confusion about basic Kahler geometry












3














I am really struggling to understand the basics of Kahler geometry and hope someone can give me some guidance. Suppose we have a complex manifold with some complex structure $J$ and let $g$ be a Hermitian metric. That is, $g$ is a Riemannian metric and satisfies the extra condition $g(JX,JY)=g(X,Y)$. So we have a Riemannian manifold $(M^{2n},g)$. Why do we all of a sudden start considering the complexified tangent space? What good does that do? The metric $g$ is of the form $$g=sum_{i,j=1}^{2n}g_{ij}E^iotimes E^j$$ where ${E^1,dots, E^{2n}}={dx^1,dy^1,dots, dx^n, dy^n}$. We introduce $frac{partial}{partial z_k}$ and $frac{partial}{partial overline z_k}$ and $dz_k$ and $doverline z_k$ and then extend the metric to $TM^mathbb{C}$ and start doing all these computations. But by doing this we've completely changed the domain of the metric. One metric is in $Omega^2_mathbb{C}(M)$ and the other is in $Omega^2(M)$. How is this new metric telling us stuff about the original one? What is the relationship between the matrix $g_{koverline j}$, where $1leq j,kleq n$ and $g_{jk}$, where $1leq j,kleq 2n$? We introduce the 'extended' connection and curvature but again, I don't see how this new information tells us something about the original manifold. For example, apparently the 'extended' definition of Ricci curvature on $TM^mathbb{C}$ is the same as the Ricci curvature on $(M^{2n},g)$. How can these two things be the same, they are functions on completely different spaces?? Another example I am struggling with is how the Kahler form $omega$ satisfies $frac{omega^n}{n!}=text{vol}_M$. We have $omega=sqrt{-1}g_{joverline k}dz^jwedge doverline z^k$ an element of $Omega^2_mathbb{C}(M)$. How can this be raised to powers to give an element of
$Omega^{2n}(M)$??



Really any help would be greatly appreciated!










share|cite|improve this question
























  • Just to address your final question quickly: Raising to the $n$th power means wedging it with itself $n$ times. Try computing $omega^2$ if $omega=sqrt{-1}(dzwedge dbar z + dwwedge dbar w)$, for example.
    – Ted Shifrin
    Feb 20 '15 at 3:38












  • One adventage of using complex variable is that when $M$ is Kahler, the Riemann Curavture tensor has only type (1, 1) component. You can take a look at Gang Tian's "Canonical metric on Kahler manifolds". Also, some other geometric operator (eg $d^*$ and $Delta_g$) has a neat formula when written using complex coordinate).
    – user99914
    Feb 20 '15 at 7:43
















3














I am really struggling to understand the basics of Kahler geometry and hope someone can give me some guidance. Suppose we have a complex manifold with some complex structure $J$ and let $g$ be a Hermitian metric. That is, $g$ is a Riemannian metric and satisfies the extra condition $g(JX,JY)=g(X,Y)$. So we have a Riemannian manifold $(M^{2n},g)$. Why do we all of a sudden start considering the complexified tangent space? What good does that do? The metric $g$ is of the form $$g=sum_{i,j=1}^{2n}g_{ij}E^iotimes E^j$$ where ${E^1,dots, E^{2n}}={dx^1,dy^1,dots, dx^n, dy^n}$. We introduce $frac{partial}{partial z_k}$ and $frac{partial}{partial overline z_k}$ and $dz_k$ and $doverline z_k$ and then extend the metric to $TM^mathbb{C}$ and start doing all these computations. But by doing this we've completely changed the domain of the metric. One metric is in $Omega^2_mathbb{C}(M)$ and the other is in $Omega^2(M)$. How is this new metric telling us stuff about the original one? What is the relationship between the matrix $g_{koverline j}$, where $1leq j,kleq n$ and $g_{jk}$, where $1leq j,kleq 2n$? We introduce the 'extended' connection and curvature but again, I don't see how this new information tells us something about the original manifold. For example, apparently the 'extended' definition of Ricci curvature on $TM^mathbb{C}$ is the same as the Ricci curvature on $(M^{2n},g)$. How can these two things be the same, they are functions on completely different spaces?? Another example I am struggling with is how the Kahler form $omega$ satisfies $frac{omega^n}{n!}=text{vol}_M$. We have $omega=sqrt{-1}g_{joverline k}dz^jwedge doverline z^k$ an element of $Omega^2_mathbb{C}(M)$. How can this be raised to powers to give an element of
$Omega^{2n}(M)$??



