What is the geometrical meaning of higher Chern forms and classes?GAGA and Chern classesProving the basic identity which implies the Chern-Weil theorem Which torsion classes in integral cohomology are Chern classes of flat bundles?Why is the integral of the second chern class an integer?Schur polynomials in the Chern classes as direct imagesHow does one go from Chern--Weil to cohomology classes on BGL(n,C)?Atiyah classes of holomorphic vector bundles with trivial Chern classesChern classes and singular hermitian metrics on vector bundlesChern Classes: two approachesWhat is $SL(2,mathbbR)$-Chern-SImons Theory?

What is the geometrical meaning of higher Chern forms and classes?


GAGA and Chern classesProving the basic identity which implies the Chern-Weil theorem Which torsion classes in integral cohomology are Chern classes of flat bundles?Why is the integral of the second chern class an integer?Schur polynomials in the Chern classes as direct imagesHow does one go from Chern--Weil to cohomology classes on BGL(n,C)?Atiyah classes of holomorphic vector bundles with trivial Chern classesChern classes and singular hermitian metrics on vector bundlesChern Classes: two approachesWhat is $SL(2,mathbbR)$-Chern-SImons Theory?













7












$begingroup$


Let $M$ be a complex manifold, $R^nabla$ be the curvature operator for connections $nabla$.
Consider a polynomial function $f:operatorname M_n(mathbbC)tomathbbC$. For the gauge group $operatornameGL_n(mathbbC)$,
if $f(A)=f(gAg^-1)$
where $A in operatorname M_n(mathbbC)$ and $g in operatornameGL_n(mathbbC)$, then $f$ is said to be an invariant polynomial function.
Let $I^k(operatorname M_n(mathbbC))$ be the set of all such polynomials of degree $k$.
Also, $bigoplus_kgeq0I^k(operatorname M_n(mathbbC))=I(operatorname M_n(mathbbC))$.



We shall need $phi_n(A)=det(A)$, $n>1$, and $phi_1(A)=Tr(A)$.



Define a global differential form ($2k$-form)
$f(R^nabla) in mathbbA^2k(M)$.
If we have the de Rham cohomology group $H^2k(M)$, then the Weil homomorphism is defined as the map
$omega:I(operatorname M_n(mathbbC))to bigoplus_kgeq 0H^2k(M)$.



The Chern forms $c_i(R^nabla)=phi_i(fracsqrt-12piR^nabla)$.



For the complex vector bundle $(mathbbE,pi,M)$, where $mathbbE$ is the total space, the Chern classes are defined as $c_i(mathbbE) in H^2k(M)$.



Therefore $c_i(mathbbE)mathrel:=omega(c_i(R^nabla))=[c_i(R^nabla)]$ (de Rham cohomology class).



The Chern forms $c_i(R^nabla) in mathbbA^2i(mathbbE)$.



$mathbbA^2i(mathbbE)$ is sheaf of smooth $mathbbE$-valued $2i$ forms on $M$.



Cohomology groups are very important in geometry for understanding the invariants that can be defined on manifolds. That is, the transformations that keep some special properties of the manifold and which analogous to the gauge transformations in physics.



Chern classes are special type of cohomology classes. If the first Chern class vanishes for a particular manifold, then it must be a Ricci-flat manifold. For example the Calabi–Yau manifolds (they have lots of other special properties, e.g., trivial canonical bundle, etc.).



But what do the higher Chern classes mean?
What uses are those higher cohomology classes corresponding to the higher Chern classes of?










share|cite|improve this question











$endgroup$











  • $begingroup$
    One answer is that if $mathbb E$ is a trivial vector bundle, then all of its Chern classes vanish. Therefore higher Chern classes are obstructions to a bundle being trivial.
    $endgroup$
    – Arun Debray
    Apr 20 at 13:21










  • $begingroup$
    I have edited to fix TeX (for example, use in $in$ rather than epsilon $epsilon$ for set membership), but some of it confuses me. For example, $mathbb A^2i(mathbb E)$ is the sheaf of smooth, $mathbb E$-valued, $2i$ … what? Forms?
    $endgroup$
    – LSpice
    Apr 20 at 18:31










