Mixing
Role of Dirac operators in Index Theorems
$begingroup$
I'm trying to approach the Atiyah-Singer Index Theorem by getting an overview of the area.
One thing that confuses me a lot is that some treatments give (and hence prove) the theorem for Dirac operators, while other sources not even mention Dirac operators and work just with general elliptic operators.
From what I read seems that the Index theorem for Dirac operators implies the theorem for general elliptic operators. Indeed the Dirac operator is some sense "the" elliptic operator. I've heard this claim can be justified with some K-theory (which I know nothing about at the moment).
Can someone explain in simple (i.e. vague) terms this idea? Why the Dirac operator is "the" elliptic operator?
Apart from the K-theoretical justification, are there any methods of showing this? Can this be seen from the Analysis side?
Thanks in advance
differential-geometry algebraic-topology differential-topology topological-k-theory elliptic-operators
asked Jan 11 at 23:27
EmarJEmarJ
411214
$endgroup$
add a comment |
$begingroup$
I'm trying to approach the Atiyah-Singer Index Theorem by getting an overview of the area.
One thing that confuses me a lot is that some treatments give (and hence prove) the theorem for Dirac operators, while other sources not even mention Dirac operators and work just with general elliptic operators.
From what I read seems that the Index theorem for Dirac operators implies the theorem for general elliptic operators. Indeed the Dirac operator is some sense "the" elliptic operator. I've heard this claim can be justified with some K-theory (which I know nothing about at the moment).
Can someone explain in simple (i.e. vague) terms this idea? Why the Dirac operator is "the" elliptic operator?
Apart from the K-theoretical justification, are there any methods of showing this? Can this be seen from the Analysis side?
Thanks in advance
differential-geometry algebraic-topology differential-topology topological-k-theory elliptic-operators
asked Jan 11 at 23:27
EmarJEmarJ
411214
$endgroup$
add a comment |
$begingroup$
I'm trying to approach the Atiyah-Singer Index Theorem by getting an overview of the area.
One thing that confuses me a lot is that some treatments give (and hence prove) the theorem for Dirac operators, while other sources not even mention Dirac operators and work just with general elliptic operators.
From what I read seems that the Index theorem for Dirac operators implies the theorem for general elliptic operators. Indeed the Dirac operator is some sense "the" elliptic operator. I've heard this claim can be justified with some K-theory (which I know nothing about at the moment).
Can someone explain in simple (i.e. vague) terms this idea? Why the Dirac operator is "the" elliptic operator?
Apart from the K-theoretical justification, are there any methods of showing this? Can this be seen from the Analysis side?
Thanks in advance
differential-geometry algebraic-topology differential-topology topological-k-theory elliptic-operators
asked Jan 11 at 23:27
EmarJEmarJ
411214
$endgroup$
I'm trying to approach the Atiyah-Singer Index Theorem by getting an overview of the area.
One thing that confuses me a lot is that some treatments give (and hence prove) the theorem for Dirac operators, while other sources not even mention Dirac operators and work just with general elliptic operators.
From what I read seems that the Index theorem for Dirac operators implies the theorem for general elliptic operators. Indeed the Dirac operator is some sense "the" elliptic operator. I've heard this claim can be justified with some K-theory (which I know nothing about at the moment).
Can someone explain in simple (i.e. vague) terms this idea? Why the Dirac operator is "the" elliptic operator?
Apart from the K-theoretical justification, are there any methods of showing this? Can this be seen from the Analysis side?
Thanks in advance
differential-geometry algebraic-topology differential-topology topological-k-theory elliptic-operators
differential-geometry algebraic-topology differential-topology topological-k-theory elliptic-operators
asked Jan 11 at 23:27
EmarJEmarJ
411214
asked Jan 11 at 23:27
EmarJEmarJ
411214
asked Jan 11 at 23:27
EmarJEmarJ
411214
asked Jan 11 at 23:27
EmarJEmarJ
411214
asked Jan 11 at 23:27
EmarJEmarJ
411214
411214
add a comment |
add a comment |
1 Answer
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$begingroup$
This is an intricate question. First of all, it is not "the Dirac operator" but a family of Dirac operators, for which there is something usually called the "local version of the index theorem".