Really any help would be greatly appreciated!










share|cite|improve this question
























  • Just to address your final question quickly: Raising to the $n$th power means wedging it with itself $n$ times. Try computing $omega^2$ if $omega=sqrt{-1}(dzwedge dbar z + dwwedge dbar w)$, for example.
    – Ted Shifrin
    Feb 20 '15 at 3:38












  • One adventage of using complex variable is that when $M$ is Kahler, the Riemann Curavture tensor has only type (1, 1) component. You can take a look at Gang Tian's "Canonical metric on Kahler manifolds". Also, some other geometric operator (eg $d^*$ and $Delta_g$) has a neat formula when written using complex coordinate).
    – user99914
    Feb 20 '15 at 7:43














3












3








3


2





I am really struggling to understand the basics of Kahler geometry and hope someone can give me some guidance. Suppose we have a complex manifold with some complex structure $J$ and let $g$ be a Hermitian metric. That is, $g$ is a Riemannian metric and satisfies the extra condition $g(JX,JY)=g(X,Y)$. So we have a Riemannian manifold $(M^{2n},g)$. Why do we all of a sudden start considering the complexified tangent space? What good does that do? The metric $g$ is of the form $$g=sum_{i,j=1}^{2n}g_{ij}E^iotimes E^j$$ where ${E^1,dots, E^{2n}}={dx^1,dy^1,dots, dx^n, dy^n}$. We introduce $frac{partial}{partial z_k}$ and $frac{partial}{partial overline z_k}$ and $dz_k$ and $doverline z_k$ and then extend the metric to $TM^mathbb{C}$ and start doing all these computations. But by doing this we've completely changed the domain of the metric. One metric is in $Omega^2_mathbb{C}(M)$ and the other is in $Omega^2(M)$. How is this new metric telling us stuff about the original one? What is the relationship between the matrix $g_{koverline j}$, where $1leq j,kleq n$ and $g_{jk}$, where $1leq j,kleq 2n$? We introduce the 'extended' connection and curvature but again, I don't see how this new information tells us something about the original manifold. For example, apparently the 'extended' definition of Ricci curvature on $TM^mathbb{C}$ is the same as the Ricci curvature on $(M^{2n},g)$. How can these two things be the same, they are functions on completely different spaces?? Another example I am struggling with is how the Kahler form $omega$ satisfies $frac{omega^n}{n!}=text{vol}_M$. We have $omega=sqrt{-1}g_{joverline k}dz^jwedge doverline z^k$ an element of $Omega^2_mathbb{C}(M)$. How can this be raised to powers to give an element of
$Omega^{2n}(M)$??



Really any help would be greatly appreciated!










share|cite|improve this question















I am really struggling to understand the basics of Kahler geometry and hope someone can give me some guidance. Suppose we have a complex manifold with some complex structure $J$ and let $g$ be a Hermitian metric. That is, $g$ is a Riemannian metric and satisfies the extra condition $g(JX,JY)=g(X,Y)$. So we have a Riemannian manifold $(M^{2n},g)$. Why do we all of a sudden start considering the complexified tangent space? What good does that do? The metric $g$ is of the form $$g=sum_{i,j=1}^{2n}g_{ij}E^iotimes E^j$$ where ${E^1,dots, E^{2n}}={dx^1,dy^1,dots, dx^n, dy^n}$. We introduce $frac{partial}{partial z_k}$ and $frac{partial}{partial overline z_k}$ and $dz_k$ and $doverline z_k$ and then extend the metric to $TM^mathbb{C}$ and start doing all these computations. But by doing this we've completely changed the domain of the metric. One metric is in $Omega^2_mathbb{C}(M)$ and the other is in $Omega^2(M)$. How is this new metric telling us stuff about the original one? What is the relationship between the matrix $g_{koverline j}$, where $1leq j,kleq n$ and $g_{jk}$, where $1leq j,kleq 2n$? We introduce the 'extended' connection and curvature but again, I don't see how this new information tells us something about the original manifold. For example, apparently the 'extended' definition of Ricci curvature on $TM^mathbb{C}$ is the same as the Ricci curvature on $(M^{2n},g)$. How can these two things be the same, they are functions on completely different spaces?? Another example I am struggling with is how the Kahler form $omega$ satisfies $frac{omega^n}{n!}=text{vol}_M$. We have $omega=sqrt{-1}g_{joverline k}dz^jwedge doverline z^k$ an element of $Omega^2_mathbb{C}(M)$. How can this be raised to powers to give an element of
$Omega^{2n}(M)$??



