Pullback of a differential form by a local diffeomorphism












6















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







share|cite|improve this question



























    6















    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







    share|cite|improve this question

























      6












      6








      6


      7






      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







      share|cite|improve this question














      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






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 25 '17 at 20:02









      user143144

      1176




      1176






















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          6














          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".






          share|cite|improve this answer



















          • 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











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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".






          share|cite|improve this answer



















          • 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
















          6














          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".






          share|cite|improve this answer



















          • 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














          6












          6








          6






          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".






          share|cite|improve this answer














          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".







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 '18 at 12:05

























          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














          • 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


















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