Pullback of a differential form by a local diffeomorphism
Suppose I have to smooth oriented manifolds, $M$ and $N$ and a local diffeomorphism $f : M rightarrow N$. Let $omega$ be a differential form of maximum degree on $N$, let's say, $r$. How can I rewrite
$$int_N omega$$
in terms of the integral of the pullback of $omega$ by $f$, $f^*omega$?
So, I know that when $f$ is a diffeomorphism, then
$$int_Nomega = pm int_M f^*omega$$
depending on whether $f$ preserves the orientation or not. But that's in part due to the fact that $f$ is bijective, but that condition is removed when assuming $f$ is a local diffeomorphism. So is there a nice way to write that integral in terms of the pullback?
integration differential-geometry smooth-manifolds pullback
asked Jan 25 '17 at 20:02
user143144
1176
add a comment |
Suppose I have to smooth oriented manifolds, $M$ and $N$ and a local diffeomorphism $f : M rightarrow N$. Let $omega$ be a differential form of maximum degree on $N$, let's say, $r$. How can I rewrite
$$int_N omega$$
in terms of the integral of the pullback of $omega$ by $f$, $f^*omega$?
So, I know that when $f$ is a diffeomorphism, then
$$int_Nomega = pm int_M f^*omega$$
depending on whether $f$ preserves the orientation or not. But that's in part due to the fact that $f$ is bijective, but that condition is removed when assuming $f$ is a local diffeomorphism. So is there a nice way to write that integral in terms of the pullback?
integration differential-geometry smooth-manifolds pullback
asked Jan 25 '17 at 20:02
user143144
1176
add a comment |
Suppose I have to smooth oriented manifolds, $M$ and $N$ and a local diffeomorphism $f : M rightarrow N$. Let $omega$ be a differential form of maximum degree on $N$, let's say, $r$. How can I rewrite
$$int_N omega$$
in terms of the integral of the pullback of $omega$ by $f$, $f^*omega$?
So, I know that when $f$ is a diffeomorphism, then
$$int_Nomega = pm int_M f^*omega$$
depending on whether $f$ preserves the orientation or not. But that's in part due to the fact that $f$ is bijective, but that condition is removed when assuming $f$ is a local diffeomorphism. So is there a nice way to write that integral in terms of the pullback?
integration differential-geometry smooth-manifolds pullback
asked Jan 25 '17 at 20:02
user143144
1176
Suppose I have to smooth oriented manifolds, $M$ and $N$ and a local diffeomorphism $f : M rightarrow N$. Let $omega$ be a differential form of maximum degree on $N$, let's say, $r$. How can I rewrite
$$int_N omega$$
in terms of the integral of the pullback of $omega$ by $f$, $f^*omega$?
So, I know that when $f$ is a diffeomorphism, then
$$int_Nomega = pm int_M f^*omega$$
depending on whether $f$ preserves the orientation or not. But that's in part due to the fact that $f$ is bijective, but that condition is removed when assuming $f$ is a local diffeomorphism. So is there a nice way to write that integral in terms of the pullback?
integration differential-geometry smooth-manifolds pullback
integration differential-geometry smooth-manifolds pullback
asked Jan 25 '17 at 20:02
user143144
1176
asked Jan 25 '17 at 20:02
user143144
1176
asked Jan 25 '17 at 20:02
user143144
1176
asked Jan 25 '17 at 20:02
user143144
1176
asked Jan 25 '17 at 20:02
user143144
1176
1176
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Yeah, and you can make it work even if $f$ is not a local diffeomorphism, only a proper map. The relevant notion is that of the degree of a smooth proper map between oriented manifolds of the same dimension.
Assume for simplicity that $M,N$ are closed (compact, without boundary), non-empty and connected and let $f colon M rightarrow N$ be an arbitrary smooth map. Since $M,N$ are oriented, we can a generator $[omega_1] in H^{text{top}}(M)$ such that $omega_1$ is consistent with the orientation on $M$ and $int_M omega_1 = 1$. Choose $[omega_2]$ similarly for $N$. The map $f$ induces a map $f^{*} colon H^{text{top}}(N) rightarrow H^{text{top}}(M)$ on cohomology and since the top cohomology groups are one-dimensional, we must have $f^{*}([omega_2]) = c [omega_1]$ for some $c in mathbb{R}$. This $c = deg(f)$ is called the degree of $f$ and is a priori a real number. However, it can be shown that $c$ is in fact an integer which can be computed by counting the number of preimages of a regular value $p in N$ with appropriate signs which take the orientations of $M,N$ into consideration.