Really any help would be greatly appreciated!







differential-geometry kahler-manifolds






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edited Feb 19 '15 at 19:14

























asked Feb 19 '15 at 17:42









user162520

670315




670315












  • Just to address your final question quickly: Raising to the $n$th power means wedging it with itself $n$ times. Try computing $omega^2$ if $omega=sqrt{-1}(dzwedge dbar z + dwwedge dbar w)$, for example.
    – Ted Shifrin
    Feb 20 '15 at 3:38












  • One adventage of using complex variable is that when $M$ is Kahler, the Riemann Curavture tensor has only type (1, 1) component. You can take a look at Gang Tian's "Canonical metric on Kahler manifolds". Also, some other geometric operator (eg $d^*$ and $Delta_g$) has a neat formula when written using complex coordinate).
    – user99914
    Feb 20 '15 at 7:43


















  • Just to address your final question quickly: Raising to the $n$th power means wedging it with itself $n$ times. Try computing $omega^2$ if $omega=sqrt{-1}(dzwedge dbar z + dwwedge dbar w)$, for example.
    – Ted Shifrin
    Feb 20 '15 at 3:38












  • One adventage of using complex variable is that when $M$ is Kahler, the Riemann Curavture tensor has only type (1, 1) component. You can take a look at Gang Tian's "Canonical metric on Kahler manifolds". Also, some other geometric operator (eg $d^*$ and $Delta_g$) has a neat formula when written using complex coordinate).
    – user99914
    Feb 20 '15 at 7:43
















Just to address your final question quickly: Raising to the $n$th power means wedging it with itself $n$ times. Try computing $omega^2$ if $omega=sqrt{-1}(dzwedge dbar z + dwwedge dbar w)$, for example.
– Ted Shifrin
Feb 20 '15 at 3:38






Just to address your final question quickly: Raising to the $n$th power means wedging it with itself $n$ times. Try computing $omega^2$ if $omega=sqrt{-1}(dzwedge dbar z + dwwedge dbar w)$, for example.
– Ted Shifrin
Feb 20 '15 at 3:38














One adventage of using complex variable is that when $M$ is Kahler, the Riemann Curavture tensor has only type (1, 1) component. You can take a look at Gang Tian's "Canonical metric on Kahler manifolds". Also, some other geometric operator (eg $d^*$ and $Delta_g$) has a neat formula when written using complex coordinate).
– user99914
Feb 20 '15 at 7:43




One adventage of using complex variable is that when $M$ is Kahler, the Riemann Curavture tensor has only type (1, 1) component. You can take a look at Gang Tian's "Canonical metric on Kahler manifolds". Also, some other geometric operator (eg $d^*$ and $Delta_g$) has a neat formula when written using complex coordinate).
– user99914
Feb 20 '15 at 7:43










2 Answers
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active

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3














Kähler manifolds are particularly nice examples of manifolds where the complex structure and the Riemannian structure coalesce perfectly. Every complex manifold may be viewed as a Riemannian manifold (since holomorphic charts are automatically smooth), but it is not necessarily the case that these two structures will be compatible. This compatibility can be expressed in a number of ways:




  • We have a notion of parallel transport on Riemannian manifolds. A certain compatibility criterion between the complex and Riemannian structures would be to require that parallel transport commutes with $J$.


  • On the tangent bundle of a complex manifold, we have two choices of canonical connection, the Levi--Civita connection from Riemannian geometry and the Chern connection from complex geometry. These two connections do not necessarily coincide on an arbitrary complex manifold.


  • On a Riemannian manifold we may diagonalise the metric at a point which gives us the Riemannian Normal Coordinates. On an a complex manifold we have these coordinates also (of course), but these coordinates are not necessarily holomorphic.



PUNCHLINE: On a Kähler manifold all of these conditions are satisfied! In fact, the Kähler condition $domega =0$ is equivalent to the three conditions above (any of which may be taken as a definition).



Closing remarks: It is worth noting that the condition $g(JX, JY) = g(X,Y)$ may be stated as requiring the complex structure $J$ to be an orthogonal transformation on each tangent space.



