# The Unapologetic Mathematician

## Tensor and Symmetric Algebras

There are a few graded algebras we can construct with our symmetric and antisymmetric tensors, and at least one of them will be useful. Remember that we also have symmetric and alternating multilinear functionals in play, so the same constructions will give rise to even more algebras.

First and easiest we have the tensor algebra on $V$. This just takes all the tensor powers of $V$ and direct sums them up

$\displaystyle T(V)=\bigoplus\limits_{n=0}^\infty V^{\otimes n}$

This gives us a big vector space — an infinite-dimensional one, in fact — but it’s not an algebra until we define a bilinear multiplication. For this one, we’ll just define the multiplication by the tensor product itself. That is, if $\mu\in V^m$ and $\nu\in V^n$ are two tensors, their product will be $\mu\otimes\nu\in V^{m+n}$, which is by definition bilinear. This algebra has an obvious grading by the number of tensorands.

This is exactly the free algebra on a vector space, and it’s just like we built the free ring on an abelian group. If we perform the construction on the dual space $V^*$ we get an algebra of functions. If $V$ has dimension $d$, then this is isomorphic to the algebra $T(V^*)\cong\mathbb{F}\{X^1,\dots,X^d\}$ of noncommutative polynomials in $d$ variables.

Next we consider the symmetric algebra on $V$, which consists of the direct sum of all the spaces of symmetric tensors

$\displaystyle S(V)=\bigoplus\limits_{n=0}^\infty S^n(V)$

with a grading again given by the number of tensorands.

Now, despite the fact that each $S^n(V)$ is a subspace of the tensor space $T^{\otimes n}$, this is not a subalgebra of $T(V)$. This is because the tensor product of two symmetric tensors may well not be symmetric itself. Instead, we will take the tensor product of $\mu\in S^m(V)$ and $\nu\in S^n(V)$, and then symmetrize it, to give $\mu\odot\nu\in S^{m+n}(V)$. This will be bilinear, and it will work with our choice of grading, but will it be associative?

If we have three symmetric tensors $\lambda\in S^l(V)$, $\mu\in S^m(V)$, and $\nu\in S^n(V)$, then we could multiply them by $(\lambda\odot\mu)\odot\nu$ or by $\lambda\odot(\mu\odot\nu)$. To get the first of these, we tensor $\lambda$ and $\mu$, symmetrize the result, then tensor with $\nu$ and symmetrize that. But since symmetrizing $\lambda\otimes\mu$ consists of adding up a number of shuffled versions of this tensor, we could tensor with $\nu$ first and then symmetrize only the first $l+m$ tensorands, before finally tensoring the entire thing. I assert that these two symmetrizations — the first one on only part of the whole term — are equivalent to simply symmetrizing the whole thing. Similarly, symmetrizing the last $m+n$ tensorands followed by symmetrizing the whole thing is equivalent to just symmetrizing the whole thing. And so both orders of multiplication are the same, and the operation $\odot$ indeed defines an associative multiplication.

To see this, remember that symmetrizing the whole term involves a sum over the symmetric group $S_{l+m+n}$, while symmetrizing over the beginning involves a sum over the subgroup $S_{l+m}\subseteq S_{l+m+n}$ consisting of those permutations acting on only the first $l+m$ places. This will be key to our proof. We consider the collection of left cosets of $S_{l+m}$ within $S_{l+m+n}$. For each one, we can pick a representative element (this is no trouble since there are only a finite number of cosets with a finite number of elements each) and collect these representatives into a set $C$. Then the whole group $S_{l+m+n}$ is the disjoint union

$\displaystyle S_{l+m+n}=\biguplus\limits_{\gamma\in\Gamma}\gamma S_{l+m}$

This will let us rewrite the symmetrizer in such a way as to make our point. So let’s write down the product of the two group algebra elements we’re interested in