A basic fact related to the index theorem is that the index of an elliptic operator is a rather robust quantity. In particular, the index depends only on the principal symbol of the operator, which is the first step towards encoding the initial operator into a topological object (related to K-theory) from which the index can be computed. So you can initially say which part of the symbol you have to know in order to determine the index. In particular, you can realize the same principal symbol as the symbol of many different operators and also of operators of different order. Passing to pseudodifferential operators, you can use operators of order zero, which is very convenient from the point of view of analsis.
What this boils down to is that each symbol determines a class in a certain abelian group, which is defined topologically (and the whole group is obtained in that way). For all symbols in a class, the corresponding operators have the same index, so this defines a homomorphism from that group to $mathbb Z$ (usually called the "analytical index"). Using pure topology (characteristic classes, Chern character,etc.), one defines another homomorphism (calles the "topological index") between the same groups and the index theorem boils down to showing that these two homomorphisms agree.
But to prove this, it suffices to check that the two homomorphisms agree on a set of generators of this group. (And also the classical proof of the index theorem proceeds by using invariance properties in order to show that computing the index for a certain set of operators suffices to establish the general theorem.) Now Dirac type operators form a family for which the index can ba computed using analysis and geometry. Here one starts with a Riemannian spin manifold and a bundle induced by a representation of the Clifford algebra. A Dirac operator on this is a first order operator whose symbol equals Clifford multiplication (which is derived from the representation of the Clifford algebra). The square of such an operator is a Laplace-type operator and one uses its heat kernel to compute the index of the operator in the form of an integral over local invariant tensors associated to the Riemann manifold (so these are derived from the Riemann curvature). Via Chern-Weil theory, such integrals can be related to characteristic classes, which allows one to prove that the analytical and topological indices agree on the symbols of Dirac-type operators. As a final step you then show that the abelian group discussed above is generated by the symbols of Dirac type operators, which closes the circle.
I don't really think that you can avoid topology in the proof of the fact that it suffices to understand the index of Dirac operators in order to verify the general index theorem. On the other hand, if you are more analytically oriented, this is a fact that you may simply take as being given at a first go.
answered Jan 12 at 12:32
Andreas CapAndreas Cap
11.2k923
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is an intricate question. First of all, it is not "the Dirac operator" but a family of Dirac operators, for which there is something usually called the "local version of the index theorem".
A basic fact related to the index theorem is that the index of an elliptic operator is a rather robust quantity. In particular, the index depends only on the principal symbol of the operator, which is the first step towards encoding the initial operator into a topological object (related to K-theory) from which the index can be computed. So you can initially say which part of the symbol you have to know in order to determine the index. In particular, you can realize the same principal symbol as the symbol of many different operators and also of operators of different order. Passing to pseudodifferential operators, you can use operators of order zero, which is very convenient from the point of view of analsis.
What this boils down to is that each symbol determines a class in a certain abelian group, which is defined topologically (and the whole group is obtained in that way). For all symbols in a class, the corresponding operators have the same index, so this defines a homomorphism from that group to $mathbb Z$ (usually called the "analytical index"). Using pure topology (characteristic classes, Chern character,etc.), one defines another homomorphism (calles the "topological index") between the same groups and the index theorem boils down to showing that these two homomorphisms agree.
But to prove this, it suffices to check that the two homomorphisms agree on a set of generators of this group. (And also the classical proof of the index theorem proceeds by using invariance properties in order to show that computing the index for a certain set of operators suffices to establish the general theorem.) Now Dirac type operators form a family for which the index can ba computed using analysis and geometry. Here one starts with a Riemannian spin manifold and a bundle induced by a representation of the Clifford algebra. A Dirac operator on this is a first order operator whose symbol equals Clifford multiplication (which is derived from the representation of the Clifford algebra). The square of such an operator is a Laplace-type operator and one uses its heat kernel to compute the index of the operator in the form of an integral over local invariant tensors associated to the Riemann manifold (so these are derived from the Riemann curvature). Via Chern-Weil theory, such integrals can be related to characteristic classes, which allows one to prove that the analytical and topological indices agree on the symbols of Dirac-type operators. As a final step you then show that the abelian group discussed above is generated by the symbols of Dirac type operators, which closes the circle.
