Elements in a symmetric algebra If $V$ is a vector space of dimension $n$, what do elements in Sym$^k(V)$ look like? I've seen the "mod out by" definition, but I am still unclear as to what this vector space looks like. I have read somewhere that they resemble polynomials.
 A: If you fix a basis $(v_1, \ldots, v_n)$ for $V$, then elements in $\mathrm{Sym}^k(V)$ are $\mathbb{F}$-linear combinations of elements of the form $v_{i_1} v_{i_2} \dots v_{i_k}$ where each $i_j$ satisfies $1 \leq i_j \leq n$ so for example $v_1^2, v_1v_2$ and $2v_1^2 + 3v_2v_3$ belong to $\mathrm{Sym}^2(V)$.
Written using multi-index notation, the elements
$$ \mathbf{v}^{\alpha} = v_1^{\alpha_1} \dots v_n^{\alpha_n} $$
where $\alpha = (\alpha_1, \ldots, \alpha_n)$ is a multi-index of degree
$$ |\alpha| = \sum_{i=1}^n \alpha_i = k $$
form a basis of $n + k + 1 \choose k$ elements for $\mathrm{Sym}^k(V)$. Note that the notation $v_{i_1} \dots v_{i_k}$ is a notation for the equivalence class of $v_{i_1} \otimes \dots \otimes v_{i_k}$ in $\mathrm{Sym}^k(V)$ so for example, we have $v_1v_2 = v_2 v_1$.
If $V = \left( \mathbb{F}^n \right)^{*}$ and we choose the basis for $V$ to be the dual basis $(x_1, \ldots, x_n)$ of the standard basis (so $x_i(e_j) = \delta_{ij}$), then elements of $\mathrm{Sym}^k(V)$ are expressions of the form
$$ \sum_{|\alpha| = k} a_{\alpha} \mathbf{x}^{\alpha} = \sum_{|\alpha| = k} a_{\alpha} x_1^{\alpha_1} \dots x_n^{\alpha_n}$$
and so are naturally identified with homogeneous polynomials of degree $k$ on $\mathbb{F}^n$. More generally, $\mathrm{Sym}^k(V^{*})$ can be taken to be a coordinate independent definition of the space of homogeneous polynomials on $V$ of degree $k$. 