  • $begingroup$
    LSpice , thanks for the corrections in the latex font . And Yes , those are $2i$ forms which I have just edited . Sorry for creating confusions for I did not note those mistakes.
    $endgroup$
    – lap top
    Apr 23 at 6:44















7












$begingroup$


Let $M$ be a complex manifold, $R^nabla$ be the curvature operator for connections $nabla$.
Consider a polynomial function $f:operatorname M_n(mathbbC)tomathbbC$. For the gauge group $operatornameGL_n(mathbbC)$,
if $f(A)=f(gAg^-1)$
where $A in operatorname M_n(mathbbC)$ and $g in operatornameGL_n(mathbbC)$, then $f$ is said to be an invariant polynomial function.
Let $I^k(operatorname M_n(mathbbC))$ be the set of all such polynomials of degree $k$.
Also, $bigoplus_kgeq0I^k(operatorname M_n(mathbbC))=I(operatorname M_n(mathbbC))$.



We shall need $phi_n(A)=det(A)$, $n>1$, and $phi_1(A)=Tr(A)$.



Define a global differential form ($2k$-form)
$f(R^nabla) in mathbbA^2k(M)$.
If we have the de Rham cohomology group $H^2k(M)$, then the Weil homomorphism is defined as the map
$omega:I(operatorname M_n(mathbbC))to bigoplus_kgeq 0H^2k(M)$.



The Chern forms $c_i(R^nabla)=phi_i(fracsqrt-12piR^nabla)$.



For the complex vector bundle $(mathbbE,pi,M)$, where $mathbbE$ is the total space, the Chern classes are defined as $c_i(mathbbE) in H^2k(M)$.



Therefore $c_i(mathbbE)mathrel:=omega(c_i(R^nabla))=[c_i(R^nabla)]$ (de Rham cohomology class).



The Chern forms $c_i(R^nabla) in mathbbA^2i(mathbbE)$.



$mathbbA^2i(mathbbE)$ is sheaf of smooth $mathbbE$-valued $2i$ forms on $M$.



Cohomology groups are very important in geometry for understanding the invariants that can be defined on manifolds. That is, the transformations that keep some special properties of the manifold and which analogous to the gauge transformations in physics.



Chern classes are special type of cohomology classes. If the first Chern class vanishes for a particular manifold, then it must be a Ricci-flat manifold. For example the Calabi–Yau manifolds (they have lots of other special properties, e.g., trivial canonical bundle, etc.).



But what do the higher Chern classes mean?
What uses are those higher cohomology classes corresponding to the higher Chern classes of?










share|cite|improve this question











$endgroup$











  • $begingroup$
    One answer is that if $mathbb E$ is a trivial vector bundle, then all of its Chern classes vanish. Therefore higher Chern classes are obstructions to a bundle being trivial.
    $endgroup$
    – Arun Debray
    Apr 20 at 13:21










  • $begingroup$
    I have edited to fix TeX (for example, use in $in$ rather than epsilon $epsilon$ for set membership), but some of it confuses me. For example, $mathbb A^2i(mathbb E)$ is the sheaf of smooth, $mathbb E$-valued, $2i$ … what? Forms?
    $endgroup$
    – LSpice
    Apr 20 at 18:31










  • $begingroup$
    LSpice , thanks for the corrections in the latex font . And Yes , those are $2i$ forms which I have just edited . Sorry for creating confusions for I did not note those mistakes.
    $endgroup$
    – lap top
    Apr 23 at 6:44













7












7








7


2



$begingroup$


Let $M$ be a complex manifold, $R^nabla$ be the curvature operator for connections $nabla$.
Consider a polynomial function $f:operatorname M_n(mathbbC)tomathbbC$. For the gauge group $operatornameGL_n(mathbbC)$,
if $f(A)=f(gAg^-1)$
where $A in operatorname M_n(mathbbC)$ and $g in operatornameGL_n(mathbbC)$, then $f$ is said to be an invariant polynomial function.
Let $I^k(operatorname M_n(mathbbC))$ be the set of all such polynomials of degree $k$.
Also, $bigoplus_kgeq0I^k(operatorname M_n(mathbbC))=I(operatorname M_n(mathbbC))$.



