Defining Bott Class by relative $K$-theory
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I am really confused with this construction of Bott Class in Page 127, Example 8.4.12
If $V$ is a complex vector space of dimension $n$, we form the complex
$$ 0 rightarrow wedge^0 V rightarrow wedge^1 V rightarrow cdots rightarrow wedge ^nV rightarrow 0 $$
of trivial vector bundles over $V$.
What does this mean? So we have $n$ trivial vector bundles, $V times Bbb C^{binom{n}{k}}$?
The map $wedge^pV rightarrow wedge^{p+1}V$ is given by $w mapsto v wedge w$. Now we pass to complex conjugate bundles. IT is clear that if $f:V rightarrow W$ is a $Bbb C$ linear map then $f:bar{V} rightarrow bar{W}$.
What does $bar{V}$ mean? Where do we use "passing to the complex conjugate"?
Denote the resulting class $b_V in K^0(DV,DS)$.
(i) How is $b_V$ induced from all the things above? Is this by the wrapping defined on pg 126?
(ii) how is $b_V$ dependent on $f$ or $W$.
algebraic-topology differential-topology k-theory topological-k-theory
asked Dec 9 '18 at 21:49
CL.CL.
2,2152825
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add a comment |
$begingroup$
I am really confused with this construction of Bott Class in Page 127, Example 8.4.12
If $V$ is a complex vector space of dimension $n$, we form the complex
$$ 0 rightarrow wedge^0 V rightarrow wedge^1 V rightarrow cdots rightarrow wedge ^nV rightarrow 0 $$
of trivial vector bundles over $V$.
What does this mean? So we have $n$ trivial vector bundles, $V times Bbb C^{binom{n}{k}}$?
The map $wedge^pV rightarrow wedge^{p+1}V$ is given by $w mapsto v wedge w$. Now we pass to complex conjugate bundles. IT is clear that if $f:V rightarrow W$ is a $Bbb C$ linear map then $f:bar{V} rightarrow bar{W}$.
What does $bar{V}$ mean? Where do we use "passing to the complex conjugate"?
Denote the resulting class $b_V in K^0(DV,DS)$.
(i) How is $b_V$ induced from all the things above? Is this by the wrapping defined on pg 126?
(ii) how is $b_V$ dependent on $f$ or $W$.
algebraic-topology differential-topology k-theory topological-k-theory
asked Dec 9 '18 at 21:49
CL.CL.
2,2152825
$endgroup$
$begingroup$
Although I do not see the purpose of the bundle $wedge^i V$, it is obvious that it stands for the product bundle $V times wedge^i V$. The map $wedge^i V to wedge^{i+1} V$ is defined fiberwise by $omega mapsto v wedge omega$ for all $v in V$. Any complex vector space $V$ can be given the complex conjugate structure by defining a new scalar multiplication via $lambda * v = overline{lambda} cdot v$. This vector space is denoted by $overline{V}$. The same construction applies to vector bundles (fiberwise). But it is not clear to me what the "resulting class" $b_V$ should be.
$endgroup$
– Paul Frost
Dec 10 '18 at 10:33
$begingroup$
math.stackexchange.com/questions/642167/… $bar{V}$ is the complex conjugate of a vector space, you can look it up on Wikipedia. Please note that there is a typo: $DS$ should be $SV$. The sequence is exact everywhere outside zero, so gives a class in $(V,Vsmallsetminus {0})cong(DV,SV)$. The passage to conjugate is related to the fact the the Bott generator is defined of index $-1$, so in this picture the section $zmapsto bar{z}$. This answers your (i). Question (ii): $f$ and $W$ were only examples
$endgroup$
– vap
Dec 12 '18 at 18:40
add a comment |
$begingroup$
I am really confused with this construction of Bott Class in Page 127, Example 8.4.12
If $V$ is a complex vector space of dimension $n$, we form the complex
$$ 0 rightarrow wedge^0 V rightarrow wedge^1 V rightarrow cdots rightarrow wedge ^nV rightarrow 0 $$
of trivial vector bundles over $V$.
What does this mean? So we have $n$ trivial vector bundles, $V times Bbb C^{binom{n}{k}}$?
