My question is motivated by the following observation:

If one considers a representation of $S_n$ on $\mathbb R^n$ given by permutation of co-ordinates, we can see that there are two invariant subspaces, generated by $$x_1+...x_n=0$$ and its orthogonal complement, spanned by the single vector $(1,...,1)$.

The definition for the symmetric algebra on $S^2A:=\{a_1 \otimes a_2+a_2 \otimes a_1=0\}$ which is also invariant under permutation of indices etc. (action of $S_2$)

I'm not used to the symmetric algebra, but one can maybe expect in general to find a similar relationship between $S_n$ and $S A$ on one hand, but what is less clear is what the orthogonal complement of $S^nA$ looks like (is it still just the collection of tensor elements that commute? This doesn't seem likely.)

My question: is there anyway to build up an analogy between polynomials and the symmetric algebra? What would the indeterminates "look like"? I expect degree $\dim V^n$ with indeterminates in the basis elements for the tensor product, but I'm not so sure.


1 Answer 1


Indeed are the same algebra, the dimension of the vectorial space is equal to the number of variables of the polynomial algebra, i.e. $\mathbb{K}[x]$ is obtained by the symmetric algebra of the 1-dimensional vector space over the field $\mathbb{K}$, $\mathbb{K}[x,y]$ is obtained by the symmetric algebra of the 2-dimensional vectorial space, etc..

I will use the notation of generators and relations to outline the thing. If it's not something you're used to, just tell me and I will change notation to a more familiar one.

Let's do an example with 2 variables and then you can generalize easily. In this case $V$ has to be a 2 dimension space. Let's call the vector of the base {$x,y$}. Then the simmetric algebra is obtained by $$S\left(V\right)=T\left(V\right)/\left\langle v{\scriptstyle \otimes}w-w{\scriptstyle \otimes}v\right\rangle,$$ where $T(V)$ is the usual tensorial algebra of tensors of any degree $T\left(V\right)\cong\mathbb{K}\oplus V\oplus\left(V\otimes V\right)\oplus\left(V\otimes V\otimes V\right)\oplus\ldots\,\,\,$. Using the base ${x,y}$ then any element in $S(V)$ is of the form $$p=a_1+a_2x+a_3y+a_4x\otimes x+a_5x\otimes y+a_6y\otimes y+....$$ or using another notation $p$ il the polynomial $$p=a_1+a_2x+a_3y+a_4x^2+a_5xy+a_6y^2+....$$ So in general if you have an $n$-dimensional vector space $V$ over $\mathbb{K}$ with basis {$x_1,x_2,....x_n$}, then the symmetric algebra is isomorfic to the algebra of polynomials with n variables $$S\left(V\right)\cong\mathbb{K}[x_1,x_2,....x_n]$$

  • $\begingroup$ Wow this is a much stronger result than I anticipated. Do you have a reference for the last paragraph? Is this a well known relationship? Thanks! $\endgroup$ Feb 18, 2017 at 20:04
  • $\begingroup$ I didn't see it anywhere specifically, but I think is a well known relationship. You can find something in this spirit in "Introduction to representation theory" of P. Etingof. $\endgroup$
    – Dac0
    Feb 18, 2017 at 22:08
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    $\begingroup$ Follow up. Wikipedia says: "If B is a basis of V, the symmetric algebra S(V) can be identified, through a canonical isomorphism, to the polynomial ring K[B], where the elements of B are considered as indeterminates. Therefore, the symmetric algebra over V can be viewed as a ‚coordinate free’ polynomial ring over V." $\endgroup$
    – M.C.
    Jul 24, 2020 at 13:52

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