Changing coefficients of cohomology and pullbacks












1














If I have a compact complex manifold $M$ and a map $f:Mto M$. Let $mathbb F$ a field. Then $H^k(M;mathbb F) $ is a $n$-dimensional vector space and chosing a basis, $f^*_{mathbb F}$ defines a matrix.



Let's say I already know how the matrix $f^*_mathbb F$ looks for $mathbb F = mathbb R$. How do I obtain $f^*_mathbb C$ from it?

Is it just the same matrix but we think of the entries as complex numbers with vanishing imaginary part?



What about the other way around? If I know $f^*_mathbb C$, how do I get $f^*_mathbb R$?

I know that there is a standard way to make a real matrix out of a compley one by replacing every entry $x=a+ ib$ by a block $begin{pmatrix} a & -b \ b & a end{pmatrix} $. But this would change the size of the matrix, so it cannot be right.










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    1














    If I have a compact complex manifold $M$ and a map $f:Mto M$. Let $mathbb F$ a field. Then $H^k(M;mathbb F) $ is a $n$-dimensional vector space and chosing a basis, $f^*_{mathbb F}$ defines a matrix.



    Let's say I already know how the matrix $f^*_mathbb F$ looks for $mathbb F = mathbb R$. How do I obtain $f^*_mathbb C$ from it?

    Is it just the same matrix but we think of the entries as complex numbers with vanishing imaginary part?



    What about the other way around? If I know $f^*_mathbb C$, how do I get $f^*_mathbb R$?

    I know that there is a standard way to make a real matrix out of a compley one by replacing every entry $x=a+ ib$ by a block $begin{pmatrix} a & -b \ b & a end{pmatrix} $. But this would change the size of the matrix, so it cannot be right.










    share|cite|improve this question

























      1












      1








      1







      If I have a compact complex manifold $M$ and a map $f:Mto M$. Let $mathbb F$ a field. Then $H^k(M;mathbb F) $ is a $n$-dimensional vector space and chosing a basis, $f^*_{mathbb F}$ defines a matrix.



      Let's say I already know how the matrix $f^*_mathbb F$ looks for $mathbb F = mathbb R$. How do I obtain $f^*_mathbb C$ from it?

      Is it just the same matrix but we think of the entries as complex numbers with vanishing imaginary part?



      What about the other way around? If I know $f^*_mathbb C$, how do I get $f^*_mathbb R$?

      I know that there is a standard way to make a real matrix out of a compley one by replacing every entry $x=a+ ib$ by a block $begin{pmatrix} a & -b \ b & a end{pmatrix} $. But this would change the size of the matrix, so it cannot be right.










      share|cite|improve this question













      If I have a compact complex manifold $M$ and a map $f:Mto M$. Let $mathbb F$ a field. Then $H^k(M;mathbb F) $ is a $n$-dimensional vector space and chosing a basis, $f^*_{mathbb F}$ defines a matrix.



      Let's say I already know how the matrix $f^*_mathbb F$ looks for $mathbb F = mathbb R$. How do I obtain $f^*_mathbb C$ from it?

      Is it just the same matrix but we think of the entries as complex numbers with vanishing imaginary part?



      What about the other way around? If I know $f^*_mathbb C$, how do I get $f^*_mathbb R$?

      I know that there is a standard way to make a real matrix out of a compley one by replacing every entry $x=a+ ib$ by a block $begin{pmatrix} a & -b \ b & a end{pmatrix} $. But this would change the size of the matrix, so it cannot be right.







      abstract-algebra matrices differential-geometry algebraic-topology homology-cohomology






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      asked yesterday









      J.Doe

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          Part 1: Relating cohomology with complex and real coefficients



          Let's take a look at the universal coefficient theorem for cohomology, which says that for a PID $R$ and $R$-module $M$, there is a natural short exact sequence
          $$0tonewcommandExt{operatorname{Ext}}Ext^1_R(H_{i-1}(X;R),M)to H^i(X;M)to newcommandHom{operatorname{Hom}}Hom_R(H_i(X;R),M)to 0,$$
          and taking $R=newcommandRR{Bbb{R}}RR$, we see $H_{i-1}(X;R)$ is free, so $Ext$ vanishes, so we get natural isomorphisms,
          $$H^i(X;M)simeq Hom_RR(H_i(X;RR),M)$$



          Now apply this to both $M=RR$ and $M=newcommandCC{Bbb{C}}CC$, to get $H^i(X;RR)simeq H_i(X;RR)^*$, where $*$ denotes taking the dual $RR$-vector space and $$H^i(X;CC) simeq Hom_RR(H_i(X;RR),CC)simeq H_i(X;RR)^*otimes CCsimeq H^i(X;RR)otimes CC,$$
          where this last natural isomorphism follows from this question for example.