The map $wedge^pV rightarrow wedge^{p+1}V$ is given by $w mapsto v wedge w$. Now we pass to complex conjugate bundles. IT is clear that if $f:V rightarrow W$ is a $Bbb C$ linear map then $f:bar{V} rightarrow bar{W}$.
What does $bar{V}$ mean? Where do we use "passing to the complex conjugate"?
Denote the resulting class $b_V in K^0(DV,DS)$.
(i) How is $b_V$ induced from all the things above? Is this by the wrapping defined on pg 126?
(ii) how is $b_V$ dependent on $f$ or $W$.
algebraic-topology differential-topology k-theory topological-k-theory
asked Dec 9 '18 at 21:49
CL.CL.
2,2152825
$endgroup$
I am really confused with this construction of Bott Class in Page 127, Example 8.4.12
If $V$ is a complex vector space of dimension $n$, we form the complex
$$ 0 rightarrow wedge^0 V rightarrow wedge^1 V rightarrow cdots rightarrow wedge ^nV rightarrow 0 $$
of trivial vector bundles over $V$.
What does this mean? So we have $n$ trivial vector bundles, $V times Bbb C^{binom{n}{k}}$?
The map $wedge^pV rightarrow wedge^{p+1}V$ is given by $w mapsto v wedge w$. Now we pass to complex conjugate bundles. IT is clear that if $f:V rightarrow W$ is a $Bbb C$ linear map then $f:bar{V} rightarrow bar{W}$.
What does $bar{V}$ mean? Where do we use "passing to the complex conjugate"?
Denote the resulting class $b_V in K^0(DV,DS)$.
(i) How is $b_V$ induced from all the things above? Is this by the wrapping defined on pg 126?
(ii) how is $b_V$ dependent on $f$ or $W$.
algebraic-topology differential-topology k-theory topological-k-theory
algebraic-topology differential-topology k-theory topological-k-theory
asked Dec 9 '18 at 21:49
CL.CL.
2,2152825
asked Dec 9 '18 at 21:49
CL.CL.
2,2152825
edited Dec 9 '18 at 22:09
CL.
asked Dec 9 '18 at 21:49
CL.CL.
2,2152825
asked Dec 9 '18 at 21:49
CL.CL.
2,2152825
asked Dec 9 '18 at 21:49
CL.CL.
2,2152825
2,2152825
$begingroup$
Although I do not see the purpose of the bundle $wedge^i V$, it is obvious that it stands for the product bundle $V times wedge^i V$. The map $wedge^i V to wedge^{i+1} V$ is defined fiberwise by $omega mapsto v wedge omega$ for all $v in V$. Any complex vector space $V$ can be given the complex conjugate structure by defining a new scalar multiplication via $lambda * v = overline{lambda} cdot v$. This vector space is denoted by $overline{V}$. The same construction applies to vector bundles (fiberwise). But it is not clear to me what the "resulting class" $b_V$ should be.
$endgroup$
– Paul Frost
Dec 10 '18 at 10:33
$begingroup$
math.stackexchange.com/questions/642167/… $bar{V}$ is the complex conjugate of a vector space, you can look it up on Wikipedia. Please note that there is a typo: $DS$ should be $SV$. The sequence is exact everywhere outside zero, so gives a class in $(V,Vsmallsetminus {0})cong(DV,SV)$. The passage to conjugate is related to the fact the the Bott generator is defined of index $-1$, so in this picture the section $zmapsto bar{z}$. This answers your (i). Question (ii): $f$ and $W$ were only examples
$endgroup$
– vap
Dec 12 '18 at 18:40
add a comment |
$begingroup$
Although I do not see the purpose of the bundle $wedge^i V$, it is obvious that it stands for the product bundle $V times wedge^i V$. The map $wedge^i V to wedge^{i+1} V$ is defined fiberwise by $omega mapsto v wedge omega$ for all $v in V$. Any complex vector space $V$ can be given the complex conjugate structure by defining a new scalar multiplication via $lambda * v = overline{lambda} cdot v$. This vector space is denoted by $overline{V}$. The same construction applies to vector bundles (fiberwise). But it is not clear to me what the "resulting class" $b_V$ should be.
