Mixing
Isomorphism of bundles , Madsen's Calculus to Cohomology
$begingroup$
Let $xi=(E,M,p)$ be a smooth bundle over $M$. Denote $Omega^0(xi)$ the smooth sections.
We also define $Omega^i(M):= Omega^0(wedge^i T^*M)$, differential $i$ forms, and $Omega^0(M)$ is then smooth functions on $M$ to $Bbb R$.
Then in Madsen's book Calculus to Cohomology, page 171, it states
$$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^2(M) otimes_{Omega^0(M)} Omega^0(xi)) cong operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^0(xi)) otimes _{Omega^0(M)} Omega^2(M)$$
I don't understand why this is true. In fact, if we relabel everything,
$$operatorname{Hom}_R(A,B otimes_RS) cong operatorname{Hom}_R(A,B) otimes_R S$$
is what we have to show. But there is only a map clear from one direction, from RHS to LHS,
$$ (f otimes s ) (a):= f(a) otimes s $$
How does one prove the isomorphism?
algebraic-topology definition differential-topology vector-bundles characteristic-classes
edited Jan 15 at 2:47
amWhy
1
asked Jan 13 at 16:22
CL.CL.
2,2432825
$endgroup$
|
show 2 more comments
$begingroup$
Let $xi=(E,M,p)$ be a smooth bundle over $M$. Denote $Omega^0(xi)$ the smooth sections.
We also define $Omega^i(M):= Omega^0(wedge^i T^*M)$, differential $i$ forms, and $Omega^0(M)$ is then smooth functions on $M$ to $Bbb R$.
Then in Madsen's book Calculus to Cohomology, page 171, it states
$$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^2(M) otimes_{Omega^0(M)} Omega^0(xi)) cong operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^0(xi)) otimes _{Omega^0(M)} Omega^2(M)$$
I don't understand why this is true. In fact, if we relabel everything,
$$operatorname{Hom}_R(A,B otimes_RS) cong operatorname{Hom}_R(A,B) otimes_R S$$
is what we have to show. But there is only a map clear from one direction, from RHS to LHS,
$$ (f otimes s ) (a):= f(a) otimes s $$
How does one prove the isomorphism?
algebraic-topology definition differential-topology vector-bundles characteristic-classes
edited Jan 15 at 2:47
amWhy
1
asked Jan 13 at 16:22
CL.CL.
2,2432825
$endgroup$
$begingroup$
I don't think this is true. Take $A=B=mathbb{Q}$ and $R=S=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 13 at 23:47
$begingroup$
You are right that the generalization does not hold. But there is something different about the case stated. Fiber wise, we have an isomorphism of vector spaces.
$endgroup$
– CL.
Jan 14 at 7:48
$begingroup$
I believe that the map $nu$ defined by $nu(fotimes c)(a)=f(a)otimes c$ is an isomorphism $nu:Hom_R(A,B)otimes_RCcong Hom_R(A,Botimes_RC)$ when either $A$ or $C$ is a finitely generated projective $R$-module. (I don't see the problem with Hempelicious's example above). Then the statement should hold since the sections of $E$ form a locally-free $C^infty(M)$ module of dimension equal to the rank of $E$. I don't have a good reference to hand, so hopefully you'll be able to track one down.
$endgroup$
– Tyrone
Jan 14 at 10:19
1
$begingroup$
@Hempelicious You have $mathbb{Q}otimes_mathbb{Z}mathbb{Z}congmathbb{Q}$ and $Hom_mathbb{Z}(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})$. Therefore $Hom_mathbb{Z}(mathbb{Q},mathbb{Q}otimes_mathbb{Z}mathbb{Z})cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})cong Hom(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}$, so it holds. Or have I made a mistake?
$endgroup$
– Tyrone
Jan 14 at 12:59
1
$begingroup$
@Tyrone: No, the mistake is mine! I meant $A=S=mathbb{Q}$ and $R=B=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 14 at 13:45
|
show 2 more comments
$begingroup$
Let $xi=(E,M,p)$ be a smooth bundle over $M$. Denote $Omega^0(xi)$ the smooth sections.
We also define $Omega^i(M):= Omega^0(wedge^i T^*M)$, differential $i$ forms, and $Omega^0(M)$ is then smooth functions on $M$ to $Bbb R$.