\displaystyle\begin{aligned}\left(\frac{1}{(l+m+n)!}\sum\limits_{\pi\in S_{l+m+n}}\pi\right)\left(\frac{1}{(l+m)!}\sum\limits_{\hat{\pi}\in S_{l+m}}\hat{\pi}\right)&=\left(\frac{1}{(l+m+n)!}\sum\limits_{\gamma\in\Gamma}\sum\limits_{\pi\in\gamma S_{l+m}}\pi\right)\left(\frac{1}{(l+m)!}\sum\limits_{\hat{\pi}\in S_{l+m}}\hat{\pi}\right)\\&=\left(\frac{1}{(l+m+n)!}\sum\limits_{\gamma\in\Gamma}\sum\limits_{\pi\in S_{l+m}}\gamma\pi\right)\left(\frac{1}{(l+m)!}\sum\limits_{\hat{\pi}\in S_{l+m}}\hat{\pi}\right)\\&=\left(\frac{1}{(l+m+n)!}\left(\sum\limits_{\gamma\in\Gamma}\gamma\right)\left(\sum\limits_{\pi\in S_{l+m}}\pi\right)\right)\left(\frac{1}{(l+m)!}\sum\limits_{\hat{\pi}\in S_{l+m}}\hat{\pi}\right)\\&=\left(\frac{1}{(l+m+n)!}\sum\limits_{\gamma\in\Gamma}\gamma\right)\left(\frac{1}{(l+m)!}\sum\limits_{\pi\in S_{l+m}}\sum\limits_{\hat{\pi}\in S_{l+m}}\pi\hat{\pi}\right)\\&=\frac{1}{(l+m+n)!}\left(\sum\limits_{\gamma\in\Gamma}\gamma\right)\left(\sum\limits_{\pi\in S_{l+m}}\pi\right)\\&=\frac{1}{(l+m+n)!}\sum\limits_{\pi\in S_{l+m+n}}\pi\end{aligned}

Essentially, because the symmetrization of the whole term subsumes symmetrization of the first $l+m$ tensorands, the smaller symmetrization can be folded in, and the resulting sum counts the whole sum exactly $(l+m)!$ times, which cancels out the normalization factor. And this proves that the multiplication is, indeed, associative.

This multiplication is also commutative. Indeed, given $\mu\in S^m(V)$ and $\nu\in S^n(V)$, we can let $\tau_{m,n}$ be the permutation which moves the last $n$ slots to the beginning of the term and the first $m$ slots to the end. Then we write

\displaystyle\begin{aligned}\mu\odot\nu&=\left(\frac{1}{(m+n)!}\sum\limits_{\pi\in S_{m+n}}\pi\right)(\mu\otimes\nu)\\&=\left(\frac{1}{(m+n)!}\sum\limits_{\pi\in S_{m+n}}\pi\tau_{m,n}\right)(\mu\otimes\nu)\\&=\left(\frac{1}{(m+n)!}\sum\limits_{\pi\in S_{m+n}}\pi\right)\left(\tau_{m,n}(\mu\otimes\nu)\right)\\&=\left(\frac{1}{(m+n)!}\sum\limits_{\pi\in S_{m+n}}\pi\right)(\nu\otimes\mu)\\&=\nu\odot\mu\end{aligned}

because right-multiplication by $\tau_{m,n}$ just shuffles around the order of the sum.

The symmetric algebra $S(V)$ is the free commutative algebra on the vector space $V$. And so it should be no surprise that the symmetric algebra on the dual space is isomorphic to the algebra of polynomial functions on $V$, where the grading is the total degree of a monomial. If $V$ has finite dimension $d$, we have $S(V^*)\cong\mathbb{F}[X^1,\dots,X^d]$.

October 26, 2009 - Posted by | Algebra, Linear Algebra

1. Those direct sums should start at n=0 if you want the algebra to have an identity.

Comment by Qiaochu Yuan | October 26, 2009 | Reply

2. Thanks for catching that.

Comment by John Armstrong | October 26, 2009 | Reply

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6. Is it possible to provide an explicit example as given in https://unapologetic.wordpress.com/2008/12/22/symmetric-tensors/ ?
Suppose we have two symmetric tensors a = [ a_1 b_1] and b = [a_2 b_2] (written in matrix form).
b_1 c_1 b_2 c_2

What would get on symmetrizing a \otimes b ?

Comment by Datta | March 12, 2012 | Reply

7. Sorry, in the above I meant the symmetric tensors a = [a_1, b_1, b_1, c_1] and b = [a_2, b_2, b_2, c_2] which can also be written as 2 by 2 symmetric matrices.

Comment by Datta | March 12, 2012 | Reply