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A natural isomorphism between the dual of k-th exterior power and the k-th exterior power of the dual?
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On the wikipedia page of Exterior Algebra under the section on alternating multilinear forms, it says:
“... By the universal property of the exterior power, the space of alternating forms of degree $k$ on $V$ is naturally isomorphic with the dual vector space $left( Lambda^k V right)^* $. If $V$ is finite-dimensional, then the latter is naturally isomorphic to $ Lambda ^k left(V^*right) $. ...â€Â
I understand the first part, but I don’t see how the dual space of the $k$-th exterior power of $V$ is naturally isomorphic to the $k$-th exterior power of the dual of $V$? I get that they have the same dimension, but what’s the specific natural isomorphism?
I want to know this because I don’t see how alternating multilinear forms correspond to exterior products of covectors.
differential-geometry differential-forms
asked Jul 16 at 7:56
user577413
838
add a comment |Â
up vote
1
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On the wikipedia page of Exterior Algebra under the section on alternating multilinear forms, it says:
“... By the universal property of the exterior power, the space of alternating forms of degree $k$ on $V$ is naturally isomorphic with the dual vector space $left( Lambda^k V right)^* $. If $V$ is finite-dimensional, then the latter is naturally isomorphic to $ Lambda ^k left(V^*right) $. ...â€Â
I understand the first part, but I don’t see how the dual space of the $k$-th exterior power of $V$ is naturally isomorphic to the $k$-th exterior power of the dual of $V$? I get that they have the same dimension, but what’s the specific natural isomorphism?
I want to know this because I don’t see how alternating multilinear forms correspond to exterior products of covectors.
differential-geometry differential-forms
asked Jul 16 at 7:56
user577413
838
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 7:57
@JoséCarlosSantos Thanks very much!
– user577413
Jul 16 at 8:11
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
On the wikipedia page of Exterior Algebra under the section on alternating multilinear forms, it says:
“... By the universal property of the exterior power, the space of alternating forms of degree $k$ on $V$ is naturally isomorphic with the dual vector space $left( Lambda^k V right)^* $. If $V$ is finite-dimensional, then the latter is naturally isomorphic to $ Lambda ^k left(V^*right) $. ...â€Â
I understand the first part, but I don’t see how the dual space of the $k$-th exterior power of $V$ is naturally isomorphic to the $k$-th exterior power of the dual of $V$? I get that they have the same dimension, but what’s the specific natural isomorphism?
I want to know this because I don’t see how alternating multilinear forms correspond to exterior products of covectors.
differential-geometry differential-forms
asked Jul 16 at 7:56
user577413
838
On the wikipedia page of Exterior Algebra under the section on alternating multilinear forms, it says:
“... By the universal property of the exterior power, the space of alternating forms of degree $k$ on $V$ is naturally isomorphic with the dual vector space $left( Lambda^k V right)^* $. If $V$ is finite-dimensional, then the latter is naturally isomorphic to $ Lambda ^k left(V^*right) $. ...â€Â
I understand the first part, but I don’t see how the dual space of the $k$-th exterior power of $V$ is naturally isomorphic to the $k$-th exterior power of the dual of $V$? I get that they have the same dimension, but what’s the specific natural isomorphism?
I want to know this because I don’t see how alternating multilinear forms correspond to exterior products of covectors.
differential-geometry differential-forms
asked Jul 16 at 7:56
user577413
838
edited Jul 16 at 8:15
asked Jul 16 at 7:56
user577413
838
asked Jul 16 at 7:56
user577413
838
asked Jul 16 at 7:56
user577413
838
838
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 7:57
@JoséCarlosSantos Thanks very much!
– user577413
Jul 16 at 8:11
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 7:57
@JoséCarlosSantos Thanks very much!
– user577413
Jul 16 at 8:11
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 7:57
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 7:57
@JoséCarlosSantos Thanks very much!
– user577413
Jul 16 at 8:11
@JoséCarlosSantos Thanks very much!