Commuting with the complex structure $J$ is desirable for the same reason that holomorphic functions are desirable relative to smooth functions. That is, a function $f : mathbb{C} longrightarrow mathbb{C}$ is holomorphic if its Jacobian matrix commutes with $J = begin{bmatrix}
0 & -1 \
1 & 0
end{bmatrix}$
.



Notice that this also alludes to why holomorphic functions are such starkly different beasts when compared with smooth functions. This requirement on the derivative is quite rigid.



References: Griffiths and Harris, Principles of Algebraic Geometry, Andrei Moroianu Lecture on Kähler Geometry, Werner Ballmann Lectures on Kähler manifolds, Gabór Szekelyhidi Introduction to Extremal Metrics.



This is also nice: Key differences between almost complex manifolds and complex manifolds



Updated: I have recently been made aware of Daniele Angella's text Cohomological Aspects in complex Non-Kähler geometry. As the name suggests, the book focuses on complex manifolds which are not Kähler, but I would highly recommend this well-written text. It has a nice treatment of almost complex manifolds, and it is often insightful to look at the behaviour of the objects which do not satisfy such a nice condition as being Kähler.






share|cite|improve this answer































    1














    Customary we write Kahler metric $$ ds^2=2sum_{a,b}
    g_{aoverline{b}}dz_aotimes doverline{z}_b
    $$



    Here $g$ is a Riemannian metric. If $E_a=frac{partial }{partial
    x_a}, E_{n+a}=frac{partial }{partial x_{n+a}}$ are coordinate
    field, then we have dual $dE_a, dE_{n+a}$.



    In further, define $$ frac{partial
    }{partial z_a} =frac{partial }{partial x_a} -ifrac{partial
    }{partial x_{n+a}} $$ so that its dual is $dz_a=frac{1}{2}{ dE_a+
    idE_{n+a} }$.



    If $g_{aoverline{b}} :=g(frac{partial }{partial
    z_a},overline{frac{partial }{partial z_b}} )$, then by a direct
    computation considering $J$, we have $$ g=2{rm Re} sum_{a,b}
    g_{aoverline{b}}dz_aotimes doverline{z}_b = {rm Re} ds^2 $$



    In further, fundamental form is $Phi (X,Y)=g(X,JY)$. By
    computation, we have $Phi = {rm Im} ds^2 $.






    share|cite|improve this answer





















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      2 Answers
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      Kähler manifolds are particularly nice examples of manifolds where the complex structure and the Riemannian structure coalesce perfectly. Every complex manifold may be viewed as a Riemannian manifold (since holomorphic charts are automatically smooth), but it is not necessarily the case that these two structures will be compatible. This compatibility can be expressed in a number of ways:




      • We have a notion of parallel transport on Riemannian manifolds. A certain compatibility criterion between the complex and Riemannian structures would be to require that parallel transport commutes with $J$.


      • On the tangent bundle of a complex manifold, we have two choices of canonical connection, the Levi--Civita connection from Riemannian geometry and the Chern connection from complex geometry. These two connections do not necessarily coincide on an arbitrary complex manifold.


      • On a Riemannian manifold we may diagonalise the metric at a point which gives us the Riemannian Normal Coordinates. On an a complex manifold we have these coordinates also (of course), but these coordinates are not necessarily holomorphic.



      PUNCHLINE: On a Kähler manifold all of these conditions are satisfied! In fact, the Kähler condition $domega =0$ is equivalent to the three conditions above (any of which may be taken as a definition).



      Closing remarks: It is worth noting that the condition $g(JX, JY) = g(X,Y)$ may be stated as requiring the complex structure $J$ to be an orthogonal transformation on each tangent space.



      Commuting with the complex structure $J$ is desirable for the same reason that holomorphic functions are desirable relative to smooth functions. That is, a function $f : mathbb{C} longrightarrow mathbb{C}$ is holomorphic if its Jacobian matrix commutes with $J = begin{bmatrix}
      0 & -1 \
      1 & 0
      end{bmatrix}$
      .



      Notice that this also alludes to why holomorphic functions are such starkly different beasts when compared with smooth functions. This requirement on the derivative is quite rigid.



      References: Griffiths and Harris, Principles of Algebraic Geometry, Andrei Moroianu Lecture on Kähler Geometry, Werner Ballmann Lectures on Kähler manifolds, Gabór Szekelyhidi Introduction to Extremal Metrics.