We shall need $phi_n(A)=det(A)$, $n>1$, and $phi_1(A)=Tr(A)$.



Define a global differential form ($2k$-form)
$f(R^nabla) in mathbbA^2k(M)$.
If we have the de Rham cohomology group $H^2k(M)$, then the Weil homomorphism is defined as the map
$omega:I(operatorname M_n(mathbbC))to bigoplus_kgeq 0H^2k(M)$.



The Chern forms $c_i(R^nabla)=phi_i(fracsqrt-12piR^nabla)$.



For the complex vector bundle $(mathbbE,pi,M)$, where $mathbbE$ is the total space, the Chern classes are defined as $c_i(mathbbE) in H^2k(M)$.



Therefore $c_i(mathbbE)mathrel:=omega(c_i(R^nabla))=[c_i(R^nabla)]$ (de Rham cohomology class).



The Chern forms $c_i(R^nabla) in mathbbA^2i(mathbbE)$.



$mathbbA^2i(mathbbE)$ is sheaf of smooth $mathbbE$-valued $2i$ forms on $M$.



Cohomology groups are very important in geometry for understanding the invariants that can be defined on manifolds. That is, the transformations that keep some special properties of the manifold and which analogous to the gauge transformations in physics.



Chern classes are special type of cohomology classes. If the first Chern class vanishes for a particular manifold, then it must be a Ricci-flat manifold. For example the Calabi–Yau manifolds (they have lots of other special properties, e.g., trivial canonical bundle, etc.).



But what do the higher Chern classes mean?
What uses are those higher cohomology classes corresponding to the higher Chern classes of?










share|cite|improve this question











$endgroup$




Let $M$ be a complex manifold, $R^nabla$ be the curvature operator for connections $nabla$.
Consider a polynomial function $f:operatorname M_n(mathbbC)tomathbbC$. For the gauge group $operatornameGL_n(mathbbC)$,
if $f(A)=f(gAg^-1)$
where $A in operatorname M_n(mathbbC)$ and $g in operatornameGL_n(mathbbC)$, then $f$ is said to be an invariant polynomial function.
Let $I^k(operatorname M_n(mathbbC))$ be the set of all such polynomials of degree $k$.
Also, $bigoplus_kgeq0I^k(operatorname M_n(mathbbC))=I(operatorname M_n(mathbbC))$.



We shall need $phi_n(A)=det(A)$, $n>1$, and $phi_1(A)=Tr(A)$.



Define a global differential form ($2k$-form)
$f(R^nabla) in mathbbA^2k(M)$.
If we have the de Rham cohomology group $H^2k(M)$, then the Weil homomorphism is defined as the map
$omega:I(operatorname M_n(mathbbC))to bigoplus_kgeq 0H^2k(M)$.



The Chern forms $c_i(R^nabla)=phi_i(fracsqrt-12piR^nabla)$.



For the complex vector bundle $(mathbbE,pi,M)$, where $mathbbE$ is the total space, the Chern classes are defined as $c_i(mathbbE) in H^2k(M)$.



Therefore $c_i(mathbbE)mathrel:=omega(c_i(R^nabla))=[c_i(R^nabla)]$ (de Rham cohomology class).



The Chern forms $c_i(R^nabla) in mathbbA^2i(mathbbE)$.



$mathbbA^2i(mathbbE)$ is sheaf of smooth $mathbbE$-valued $2i$ forms on $M$.



Cohomology groups are very important in geometry for understanding the invariants that can be defined on manifolds. That is, the transformations that keep some special properties of the manifold and which analogous to the gauge transformations in physics.



Chern classes are special type of cohomology classes. If the first Chern class vanishes for a particular manifold, then it must be a Ricci-flat manifold. For example the Calabi–Yau manifolds (they have lots of other special properties, e.g., trivial canonical bundle, etc.).



