Mixing
Clarification on representing K-theory by vector bundles.
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In the Vector bundles and K-theory text by Hatcher, on page 40 under Ring structures it says:
'For elements of $K(X)$ represented by vector bundles $E_1$ and $E_2$ their product in $K(X)$ will be represented by the bundle $E_1 times E_2$, so for arbitrary elements of $K(X)$ represented by differences of vector bundles, their product in $K(X)$ is defined by the formula
$$(E_1 - E_1')(E_2 - E_2') = E_1 otimes E_2 - E_1 otimes E_2' - E_1' otimes E_2 + E_1' otimes E_2'$$
The second statement I get but I am a bit confused when representing elements of $K(X)$ by vector bundles, as opposed to differences of vector bundles. I'd have guessed that given a representative of $K(X)$ by a vector bundle $E_1$, this is equivalent to the difference $E_1 - 0$ where zero denotes the trivial bundle of zero, as only then do the two multiplications agree with one another. However not every difference $E - F$ can be represented as a $H - 0$, (just take rank(F)>rank(E)), so am I interpreting it wrong or when we are representing elements by vector bundles are we restricting to only the ones with a difference equivalence. Can anyone clarify what is meant?
topological-k-theory
asked Jan 30 at 1:15
jojojojo
6217
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$begingroup$
In the Vector bundles and K-theory text by Hatcher, on page 40 under Ring structures it says:
'For elements of $K(X)$ represented by vector bundles $E_1$ and $E_2$ their product in $K(X)$ will be represented by the bundle $E_1 times E_2$, so for arbitrary elements of $K(X)$ represented by differences of vector bundles, their product in $K(X)$ is defined by the formula
$$(E_1 - E_1')(E_2 - E_2') = E_1 otimes E_2 - E_1 otimes E_2' - E_1' otimes E_2 + E_1' otimes E_2'$$
The second statement I get but I am a bit confused when representing elements of $K(X)$ by vector bundles, as opposed to differences of vector bundles. I'd have guessed that given a representative of $K(X)$ by a vector bundle $E_1$, this is equivalent to the difference $E_1 - 0$ where zero denotes the trivial bundle of zero, as only then do the two multiplications agree with one another. However not every difference $E - F$ can be represented as a $H - 0$, (just take rank(F)>rank(E)), so am I interpreting it wrong or when we are representing elements by vector bundles are we restricting to only the ones with a difference equivalence. Can anyone clarify what is meant?
topological-k-theory
asked Jan 30 at 1:15
jojojojo
6217
$endgroup$
add a comment |
$begingroup$
In the Vector bundles and K-theory text by Hatcher, on page 40 under Ring structures it says:
'For elements of $K(X)$ represented by vector bundles $E_1$ and $E_2$ their product in $K(X)$ will be represented by the bundle $E_1 times E_2$, so for arbitrary elements of $K(X)$ represented by differences of vector bundles, their product in $K(X)$ is defined by the formula
$$(E_1 - E_1')(E_2 - E_2') = E_1 otimes E_2 - E_1 otimes E_2' - E_1' otimes E_2 + E_1' otimes E_2'$$
The second statement I get but I am a bit confused when representing elements of $K(X)$ by vector bundles, as opposed to differences of vector bundles. I'd have guessed that given a representative of $K(X)$ by a vector bundle $E_1$, this is equivalent to the difference $E_1 - 0$ where zero denotes the trivial bundle of zero, as only then do the two multiplications agree with one another. However not every difference $E - F$ can be represented as a $H - 0$, (just take rank(F)>rank(E)), so am I interpreting it wrong or when we are representing elements by vector bundles are we restricting to only the ones with a difference equivalence. Can anyone clarify what is meant?
topological-k-theory
asked Jan 30 at 1:15
jojojojo
6217
$endgroup$
In the Vector bundles and K-theory text by Hatcher, on page 40 under Ring structures it says:
'For elements of $K(X)$ represented by vector bundles $E_1$ and $E_2$ their product in $K(X)$ will be represented by the bundle $E_1 times E_2$, so for arbitrary elements of $K(X)$ represented by differences of vector bundles, their product in $K(X)$ is defined by the formula
$$(E_1 - E_1')(E_2 - E_2') = E_1 otimes E_2 - E_1 otimes E_2' - E_1' otimes E_2 + E_1' otimes E_2'$$
The second statement I get but I am a bit confused when representing elements of $K(X)$ by vector bundles, as opposed to differences of vector bundles. I'd have guessed that given a representative of $K(X)$ by a vector bundle $E_1$, this is equivalent to the difference $E_1 - 0$ where zero denotes the trivial bundle of zero, as only then do the two multiplications agree with one another. However not every difference $E - F$ can be represented as a $H - 0$, (just take rank(F)>rank(E)), so am I interpreting it wrong or when we are representing elements by vector bundles are we restricting to only the ones with a difference equivalence. Can anyone clarify what is meant?
topological-k-theory
topological-k-theory
asked Jan 30 at 1:15
jojojojo
6217
asked Jan 30 at 1:15
jojojojo
6217
asked Jan 30 at 1:15
jojojojo
6217
asked Jan 30 at 1:15
jojojojo
6217
asked Jan 30 at 1:15
jojojojo
6217
6217
add a comment |
add a comment |
1 Answer
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To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.
answered Jan 30 at 1:27
Tsemo AristideTsemo Aristide
60.1k11446
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1 Answer
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1 Answer
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$begingroup$
To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.
answered Jan 30 at 1:27
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.
answered Jan 30 at 1:27
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.
answered Jan 30 at 1:27
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
To construct $K(X)$ one starts with the set of vector bundles over $X$ which is a additive semi-group when endowed with the Whitney sum. Then one formally completes this semi-group by adding the inverse. This exactly the same type of operation when one constructs $mathbb{Z}$ from $mathbb{N}$. With this in mind the generalization of the multiplication on $K(X)$ is similar to define the multiplication of $mathbb{Z}$ knowing the multiplication of $mathbb{N}$.
answered Jan 30 at 1:27
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 1:27
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 1:27
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 1:27
Tsemo AristideTsemo Aristide
60.1k11446
60.1k11446
add a comment |
add a comment |
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