      This is also nice: Key differences between almost complex manifolds and complex manifolds



      Updated: I have recently been made aware of Daniele Angella's text Cohomological Aspects in complex Non-Kähler geometry. As the name suggests, the book focuses on complex manifolds which are not Kähler, but I would highly recommend this well-written text. It has a nice treatment of almost complex manifolds, and it is often insightful to look at the behaviour of the objects which do not satisfy such a nice condition as being Kähler.






      share|cite|improve this answer




























        3














        Kähler manifolds are particularly nice examples of manifolds where the complex structure and the Riemannian structure coalesce perfectly. Every complex manifold may be viewed as a Riemannian manifold (since holomorphic charts are automatically smooth), but it is not necessarily the case that these two structures will be compatible. This compatibility can be expressed in a number of ways:




        • We have a notion of parallel transport on Riemannian manifolds. A certain compatibility criterion between the complex and Riemannian structures would be to require that parallel transport commutes with $J$.


        • On the tangent bundle of a complex manifold, we have two choices of canonical connection, the Levi--Civita connection from Riemannian geometry and the Chern connection from complex geometry. These two connections do not necessarily coincide on an arbitrary complex manifold.


        • On a Riemannian manifold we may diagonalise the metric at a point which gives us the Riemannian Normal Coordinates. On an a complex manifold we have these coordinates also (of course), but these coordinates are not necessarily holomorphic.



        PUNCHLINE: On a Kähler manifold all of these conditions are satisfied! In fact, the Kähler condition $domega =0$ is equivalent to the three conditions above (any of which may be taken as a definition).



        Closing remarks: It is worth noting that the condition $g(JX, JY) = g(X,Y)$ may be stated as requiring the complex structure $J$ to be an orthogonal transformation on each tangent space.



        Commuting with the complex structure $J$ is desirable for the same reason that holomorphic functions are desirable relative to smooth functions. That is, a function $f : mathbb{C} longrightarrow mathbb{C}$ is holomorphic if its Jacobian matrix commutes with $J = begin{bmatrix}
        0 & -1 \
        1 & 0
        end{bmatrix}$
        .



        Notice that this also alludes to why holomorphic functions are such starkly different beasts when compared with smooth functions. This requirement on the derivative is quite rigid.



        References: Griffiths and Harris, Principles of Algebraic Geometry, Andrei Moroianu Lecture on Kähler Geometry, Werner Ballmann Lectures on Kähler manifolds, Gabór Szekelyhidi Introduction to Extremal Metrics.



        This is also nice: Key differences between almost complex manifolds and complex manifolds



        Updated: I have recently been made aware of Daniele Angella's text Cohomological Aspects in complex Non-Kähler geometry. As the name suggests, the book focuses on complex manifolds which are not Kähler, but I would highly recommend this well-written text. It has a nice treatment of almost complex manifolds, and it is often insightful to look at the behaviour of the objects which do not satisfy such a nice condition as being Kähler.






        share|cite|improve this answer


























          3












          3








          3






          Kähler manifolds are particularly nice examples of manifolds where the complex structure and the Riemannian structure coalesce perfectly. Every complex manifold may be viewed as a Riemannian manifold (since holomorphic charts are automatically smooth), but it is not necessarily the case that these two structures will be compatible. This compatibility can be expressed in a number of ways:




          • We have a notion of parallel transport on Riemannian manifolds. A certain compatibility criterion between the complex and Riemannian structures would be to require that parallel transport commutes with $J$.


          • On the tangent bundle of a complex manifold, we have two choices of canonical connection, the Levi--Civita connection from Riemannian geometry and the Chern connection from complex geometry. These two connections do not necessarily coincide on an arbitrary complex manifold.


          • On a Riemannian manifold we may diagonalise the metric at a point which gives us the Riemannian Normal Coordinates. On an a complex manifold we have these coordinates also (of course), but these coordinates are not necessarily holomorphic.



          PUNCHLINE: On a Kähler manifold all of these conditions are satisfied! In fact, the Kähler condition $domega =0$ is equivalent to the three conditions above (any of which may be taken as a definition).



          Closing remarks: It is worth noting that the condition $g(JX, JY) = g(X,Y)$ may be stated as requiring the complex structure $J$ to be an orthogonal transformation on each tangent space.



