Pull back of wedge of differential forms
up vote
1
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Suppose that $phi : M rightarrow N$ is a smooth map between differential manifolds. Let $omega, eta$ be forms on $N$. Is there an easy proof for the fact that
$$phi^*(omega wedge eta) = phi^* omega wedge phi^* eta?$$
In other words, $phi^* : Omega(N) rightarrow Omega(M)$ is a graded algebra homomorphism. (By easy, I mean does not require expanding both sides using the definition.)
differential-geometry differential-forms
edited Mar 19 '17 at 0:48
Paul
15.9k33666
asked Mar 18 '17 at 23:33
An Hoa
1,050511
add a comment |
up vote
1
down vote
favorite
Suppose that $phi : M rightarrow N$ is a smooth map between differential manifolds. Let $omega, eta$ be forms on $N$. Is there an easy proof for the fact that
$$phi^*(omega wedge eta) = phi^* omega wedge phi^* eta?$$
In other words, $phi^* : Omega(N) rightarrow Omega(M)$ is a graded algebra homomorphism. (By easy, I mean does not require expanding both sides using the definition.)
differential-geometry differential-forms
edited Mar 19 '17 at 0:48
Paul
15.9k33666
asked Mar 18 '17 at 23:33
An Hoa
1,050511
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that $phi : M rightarrow N$ is a smooth map between differential manifolds. Let $omega, eta$ be forms on $N$. Is there an easy proof for the fact that
$$phi^*(omega wedge eta) = phi^* omega wedge phi^* eta?$$
In other words, $phi^* : Omega(N) rightarrow Omega(M)$ is a graded algebra homomorphism. (By easy, I mean does not require expanding both sides using the definition.)
differential-geometry differential-forms
edited Mar 19 '17 at 0:48
Paul
15.9k33666
asked Mar 18 '17 at 23:33
An Hoa
1,050511
Suppose that $phi : M rightarrow N$ is a smooth map between differential manifolds. Let $omega, eta$ be forms on $N$. Is there an easy proof for the fact that
$$phi^*(omega wedge eta) = phi^* omega wedge phi^* eta?$$
In other words, $phi^* : Omega(N) rightarrow Omega(M)$ is a graded algebra homomorphism. (By easy, I mean does not require expanding both sides using the definition.)
differential-geometry differential-forms
differential-geometry differential-forms
edited Mar 19 '17 at 0:48
Paul
15.9k33666
asked Mar 18 '17 at 23:33
An Hoa
1,050511
edited Mar 19 '17 at 0:48
Paul
15.9k33666
asked Mar 18 '17 at 23:33
An Hoa
1,050511
edited Mar 19 '17 at 0:48
Paul
15.9k33666
edited Mar 19 '17 at 0:48
Paul
15.9k33666
edited Mar 19 '17 at 0:48
Paul
15.9k33666
15.9k33666
asked Mar 18 '17 at 23:33
An Hoa
1,050511
asked Mar 18 '17 at 23:33
An Hoa
1,050511
asked Mar 18 '17 at 23:33
An Hoa
1,050511
1,050511
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
I've read your question wrong, here is a more general answer you want, I guess.
We will use this following proprieties:
$$phi^*(omega_1+omega_2)= phi^*(omega_1)+phi^*(omega_2)$$
$$omega wedge eta = frac {(m+l)!}{m!l!}Alt(omega otimes eta) $$
$$Alt(omega otimes eta)=frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma $$
By that, $$phi^*(omega wedge eta)= frac {(m+l)!}{m!l!}phi^*(Alt(omega otimes eta)) = frac {(m+l)!}{m!l!}phi^*(frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma) =\ =frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma $$
Now if we prove that $phi^*(omega otimes eta)circsigma = (phi^*omega otimes phi^*eta)circsigma $, we have that
$$frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma = frac {1}{m!l!} sum_{sigma in S_ {m+l }} (phi^*omega otimesphi^*eta)circsigma = phi^*(omega)wedge phi^*(eta)$$
Proving that $phi^*(omega otimes eta)=(phi^*omega otimes phi^*eta)$ is easy.