Knowing that, given $omega in Omega^{text{top}}(N)$, write $[omega] = c[omega_2]$ for some $c in mathbb{R}$. Then
$$ [f^{*}(omega)] = f^{*}([omega]) = f^{*}(c[omega_2]) = c deg(f) [omega_1] $$
so
$$ int_M f^{*}(omega) = deg(f) c int_N omega_1 = deg(f) c = deg(f) cint_N omega_2 = deg(f) int_N omega. $$
In particular, if $f$ is a diffeomorphism, $deg(f) = pm 1$ so this generalizes your starting point. If $f$ is a local diffeomorphism then $f colon M rightarrow N$ is a covering map and $deg(f)$ will be the number of points in an arbirary fiber, counted with appropriate signs. For much more details and proofs, see the book "Differential Forms in Algebraic Topology".
answered Jan 25 '17 at 21:00
levap
47k23273
3
For your particular case, you can avoid mentioning cohomology by showing first (assuming $M,N$ are compact) that $f$ is a covering map. Then, over a basic neighborhood $U subseteq N$, we have $f^{-1}(U) = sqcup {U_i}$ where $f|_{U_i} colon U_i rightarrow U$ is a diffeomorphism which reverses or preserves orientation. Hence $int_{f^{-1}(U)} f^{*}(omega) = sum pm int_{N} omega$. Then you can get the global result by a partition of unity after showing that "the signs stay constant" on the intersections so there isn't any unwanted cancellations.
– levap
Jan 25 '17 at 21:19
This is super cool! Just one thing, when using the fact that local diffeomorphism is a covering map, don't I need to assume at least that $f$ is surjective or $N$ is connected? And that would also assure that the signs stay constant on the intersections, right?
– user143144
Jan 26 '17 at 9:04
Yeah, you are definitely right. I somehow tend to think of my manifolds as always connected, but you're right. I've edited my answer.
– levap
Jan 26 '17 at 18:34
add a comment |
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Yeah, and you can make it work even if $f$ is not a local diffeomorphism, only a proper map. The relevant notion is that of the degree of a smooth proper map between oriented manifolds of the same dimension.
Assume for simplicity that $M,N$ are closed (compact, without boundary), non-empty and connected and let $f colon M rightarrow N$ be an arbitrary smooth map. Since $M,N$ are oriented, we can a generator $[omega_1] in H^{text{top}}(M)$ such that $omega_1$ is consistent with the orientation on $M$ and $int_M omega_1 = 1$. Choose $[omega_2]$ similarly for $N$. The map $f$ induces a map $f^{*} colon H^{text{top}}(N) rightarrow H^{text{top}}(M)$ on cohomology and since the top cohomology groups are one-dimensional, we must have $f^{*}([omega_2]) = c [omega_1]$ for some $c in mathbb{R}$. This $c = deg(f)$ is called the degree of $f$ and is a priori a real number. However, it can be shown that $c$ is in fact an integer which can be computed by counting the number of preimages of a regular value $p in N$ with appropriate signs which take the orientations of $M,N$ into consideration.
Knowing that, given $omega in Omega^{text{top}}(N)$, write $[omega] = c[omega_2]$ for some $c in mathbb{R}$. Then
$$ [f^{*}(omega)] = f^{*}([omega]) = f^{*}(c[omega_2]) = c deg(f) [omega_1] $$
so
$$ int_M f^{*}(omega) = deg(f) c int_N omega_1 = deg(f) c = deg(f) cint_N omega_2 = deg(f) int_N omega. $$
In particular, if $f$ is a diffeomorphism, $deg(f) = pm 1$ so this generalizes your starting point. If $f$ is a local diffeomorphism then $f colon M rightarrow N$ is a covering map and $deg(f)$ will be the number of points in an arbirary fiber, counted with appropriate signs. For much more details and proofs, see the book "Differential Forms in Algebraic Topology".