          Thus we see that cohomology with complex coefficients is the complexification of cohomology with real coefficients.



          Part 2: Converting a matrix for $f^*_RR$ to a matrix for $f^*_CC$.



          Since cohomology with complex coefficients is just the complexification of cohomology with real coefficients, the matrix of the pullback for complex coefficients will be the exact same matrix as the matrix for the pullback with real coefficients, but where we regard the real matrix as now being a complex matrix in the obvious way. (I.e. exactly what you suggested first)



          Part 3: The other direction



          To go the other way, it's a bit more complicated, and I'll see if I can give the general story for complexifications overall.



          Suppose we have a real vector space $V$, and its complexification, $V_CC=VotimesCC$, and another real vector space $W$, and its complexification $W_CC$, and a map $T : Vto W$, which induces a map $S:V_CCto W_CC$.



          Now if we have bases ${e_i}$ and ${f_j}$ for $V$ and $W$ respectively, and we compute the matrix of $S$ with respect to these (now regarded as bases for $V_CC$ and $W_CC$), then you can check that the resulting matrix for $S$ will be real, and will equal the matrix for $T$ computed with respect to ${e_i}$ and ${f_j}$.



          The difficulty arises when we have the matrix of $S$ with respect to what you can think of as "non-real" basis vectors, since then we don't have obvious bases for $V$ and $W$ to compute $T$ with respect to. It's then a bit complicated, so I'll leave my answer here.






          share|cite|improve this answer























          • Very nice, thank you
            – J.Doe
            yesterday










          • $left{frac{1}{2}(e_i+tau e_i)right}$ does not form a basis in general. For example, take $V=mathbb{R}$ so that $V_{mathbb{C}}=mathbb{C}$. This is a dimension 1 vector space. Consider the basis ${i}$. Then ${i,tau i}$ is not a $mathbb{R}$-basis of $V_{mathbb{C}}$ and $frac{1}{2}(i+tau i)$ is not a basis of $V$.
            – Roland
            yesterday










          • @Roland Yes, I realized that as well, I've decided to delete that portion of my answer, and just leave it where it is now.
            – jgon
            12 hours ago











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          2














          Part 1: Relating cohomology with complex and real coefficients



          Let's take a look at the universal coefficient theorem for cohomology, which says that for a PID $R$ and $R$-module $M$, there is a natural short exact sequence
          $$0tonewcommandExt{operatorname{Ext}}Ext^1_R(H_{i-1}(X;R),M)to H^i(X;M)to newcommandHom{operatorname{Hom}}Hom_R(H_i(X;R),M)to 0,$$
          and taking $R=newcommandRR{Bbb{R}}RR$, we see $H_{i-1}(X;R)$ is free, so $Ext$ vanishes, so we get natural isomorphisms,
          $$H^i(X;M)simeq Hom_RR(H_i(X;RR),M)$$



          Now apply this to both $M=RR$ and $M=newcommandCC{Bbb{C}}CC$, to get $H^i(X;RR)simeq H_i(X;RR)^*$, where $*$ denotes taking the dual $RR$-vector space and $$H^i(X;CC) simeq Hom_RR(H_i(X;RR),CC)simeq H_i(X;RR)^*otimes CCsimeq H^i(X;RR)otimes CC,$$
          where this last natural isomorphism follows from this question for example.



          Thus we see that cohomology with complex coefficients is the complexification of cohomology with real coefficients.



          Part 2: Converting a matrix for $f^*_RR$ to a matrix for $f^*_CC$.



          Since cohomology with complex coefficients is just the complexification of cohomology with real coefficients, the matrix of the pullback for complex coefficients will be the exact same matrix as the matrix for the pullback with real coefficients, but where we regard the real matrix as now being a complex matrix in the obvious way. (I.e. exactly what you suggested first)



          Part 3: The other direction



          To go the other way, it's a bit more complicated, and I'll see if I can give the general story for complexifications overall.