edited Nov 22 at 15:30
Community♦
1
answered Mar 19 '17 at 5:14
Danuso Rocha
1367
add a comment |
up vote
2
down vote
Let $p$ be a point $M$, $v,w in T_pM$ and $omega_1, omega_2, omega$ and $eta $ be 1-forms of $M$ . We have to use that:
$$(omega_1 wedge omega_2)_p(v,w) =
begin{vmatrix}
{omega_1}_p(v) & {omega_1}_p(w) \
{omega_2}_p(v) & {omega_2}_p(w)
end{vmatrix}$$
$phi^*(omega wedgeeta)_p(v,w)
= (omega wedge eta)_{phi (p)}(d phi_p(v),d phi_p(w)) = begin{vmatrix}
{(omega)}_{phi(p)}(d phi_p(v)) & {(omega)}_{phi(p)}(d phi_p(w)) \
{(eta)}_{phi(p)}(d phi_p(v) & {(eta)}_{phi(p)}(d phi_p(w))
end{vmatrix} = begin{vmatrix}
{(phi^*omega)}_p(v) & {(phi^*omega)}_p(w) \
{(phi^*eta)}_p(v) & {(phi^*eta)}_p(w)
end{vmatrix} = (phi^*omega) wedge (phi^*eta)_p(v,w) $.
So we have $phi^*( omega wedge eta) = phi^*(omega) wedge phi^*(eta).$
answered Mar 19 '17 at 2:52
Danuso Rocha
1367
1
I see what you are trying to do, I like you aprroach. But what if $omega=sum f_{i_1,ldots, i_k} dx_{i_1}wedgecdotswedge dx_{i_k}in Gamma^infty Big(Lambda^kbig( T^* M big)Big) $ and $eta=sum g_{j_1,ldots, j_ell} dx_{j_1}wedgecdotswedge dx_{j_ell}in Gamma^infty Big(Lambda^ellbig( T^* M big)Big)$ are arbitrary forms ( of degree $k$ and $ell$ respectively) ?
– EternalBlood
Mar 19 '17 at 3:05
Maybe using this first case, linearity and induction.
– Danuso Rocha
Mar 19 '17 at 3:18
Thanks for alert me, @EternalBlood. I gave the general answer. :)
– Danuso Rocha
Mar 19 '17 at 5:16
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
oldest
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active
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up vote
0
down vote
accepted
I've read your question wrong, here is a more general answer you want, I guess.
We will use this following proprieties:
$$phi^*(omega_1+omega_2)= phi^*(omega_1)+phi^*(omega_2)$$
$$omega wedge eta = frac {(m+l)!}{m!l!}Alt(omega otimes eta) $$
$$Alt(omega otimes eta)=frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma $$
By that, $$phi^*(omega wedge eta)= frac {(m+l)!}{m!l!}phi^*(Alt(omega otimes eta)) = frac {(m+l)!}{m!l!}phi^*(frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma) =\ =frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma $$
Now if we prove that $phi^*(omega otimes eta)circsigma = (phi^*omega otimes phi^*eta)circsigma $, we have that
$$frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma = frac {1}{m!l!} sum_{sigma in S_ {m+l }} (phi^*omega otimesphi^*eta)circsigma = phi^*(omega)wedge phi^*(eta)$$
Proving that $phi^*(omega otimes eta)=(phi^*omega otimes phi^*eta)$ is easy.
edited Nov 22 at 15:30
Community♦
1
answered Mar 19 '17 at 5:14
Danuso Rocha
1367
add a comment |
up vote
0
down vote
accepted
I've read your question wrong, here is a more general answer you want, I guess.
We will use this following proprieties:
$$phi^*(omega_1+omega_2)= phi^*(omega_1)+phi^*(omega_2)$$
$$omega wedge eta = frac {(m+l)!}{m!l!}Alt(omega otimes eta) $$
$$Alt(omega otimes eta)=frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma $$
By that, $$phi^*(omega wedge eta)= frac {(m+l)!}{m!l!}phi^*(Alt(omega otimes eta)) = frac {(m+l)!}{m!l!}phi^*(frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma) =\ =frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma $$
Now if we prove that $phi^*(omega otimes eta)circsigma = (phi^*omega otimes phi^*eta)circsigma $, we have that
$$frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma = frac {1}{m!l!} sum_{sigma in S_ {m+l }} (phi^*omega otimesphi^*eta)circsigma = phi^*(omega)wedge phi^*(eta)$$
Proving that $phi^*(omega otimes eta)=(phi^*omega otimes phi^*eta)$ is easy.
edited Nov 22 at 15:30
Community♦
1
answered Mar 19 '17 at 5:14
Danuso Rocha
1367
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I've read your question wrong, here is a more general answer you want, I guess.