answered Jan 25 '17 at 21:00
levap
47k23273
3
For your particular case, you can avoid mentioning cohomology by showing first (assuming $M,N$ are compact) that $f$ is a covering map. Then, over a basic neighborhood $U subseteq N$, we have $f^{-1}(U) = sqcup {U_i}$ where $f|_{U_i} colon U_i rightarrow U$ is a diffeomorphism which reverses or preserves orientation. Hence $int_{f^{-1}(U)} f^{*}(omega) = sum pm int_{N} omega$. Then you can get the global result by a partition of unity after showing that "the signs stay constant" on the intersections so there isn't any unwanted cancellations.
– levap
Jan 25 '17 at 21:19
This is super cool! Just one thing, when using the fact that local diffeomorphism is a covering map, don't I need to assume at least that $f$ is surjective or $N$ is connected? And that would also assure that the signs stay constant on the intersections, right?
– user143144
Jan 26 '17 at 9:04
Yeah, you are definitely right. I somehow tend to think of my manifolds as always connected, but you're right. I've edited my answer.
– levap
Jan 26 '17 at 18:34
add a comment |
Yeah, and you can make it work even if $f$ is not a local diffeomorphism, only a proper map. The relevant notion is that of the degree of a smooth proper map between oriented manifolds of the same dimension.
Assume for simplicity that $M,N$ are closed (compact, without boundary), non-empty and connected and let $f colon M rightarrow N$ be an arbitrary smooth map. Since $M,N$ are oriented, we can a generator $[omega_1] in H^{text{top}}(M)$ such that $omega_1$ is consistent with the orientation on $M$ and $int_M omega_1 = 1$. Choose $[omega_2]$ similarly for $N$. The map $f$ induces a map $f^{*} colon H^{text{top}}(N) rightarrow H^{text{top}}(M)$ on cohomology and since the top cohomology groups are one-dimensional, we must have $f^{*}([omega_2]) = c [omega_1]$ for some $c in mathbb{R}$. This $c = deg(f)$ is called the degree of $f$ and is a priori a real number. However, it can be shown that $c$ is in fact an integer which can be computed by counting the number of preimages of a regular value $p in N$ with appropriate signs which take the orientations of $M,N$ into consideration.
Knowing that, given $omega in Omega^{text{top}}(N)$, write $[omega] = c[omega_2]$ for some $c in mathbb{R}$. Then
$$ [f^{*}(omega)] = f^{*}([omega]) = f^{*}(c[omega_2]) = c deg(f) [omega_1] $$
so
$$ int_M f^{*}(omega) = deg(f) c int_N omega_1 = deg(f) c = deg(f) cint_N omega_2 = deg(f) int_N omega. $$
In particular, if $f$ is a diffeomorphism, $deg(f) = pm 1$ so this generalizes your starting point. If $f$ is a local diffeomorphism then $f colon M rightarrow N$ is a covering map and $deg(f)$ will be the number of points in an arbirary fiber, counted with appropriate signs. For much more details and proofs, see the book "Differential Forms in Algebraic Topology".
answered Jan 25 '17 at 21:00
levap
47k23273
3
For your particular case, you can avoid mentioning cohomology by showing first (assuming $M,N$ are compact) that $f$ is a covering map. Then, over a basic neighborhood $U subseteq N$, we have $f^{-1}(U) = sqcup {U_i}$ where $f|_{U_i} colon U_i rightarrow U$ is a diffeomorphism which reverses or preserves orientation. Hence $int_{f^{-1}(U)} f^{*}(omega) = sum pm int_{N} omega$. Then you can get the global result by a partition of unity after showing that "the signs stay constant" on the intersections so there isn't any unwanted cancellations.
– levap
Jan 25 '17 at 21:19
This is super cool! Just one thing, when using the fact that local diffeomorphism is a covering map, don't I need to assume at least that $f$ is surjective or $N$ is connected? And that would also assure that the signs stay constant on the intersections, right?
– user143144
Jan 26 '17 at 9:04
Yeah, you are definitely right. I somehow tend to think of my manifolds as always connected, but you're right. I've edited my answer.
