# Relationship between polynomials and the symmetric algebra.

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.

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 $$S\left(V\right)\cong\mathbb{K}[x_1,x_2,....x_n]$$

• 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! Feb 18, 2017 at 20:04
• 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.
– Dac0
Feb 18, 2017 at 22:08
• 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."
– M.C.
Jul 24, 2020 at 13:52