We will use this following proprieties:
$$phi^*(omega_1+omega_2)= phi^*(omega_1)+phi^*(omega_2)$$
$$omega wedge eta = frac {(m+l)!}{m!l!}Alt(omega otimes eta) $$
$$Alt(omega otimes eta)=frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma $$
By that, $$phi^*(omega wedge eta)= frac {(m+l)!}{m!l!}phi^*(Alt(omega otimes eta)) = frac {(m+l)!}{m!l!}phi^*(frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma) =\ =frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma $$
Now if we prove that $phi^*(omega otimes eta)circsigma = (phi^*omega otimes phi^*eta)circsigma $, we have that
$$frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma = frac {1}{m!l!} sum_{sigma in S_ {m+l }} (phi^*omega otimesphi^*eta)circsigma = phi^*(omega)wedge phi^*(eta)$$
Proving that $phi^*(omega otimes eta)=(phi^*omega otimes phi^*eta)$ is easy.
edited Nov 22 at 15:30
Community♦
1
answered Mar 19 '17 at 5:14
Danuso Rocha
1367
I've read your question wrong, here is a more general answer you want, I guess.
We will use this following proprieties:
$$phi^*(omega_1+omega_2)= phi^*(omega_1)+phi^*(omega_2)$$
$$omega wedge eta = frac {(m+l)!}{m!l!}Alt(omega otimes eta) $$
$$Alt(omega otimes eta)=frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma $$
By that, $$phi^*(omega wedge eta)= frac {(m+l)!}{m!l!}phi^*(Alt(omega otimes eta)) = frac {(m+l)!}{m!l!}phi^*(frac {1}{(m+l)!} sum_{sigma in S_ {m+l }} (omega otimes eta)circsigma) =\ =frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma $$
Now if we prove that $phi^*(omega otimes eta)circsigma = (phi^*omega otimes phi^*eta)circsigma $, we have that
$$frac {1}{m!l!} sum_{sigma in S_ {m+l }}phi^* (omega otimes eta)circsigma = frac {1}{m!l!} sum_{sigma in S_ {m+l }} (phi^*omega otimesphi^*eta)circsigma = phi^*(omega)wedge phi^*(eta)$$
Proving that $phi^*(omega otimes eta)=(phi^*omega otimes phi^*eta)$ is easy.
edited Nov 22 at 15:30
Community♦
1
answered Mar 19 '17 at 5:14
Danuso Rocha
1367
edited Nov 22 at 15:30
Community♦
1
edited Nov 22 at 15:30
Community♦
1
edited Nov 22 at 15:30
Community♦
1
1
answered Mar 19 '17 at 5:14
Danuso Rocha
1367
answered Mar 19 '17 at 5:14
Danuso Rocha
1367
answered Mar 19 '17 at 5:14
Danuso Rocha
1367
1367
add a comment |
add a comment |
up vote
2
down vote
Let $p$ be a point $M$, $v,w in T_pM$ and $omega_1, omega_2, omega$ and $eta $ be 1-forms of $M$ . We have to use that:
$$(omega_1 wedge omega_2)_p(v,w) =
begin{vmatrix}
{omega_1}_p(v) & {omega_1}_p(w) \
{omega_2}_p(v) & {omega_2}_p(w)
end{vmatrix}$$
$phi^*(omega wedgeeta)_p(v,w)
= (omega wedge eta)_{phi (p)}(d phi_p(v),d phi_p(w)) = begin{vmatrix}
{(omega)}_{phi(p)}(d phi_p(v)) & {(omega)}_{phi(p)}(d phi_p(w)) \
{(eta)}_{phi(p)}(d phi_p(v) & {(eta)}_{phi(p)}(d phi_p(w))
end{vmatrix} = begin{vmatrix}
{(phi^*omega)}_p(v) & {(phi^*omega)}_p(w) \
{(phi^*eta)}_p(v) & {(phi^*eta)}_p(w)
end{vmatrix} = (phi^*omega) wedge (phi^*eta)_p(v,w) $.
So we have $phi^*( omega wedge eta) = phi^*(omega) wedge phi^*(eta).$
answered Mar 19 '17 at 2:52
Danuso Rocha
1367
1
I see what you are trying to do, I like you aprroach. But what if $omega=sum f_{i_1,ldots, i_k} dx_{i_1}wedgecdotswedge dx_{i_k}in Gamma^infty Big(Lambda^kbig( T^* M big)Big) $ and $eta=sum g_{j_1,ldots, j_ell} dx_{j_1}wedgecdotswedge dx_{j_ell}in Gamma^infty Big(Lambda^ellbig( T^* M big)Big)$ are arbitrary forms ( of degree $k$ and $ell$ respectively) ?
