Existence of Thom Class
$begingroup$
In page 133, Theorem 8.5.5.
(The Thom isomoprhism theorem) Let $pi:V rightarrow X$ be a complex vector bundle of rank $n$, over al locally comapct space $X$. Let
$$ 0 rightarrow pi^* wedge ^0 V rightarrow pi^* wedge ^1 V rightarrow cdots rightarrow pi^* wedge ^n V rightarrow0 $$
we the chain complex, the map $pi^* wedge ^pV rightarrow pi^* wedge ^{p+1} V$ is given over $v in V$ by taking the exterior product with $v$. The Thom class $t_V in K_c(V)$ is the complex conjugate of the wrap of this complex. Then the map
$$ th^K:K_c(X) rightarrow K_c(V), : , x mapsto pi^*x # t_V $$
is an isomoprhism.
The remark says:
We do not need to know the Thom isomoprhism theorem, but only the existence of the Thom Class.
I'm confused with this statement. Is there something needed to prove for the existence?
I think what we have to prove is that $t_V$ is indeed an element of $K_c(V)$? How is this justified?
algebraic-topology vector-bundles k-theory topological-k-theory
$endgroup$
add a comment |
$begingroup$
In page 133, Theorem 8.5.5.
(The Thom isomoprhism theorem) Let $pi:V rightarrow X$ be a complex vector bundle of rank $n$, over al locally comapct space $X$. Let
$$ 0 rightarrow pi^* wedge ^0 V rightarrow pi^* wedge ^1 V rightarrow cdots rightarrow pi^* wedge ^n V rightarrow0 $$
we the chain complex, the map $pi^* wedge ^pV rightarrow pi^* wedge ^{p+1} V$ is given over $v in V$ by taking the exterior product with $v$. The Thom class $t_V in K_c(V)$ is the complex conjugate of the wrap of this complex. Then the map
$$ th^K:K_c(X) rightarrow K_c(V), : , x mapsto pi^*x # t_V $$
is an isomoprhism.
The remark says:
We do not need to know the Thom isomoprhism theorem, but only the existence of the Thom Class.
I'm confused with this statement. Is there something needed to prove for the existence?
I think what we have to prove is that $t_V$ is indeed an element of $K_c(V)$? How is this justified?
algebraic-topology vector-bundles k-theory topological-k-theory
$endgroup$
add a comment |
$begingroup$
In page 133, Theorem 8.5.5.
(The Thom isomoprhism theorem) Let $pi:V rightarrow X$ be a complex vector bundle of rank $n$, over al locally comapct space $X$. Let
$$ 0 rightarrow pi^* wedge ^0 V rightarrow pi^* wedge ^1 V rightarrow cdots rightarrow pi^* wedge ^n V rightarrow0 $$
we the chain complex, the map $pi^* wedge ^pV rightarrow pi^* wedge ^{p+1} V$ is given over $v in V$ by taking the exterior product with $v$. The Thom class $t_V in K_c(V)$ is the complex conjugate of the wrap of this complex. Then the map
$$ th^K:K_c(X) rightarrow K_c(V), : , x mapsto pi^*x # t_V $$
is an isomoprhism.
The remark says:
We do not need to know the Thom isomoprhism theorem, but only the existence of the Thom Class.
I'm confused with this statement. Is there something needed to prove for the existence?
I think what we have to prove is that $t_V$ is indeed an element of $K_c(V)$? How is this justified?
algebraic-topology vector-bundles k-theory topological-k-theory
$endgroup$
In page 133, Theorem 8.5.5.
(The Thom isomoprhism theorem) Let $pi:V rightarrow X$ be a complex vector bundle of rank $n$, over al locally comapct space $X$. Let
$$ 0 rightarrow pi^* wedge ^0 V rightarrow pi^* wedge ^1 V rightarrow cdots rightarrow pi^* wedge ^n V rightarrow0 $$
we the chain complex, the map $pi^* wedge ^pV rightarrow pi^* wedge ^{p+1} V$ is given over $v in V$ by taking the exterior product with $v$. The Thom class $t_V in K_c(V)$ is the complex conjugate of the wrap of this complex. Then the map
$$ th^K:K_c(X) rightarrow K_c(V), : , x mapsto pi^*x # t_V $$
is an isomoprhism.
The remark says:
We do not need to know the Thom isomoprhism theorem, but only the existence of the Thom Class.
I'm confused with this statement. Is there something needed to prove for the existence?
I think what we have to prove is that $t_V$ is indeed an element of $K_c(V)$? How is this justified?
algebraic-topology vector-bundles k-theory topological-k-theory
algebraic-topology vector-bundles k-theory topological-k-theory
edited Dec 19 '18 at 15:35
CL.
asked Dec 10 '18 at 1:56
CL.CL.
2,2482825
2,2482825
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