          Commuting with the complex structure $J$ is desirable for the same reason that holomorphic functions are desirable relative to smooth functions. That is, a function $f : mathbb{C} longrightarrow mathbb{C}$ is holomorphic if its Jacobian matrix commutes with $J = begin{bmatrix}
          0 & -1 \
          1 & 0
          end{bmatrix}$
          .



          Notice that this also alludes to why holomorphic functions are such starkly different beasts when compared with smooth functions. This requirement on the derivative is quite rigid.



          References: Griffiths and Harris, Principles of Algebraic Geometry, Andrei Moroianu Lecture on Kähler Geometry, Werner Ballmann Lectures on Kähler manifolds, Gabór Szekelyhidi Introduction to Extremal Metrics.



          This is also nice: Key differences between almost complex manifolds and complex manifolds



          Updated: I have recently been made aware of Daniele Angella's text Cohomological Aspects in complex Non-Kähler geometry. As the name suggests, the book focuses on complex manifolds which are not Kähler, but I would highly recommend this well-written text. It has a nice treatment of almost complex manifolds, and it is often insightful to look at the behaviour of the objects which do not satisfy such a nice condition as being Kähler.






          share|cite|improve this answer














          Kähler manifolds are particularly nice examples of manifolds where the complex structure and the Riemannian structure coalesce perfectly. Every complex manifold may be viewed as a Riemannian manifold (since holomorphic charts are automatically smooth), but it is not necessarily the case that these two structures will be compatible. This compatibility can be expressed in a number of ways:




          • We have a notion of parallel transport on Riemannian manifolds. A certain compatibility criterion between the complex and Riemannian structures would be to require that parallel transport commutes with $J$.


          • On the tangent bundle of a complex manifold, we have two choices of canonical connection, the Levi--Civita connection from Riemannian geometry and the Chern connection from complex geometry. These two connections do not necessarily coincide on an arbitrary complex manifold.


          • On a Riemannian manifold we may diagonalise the metric at a point which gives us the Riemannian Normal Coordinates. On an a complex manifold we have these coordinates also (of course), but these coordinates are not necessarily holomorphic.



          PUNCHLINE: On a Kähler manifold all of these conditions are satisfied! In fact, the Kähler condition $domega =0$ is equivalent to the three conditions above (any of which may be taken as a definition).



          Closing remarks: It is worth noting that the condition $g(JX, JY) = g(X,Y)$ may be stated as requiring the complex structure $J$ to be an orthogonal transformation on each tangent space.



          Commuting with the complex structure $J$ is desirable for the same reason that holomorphic functions are desirable relative to smooth functions. That is, a function $f : mathbb{C} longrightarrow mathbb{C}$ is holomorphic if its Jacobian matrix commutes with $J = begin{bmatrix}
          0 & -1 \
          1 & 0
          end{bmatrix}$
          .



          Notice that this also alludes to why holomorphic functions are such starkly different beasts when compared with smooth functions. This requirement on the derivative is quite rigid.



          References: Griffiths and Harris, Principles of Algebraic Geometry, Andrei Moroianu Lecture on Kähler Geometry, Werner Ballmann Lectures on Kähler manifolds, Gabór Szekelyhidi Introduction to Extremal Metrics.



          This is also nice: Key differences between almost complex manifolds and complex manifolds



          Updated: I have recently been made aware of Daniele Angella's text Cohomological Aspects in complex Non-Kähler geometry. As the name suggests, the book focuses on complex manifolds which are not Kähler, but I would highly recommend this well-written text. It has a nice treatment of almost complex manifolds, and it is often insightful to look at the behaviour of the objects which do not satisfy such a nice condition as being Kähler.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 2 '18 at 21:11

























          answered Jun 14 '18 at 22:18









          AmorFati

          396529




          396529























              1














              Customary we write Kahler metric $$ ds^2=2sum_{a,b}
              g_{aoverline{b}}dz_aotimes doverline{z}_b
              $$



              Here $g$ is a Riemannian metric. If $E_a=frac{partial }{partial
              x_a}, E_{n+a}=frac{partial }{partial x_{n+a}}$ are coordinate
              field, then we have dual $dE_a, dE_{n+a}$.