$endgroup$
– Paul Frost
Dec 10 '18 at 10:33
$begingroup$
math.stackexchange.com/questions/642167/… $bar{V}$ is the complex conjugate of a vector space, you can look it up on Wikipedia. Please note that there is a typo: $DS$ should be $SV$. The sequence is exact everywhere outside zero, so gives a class in $(V,Vsmallsetminus {0})cong(DV,SV)$. The passage to conjugate is related to the fact the the Bott generator is defined of index $-1$, so in this picture the section $zmapsto bar{z}$. This answers your (i). Question (ii): $f$ and $W$ were only examples
$endgroup$
– vap
Dec 12 '18 at 18:40
$begingroup$
Although I do not see the purpose of the bundle $wedge^i V$, it is obvious that it stands for the product bundle $V times wedge^i V$. The map $wedge^i V to wedge^{i+1} V$ is defined fiberwise by $omega mapsto v wedge omega$ for all $v in V$. Any complex vector space $V$ can be given the complex conjugate structure by defining a new scalar multiplication via $lambda * v = overline{lambda} cdot v$. This vector space is denoted by $overline{V}$. The same construction applies to vector bundles (fiberwise). But it is not clear to me what the "resulting class" $b_V$ should be.
$endgroup$
– Paul Frost
Dec 10 '18 at 10:33
$begingroup$
Although I do not see the purpose of the bundle $wedge^i V$, it is obvious that it stands for the product bundle $V times wedge^i V$. The map $wedge^i V to wedge^{i+1} V$ is defined fiberwise by $omega mapsto v wedge omega$ for all $v in V$. Any complex vector space $V$ can be given the complex conjugate structure by defining a new scalar multiplication via $lambda * v = overline{lambda} cdot v$. This vector space is denoted by $overline{V}$. The same construction applies to vector bundles (fiberwise). But it is not clear to me what the "resulting class" $b_V$ should be.
$endgroup$
– Paul Frost
Dec 10 '18 at 10:33
$begingroup$
math.stackexchange.com/questions/642167/… $bar{V}$ is the complex conjugate of a vector space, you can look it up on Wikipedia. Please note that there is a typo: $DS$ should be $SV$. The sequence is exact everywhere outside zero, so gives a class in $(V,Vsmallsetminus {0})cong(DV,SV)$. The passage to conjugate is related to the fact the the Bott generator is defined of index $-1$, so in this picture the section $zmapsto bar{z}$. This answers your (i). Question (ii): $f$ and $W$ were only examples
$endgroup$
– vap
Dec 12 '18 at 18:40
$begingroup$
math.stackexchange.com/questions/642167/… $bar{V}$ is the complex conjugate of a vector space, you can look it up on Wikipedia. Please note that there is a typo: $DS$ should be $SV$. The sequence is exact everywhere outside zero, so gives a class in $(V,Vsmallsetminus {0})cong(DV,SV)$. The passage to conjugate is related to the fact the the Bott generator is defined of index $-1$, so in this picture the section $zmapsto bar{z}$. This answers your (i). Question (ii): $f$ and $W$ were only examples
$endgroup$
– vap
Dec 12 '18 at 18:40
add a comment |
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$begingroup$
Although I do not see the purpose of the bundle $wedge^i V$, it is obvious that it stands for the product bundle $V times wedge^i V$. The map $wedge^i V to wedge^{i+1} V$ is defined fiberwise by $omega mapsto v wedge omega$ for all $v in V$. Any complex vector space $V$ can be given the complex conjugate structure by defining a new scalar multiplication via $lambda * v = overline{lambda} cdot v$. This vector space is denoted by $overline{V}$. The same construction applies to vector bundles (fiberwise). But it is not clear to me what the "resulting class" $b_V$ should be.
$endgroup$
– Paul Frost
Dec 10 '18 at 10:33
$begingroup$
math.stackexchange.com/questions/642167/… $bar{V}$ is the complex conjugate of a vector space, you can look it up on Wikipedia. Please note that there is a typo: $DS$ should be $SV$. The sequence is exact everywhere outside zero, so gives a class in $(V,Vsmallsetminus {0})cong(DV,SV)$. The passage to conjugate is related to the fact the the Bott generator is defined of index $-1$, so in this picture the section $zmapsto bar{z}$. This answers your (i). Question (ii): $f$ and $W$ were only examples
$endgroup$
– vap
Dec 12 '18 at 18:40