Properties of forms and pullbacks
$begingroup$
Let $j: mathbb{S}^2 rightarrow mathbb{R}^3setminus{0}$ be the canonical injection and $alpha$ a k-form over $mathbb{R}^3setminus{0}$.
- If $alpha$ is closed or exact, is it the same for $j^*alpha$?
- If yes, can this be generalized to any regular map: $f: M rightarrow N$ and any $k$-form over $N$?
differential-geometry smooth-manifolds exterior-derivative
$endgroup$
add a comment |
$begingroup$
Let $j: mathbb{S}^2 rightarrow mathbb{R}^3setminus{0}$ be the canonical injection and $alpha$ a k-form over $mathbb{R}^3setminus{0}$.
- If $alpha$ is closed or exact, is it the same for $j^*alpha$?
- If yes, can this be generalized to any regular map: $f: M rightarrow N$ and any $k$-form over $N$?
differential-geometry smooth-manifolds exterior-derivative
$endgroup$
add a comment |
$begingroup$
Let $j: mathbb{S}^2 rightarrow mathbb{R}^3setminus{0}$ be the canonical injection and $alpha$ a k-form over $mathbb{R}^3setminus{0}$.
- If $alpha$ is closed or exact, is it the same for $j^*alpha$?
- If yes, can this be generalized to any regular map: $f: M rightarrow N$ and any $k$-form over $N$?
differential-geometry smooth-manifolds exterior-derivative
$endgroup$
Let $j: mathbb{S}^2 rightarrow mathbb{R}^3setminus{0}$ be the canonical injection and $alpha$ a k-form over $mathbb{R}^3setminus{0}$.
- If $alpha$ is closed or exact, is it the same for $j^*alpha$?
- If yes, can this be generalized to any regular map: $f: M rightarrow N$ and any $k$-form over $N$?
differential-geometry smooth-manifolds exterior-derivative
differential-geometry smooth-manifolds exterior-derivative
edited Dec 25 '18 at 0:44
Travis
62.8k767149
62.8k767149
asked Dec 25 '18 at 0:26
PerelManPerelMan
661313
661313
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2 Answers
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$begingroup$
The key fact that you need is the naturality of the exterior derivative:
If $F : M to N$ is a smooth map between smooth manifolds, the pullback map $F^* : Gamma(bigwedge^k T^*N) to Gamma(bigwedge^k T^*M)$ commutes with the exterior derivative, that is, for any smooth $k$-form $alpha in Gamma(bigwedge^k T^*N)$ on $N$, $$boxed{F^* dalpha = d(F^* alpha)} .$$
For more, see, e.g., $S$ 12 of Lee's Introduction to Smooth Manifolds, where this fact is labeled Lemma 12.16.
$endgroup$
add a comment |
$begingroup$
As @Travis pointed out, you should use the fact that the pullback of differential forms commutes with the differential.
Moreover, if $f: M rightarrow N$ is a smooth map, then the induced map $f^{*}$ takes closed form to the closed forms and exact forms to the exact forms, which descends to the linear map between cohomology groups of $M$ and $N$
$$f^{*}: H^{n}(M) rightarrow H^{n}(N)$$
Indeed, let $omega$ be an exact form, then there exists a form $eta$ such that $w = d eta$, then $$f^{*} (omega) = f^{*} d eta = d (G^{*} eta)$$ whilst the latter implies that that the $f^{*}(omega)$ is also exact.
Now let $omega$ be a closed form, hence, since the derivative commutes with the pullback
$$d(f^{*} omega) = f^{*} (d omega) = f^{*} 0 = 0$$
So, if $[omega] = [omega']$, then $f^{*}[omega] = f^{*}[omega']$, as desired.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The key fact that you need is the naturality of the exterior derivative:
If $F : M to N$ is a smooth map between smooth manifolds, the pullback map $F^* : Gamma(bigwedge^k T^*N) to Gamma(bigwedge^k T^*M)$ commutes with the exterior derivative, that is, for any smooth $k$-form $alpha in Gamma(bigwedge^k T^*N)$ on $N$, $$boxed{F^* dalpha = d(F^* alpha)} .$$
For more, see, e.g., $S$ 12 of Lee's Introduction to Smooth Manifolds, where this fact is labeled Lemma 12.16.
$endgroup$
add a comment |
$begingroup$
The key fact that you need is the naturality of the exterior derivative:
If $F : M to N$ is a smooth map between smooth manifolds, the pullback map $F^* : Gamma(bigwedge^k T^*N) to Gamma(bigwedge^k T^*M)$ commutes with the exterior derivative, that is, for any smooth $k$-form $alpha in Gamma(bigwedge^k T^*N)$ on $N$, $$boxed{F^* dalpha = d(F^* alpha)} .$$
For more, see, e.g., $S$ 12 of Lee's Introduction to Smooth Manifolds, where this fact is labeled Lemma 12.16.
