Exterior powers of a vector space and its dual space

[Jakob1]

Hello.

I've just read about natural identifications of exterior powers with spaces of alternating maps, etc here: Some Natural Identifications

However, I have problems showing that the following operations give the same space:

$$V \rightarrow \Lambda_p V \rightarrow (\Lambda_p V)^* \cong \Lambda_p V^*$$

$$V \rightarrow V^* \rightarrow \Lambda_p V ^*$$

I know that $$(\Lambda_p V)^* \cong \Lambda_p V^*$$, because $$(\Lambda_p V)^* \cong \mathcal{A}_p(V)$$, and $$\mathcal{A}_p(V) \ni f \rightarrow L_f \in (\Lambda_p V)^*$$, where $$L_f$$ is the only linear map which makes the universal factorization diagram commute, is an isomorphism.

And $$\mathcal{A}_p(V) \cong \Lambda_p V^*$$, and here we consider this map:

$$V^* \times ... \times V^* \ni (f_1, ..., f_p) \rightarrow (V^p \ni (v_1, ..., v_p) \rightarrow \det [f_i(v_j)] \in \mathbb{K}) \in \mathcal{A}_p(V)$$ which is $p$-linear and antisymmetyric and together with the exterior power map $$V^* \times ... \times V^* \rightarrow \Lambda_p V$$ we get the universal factorization diagram and
the isomorphism we are looking for is the only linear map which makes the diagram commute.

This is all I know at the moment.

Could you help me with this problem?

Thank you.

 

[ThePerfectHacker]

What does "isomorphic" mean here? Does it mean isomorphic as vector spaces?

If you want to show that $(\Lambda^p V)^*$ is isomorphic to $\Lambda^p (V^*)$ as vector spaces all you do is a simple dimension count. The dimension of $V$ and $V^*$ are equal call it $n$ and so the dimensions of the exterior is ${p\choose n}$.

Do you want to show they are isomorphic as representations instead? But if so that would be wrong, I think you need $(\Lambda^p V)^* \simeq \Lambda^{p-n} V^*$.

 

[Jakob1]

Yes, I mean isomorphic as vector spaces.
I'm sorry, my question wasn't clear.

Actually, I need to show that there are two equivalent ways of introducing dot product on $$\Lambda_p (V^*)$$.

Let $$\beta: V \ni v \rightarrow <v | \cdot> \in V^*$$ - here $$< \cdot | \cdot>$$ is the inner product in $$V$$ and $$\beta$$ is an isomorphism.

For $$f, g \in V^*$$ we define $$<f | g> = <\beta ^{-1} (f) | \beta ^{-1} (g)>$$We need to find an inner product on $$\Lambda_p V$$ for which we will have $$\Lambda ^p \beta : \Lambda_p V \ni \xi \rightarrow <\xi | \cdot>_{\Lambda_pV} \in (\Lambda _p V)^*$$.

$$\xi = v_1 \wedge ... \wedge v_p, \ \ \eta = w_1 \wedge ... \wedge w_p$$

And the inner product on $$\Lambda_pV$$ is $$<\xi | \eta> = ((\Lambda^p \beta)(\xi))(\eta) = (\beta(v_1) \wedge ... \wedge \beta (v_p))(w_1 \wedge ... \wedge w_p) = \det [(\beta(v_i))(w_j)]_{i,j=1,...,p}$$.

Then I suppose we could use the first procedure again to find the inner product on $$(\Lambda_pV)^* \cong \Lambda_p(V)^*$$

So my question is will we get the same if we
1) first use the second procedure to find the inner product on $$\Lambda_pV$$ and then the first one to get the inner product on $$(\Lambda_pV)^* \cong \Lambda_p(V)^*$$

2) first find the inner product on $$V^*$$ and then on its exterior power.