I don't really think that you can avoid topology in the proof of the fact that it suffices to understand the index of Dirac operators in order to verify the general index theorem. On the other hand, if you are more analytically oriented, this is a fact that you may simply take as being given at a first go.
answered Jan 12 at 12:32
Andreas CapAndreas Cap
11.2k923
$endgroup$
add a comment |
$begingroup$
This is an intricate question. First of all, it is not "the Dirac operator" but a family of Dirac operators, for which there is something usually called the "local version of the index theorem".
A basic fact related to the index theorem is that the index of an elliptic operator is a rather robust quantity. In particular, the index depends only on the principal symbol of the operator, which is the first step towards encoding the initial operator into a topological object (related to K-theory) from which the index can be computed. So you can initially say which part of the symbol you have to know in order to determine the index. In particular, you can realize the same principal symbol as the symbol of many different operators and also of operators of different order. Passing to pseudodifferential operators, you can use operators of order zero, which is very convenient from the point of view of analsis.
What this boils down to is that each symbol determines a class in a certain abelian group, which is defined topologically (and the whole group is obtained in that way). For all symbols in a class, the corresponding operators have the same index, so this defines a homomorphism from that group to $mathbb Z$ (usually called the "analytical index"). Using pure topology (characteristic classes, Chern character,etc.), one defines another homomorphism (calles the "topological index") between the same groups and the index theorem boils down to showing that these two homomorphisms agree.
But to prove this, it suffices to check that the two homomorphisms agree on a set of generators of this group. (And also the classical proof of the index theorem proceeds by using invariance properties in order to show that computing the index for a certain set of operators suffices to establish the general theorem.) Now Dirac type operators form a family for which the index can ba computed using analysis and geometry. Here one starts with a Riemannian spin manifold and a bundle induced by a representation of the Clifford algebra. A Dirac operator on this is a first order operator whose symbol equals Clifford multiplication (which is derived from the representation of the Clifford algebra). The square of such an operator is a Laplace-type operator and one uses its heat kernel to compute the index of the operator in the form of an integral over local invariant tensors associated to the Riemann manifold (so these are derived from the Riemann curvature). Via Chern-Weil theory, such integrals can be related to characteristic classes, which allows one to prove that the analytical and topological indices agree on the symbols of Dirac-type operators. As a final step you then show that the abelian group discussed above is generated by the symbols of Dirac type operators, which closes the circle.
I don't really think that you can avoid topology in the proof of the fact that it suffices to understand the index of Dirac operators in order to verify the general index theorem. On the other hand, if you are more analytically oriented, this is a fact that you may simply take as being given at a first go.
answered Jan 12 at 12:32
Andreas CapAndreas Cap
11.2k923
$endgroup$
add a comment |
$begingroup$
This is an intricate question. First of all, it is not "the Dirac operator" but a family of Dirac operators, for which there is something usually called the "local version of the index theorem".
A basic fact related to the index theorem is that the index of an elliptic operator is a rather robust quantity. In particular, the index depends only on the principal symbol of the operator, which is the first step towards encoding the initial operator into a topological object (related to K-theory) from which the index can be computed. So you can initially say which part of the symbol you have to know in order to determine the index. In particular, you can realize the same principal symbol as the symbol of many different operators and also of operators of different order. Passing to pseudodifferential operators, you can use operators of order zero, which is very convenient from the point of view of analsis.
What this boils down to is that each symbol determines a class in a certain abelian group, which is defined topologically (and the whole group is obtained in that way). For all symbols in a class, the corresponding operators have the same index, so this defines a homomorphism from that group to $mathbb Z$ (usually called the "analytical index"). Using pure topology (characteristic classes, Chern character,etc.), one defines another homomorphism (calles the "topological index") between the same groups and the index theorem boils down to showing that these two homomorphisms agree.
