamma_{i_0,j_0}\langle x,v_{i_0}\rangle_1\langle y,w_{j_0}\rangle_2=\sum_{\substack{1\leqslant i\leqslant n\\ 1\leqslant j\leqslant m}}\gamma_{i,j}\left(v_i\otimes w_j\right)(x,y)=0$

and since $\langle x,v_{i_0}\rangle_1$ and $\langle y,w_{j_0}\rangle_2$ are non-zero we may conclude that $\gamma_{i_0,j_0}=0$. Since $i_0\in [n]$ and $j_0\in[m]$ was arbitrary the conclusion follows. $\blacksquare$

It follows then that $\text{AntiBil}\left(\mathscr{V},\mathscr{W}\right)$ can be considered as an actual tensor product. From now on out we shall denote $\text{AntiBil}\left(\mathscr{V},\mathscr{W}\right)$  as $\mathscr{V}\otimes\mathscr{W}$ (when discussing the topic of rep. theory) and $\otimes$ shall always denote as it did in the above.

We now wish to endow $\mathscr{V}\otimes\mathscr{W}$ with an inner product structure so we can start discussing the tensor product of representations. We begin with a theorem

Theorem: There exists a unique inner product $\langle\cdot,\cdot\rangle$ on $\mathscr{V}\otimes\mathscr{W}$ such that $\langle A,u\otimes v\rangle= A(u,v)$.

Proof: While clear since this defines it on a basis we show a proof nonetheless. Suppose that $\langle\cdot,\cdot\rangle$ were such an inner product and $\left\{v_1,\cdots,v_n\right\}$ and $\{w_1,\cdots,w_m\}$ bases for $\mathscr{V}$ and $\mathscr{W}$. Then,  we’d then see that the following would necessarily be true:

$\displaystyle \left\langle\sum_{\substack{1\leqslant i\leqslant n\\ 1\leqslant j\leqslant m}}\alpha_{i,j}\left(v_i\otimes w_j\right),\sum_{\substack{1\leqslant k\leqslant n\\ 1\leqslant \ell\leqslant m}}\beta_{k,\ell}\left(v_k\otimes w_{\ell}\right)\right\rangle=\sum_{\substack{1\leqslant i\leqslant n\\ 1\leqslant j\leqslant m}}\sum_{\substack{1\leqslant k\leqslant n\\ 1\leqslant \ell\leqslant m}}\alpha_{i,j}\overline{\beta_{k,\ell}}\left\langle v_i,v_k\right\rangle_1\left\langle w_j,w_{\ell}\right\rangle_2$

and it’s evident from this formulation that the right hand side of the above is, in fact, an inner product. Thus, it remains to show that this inner product satisfies the desired quality. To do this we let $\displaystyle v=\sum_{i=1}^{n}\alpha_i v_i\in\mathscr{V}$, $\displaystyle w=\sum_{j=1}^{m}\beta_j w_j\in\mathscr{W}$ and $\displaystyle \sum_{\substack{1\leqslant k\leqslant n\\ 1\leqslant \ell\leqslant m}}\gamma_{k,\ell}(v_k\otimes w_\ell)\in\mathscr{V}\otimes\mathscr{W}$. Then,

\displaystyle \begin{aligned}\left(\sum_{\substack{1\leqslant k\leqslant n\\ 1\leqslant \ell\leqslant m}}\gamma_{k,\ell}(v_k\otimes w_\ell)\right)\left(\sum_{i=1}^{n}\alpha_i v_i,\sum_{j=1}^{m}\beta_j w_j\right) &= \sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{\substack{1\leqslant k\leqslant n\\ 1\leqslant \ell\leqslant m}}\overline{\alpha_i\beta_j}\gamma_{k,\ell}\left(v_k\otimes w_\ell\right)(v_i,w_j)\\ &= \sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{\substack{1\leqslant k\leqslant n\\ 1\leqslant \ell\leqslant m}}\overline{\alpha_i\beta_j}\gamma_{k,\ell}\left\langle v_k,v_i\right\rangle_1\left\langle w_\ell,w_j\right\rangle_2\end{aligned}

But, with equal validity

\displaystyle \begin{aligned}\left\langle \sum_{\substack{1\leqslant k\leqslant n\\ 1\leqslant \ell\leqslant m}}\gamma_{k,\ell}\left(v_k\otimes v_\ell\right),\left(\sum_{i=1}^{n}\alpha_i v_i\right)\otimes\left(\sum_{j=1}^{m}\beta_j w_j\right)\right\rangle &= \left\langle \sum_{\substack{1\leqslant k\leqslant n\\ 1\leqslant \ell\leqslant m}}\gamma_{k,\ell}\left(v_k\otimes w_\ell\right),\sum_{i=1}^{n}\sum_{j=1}^{m}\alpha_i\beta_j \left(v_i\otimes w_j\right)\right\rangle\\ &= \sum_{\substack{1\leqslant k\leqslant n\\ 1\leqslant \ell\leqslant m}}\gamma_{i,j}\overline{\alpha_i\beta_j}\left\langle v_k,v_i\right\rangle_1\left\langle w_\ell,w_j\right\rangle_2\end{aligned}

from where the conclusion follows. $\blacksquare$

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print.

February 14, 2011 -

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