– user577413
Jul 16 at 8:11
add a comment |Â
1 Answer
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Usually the natural isomorphism is given via a perfect pairing $$Lambda^k V^ast times Lambda^k V rightarrow k,~ (phi_1wedge cdots wedge phi_k, v_1wedgecdots wedge v_k)mapsto det(phi_i(v_j))$$
which gives rise to the desired isomorphism. The naturality of this construction is easily verified.
answered Jul 16 at 9:10
asdq
1,4651417
I see. Please excuse a further question: what’s the motivation behind this pairing? Correct me if I’m wrong, but this seems to be related to the inner product on k-vectors from an inner product on V.
– user577413
Jul 16 at 11:25
You are correct: Precompose the above map by $Lambda^k (cdot,cdot)times Lambda^k rmid$ where $(cdot,cdot)colon Vrightarrow V^ast$ is the isomorphism induced by the inner product, and you get the inner product of the wedge product.
– asdq
Jul 16 at 11:33
As for motivation, I think it is a good idea to get a geometric intuition about this. There are ways to visualize the wedge product in low dimensions, I think you find something about this on Wikipedia.
– asdq
Jul 16 at 11:37
Great! Now my question reduces to why the inner product on wedges ought to be a determinant.
– user577413
Jul 16 at 13:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Usually the natural isomorphism is given via a perfect pairing $$Lambda^k V^ast times Lambda^k V rightarrow k,~ (phi_1wedge cdots wedge phi_k, v_1wedgecdots wedge v_k)mapsto det(phi_i(v_j))$$
which gives rise to the desired isomorphism. The naturality of this construction is easily verified.
answered Jul 16 at 9:10
asdq
1,4651417
I see. Please excuse a further question: what’s the motivation behind this pairing? Correct me if I’m wrong, but this seems to be related to the inner product on k-vectors from an inner product on V.
– user577413
Jul 16 at 11:25
You are correct: Precompose the above map by $Lambda^k (cdot,cdot)times Lambda^k rmid$ where $(cdot,cdot)colon Vrightarrow V^ast$ is the isomorphism induced by the inner product, and you get the inner product of the wedge product.
– asdq
Jul 16 at 11:33
As for motivation, I think it is a good idea to get a geometric intuition about this. There are ways to visualize the wedge product in low dimensions, I think you find something about this on Wikipedia.
– asdq
Jul 16 at 11:37
Great! Now my question reduces to why the inner product on wedges ought to be a determinant.
– user577413
Jul 16 at 13:57
add a comment |Â
up vote
0
down vote
Usually the natural isomorphism is given via a perfect pairing $$Lambda^k V^ast times Lambda^k V rightarrow k,~ (phi_1wedge cdots wedge phi_k, v_1wedgecdots wedge v_k)mapsto det(phi_i(v_j))$$
which gives rise to the desired isomorphism. The naturality of this construction is easily verified.
answered Jul 16 at 9:10
asdq
1,4651417
I see. Please excuse a further question: what’s the motivation behind this pairing? Correct me if I’m wrong, but this seems to be related to the inner product on k-vectors from an inner product on V.
– user577413
Jul 16 at 11:25
You are correct: Precompose the above map by $Lambda^k (cdot,cdot)times Lambda^k rmid$ where $(cdot,cdot)colon Vrightarrow V^ast$ is the isomorphism induced by the inner product, and you get the inner product of the wedge product.
– asdq
Jul 16 at 11:33
As for motivation, I think it is a good idea to get a geometric intuition about this. There are ways to visualize the wedge product in low dimensions, I think you find something about this on Wikipedia.
– asdq
Jul 16 at 11:37
Great! Now my question reduces to why the inner product on wedges ought to be a determinant.