          Suppose we have a real vector space $V$, and its complexification, $V_CC=VotimesCC$, and another real vector space $W$, and its complexification $W_CC$, and a map $T : Vto W$, which induces a map $S:V_CCto W_CC$.



          Now if we have bases ${e_i}$ and ${f_j}$ for $V$ and $W$ respectively, and we compute the matrix of $S$ with respect to these (now regarded as bases for $V_CC$ and $W_CC$), then you can check that the resulting matrix for $S$ will be real, and will equal the matrix for $T$ computed with respect to ${e_i}$ and ${f_j}$.



          The difficulty arises when we have the matrix of $S$ with respect to what you can think of as "non-real" basis vectors, since then we don't have obvious bases for $V$ and $W$ to compute $T$ with respect to. It's then a bit complicated, so I'll leave my answer here.






          share|cite|improve this answer























          • Very nice, thank you
            – J.Doe
            yesterday










          • $left{frac{1}{2}(e_i+tau e_i)right}$ does not form a basis in general. For example, take $V=mathbb{R}$ so that $V_{mathbb{C}}=mathbb{C}$. This is a dimension 1 vector space. Consider the basis ${i}$. Then ${i,tau i}$ is not a $mathbb{R}$-basis of $V_{mathbb{C}}$ and $frac{1}{2}(i+tau i)$ is not a basis of $V$.
            – Roland
            yesterday










          • @Roland Yes, I realized that as well, I've decided to delete that portion of my answer, and just leave it where it is now.
            – jgon
            12 hours ago
















          2














          Part 1: Relating cohomology with complex and real coefficients



          Let's take a look at the universal coefficient theorem for cohomology, which says that for a PID $R$ and $R$-module $M$, there is a natural short exact sequence
          $$0tonewcommandExt{operatorname{Ext}}Ext^1_R(H_{i-1}(X;R),M)to H^i(X;M)to newcommandHom{operatorname{Hom}}Hom_R(H_i(X;R),M)to 0,$$
          and taking $R=newcommandRR{Bbb{R}}RR$, we see $H_{i-1}(X;R)$ is free, so $Ext$ vanishes, so we get natural isomorphisms,
          $$H^i(X;M)simeq Hom_RR(H_i(X;RR),M)$$



          Now apply this to both $M=RR$ and $M=newcommandCC{Bbb{C}}CC$, to get $H^i(X;RR)simeq H_i(X;RR)^*$, where $*$ denotes taking the dual $RR$-vector space and $$H^i(X;CC) simeq Hom_RR(H_i(X;RR),CC)simeq H_i(X;RR)^*otimes CCsimeq H^i(X;RR)otimes CC,$$
          where this last natural isomorphism follows from this question for example.



          Thus we see that cohomology with complex coefficients is the complexification of cohomology with real coefficients.



          Part 2: Converting a matrix for $f^*_RR$ to a matrix for $f^*_CC$.



          Since cohomology with complex coefficients is just the complexification of cohomology with real coefficients, the matrix of the pullback for complex coefficients will be the exact same matrix as the matrix for the pullback with real coefficients, but where we regard the real matrix as now being a complex matrix in the obvious way. (I.e. exactly what you suggested first)



          Part 3: The other direction



          To go the other way, it's a bit more complicated, and I'll see if I can give the general story for complexifications overall.



          Suppose we have a real vector space $V$, and its complexification, $V_CC=VotimesCC$, and another real vector space $W$, and its complexification $W_CC$, and a map $T : Vto W$, which induces a map $S:V_CCto W_CC$.



          Now if we have bases ${e_i}$ and ${f_j}$ for $V$ and $W$ respectively, and we compute the matrix of $S$ with respect to these (now regarded as bases for $V_CC$ and $W_CC$), then you can check that the resulting matrix for $S$ will be real, and will equal the matrix for $T$ computed with respect to ${e_i}$ and ${f_j}$.