              In further, define $$ frac{partial
              }{partial z_a} =frac{partial }{partial x_a} -ifrac{partial
              }{partial x_{n+a}} $$ so that its dual is $dz_a=frac{1}{2}{ dE_a+
              idE_{n+a} }$.



              If $g_{aoverline{b}} :=g(frac{partial }{partial
              z_a},overline{frac{partial }{partial z_b}} )$, then by a direct
              computation considering $J$, we have $$ g=2{rm Re} sum_{a,b}
              g_{aoverline{b}}dz_aotimes doverline{z}_b = {rm Re} ds^2 $$



              In further, fundamental form is $Phi (X,Y)=g(X,JY)$. By
              computation, we have $Phi = {rm Im} ds^2 $.






              share|cite|improve this answer


























                1














                Customary we write Kahler metric $$ ds^2=2sum_{a,b}
                g_{aoverline{b}}dz_aotimes doverline{z}_b
                $$



                Here $g$ is a Riemannian metric. If $E_a=frac{partial }{partial
                x_a}, E_{n+a}=frac{partial }{partial x_{n+a}}$ are coordinate
                field, then we have dual $dE_a, dE_{n+a}$.



                In further, define $$ frac{partial
                }{partial z_a} =frac{partial }{partial x_a} -ifrac{partial
                }{partial x_{n+a}} $$ so that its dual is $dz_a=frac{1}{2}{ dE_a+
                idE_{n+a} }$.



                If $g_{aoverline{b}} :=g(frac{partial }{partial
                z_a},overline{frac{partial }{partial z_b}} )$, then by a direct
                computation considering $J$, we have $$ g=2{rm Re} sum_{a,b}
                g_{aoverline{b}}dz_aotimes doverline{z}_b = {rm Re} ds^2 $$



                In further, fundamental form is $Phi (X,Y)=g(X,JY)$. By
                computation, we have $Phi = {rm Im} ds^2 $.






                share|cite|improve this answer
























                  1












                  1








                  1






                  Customary we write Kahler metric $$ ds^2=2sum_{a,b}
                  g_{aoverline{b}}dz_aotimes doverline{z}_b
                  $$



                  Here $g$ is a Riemannian metric. If $E_a=frac{partial }{partial
                  x_a}, E_{n+a}=frac{partial }{partial x_{n+a}}$ are coordinate
                  field, then we have dual $dE_a, dE_{n+a}$.



                  In further, define $$ frac{partial
                  }{partial z_a} =frac{partial }{partial x_a} -ifrac{partial
                  }{partial x_{n+a}} $$ so that its dual is $dz_a=frac{1}{2}{ dE_a+
                  idE_{n+a} }$.



                  If $g_{aoverline{b}} :=g(frac{partial }{partial
                  z_a},overline{frac{partial }{partial z_b}} )$, then by a direct
                  computation considering $J$, we have $$ g=2{rm Re} sum_{a,b}
                  g_{aoverline{b}}dz_aotimes doverline{z}_b = {rm Re} ds^2 $$



                  In further, fundamental form is $Phi (X,Y)=g(X,JY)$. By
                  computation, we have $Phi = {rm Im} ds^2 $.






                  share|cite|improve this answer












                  Customary we write Kahler metric $$ ds^2=2sum_{a,b}
                  g_{aoverline{b}}dz_aotimes doverline{z}_b
                  $$



                  Here $g$ is a Riemannian metric. If $E_a=frac{partial }{partial
                  x_a}, E_{n+a}=frac{partial }{partial x_{n+a}}$ are coordinate
                  field, then we have dual $dE_a, dE_{n+a}$.



                  In further, define $$ frac{partial
                  }{partial z_a} =frac{partial }{partial x_a} -ifrac{partial
                  }{partial x_{n+a}} $$ so that its dual is $dz_a=frac{1}{2}{ dE_a+
                  idE_{n+a} }$.



                  If $g_{aoverline{b}} :=g(frac{partial }{partial
                  z_a},overline{frac{partial }{partial z_b}} )$, then by a direct
                  computation considering $J$, we have $$ g=2{rm Re} sum_{a,b}
                  g_{aoverline{b}}dz_aotimes doverline{z}_b = {rm Re} ds^2 $$



                  In further, fundamental form is $Phi (X,Y)=g(X,JY)$. By
                  computation, we have $Phi = {rm Im} ds^2 $.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 27 '18 at 4:05









                  HK Lee

                  13.8k52057




                  13.8k52057






























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