Relative $K$-theory, definition
On pg125 on proving that that the concordance classes quotiented by the acyclic ones froms a group, there is a lemma:
Lemma 8.4.5 Let $(E,F,f)$ and $(E,F,g)$ be two $K$-cycles on $(X,Y)$. Assume that
$$f^*g +g^*f ge 0 text{ or } fg^*+gf^* ge 0$$
holds over $Y$. Then these two cycles are homotopic.
I do not understand this statement, where does the equality come from? At best I know locally we can regard the bundle morphism $f$ as a map $f:U rightarrow GL_n(Bbb C)$.
What is $f^*$? Is it the conjugate?
vector-bundles k-theory topological-k-theory
asked Dec 9 at 11:53
CL.
2,1272822
add a comment |
On pg125 on proving that that the concordance classes quotiented by the acyclic ones froms a group, there is a lemma:
Lemma 8.4.5 Let $(E,F,f)$ and $(E,F,g)$ be two $K$-cycles on $(X,Y)$. Assume that
$$f^*g +g^*f ge 0 text{ or } fg^*+gf^* ge 0$$
holds over $Y$. Then these two cycles are homotopic.
I do not understand this statement, where does the equality come from? At best I know locally we can regard the bundle morphism $f$ as a map $f:U rightarrow GL_n(Bbb C)$.
What is $f^*$? Is it the conjugate?
vector-bundles k-theory topological-k-theory
asked Dec 9 at 11:53
CL.
2,1272822
The only reasonable explanation is that $f^*$ is the adjoint bundle morphism (see e.g. en.wikipedia.org/wiki/Hermitian_adjoint). This is defined fiberwise by taking adjoints of linear maps (which requires the we have inner products on the vector bundles). Then $ge 0$ means positive semidefinite.
– Paul Frost
Dec 10 at 10:54
add a comment |
On pg125 on proving that that the concordance classes quotiented by the acyclic ones froms a group, there is a lemma:
Lemma 8.4.5 Let $(E,F,f)$ and $(E,F,g)$ be two $K$-cycles on $(X,Y)$. Assume that
$$f^*g +g^*f ge 0 text{ or } fg^*+gf^* ge 0$$
holds over $Y$. Then these two cycles are homotopic.
I do not understand this statement, where does the equality come from? At best I know locally we can regard the bundle morphism $f$ as a map $f:U rightarrow GL_n(Bbb C)$.
What is $f^*$? Is it the conjugate?
vector-bundles k-theory topological-k-theory
asked Dec 9 at 11:53
CL.
2,1272822
On pg125 on proving that that the concordance classes quotiented by the acyclic ones froms a group, there is a lemma:
Lemma 8.4.5 Let $(E,F,f)$ and $(E,F,g)$ be two $K$-cycles on $(X,Y)$. Assume that
$$f^*g +g^*f ge 0 text{ or } fg^*+gf^* ge 0$$
holds over $Y$. Then these two cycles are homotopic.
I do not understand this statement, where does the equality come from? At best I know locally we can regard the bundle morphism $f$ as a map $f:U rightarrow GL_n(Bbb C)$.
What is $f^*$? Is it the conjugate?
vector-bundles k-theory topological-k-theory
vector-bundles k-theory topological-k-theory
asked Dec 9 at 11:53
CL.
2,1272822
asked Dec 9 at 11:53
CL.
2,1272822
edited Dec 9 at 15:46
asked Dec 9 at 11:53
CL.
2,1272822
asked Dec 9 at 11:53
CL.
2,1272822
asked Dec 9 at 11:53
CL.
2,1272822
2,1272822
The only reasonable explanation is that $f^*$ is the adjoint bundle morphism (see e.g. en.wikipedia.org/wiki/Hermitian_adjoint). This is defined fiberwise by taking adjoints of linear maps (which requires the we have inner products on the vector bundles). Then $ge 0$ means positive semidefinite.
– Paul Frost
Dec 10 at 10:54
add a comment |
The only reasonable explanation is that $f^*$ is the adjoint bundle morphism (see e.g. en.wikipedia.org/wiki/Hermitian_adjoint). This is defined fiberwise by taking adjoints of linear maps (which requires the we have inner products on the vector bundles). Then $ge 0$ means positive semidefinite.
– Paul Frost
Dec 10 at 10:54
The only reasonable explanation is that $f^*$ is the adjoint bundle morphism (see e.g. en.wikipedia.org/wiki/Hermitian_adjoint). This is defined fiberwise by taking adjoints of linear maps (which requires the we have inner products on the vector bundles). Then $ge 0$ means positive semidefinite.
– Paul Frost
Dec 10 at 10:54
The only reasonable explanation is that $f^*$ is the adjoint bundle morphism (see e.g. en.wikipedia.org/wiki/Hermitian_adjoint). This is defined fiberwise by taking adjoints of linear maps (which requires the we have inner products on the vector bundles). Then $ge 0$ means positive semidefinite.
– Paul Frost
Dec 10 at 10:54
add a comment |
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The only reasonable explanation is that $f^*$ is the adjoint bundle morphism (see e.g. en.wikipedia.org/wiki/Hermitian_adjoint). This is defined fiberwise by taking adjoints of linear maps (which requires the we have inner products on the vector bundles). Then $ge 0$ means positive semidefinite.
– Paul Frost
Dec 10 at 10:54