          The difficulty arises when we have the matrix of $S$ with respect to what you can think of as "non-real" basis vectors, since then we don't have obvious bases for $V$ and $W$ to compute $T$ with respect to. It's then a bit complicated, so I'll leave my answer here.






          share|cite|improve this answer























          • Very nice, thank you
            – J.Doe
            yesterday










          • $left{frac{1}{2}(e_i+tau e_i)right}$ does not form a basis in general. For example, take $V=mathbb{R}$ so that $V_{mathbb{C}}=mathbb{C}$. This is a dimension 1 vector space. Consider the basis ${i}$. Then ${i,tau i}$ is not a $mathbb{R}$-basis of $V_{mathbb{C}}$ and $frac{1}{2}(i+tau i)$ is not a basis of $V$.
            – Roland
            yesterday










          • @Roland Yes, I realized that as well, I've decided to delete that portion of my answer, and just leave it where it is now.
            – jgon
            12 hours ago














          2












          2








          2






          Part 1: Relating cohomology with complex and real coefficients



          Let's take a look at the universal coefficient theorem for cohomology, which says that for a PID $R$ and $R$-module $M$, there is a natural short exact sequence
          $$0tonewcommandExt{operatorname{Ext}}Ext^1_R(H_{i-1}(X;R),M)to H^i(X;M)to newcommandHom{operatorname{Hom}}Hom_R(H_i(X;R),M)to 0,$$
          and taking $R=newcommandRR{Bbb{R}}RR$, we see $H_{i-1}(X;R)$ is free, so $Ext$ vanishes, so we get natural isomorphisms,
          $$H^i(X;M)simeq Hom_RR(H_i(X;RR),M)$$



          Now apply this to both $M=RR$ and $M=newcommandCC{Bbb{C}}CC$, to get $H^i(X;RR)simeq H_i(X;RR)^*$, where $*$ denotes taking the dual $RR$-vector space and $$H^i(X;CC) simeq Hom_RR(H_i(X;RR),CC)simeq H_i(X;RR)^*otimes CCsimeq H^i(X;RR)otimes CC,$$
          where this last natural isomorphism follows from this question for example.



          Thus we see that cohomology with complex coefficients is the complexification of cohomology with real coefficients.



          Part 2: Converting a matrix for $f^*_RR$ to a matrix for $f^*_CC$.



          Since cohomology with complex coefficients is just the complexification of cohomology with real coefficients, the matrix of the pullback for complex coefficients will be the exact same matrix as the matrix for the pullback with real coefficients, but where we regard the real matrix as now being a complex matrix in the obvious way. (I.e. exactly what you suggested first)



          Part 3: The other direction



          To go the other way, it's a bit more complicated, and I'll see if I can give the general story for complexifications overall.



          Suppose we have a real vector space $V$, and its complexification, $V_CC=VotimesCC$, and another real vector space $W$, and its complexification $W_CC$, and a map $T : Vto W$, which induces a map $S:V_CCto W_CC$.



          Now if we have bases ${e_i}$ and ${f_j}$ for $V$ and $W$ respectively, and we compute the matrix of $S$ with respect to these (now regarded as bases for $V_CC$ and $W_CC$), then you can check that the resulting matrix for $S$ will be real, and will equal the matrix for $T$ computed with respect to ${e_i}$ and ${f_j}$.



          The difficulty arises when we have the matrix of $S$ with respect to what you can think of as "non-real" basis vectors, since then we don't have obvious bases for $V$ and $W$ to compute $T$ with respect to. It's then a bit complicated, so I'll leave my answer here.






          share|cite|improve this answer














          Part 1: Relating cohomology with complex and real coefficients



          Let's take a look at the universal coefficient theorem for cohomology, which says that for a PID $R$ and $R$-module $M$, there is a natural short exact sequence
          $$0tonewcommandExt{operatorname{Ext}}Ext^1_R(H_{i-1}(X;R),M)to H^i(X;M)to newcommandHom{operatorname{Hom}}Hom_R(H_i(X;R),M)to 0,$$
          and taking $R=newcommandRR{Bbb{R}}RR$, we see $H_{i-1}(X;R)$ is free, so $Ext$ vanishes, so we get natural isomorphisms,
          $$H^i(X;M)simeq Hom_RR(H_i(X;RR),M)$$



          Now apply this to both $M=RR$ and $M=newcommandCC{Bbb{C}}CC$, to get $H^i(X;RR)simeq H_i(X;RR)^*$, where $*$ denotes taking the dual $RR$-vector space and $$H^i(X;CC) simeq Hom_RR(H_i(X;RR),CC)simeq H_i(X;RR)^*otimes CCsimeq H^i(X;RR)otimes CC,$$
          where this last natural isomorphism follows from this question for example.