– levap
Jan 26 '17 at 18:34
add a comment |
Yeah, and you can make it work even if $f$ is not a local diffeomorphism, only a proper map. The relevant notion is that of the degree of a smooth proper map between oriented manifolds of the same dimension.
Assume for simplicity that $M,N$ are closed (compact, without boundary), non-empty and connected and let $f colon M rightarrow N$ be an arbitrary smooth map. Since $M,N$ are oriented, we can a generator $[omega_1] in H^{text{top}}(M)$ such that $omega_1$ is consistent with the orientation on $M$ and $int_M omega_1 = 1$. Choose $[omega_2]$ similarly for $N$. The map $f$ induces a map $f^{*} colon H^{text{top}}(N) rightarrow H^{text{top}}(M)$ on cohomology and since the top cohomology groups are one-dimensional, we must have $f^{*}([omega_2]) = c [omega_1]$ for some $c in mathbb{R}$. This $c = deg(f)$ is called the degree of $f$ and is a priori a real number. However, it can be shown that $c$ is in fact an integer which can be computed by counting the number of preimages of a regular value $p in N$ with appropriate signs which take the orientations of $M,N$ into consideration.
Knowing that, given $omega in Omega^{text{top}}(N)$, write $[omega] = c[omega_2]$ for some $c in mathbb{R}$. Then
$$ [f^{*}(omega)] = f^{*}([omega]) = f^{*}(c[omega_2]) = c deg(f) [omega_1] $$
so
$$ int_M f^{*}(omega) = deg(f) c int_N omega_1 = deg(f) c = deg(f) cint_N omega_2 = deg(f) int_N omega. $$
In particular, if $f$ is a diffeomorphism, $deg(f) = pm 1$ so this generalizes your starting point. If $f$ is a local diffeomorphism then $f colon M rightarrow N$ is a covering map and $deg(f)$ will be the number of points in an arbirary fiber, counted with appropriate signs. For much more details and proofs, see the book "Differential Forms in Algebraic Topology".
answered Jan 25 '17 at 21:00
levap
47k23273
Yeah, and you can make it work even if $f$ is not a local diffeomorphism, only a proper map. The relevant notion is that of the degree of a smooth proper map between oriented manifolds of the same dimension.
Assume for simplicity that $M,N$ are closed (compact, without boundary), non-empty and connected and let $f colon M rightarrow N$ be an arbitrary smooth map. Since $M,N$ are oriented, we can a generator $[omega_1] in H^{text{top}}(M)$ such that $omega_1$ is consistent with the orientation on $M$ and $int_M omega_1 = 1$. Choose $[omega_2]$ similarly for $N$. The map $f$ induces a map $f^{*} colon H^{text{top}}(N) rightarrow H^{text{top}}(M)$ on cohomology and since the top cohomology groups are one-dimensional, we must have $f^{*}([omega_2]) = c [omega_1]$ for some $c in mathbb{R}$. This $c = deg(f)$ is called the degree of $f$ and is a priori a real number. However, it can be shown that $c$ is in fact an integer which can be computed by counting the number of preimages of a regular value $p in N$ with appropriate signs which take the orientations of $M,N$ into consideration.
Knowing that, given $omega in Omega^{text{top}}(N)$, write $[omega] = c[omega_2]$ for some $c in mathbb{R}$. Then
$$ [f^{*}(omega)] = f^{*}([omega]) = f^{*}(c[omega_2]) = c deg(f) [omega_1] $$
so
$$ int_M f^{*}(omega) = deg(f) c int_N omega_1 = deg(f) c = deg(f) cint_N omega_2 = deg(f) int_N omega. $$
In particular, if $f$ is a diffeomorphism, $deg(f) = pm 1$ so this generalizes your starting point. If $f$ is a local diffeomorphism then $f colon M rightarrow N$ is a covering map and $deg(f)$ will be the number of points in an arbirary fiber, counted with appropriate signs. For much more details and proofs, see the book "Differential Forms in Algebraic Topology".
answered Jan 25 '17 at 21:00
levap
47k23273
edited Nov 28 '18 at 12:05
answered Jan 25 '17 at 21:00
levap
47k23273
answered Jan 25 '17 at 21:00
levap
47k23273
answered Jan 25 '17 at 21:00
levap
47k23273
47k23273
3
For your particular case, you can avoid mentioning cohomology by showing first (assuming $M,N$ are compact) that $f$ is a covering map. Then, over a basic neighborhood $U subseteq N$, we have $f^{-1}(U) = sqcup {U_i}$ where $f|_{U_i} colon U_i rightarrow U$ is a diffeomorphism which reverses or preserves orientation. Hence $int_{f^{-1}(U)} f^{*}(omega) = sum pm int_{N} omega$. Then you can get the global result by a partition of unity after showing that "the signs stay constant" on the intersections so there isn't any unwanted cancellations.
