Wedge products and Basis for the set of antisymmetric tensors

Question

I just started studying Antisymmetric tensors. After introducing the antisymmetric tensors, the wedge product of dual vectors was defined. Then it was said that the set $\{e^i \wedge e^j\}_{i<j}$ is linearly independent and spans the space of all antisymmetric (r,0) tensors $\Lambda^2V^*$. The proof was left as an exercise in the book. The hint given in the book is - evaluate an arbitrary linear combination on an arbitrary tensor product $e_k \otimes e_l$.

To my understanding: $e_k \otimes e_l$ is a basis vector for $V \otimes V$, which would turn out to be the set of all (0,2) tensors. I can write a linear combination like this: \begin{equation} \sum\limits_{i,j}c_{ij}e^i \wedge e^j \end{equation} But I'm not really sure how to do evaluate this on the tensor product $e_k \otimes e_l$ since the wedge product itself is made of tensor products. As far as I know, for $V\otimes W$, \begin{equation} (v\otimes w)(h,g)=h(v)g(w) \forall h \in V^*, g \in W^*. \end{equation} Evaluating a tensor product on another tensor product doesn't make sense to me. The book has mentioned no such thing.

So, please state the proof and explain these things in a way that I can understand. (I'm a physics student and not actually a math one.)

Answer 1

I will address:

But I'm not really sure how to do evaluate this on the tensor product $e_k \otimes e_l$ since the wedge product itself is made of tensor products. As far as I know, for $V\otimes W$, \begin{equation} (v\otimes w)(h,g)=h(v)g(w) \forall h \in V^*, g \in W^*. \end{equation} Evaluating a tensor product on another tensor product doesn't make sense to me. The book has mentioned no such thing.

Here you have an element of $V\otimes W$ acting as a map $v\otimes w:V^*\times W^*\to \mathbb R$. You can check that this map is bi-linear, and a bi-linear map induces a map on the tensor product $V^*\otimes W^* \to \mathbb R$ by defining $$(v\otimes w)(h\otimes g) := (v\otimes w)(h,g).$$ Note that this is well-defined because the tensor product properties exactly match up with the bi-linearity properties (e.g. $(h+k)\otimes g = h\otimes g + k\otimes g$, etc.).

(I think this "multi-linear map property" actually characterizes the tensor product -- I was looking for a specific Math SE answer which explained tensors and their universal property really well but couldn't find it, maybe someone knows what I'm talking about.) You should also play around with $(V^*)^*=V$ to get dual statements of the above.