But what do the higher Chern classes mean?
What uses are those higher cohomology classes corresponding to the higher Chern classes of?







complex-geometry cohomology vector-bundles calabi-yau chern-classes






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 23 at 6:41







lap top

















asked Apr 20 at 9:16









lap toplap top

643




643











  • $begingroup$
    One answer is that if $mathbb E$ is a trivial vector bundle, then all of its Chern classes vanish. Therefore higher Chern classes are obstructions to a bundle being trivial.
    $endgroup$
    – Arun Debray
    Apr 20 at 13:21










  • $begingroup$
    I have edited to fix TeX (for example, use in $in$ rather than epsilon $epsilon$ for set membership), but some of it confuses me. For example, $mathbb A^2i(mathbb E)$ is the sheaf of smooth, $mathbb E$-valued, $2i$ … what? Forms?
    $endgroup$
    – LSpice
    Apr 20 at 18:31










  • $begingroup$
    LSpice , thanks for the corrections in the latex font . And Yes , those are $2i$ forms which I have just edited . Sorry for creating confusions for I did not note those mistakes.
    $endgroup$
    – lap top
    Apr 23 at 6:44
















  • $begingroup$
    One answer is that if $mathbb E$ is a trivial vector bundle, then all of its Chern classes vanish. Therefore higher Chern classes are obstructions to a bundle being trivial.
    $endgroup$
    – Arun Debray
    Apr 20 at 13:21










  • $begingroup$
    I have edited to fix TeX (for example, use in $in$ rather than epsilon $epsilon$ for set membership), but some of it confuses me. For example, $mathbb A^2i(mathbb E)$ is the sheaf of smooth, $mathbb E$-valued, $2i$ … what? Forms?
    $endgroup$
    – LSpice
    Apr 20 at 18:31










  • $begingroup$
    LSpice , thanks for the corrections in the latex font . And Yes , those are $2i$ forms which I have just edited . Sorry for creating confusions for I did not note those mistakes.
    $endgroup$
    – lap top
    Apr 23 at 6:44















$begingroup$
One answer is that if $mathbb E$ is a trivial vector bundle, then all of its Chern classes vanish. Therefore higher Chern classes are obstructions to a bundle being trivial.
$endgroup$
– Arun Debray
Apr 20 at 13:21




$begingroup$
One answer is that if $mathbb E$ is a trivial vector bundle, then all of its Chern classes vanish. Therefore higher Chern classes are obstructions to a bundle being trivial.
$endgroup$
– Arun Debray
Apr 20 at 13:21












$begingroup$
I have edited to fix TeX (for example, use in $in$ rather than epsilon $epsilon$ for set membership), but some of it confuses me. For example, $mathbb A^2i(mathbb E)$ is the sheaf of smooth, $mathbb E$-valued, $2i$ … what? Forms?
$endgroup$
– LSpice
Apr 20 at 18:31




$begingroup$
I have edited to fix TeX (for example, use in $in$ rather than epsilon $epsilon$ for set membership), but some of it confuses me. For example, $mathbb A^2i(mathbb E)$ is the sheaf of smooth, $mathbb E$-valued, $2i$ … what? Forms?
$endgroup$
– LSpice
Apr 20 at 18:31












$begingroup$
LSpice , thanks for the corrections in the latex font . And Yes , those are $2i$ forms which I have just edited . Sorry for creating confusions for I did not note those mistakes.
$endgroup$
– lap top
Apr 23 at 6:44




$begingroup$
LSpice , thanks for the corrections in the latex font . And Yes , those are $2i$ forms which I have just edited . Sorry for creating confusions for I did not note those mistakes.
$endgroup$
– lap top
Apr 23 at 6:44










1 Answer
1






active

oldest

votes


















6












$begingroup$

This is a big topic, which should be covered in the union of many standard texts (Chern, Griffiths-Harris, Milnor-Stasheff...). I'll list a few answers off the top of my head.



  1. Suppose that $L$ is the tautological line bundle on the complex projective space plane, then it's clearly not trivial, and neither is $V= Loplus L^-1$. But $c_1(V)=0$, because the trace of curvature is zero for an induced connection. Of course, $V$ is certainly not trivial. So it's natural to look for higher cohomological obstructions (as Arun suggested), e.g. $c_2(V) = -c_1(L)^2not=0$ would work.

