Dimension of the symmetric\alternating k-tensor over an $n$-dimensional vector space.

Question

I want to solve this question:

Suppose $V$ is a vector of dimension $n$ over a field $F$ of characteristic not equal to 2. Calculate dim $Sym^{k}(V)$(the symmetric k tensor ).

I know that $(Sym^k(V))^*,$ where $*$ denotes the dual is isomorphic to the homogeneous polynomial of degree $k$ in $n$-variables $F[x_1, \dots, x_n]_k$. and I know that in case of $F[x_1, x_2]_k$ its dimension is $k+1$ but I do not know how to generalize its dimension when we have $n$-variables. Could someone clarify this to me please?

Also, how can I calculate dim of $\wedge^k(V)$ (skew symmetric forms)

Answer 1

For Symmetric tensors you can look at this answer . I'll provide an outline for alternating tensors as I cannot find a proper link right now

Given a basis $\{e_{1},...,e_{n}\}$ of $V$.

Define $\{E^{1},...,E^{n}\}$ the dual basis corresponding to $\{e_{1},...,e_{n}\}$.

for a multi-index $(i_{1},...,i_{k})= I$ such that $1\leq i_{j}\leq n$ for all $j=1,2,...,k$.

Define $$E^{I}(v_{1},...,v_{n})=\begin{vmatrix} E^{i_{1}}(v_{1})& E^{i_{2}}(v_{1})&\cdots& E^{i_{k}}(v_{1}) \\ \vdots&\vdots&\cdots & \vdots \\ E^{i_{1}}(v_{k})&E^{i_{2}}(v_{k})&\cdots &E^{i_{k}}(v_{k})\end{vmatrix}$$

Then you can prove that $\{E^{I}:i_{1}<i_{2}<...<i_{k}\}$ forms a basis for alternating forms .

That is you can express any alternating form as $$\sum^{\text{increasing}}_{I}c_{I}E^{I}$$ . That is you are summing over all increasing multi-indices.

And that the above set is linearly independent.

The cardinality is precisely $\dbinom{n}{k}$ as there are precisely $\dbinom{n}{k}$ many ways to pick and arrange $k$ numbers out of $n$ and arrange in increasing order.

For more details(complete proof) see John M Lee's introduction to smooth manifolds chapter 14.