          Thus we see that cohomology with complex coefficients is the complexification of cohomology with real coefficients.



          Part 2: Converting a matrix for $f^*_RR$ to a matrix for $f^*_CC$.



          Since cohomology with complex coefficients is just the complexification of cohomology with real coefficients, the matrix of the pullback for complex coefficients will be the exact same matrix as the matrix for the pullback with real coefficients, but where we regard the real matrix as now being a complex matrix in the obvious way. (I.e. exactly what you suggested first)



          Part 3: The other direction



          To go the other way, it's a bit more complicated, and I'll see if I can give the general story for complexifications overall.



          Suppose we have a real vector space $V$, and its complexification, $V_CC=VotimesCC$, and another real vector space $W$, and its complexification $W_CC$, and a map $T : Vto W$, which induces a map $S:V_CCto W_CC$.



          Now if we have bases ${e_i}$ and ${f_j}$ for $V$ and $W$ respectively, and we compute the matrix of $S$ with respect to these (now regarded as bases for $V_CC$ and $W_CC$), then you can check that the resulting matrix for $S$ will be real, and will equal the matrix for $T$ computed with respect to ${e_i}$ and ${f_j}$.



          The difficulty arises when we have the matrix of $S$ with respect to what you can think of as "non-real" basis vectors, since then we don't have obvious bases for $V$ and $W$ to compute $T$ with respect to. It's then a bit complicated, so I'll leave my answer here.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 12 hours ago

























          answered yesterday









          jgon

          13.1k21941




          13.1k21941












          • Very nice, thank you
            – J.Doe
            yesterday










          • $left{frac{1}{2}(e_i+tau e_i)right}$ does not form a basis in general. For example, take $V=mathbb{R}$ so that $V_{mathbb{C}}=mathbb{C}$. This is a dimension 1 vector space. Consider the basis ${i}$. Then ${i,tau i}$ is not a $mathbb{R}$-basis of $V_{mathbb{C}}$ and $frac{1}{2}(i+tau i)$ is not a basis of $V$.
            – Roland
            yesterday










          • @Roland Yes, I realized that as well, I've decided to delete that portion of my answer, and just leave it where it is now.
            – jgon
            12 hours ago


















          • Very nice, thank you
            – J.Doe
            yesterday










          • $left{frac{1}{2}(e_i+tau e_i)right}$ does not form a basis in general. For example, take $V=mathbb{R}$ so that $V_{mathbb{C}}=mathbb{C}$. This is a dimension 1 vector space. Consider the basis ${i}$. Then ${i,tau i}$ is not a $mathbb{R}$-basis of $V_{mathbb{C}}$ and $frac{1}{2}(i+tau i)$ is not a basis of $V$.
            – Roland
            yesterday










          • @Roland Yes, I realized that as well, I've decided to delete that portion of my answer, and just leave it where it is now.
            – jgon
            12 hours ago
















          Very nice, thank you
          – J.Doe
          yesterday




          Very nice, thank you
          – J.Doe
          yesterday












          $left{frac{1}{2}(e_i+tau e_i)right}$ does not form a basis in general. For example, take $V=mathbb{R}$ so that $V_{mathbb{C}}=mathbb{C}$. This is a dimension 1 vector space. Consider the basis ${i}$. Then ${i,tau i}$ is not a $mathbb{R}$-basis of $V_{mathbb{C}}$ and $frac{1}{2}(i+tau i)$ is not a basis of $V$.
          – Roland
          yesterday




          $left{frac{1}{2}(e_i+tau e_i)right}$ does not form a basis in general. For example, take $V=mathbb{R}$ so that $V_{mathbb{C}}=mathbb{C}$. This is a dimension 1 vector space. Consider the basis ${i}$. Then ${i,tau i}$ is not a $mathbb{R}$-basis of $V_{mathbb{C}}$ and $frac{1}{2}(i+tau i)$ is not a basis of $V$.
          – Roland
          yesterday












          @Roland Yes, I realized that as well, I've decided to delete that portion of my answer, and just leave it where it is now.
          – jgon
          12 hours ago




          @Roland Yes, I realized that as well, I've decided to delete that portion of my answer, and just leave it where it is now.
          – jgon
          12 hours ago


















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