  2. For universal bundles on Grassmanians, Chern classes have natural geometric interpretations involving Schubert cycles.

  3. Chern classes come up in formulas expressing answers to natural geometric questions: Gauss-Bonnet, Riemann-Roch, or more general index theorems.





share|cite|improve this answer











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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    This is a big topic, which should be covered in the union of many standard texts (Chern, Griffiths-Harris, Milnor-Stasheff...). I'll list a few answers off the top of my head.



    1. Suppose that $L$ is the tautological line bundle on the complex projective space plane, then it's clearly not trivial, and neither is $V= Loplus L^-1$. But $c_1(V)=0$, because the trace of curvature is zero for an induced connection. Of course, $V$ is certainly not trivial. So it's natural to look for higher cohomological obstructions (as Arun suggested), e.g. $c_2(V) = -c_1(L)^2not=0$ would work.

    2. For universal bundles on Grassmanians, Chern classes have natural geometric interpretations involving Schubert cycles.

    3. Chern classes come up in formulas expressing answers to natural geometric questions: Gauss-Bonnet, Riemann-Roch, or more general index theorems.





    share|cite|improve this answer











    $endgroup$

















      6












      $begingroup$

      This is a big topic, which should be covered in the union of many standard texts (Chern, Griffiths-Harris, Milnor-Stasheff...). I'll list a few answers off the top of my head.



      1. Suppose that $L$ is the tautological line bundle on the complex projective space plane, then it's clearly not trivial, and neither is $V= Loplus L^-1$. But $c_1(V)=0$, because the trace of curvature is zero for an induced connection. Of course, $V$ is certainly not trivial. So it's natural to look for higher cohomological obstructions (as Arun suggested), e.g. $c_2(V) = -c_1(L)^2not=0$ would work.

      2. For universal bundles on Grassmanians, Chern classes have natural geometric interpretations involving Schubert cycles.

      3. Chern classes come up in formulas expressing answers to natural geometric questions: Gauss-Bonnet, Riemann-Roch, or more general index theorems.





      share|cite|improve this answer











      $endgroup$















        6












        6








        6





        $begingroup$

        This is a big topic, which should be covered in the union of many standard texts (Chern, Griffiths-Harris, Milnor-Stasheff...). I'll list a few answers off the top of my head.



        1. Suppose that $L$ is the tautological line bundle on the complex projective space plane, then it's clearly not trivial, and neither is $V= Loplus L^-1$. But $c_1(V)=0$, because the trace of curvature is zero for an induced connection. Of course, $V$ is certainly not trivial. So it's natural to look for higher cohomological obstructions (as Arun suggested), e.g. $c_2(V) = -c_1(L)^2not=0$ would work.

        2. For universal bundles on Grassmanians, Chern classes have natural geometric interpretations involving Schubert cycles.

        3. Chern classes come up in formulas expressing answers to natural geometric questions: Gauss-Bonnet, Riemann-Roch, or more general index theorems.





        share|cite|improve this answer











        $endgroup$



        This is a big topic, which should be covered in the union of many standard texts (Chern, Griffiths-Harris, Milnor-Stasheff...). I'll list a few answers off the top of my head.



        1. Suppose that $L$ is the tautological line bundle on the complex projective space plane, then it's clearly not trivial, and neither is $V= Loplus L^-1$. But $c_1(V)=0$, because the trace of curvature is zero for an induced connection. Of course, $V$ is certainly not trivial. So it's natural to look for higher cohomological obstructions (as Arun suggested), e.g. $c_2(V) = -c_1(L)^2not=0$ would work.

        2. For universal bundles on Grassmanians, Chern classes have natural geometric interpretations involving Schubert cycles.

        3. Chern classes come up in formulas expressing answers to natural geometric questions: Gauss-Bonnet, Riemann-Roch, or more general index theorems.






        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 20 at 16:53

























        answered Apr 20 at 14:08









        Donu ArapuraDonu Arapura

        25.5k267126




        25.5k267126



























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