– EternalBlood
Mar 19 '17 at 3:05
Maybe using this first case, linearity and induction.
– Danuso Rocha
Mar 19 '17 at 3:18
Thanks for alert me, @EternalBlood. I gave the general answer. :)
– Danuso Rocha
Mar 19 '17 at 5:16
add a comment |
up vote
2
down vote
Let $p$ be a point $M$, $v,w in T_pM$ and $omega_1, omega_2, omega$ and $eta $ be 1-forms of $M$ . We have to use that:
$$(omega_1 wedge omega_2)_p(v,w) =
begin{vmatrix}
{omega_1}_p(v) & {omega_1}_p(w) \
{omega_2}_p(v) & {omega_2}_p(w)
end{vmatrix}$$
$phi^*(omega wedgeeta)_p(v,w)
= (omega wedge eta)_{phi (p)}(d phi_p(v),d phi_p(w)) = begin{vmatrix}
{(omega)}_{phi(p)}(d phi_p(v)) & {(omega)}_{phi(p)}(d phi_p(w)) \
{(eta)}_{phi(p)}(d phi_p(v) & {(eta)}_{phi(p)}(d phi_p(w))
end{vmatrix} = begin{vmatrix}
{(phi^*omega)}_p(v) & {(phi^*omega)}_p(w) \
{(phi^*eta)}_p(v) & {(phi^*eta)}_p(w)
end{vmatrix} = (phi^*omega) wedge (phi^*eta)_p(v,w) $.
So we have $phi^*( omega wedge eta) = phi^*(omega) wedge phi^*(eta).$
answered Mar 19 '17 at 2:52
Danuso Rocha
1367
1
I see what you are trying to do, I like you aprroach. But what if $omega=sum f_{i_1,ldots, i_k} dx_{i_1}wedgecdotswedge dx_{i_k}in Gamma^infty Big(Lambda^kbig( T^* M big)Big) $ and $eta=sum g_{j_1,ldots, j_ell} dx_{j_1}wedgecdotswedge dx_{j_ell}in Gamma^infty Big(Lambda^ellbig( T^* M big)Big)$ are arbitrary forms ( of degree $k$ and $ell$ respectively) ?
– EternalBlood
Mar 19 '17 at 3:05
Maybe using this first case, linearity and induction.
– Danuso Rocha
Mar 19 '17 at 3:18
Thanks for alert me, @EternalBlood. I gave the general answer. :)
– Danuso Rocha
Mar 19 '17 at 5:16
add a comment |
up vote
2
down vote
up vote
2
down vote
Let $p$ be a point $M$, $v,w in T_pM$ and $omega_1, omega_2, omega$ and $eta $ be 1-forms of $M$ . We have to use that:
$$(omega_1 wedge omega_2)_p(v,w) =
begin{vmatrix}
{omega_1}_p(v) & {omega_1}_p(w) \
{omega_2}_p(v) & {omega_2}_p(w)
end{vmatrix}$$
$phi^*(omega wedgeeta)_p(v,w)
= (omega wedge eta)_{phi (p)}(d phi_p(v),d phi_p(w)) = begin{vmatrix}
{(omega)}_{phi(p)}(d phi_p(v)) & {(omega)}_{phi(p)}(d phi_p(w)) \
{(eta)}_{phi(p)}(d phi_p(v) & {(eta)}_{phi(p)}(d phi_p(w))
end{vmatrix} = begin{vmatrix}
{(phi^*omega)}_p(v) & {(phi^*omega)}_p(w) \
{(phi^*eta)}_p(v) & {(phi^*eta)}_p(w)
end{vmatrix} = (phi^*omega) wedge (phi^*eta)_p(v,w) $.