But to prove this, it suffices to check that the two homomorphisms agree on a set of generators of this group. (And also the classical proof of the index theorem proceeds by using invariance properties in order to show that computing the index for a certain set of operators suffices to establish the general theorem.) Now Dirac type operators form a family for which the index can ba computed using analysis and geometry. Here one starts with a Riemannian spin manifold and a bundle induced by a representation of the Clifford algebra. A Dirac operator on this is a first order operator whose symbol equals Clifford multiplication (which is derived from the representation of the Clifford algebra). The square of such an operator is a Laplace-type operator and one uses its heat kernel to compute the index of the operator in the form of an integral over local invariant tensors associated to the Riemann manifold (so these are derived from the Riemann curvature). Via Chern-Weil theory, such integrals can be related to characteristic classes, which allows one to prove that the analytical and topological indices agree on the symbols of Dirac-type operators. As a final step you then show that the abelian group discussed above is generated by the symbols of Dirac type operators, which closes the circle.
I don't really think that you can avoid topology in the proof of the fact that it suffices to understand the index of Dirac operators in order to verify the general index theorem. On the other hand, if you are more analytically oriented, this is a fact that you may simply take as being given at a first go.
answered Jan 12 at 12:32
Andreas CapAndreas Cap
11.2k923
$endgroup$
This is an intricate question. First of all, it is not "the Dirac operator" but a family of Dirac operators, for which there is something usually called the "local version of the index theorem".
A basic fact related to the index theorem is that the index of an elliptic operator is a rather robust quantity. In particular, the index depends only on the principal symbol of the operator, which is the first step towards encoding the initial operator into a topological object (related to K-theory) from which the index can be computed. So you can initially say which part of the symbol you have to know in order to determine the index. In particular, you can realize the same principal symbol as the symbol of many different operators and also of operators of different order. Passing to pseudodifferential operators, you can use operators of order zero, which is very convenient from the point of view of analsis.
What this boils down to is that each symbol determines a class in a certain abelian group, which is defined topologically (and the whole group is obtained in that way). For all symbols in a class, the corresponding operators have the same index, so this defines a homomorphism from that group to $mathbb Z$ (usually called the "analytical index"). Using pure topology (characteristic classes, Chern character,etc.), one defines another homomorphism (calles the "topological index") between the same groups and the index theorem boils down to showing that these two homomorphisms agree.
But to prove this, it suffices to check that the two homomorphisms agree on a set of generators of this group. (And also the classical proof of the index theorem proceeds by using invariance properties in order to show that computing the index for a certain set of operators suffices to establish the general theorem.) Now Dirac type operators form a family for which the index can ba computed using analysis and geometry. Here one starts with a Riemannian spin manifold and a bundle induced by a representation of the Clifford algebra. A Dirac operator on this is a first order operator whose symbol equals Clifford multiplication (which is derived from the representation of the Clifford algebra). The square of such an operator is a Laplace-type operator and one uses its heat kernel to compute the index of the operator in the form of an integral over local invariant tensors associated to the Riemann manifold (so these are derived from the Riemann curvature). Via Chern-Weil theory, such integrals can be related to characteristic classes, which allows one to prove that the analytical and topological indices agree on the symbols of Dirac-type operators. As a final step you then show that the abelian group discussed above is generated by the symbols of Dirac type operators, which closes the circle.
I don't really think that you can avoid topology in the proof of the fact that it suffices to understand the index of Dirac operators in order to verify the general index theorem. On the other hand, if you are more analytically oriented, this is a fact that you may simply take as being given at a first go.
answered Jan 12 at 12:32
Andreas CapAndreas Cap
11.2k923
answered Jan 12 at 12:32
Andreas CapAndreas Cap
11.2k923
answered Jan 12 at 12:32
Andreas CapAndreas Cap
11.2k923
answered Jan 12 at 12:32
Andreas CapAndreas Cap
11.2k923
11.2k923
add a comment |
add a comment |
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