Dimension of symmetric $n$-forms on vector space of dimension $n$

Question

Let $E$ be a vector space of dimension $n$ over a field $k=\mathbb R$ or $\mathbb C$, and consider $S^n E^*$ to be the space of $n$-linear symmetric forms. What is the dimension of $S^n E^*$?

I would say 1, because I can't think of anything other than multiples of the form $(x_1, \ldots, x_n) \mapsto x_1 \ldots x_n$. But I'm not sure this is correct.

Answer 1

Abstract

Number of all $n$-linear forms on $n$-dimensional space is $n^n$. This is because $n$-linear form is determined by values it takes on set $\mathcal{A}$ of all possible $n$-tuples of vectors from basis, and clearly $\mathcal{A}$ has $n^n$ elements.

Symmetry means that every permutation of arguments in $n$-linear form $L: E^n \rightarrow \mathbb{K} $ gives the same result. Therefore for two $n$-tuples of vectors $(v_1,..,v_n), (w_1,..,w_n)\in E^n$ that are permutations of each other $L(v_1,..,v_n)=L(w_1,..,w_n)$. Now it is enough to consider only $n$-tuples of vectors from basis with nondecreasing indexes: $\mathcal{N}\subset\mathcal{A}$. The power of $\mathcal{N}$ and the answer to your question is ${2n-1}\choose{n-1}$. Below is not too short substantiation.

Introduction

Lets state the problem precisely. Consider vector space $E$ over field $\mathbb{K}$ with $dimE=n$ and basis $B=\{e_1,..,e_n\}$. Mapping $L: E^n \rightarrow \mathbb{K} $ is n-linear form when it is linear with restriction to every single variable, formally:

• For all $1\le i \le n$ and fixed vectors $x_1,..,x_{i-1},x_{i+1},..,x_n \in E$ mapping $E\ni \mathbf x \mapsto L(x_1,..,x_{i-1},\mathbf x,x_{i+1},..,x_n) \in \mathbb{K}$ is linear.

Moreover it is symmetric if its value doesn't change after swap of two arguments, i.e.:

• For all $1\le i<j \le n$: $L(x_1,..,x_i,..,x_j,..,x_n)=L(x_1,..,x_j,..,x_i,..,x_n)$

Mapping $(x_1,..,x_n) \mapsto x_1..x_n$ isn't well defined, because vector multiplication is not defined too. What you mean is probably multiplication of coordinates; $$(x_1,..,x_n)\mapsto x_{11}..x_{n1}$$ where $x_i=x_{i1}e_1+..+x_{in}e_n$ and $x_{ij}\in \mathbb{K}$. It is convenient to think of $E$ simply as $\mathbb{K}^n$ to which it is isomorphic.

Calculations

Consider simple example when $n=2$ and $E=\mathbb{R}^2$ is equipped with standard basis. Let $L:(\mathbb{R}^2)^2 \rightarrow \mathbb{R}$ be bilinear symmetric form. Then:

$$L(x_1,x_2)=L((x_{11},x_{12}),(x_{21},x_{22}))$$ $$L((x_{11},x_{12}),(x_{21},x_{22}))= L((x_{11},0),(x_{21},x_{22}))+L((0,x_{12}),(x_{21},x_{22})) = x_{11}L((1,0),(x_{21},x_{22}))+x_{12}L((0,1),(x_{21},x_{22}))= x_{11}L(e_1,(x_{21},x_{22}))+x_{12}L(e_2,(x_{21},x_{22}))$$ Analogous treatment of the second argument gives $$ L((x_{11},x_{12}),(x_{21},x_{22})) = x_{11}(x_{21}L(e_1,e_1)+x_{22}L(e_1,e_2)) + x_{12}(x_{21}L(e_2,e_1)+x_{22}L(e_2,e_2))$$ and by symmetry $$ x_{11}x_{22}L(e_1,e_2)+x_{12}x_{21}L(e_2,e_1)= (x_{11}x_{22}+x_{12}x_{21})L(e_1,e_2)$$ so finally $$ L(x_1,x_2)= x_{11}x_{21}L(e_1,e_1)+(x_{11}x_{22}+x_{12}x_{21})L(e_1,e_2)+x_{12}x_{22}L(e_2,e_2)$$ and those three summands are independent, so $\{L(e_1,e_1), L(e_1,e_2), L(e_2,e_2)\}$ is a basis of bilinear symmetric forms on $\mathbb{R}^2$. Reasoning with arbitrary $n$ and $\mathbb{K}$ is the same in principle - by symmetry the order in $L(e_{k_1},.., e_{k_n})$ can be changed to nondecreasing (or in fact to any fixed ordering) $L(e_{i_1},.., e_{i_n})$. Exact formula is somewhat messy: $$ L(x_1,..,x_n)=\sum_{I \in \mathcal{I}}(\sum_{\sigma\in P_n^{I}}x_{1\ i_{\sigma(n)}}..x_{n\ i_{\sigma(n)}})L(e_{i_1},..,e_{i_n})$$ where $I=(i_1,..,i_n),\ \ \mathcal{I}=\{(i_1,..,i_n)\in \mathbb{N}^n:1\le i_1 \le..\le i_n \le n\},\ \ $ and $P_n^I$ is subset of $n$-permutation group such that every possible reordering of $(i_1,..,i_n)$ is received by exactly one element (this set is not unique). Thus if $\sigma, \tau \in P_n^I$ and $\ \sigma\neq \tau$ then $(i_{\sigma(1)},..,i_{\sigma(n)})\neq (i_{\tau(1)},..,i_{\tau(n)})$. Final observation is $|\mathcal{I}|=|\mathcal{N}|=$ $ {2n-1}\choose{n-1}$.