So we have $phi^*( omega wedge eta) = phi^*(omega) wedge phi^*(eta).$
answered Mar 19 '17 at 2:52
Danuso Rocha
1367
Let $p$ be a point $M$, $v,w in T_pM$ and $omega_1, omega_2, omega$ and $eta $ be 1-forms of $M$ . We have to use that:
$$(omega_1 wedge omega_2)_p(v,w) =
begin{vmatrix}
{omega_1}_p(v) & {omega_1}_p(w) \
{omega_2}_p(v) & {omega_2}_p(w)
end{vmatrix}$$
$phi^*(omega wedgeeta)_p(v,w)
= (omega wedge eta)_{phi (p)}(d phi_p(v),d phi_p(w)) = begin{vmatrix}
{(omega)}_{phi(p)}(d phi_p(v)) & {(omega)}_{phi(p)}(d phi_p(w)) \
{(eta)}_{phi(p)}(d phi_p(v) & {(eta)}_{phi(p)}(d phi_p(w))
end{vmatrix} = begin{vmatrix}
{(phi^*omega)}_p(v) & {(phi^*omega)}_p(w) \
{(phi^*eta)}_p(v) & {(phi^*eta)}_p(w)
end{vmatrix} = (phi^*omega) wedge (phi^*eta)_p(v,w) $.
So we have $phi^*( omega wedge eta) = phi^*(omega) wedge phi^*(eta).$
answered Mar 19 '17 at 2:52
Danuso Rocha
1367
edited Mar 19 '17 at 3:04
answered Mar 19 '17 at 2:52
Danuso Rocha
1367
answered Mar 19 '17 at 2:52
Danuso Rocha
1367
answered Mar 19 '17 at 2:52
Danuso Rocha
1367
1367
1
I see what you are trying to do, I like you aprroach. But what if $omega=sum f_{i_1,ldots, i_k} dx_{i_1}wedgecdotswedge dx_{i_k}in Gamma^infty Big(Lambda^kbig( T^* M big)Big) $ and $eta=sum g_{j_1,ldots, j_ell} dx_{j_1}wedgecdotswedge dx_{j_ell}in Gamma^infty Big(Lambda^ellbig( T^* M big)Big)$ are arbitrary forms ( of degree $k$ and $ell$ respectively) ?
– EternalBlood
Mar 19 '17 at 3:05
Maybe using this first case, linearity and induction.
– Danuso Rocha
Mar 19 '17 at 3:18
Thanks for alert me, @EternalBlood. I gave the general answer. :)
– Danuso Rocha
Mar 19 '17 at 5:16
add a comment |
1
I see what you are trying to do, I like you aprroach. But what if $omega=sum f_{i_1,ldots, i_k} dx_{i_1}wedgecdotswedge dx_{i_k}in Gamma^infty Big(Lambda^kbig( T^* M big)Big) $ and $eta=sum g_{j_1,ldots, j_ell} dx_{j_1}wedgecdotswedge dx_{j_ell}in Gamma^infty Big(Lambda^ellbig( T^* M big)Big)$ are arbitrary forms ( of degree $k$ and $ell$ respectively) ?
– EternalBlood
Mar 19 '17 at 3:05
Maybe using this first case, linearity and induction.
– Danuso Rocha
Mar 19 '17 at 3:18
Thanks for alert me, @EternalBlood. I gave the general answer. :)
– Danuso Rocha
Mar 19 '17 at 5:16
1
1
I see what you are trying to do, I like you aprroach. But what if $omega=sum f_{i_1,ldots, i_k} dx_{i_1}wedgecdotswedge dx_{i_k}in Gamma^infty Big(Lambda^kbig( T^* M big)Big) $ and $eta=sum g_{j_1,ldots, j_ell} dx_{j_1}wedgecdotswedge dx_{j_ell}in Gamma^infty Big(Lambda^ellbig( T^* M big)Big)$ are arbitrary forms ( of degree $k$ and $ell$ respectively) ?
– EternalBlood
Mar 19 '17 at 3:05
I see what you are trying to do, I like you aprroach. But what if $omega=sum f_{i_1,ldots, i_k} dx_{i_1}wedgecdotswedge dx_{i_k}in Gamma^infty Big(Lambda^kbig( T^* M big)Big) $ and $eta=sum g_{j_1,ldots, j_ell} dx_{j_1}wedgecdotswedge dx_{j_ell}in Gamma^infty Big(Lambda^ellbig( T^* M big)Big)$ are arbitrary forms ( of degree $k$ and $ell$ respectively) ?
– EternalBlood
Mar 19 '17 at 3:05
Maybe using this first case, linearity and induction.
– Danuso Rocha
Mar 19 '17 at 3:18
Maybe using this first case, linearity and induction.
– Danuso Rocha
Mar 19 '17 at 3:18
Thanks for alert me, @EternalBlood. I gave the general answer. :)
– Danuso Rocha
Mar 19 '17 at 5:16
Thanks for alert me, @EternalBlood. I gave the general answer. :)
– Danuso Rocha
Mar 19 '17 at 5:16
add a comment |
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