$endgroup$
add a comment |
$begingroup$
The key fact that you need is the naturality of the exterior derivative:
If $F : M to N$ is a smooth map between smooth manifolds, the pullback map $F^* : Gamma(bigwedge^k T^*N) to Gamma(bigwedge^k T^*M)$ commutes with the exterior derivative, that is, for any smooth $k$-form $alpha in Gamma(bigwedge^k T^*N)$ on $N$, $$boxed{F^* dalpha = d(F^* alpha)} .$$
For more, see, e.g., $S$ 12 of Lee's Introduction to Smooth Manifolds, where this fact is labeled Lemma 12.16.
$endgroup$
The key fact that you need is the naturality of the exterior derivative:
If $F : M to N$ is a smooth map between smooth manifolds, the pullback map $F^* : Gamma(bigwedge^k T^*N) to Gamma(bigwedge^k T^*M)$ commutes with the exterior derivative, that is, for any smooth $k$-form $alpha in Gamma(bigwedge^k T^*N)$ on $N$, $$boxed{F^* dalpha = d(F^* alpha)} .$$
For more, see, e.g., $S$ 12 of Lee's Introduction to Smooth Manifolds, where this fact is labeled Lemma 12.16.
edited Dec 25 '18 at 0:55
answered Dec 25 '18 at 0:35
TravisTravis
62.8k767149
62.8k767149
add a comment |
add a comment |
$begingroup$
As @Travis pointed out, you should use the fact that the pullback of differential forms commutes with the differential.
Moreover, if $f: M rightarrow N$ is a smooth map, then the induced map $f^{*}$ takes closed form to the closed forms and exact forms to the exact forms, which descends to the linear map between cohomology groups of $M$ and $N$
$$f^{*}: H^{n}(M) rightarrow H^{n}(N)$$
Indeed, let $omega$ be an exact form, then there exists a form $eta$ such that $w = d eta$, then $$f^{*} (omega) = f^{*} d eta = d (G^{*} eta)$$ whilst the latter implies that that the $f^{*}(omega)$ is also exact.
Now let $omega$ be a closed form, hence, since the derivative commutes with the pullback
$$d(f^{*} omega) = f^{*} (d omega) = f^{*} 0 = 0$$
So, if $[omega] = [omega']$, then $f^{*}[omega] = f^{*}[omega']$, as desired.
$endgroup$
add a comment |
$begingroup$
As @Travis pointed out, you should use the fact that the pullback of differential forms commutes with the differential.
Moreover, if $f: M rightarrow N$ is a smooth map, then the induced map $f^{*}$ takes closed form to the closed forms and exact forms to the exact forms, which descends to the linear map between cohomology groups of $M$ and $N$
$$f^{*}: H^{n}(M) rightarrow H^{n}(N)$$
Indeed, let $omega$ be an exact form, then there exists a form $eta$ such that $w = d eta$, then $$f^{*} (omega) = f^{*} d eta = d (G^{*} eta)$$ whilst the latter implies that that the $f^{*}(omega)$ is also exact.
Now let $omega$ be a closed form, hence, since the derivative commutes with the pullback
$$d(f^{*} omega) = f^{*} (d omega) = f^{*} 0 = 0$$
So, if $[omega] = [omega']$, then $f^{*}[omega] = f^{*}[omega']$, as desired.
$endgroup$
add a comment |
$begingroup$
As @Travis pointed out, you should use the fact that the pullback of differential forms commutes with the differential.
Moreover, if $f: M rightarrow N$ is a smooth map, then the induced map $f^{*}$ takes closed form to the closed forms and exact forms to the exact forms, which descends to the linear map between cohomology groups of $M$ and $N$
$$f^{*}: H^{n}(M) rightarrow H^{n}(N)$$
Indeed, let $omega$ be an exact form, then there exists a form $eta$ such that $w = d eta$, then $$f^{*} (omega) = f^{*} d eta = d (G^{*} eta)$$ whilst the latter implies that that the $f^{*}(omega)$ is also exact.
Now let $omega$ be a closed form, hence, since the derivative commutes with the pullback
$$d(f^{*} omega) = f^{*} (d omega) = f^{*} 0 = 0$$
So, if $[omega] = [omega']$, then $f^{*}[omega] = f^{*}[omega']$, as desired.
$endgroup$
As @Travis pointed out, you should use the fact that the pullback of differential forms commutes with the differential.
Moreover, if $f: M rightarrow N$ is a smooth map, then the induced map $f^{*}$ takes closed form to the closed forms and exact forms to the exact forms, which descends to the linear map between cohomology groups of $M$ and $N$
$$f^{*}: H^{n}(M) rightarrow H^{n}(N)$$
Indeed, let $omega$ be an exact form, then there exists a form $eta$ such that $w = d eta$, then $$f^{*} (omega) = f^{*} d eta = d (G^{*} eta)$$ whilst the latter implies that that the $f^{*}(omega)$ is also exact.
Now let $omega$ be a closed form, hence, since the derivative commutes with the pullback
$$d(f^{*} omega) = f^{*} (d omega) = f^{*} 0 = 0$$
So, if $[omega] = [omega']$, then $f^{*}[omega] = f^{*}[omega']$, as desired.
edited Dec 25 '18 at 1:48
answered Dec 25 '18 at 0:45
hyperkahlerhyperkahler
1,487714
1,487714
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add a comment |
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