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

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)

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} 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.