Exterior powers of a vector space and its dual space

In summary, the conversation discusses the problem of showing that two operations give the same space: V \rightarrow \Lambda_p V \rightarrow (\Lambda_p V)^* \cong \Lambda_p V^* and V \rightarrow V^* \rightarrow \Lambda_p V ^*. The speaker knows that (\Lambda_p V)^* \cong \Lambda_p V^* and \mathcal{A}_p(V) \cong \Lambda_p V^*, and is trying to find an inner product on \Lambda_p V for which \Lambda ^p \beta : \Lambda_p V \rightarrow (\Lambda _p V)^*. The question is whether using the first procedure (finding the inner product on V^* and then
  • #1
Jakob1
23
0
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:

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

\(\displaystyle V \rightarrow V^* \rightarrow \Lambda_p V ^*\)

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

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

\(\displaystyle V^* \times ... \times V^* \ni (f_1, ..., f_p) \rightarrow (V^p \ni (v_1, ..., v_p) \rightarrow \det [f_i(v_j)] \in \mathbb{K}) \in \mathcal{A}_p(V)\) which is $p$-linear and antisymmetyric and together with the exterior power map \(\displaystyle V^* \times ... \times V^* \rightarrow \Lambda_p V\) 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.
 
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  • #2
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 $(\Lambda^p V)^* \simeq \Lambda^{p-n} V^*$.
 
  • #3
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 \(\displaystyle \Lambda_p (V^*)\).

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

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

\(\displaystyle \xi = v_1 \wedge ... \wedge v_p, \ \ \eta = w_1 \wedge ... \wedge w_p\)

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

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

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

2) first find the inner product on \(\displaystyle V^*\) and then on its exterior power.
 
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FAQ: Exterior powers of a vector space and its dual space

What are exterior powers of a vector space and its dual space?

Exterior powers are a type of mathematical operator that is used to define a new vector space from an existing vector space. It is a way to extend the concept of a vector space to include anti-symmetric tensors. The dual space is the set of all linear functionals on the original vector space.

How are exterior powers and dual spaces related?

Exterior powers and dual spaces are closely related as they both involve the concept of extending a vector space. The exterior power of a vector space is isomorphic to the dual space of the exterior power of its dual space. In simpler terms, the exterior power of a vector space and its dual space are dual to each other.

What is the purpose of using exterior powers and dual spaces?

The use of exterior powers and dual spaces allows for a more comprehensive understanding and analysis of vector spaces. They provide a way to study anti-symmetric tensors, which are important in areas such as differential geometry and physics. They also have practical applications in fields such as computer vision and signal processing.

How do you compute the exterior power or dual space of a vector space?

To compute the exterior power of a vector space, one can use the wedge product of vectors. The dimension of the exterior power is equal to the number of vectors in the wedge product. To compute the dual space, one can use the transpose or adjoint of a linear transformation to map the dual space to the original vector space.

Are there any limitations to using exterior powers and dual spaces?

While exterior powers and dual spaces are powerful tools in mathematics and science, they do have limitations. They are only defined for finite-dimensional vector spaces, and their calculations can become computationally intensive for higher dimensions. Additionally, interpreting the results of exterior powers and dual spaces can be challenging, as they involve abstract concepts and notations.

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