– user577413
Jul 16 at 13:57
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Usually the natural isomorphism is given via a perfect pairing $$Lambda^k V^ast times Lambda^k V rightarrow k,~ (phi_1wedge cdots wedge phi_k, v_1wedgecdots wedge v_k)mapsto det(phi_i(v_j))$$
which gives rise to the desired isomorphism. The naturality of this construction is easily verified.
answered Jul 16 at 9:10
asdq
1,4651417
Usually the natural isomorphism is given via a perfect pairing $$Lambda^k V^ast times Lambda^k V rightarrow k,~ (phi_1wedge cdots wedge phi_k, v_1wedgecdots wedge v_k)mapsto det(phi_i(v_j))$$
which gives rise to the desired isomorphism. The naturality of this construction is easily verified.
answered Jul 16 at 9:10
asdq
1,4651417
answered Jul 16 at 9:10
asdq
1,4651417
answered Jul 16 at 9:10
asdq
1,4651417
answered Jul 16 at 9:10
asdq
1,4651417
1,4651417
I see. Please excuse a further question: what’s the motivation behind this pairing? Correct me if I’m wrong, but this seems to be related to the inner product on k-vectors from an inner product on V.
– user577413
Jul 16 at 11:25
You are correct: Precompose the above map by $Lambda^k (cdot,cdot)times Lambda^k rmid$ where $(cdot,cdot)colon Vrightarrow V^ast$ is the isomorphism induced by the inner product, and you get the inner product of the wedge product.
– asdq
Jul 16 at 11:33
As for motivation, I think it is a good idea to get a geometric intuition about this. There are ways to visualize the wedge product in low dimensions, I think you find something about this on Wikipedia.
– asdq
Jul 16 at 11:37
Great! Now my question reduces to why the inner product on wedges ought to be a determinant.
– user577413
Jul 16 at 13:57
add a comment |Â
I see. Please excuse a further question: what’s the motivation behind this pairing? Correct me if I’m wrong, but this seems to be related to the inner product on k-vectors from an inner product on V.
– user577413
Jul 16 at 11:25
You are correct: Precompose the above map by $Lambda^k (cdot,cdot)times Lambda^k rmid$ where $(cdot,cdot)colon Vrightarrow V^ast$ is the isomorphism induced by the inner product, and you get the inner product of the wedge product.
– asdq
Jul 16 at 11:33
As for motivation, I think it is a good idea to get a geometric intuition about this. There are ways to visualize the wedge product in low dimensions, I think you find something about this on Wikipedia.
– asdq
Jul 16 at 11:37
Great! Now my question reduces to why the inner product on wedges ought to be a determinant.
– user577413
Jul 16 at 13:57
I see. Please excuse a further question: what’s the motivation behind this pairing? Correct me if I’m wrong, but this seems to be related to the inner product on k-vectors from an inner product on V.
– user577413
Jul 16 at 11:25
I see. Please excuse a further question: what’s the motivation behind this pairing? Correct me if I’m wrong, but this seems to be related to the inner product on k-vectors from an inner product on V.
– user577413
Jul 16 at 11:25
You are correct: Precompose the above map by $Lambda^k (cdot,cdot)times Lambda^k rmid$ where $(cdot,cdot)colon Vrightarrow V^ast$ is the isomorphism induced by the inner product, and you get the inner product of the wedge product.
– asdq
Jul 16 at 11:33
You are correct: Precompose the above map by $Lambda^k (cdot,cdot)times Lambda^k rmid$ where $(cdot,cdot)colon Vrightarrow V^ast$ is the isomorphism induced by the inner product, and you get the inner product of the wedge product.
– asdq
Jul 16 at 11:33
As for motivation, I think it is a good idea to get a geometric intuition about this. There are ways to visualize the wedge product in low dimensions, I think you find something about this on Wikipedia.
– asdq
Jul 16 at 11:37
As for motivation, I think it is a good idea to get a geometric intuition about this. There are ways to visualize the wedge product in low dimensions, I think you find something about this on Wikipedia.
– asdq
Jul 16 at 11:37
Great! Now my question reduces to why the inner product on wedges ought to be a determinant.
– user577413
Jul 16 at 13:57
Great! Now my question reduces to why the inner product on wedges ought to be a determinant.
– user577413
Jul 16 at 13:57
add a comment |Â
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 7:57
@JoséCarlosSantos Thanks very much!
– user577413
Jul 16 at 8:11