In Tu's book "Geometry" there is the following statement on page 171:
I would like to show that $span\{e_{I}^{*} \}=(\Lambda_{k}V)^{*}$. For some reason I get completely confused by all the indices. I tried the following so far, which is I think more or less the same as in the case of $V^{*}$, which is often done in a first course in linear algebra
Question: How do I get the inidices right in the following computations?
Let $\phi\in(\Lambda_{k}V)^{*}$. Then $\phi(v_{1}\wedge...\wedge v_{n})=\phi((\sum v_{i_{1}}e_{i_{1}})\wedge...\wedge(\sum v_{i_{n}}e_{i_{n}}))$. Then I would use the multilinearity of $\phi$. Then
$\phi((\sum v_{i_{1}}e_{i_{1}})\wedge...\wedge(\sum v_{i_{n}}e_{i_{n}}))=\sum_{I} v_{i_{1}}...v_{i_{n}}\phi(e_{i_{1}}\wedge...\wedge e_{i_{n}})$, $I:=(i_{1},...,i_{n})$.
This is where I get stuck, because I think that's not the right way of writing it down.
Next I would look at $\sum \phi(e_{i_{1}}\wedge...\wedge e_{i_{n}})e_{I}^{*}$ and evaluate it on $v_{1}\wedge...\wedge v_{n}$. Hence,
$\sum_{I} \phi(e_{i_{1}}\wedge...\wedge e_{i_{n}})e_{I}^{*}(v_{1}\wedge...\wedge v_{n})$
$=\sum_{I} \phi(e_{i_{1}}\wedge...\wedge e_{i_{n}})e_{I}^{*}((\sum v_{j_{1}}e_{j_{1}})\wedge...\wedge(\sum v_{j_{n}}e_{j_{n}}))$
$=\sum_{I,J} v_{j_{1}}...v_{j_{n}}\phi(e_{i_{1}}\wedge...\wedge e_{i_{n}})e_{I}^{*}(e_{j_{1}}\wedge...\wedge e_{j_{n}})$
$=\sum_{I,J} v_{j_{1}}...v_{j_{n}}\phi(e_{i_{1}}\wedge...\wedge e_{i_{n}})\delta_{I,J}$ (Again, I am not sure about the indices.)
$=\sum_{I} v_{j_{1}}...v_{j_{n}}\phi(e_{i_{1}}\wedge...\wedge e_{i_{n}})$.
From which I would conclude that
$\phi=\sum_{I} \phi(e_{i_{1}}\wedge...\wedge e_{i_{n}})e_{I}^{*} $
Thank you very much in advance!
$(e_J)_J$ Forms a basis for the space. For each $J$, $e_J^{*}$ is a linear map $\Lambda^{k}(V)^{\vee}\rightarrow \mathbb{R}$ defined by its action on the basis then extended linearly. Namely, $e_J^*(e_K)=0$ if $J=K$ and 0 otherwise.
More generally, if $V$ is finite dimensional with basis $e_1,...,e_n$, then the dual basis, $(e_j^*)_j$ are defined as above. Namely, $e_j^*(e_i)=\delta_{i,j}$. Once you show that the dual basis, in this general case, is infact a basis, then you get what the author wrote for free.