– levap
Jan 25 '17 at 21:19
This is super cool! Just one thing, when using the fact that local diffeomorphism is a covering map, don't I need to assume at least that $f$ is surjective or $N$ is connected? And that would also assure that the signs stay constant on the intersections, right?
– user143144
Jan 26 '17 at 9:04
Yeah, you are definitely right. I somehow tend to think of my manifolds as always connected, but you're right. I've edited my answer.
– levap
Jan 26 '17 at 18:34
add a comment |
3
For your particular case, you can avoid mentioning cohomology by showing first (assuming $M,N$ are compact) that $f$ is a covering map. Then, over a basic neighborhood $U subseteq N$, we have $f^{-1}(U) = sqcup {U_i}$ where $f|_{U_i} colon U_i rightarrow U$ is a diffeomorphism which reverses or preserves orientation. Hence $int_{f^{-1}(U)} f^{*}(omega) = sum pm int_{N} omega$. Then you can get the global result by a partition of unity after showing that "the signs stay constant" on the intersections so there isn't any unwanted cancellations.
– levap
Jan 25 '17 at 21:19
This is super cool! Just one thing, when using the fact that local diffeomorphism is a covering map, don't I need to assume at least that $f$ is surjective or $N$ is connected? And that would also assure that the signs stay constant on the intersections, right?
– user143144
Jan 26 '17 at 9:04
Yeah, you are definitely right. I somehow tend to think of my manifolds as always connected, but you're right. I've edited my answer.
– levap
Jan 26 '17 at 18:34
3
3
For your particular case, you can avoid mentioning cohomology by showing first (assuming $M,N$ are compact) that $f$ is a covering map. Then, over a basic neighborhood $U subseteq N$, we have $f^{-1}(U) = sqcup {U_i}$ where $f|_{U_i} colon U_i rightarrow U$ is a diffeomorphism which reverses or preserves orientation. Hence $int_{f^{-1}(U)} f^{*}(omega) = sum pm int_{N} omega$. Then you can get the global result by a partition of unity after showing that "the signs stay constant" on the intersections so there isn't any unwanted cancellations.
– levap
Jan 25 '17 at 21:19
For your particular case, you can avoid mentioning cohomology by showing first (assuming $M,N$ are compact) that $f$ is a covering map. Then, over a basic neighborhood $U subseteq N$, we have $f^{-1}(U) = sqcup {U_i}$ where $f|_{U_i} colon U_i rightarrow U$ is a diffeomorphism which reverses or preserves orientation. Hence $int_{f^{-1}(U)} f^{*}(omega) = sum pm int_{N} omega$. Then you can get the global result by a partition of unity after showing that "the signs stay constant" on the intersections so there isn't any unwanted cancellations.
– levap
Jan 25 '17 at 21:19
This is super cool! Just one thing, when using the fact that local diffeomorphism is a covering map, don't I need to assume at least that $f$ is surjective or $N$ is connected? And that would also assure that the signs stay constant on the intersections, right?
– user143144
Jan 26 '17 at 9:04
This is super cool! Just one thing, when using the fact that local diffeomorphism is a covering map, don't I need to assume at least that $f$ is surjective or $N$ is connected? And that would also assure that the signs stay constant on the intersections, right?
– user143144
Jan 26 '17 at 9:04
Yeah, you are definitely right. I somehow tend to think of my manifolds as always connected, but you're right. I've edited my answer.
– levap
Jan 26 '17 at 18:34
Yeah, you are definitely right. I somehow tend to think of my manifolds as always connected, but you're right. I've edited my answer.
– levap
Jan 26 '17 at 18:34
add a comment |
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