Then in Madsen's book Calculus to Cohomology, page 171, it states
$$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^2(M) otimes_{Omega^0(M)} Omega^0(xi)) cong operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^0(xi)) otimes _{Omega^0(M)} Omega^2(M)$$
I don't understand why this is true. In fact, if we relabel everything,
$$operatorname{Hom}_R(A,B otimes_RS) cong operatorname{Hom}_R(A,B) otimes_R S$$
is what we have to show. But there is only a map clear from one direction, from RHS to LHS,
$$ (f otimes s ) (a):= f(a) otimes s $$
How does one prove the isomorphism?
algebraic-topology definition differential-topology vector-bundles characteristic-classes
edited Jan 15 at 2:47
amWhy
1
asked Jan 13 at 16:22
CL.CL.
2,2432825
$endgroup$
Let $xi=(E,M,p)$ be a smooth bundle over $M$. Denote $Omega^0(xi)$ the smooth sections.
We also define $Omega^i(M):= Omega^0(wedge^i T^*M)$, differential $i$ forms, and $Omega^0(M)$ is then smooth functions on $M$ to $Bbb R$.
Then in Madsen's book Calculus to Cohomology, page 171, it states
$$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^2(M) otimes_{Omega^0(M)} Omega^0(xi)) cong operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^0(xi)) otimes _{Omega^0(M)} Omega^2(M)$$
I don't understand why this is true. In fact, if we relabel everything,
$$operatorname{Hom}_R(A,B otimes_RS) cong operatorname{Hom}_R(A,B) otimes_R S$$
is what we have to show. But there is only a map clear from one direction, from RHS to LHS,
$$ (f otimes s ) (a):= f(a) otimes s $$
How does one prove the isomorphism?
algebraic-topology definition differential-topology vector-bundles characteristic-classes
algebraic-topology definition differential-topology vector-bundles characteristic-classes
edited Jan 15 at 2:47
amWhy
1
asked Jan 13 at 16:22
CL.CL.
2,2432825
edited Jan 15 at 2:47
amWhy
1
asked Jan 13 at 16:22
CL.CL.
2,2432825
edited Jan 15 at 2:47
amWhy
1
edited Jan 15 at 2:47
amWhy
1
edited Jan 15 at 2:47
amWhy
1
1
asked Jan 13 at 16:22
CL.CL.
2,2432825
asked Jan 13 at 16:22
CL.CL.
2,2432825
asked Jan 13 at 16:22
CL.CL.
2,2432825
2,2432825
$begingroup$
I don't think this is true. Take $A=B=mathbb{Q}$ and $R=S=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 13 at 23:47
$begingroup$
You are right that the generalization does not hold. But there is something different about the case stated. Fiber wise, we have an isomorphism of vector spaces.
$endgroup$
– CL.
Jan 14 at 7:48
$begingroup$
I believe that the map $nu$ defined by $nu(fotimes c)(a)=f(a)otimes c$ is an isomorphism $nu:Hom_R(A,B)otimes_RCcong Hom_R(A,Botimes_RC)$ when either $A$ or $C$ is a finitely generated projective $R$-module. (I don't see the problem with Hempelicious's example above). Then the statement should hold since the sections of $E$ form a locally-free $C^infty(M)$ module of dimension equal to the rank of $E$. I don't have a good reference to hand, so hopefully you'll be able to track one down.
$endgroup$
– Tyrone
Jan 14 at 10:19
1
$begingroup$
@Hempelicious You have $mathbb{Q}otimes_mathbb{Z}mathbb{Z}congmathbb{Q}$ and $Hom_mathbb{Z}(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})$. Therefore $Hom_mathbb{Z}(mathbb{Q},mathbb{Q}otimes_mathbb{Z}mathbb{Z})cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})cong Hom(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}$, so it holds. Or have I made a mistake?
$endgroup$
– Tyrone
Jan 14 at 12:59
1
$begingroup$
@Tyrone: No, the mistake is mine! I meant $A=S=mathbb{Q}$ and $R=B=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 14 at 13:45
|
show 2 more comments
$begingroup$
I don't think this is true. Take $A=B=mathbb{Q}$ and $R=S=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 13 at 23:47
$begingroup$
You are right that the generalization does not hold. But there is something different about the case stated. Fiber wise, we have an isomorphism of vector spaces.
$endgroup$
– CL.
Jan 14 at 7:48
$begingroup$
I believe that the map $nu$ defined by $nu(fotimes c)(a)=f(a)otimes c$ is an isomorphism $nu:Hom_R(A,B)otimes_RCcong Hom_R(A,Botimes_RC)$ when either $A$ or $C$ is a finitely generated projective $R$-module. (I don't see the problem with Hempelicious's example above). Then the statement should hold since the sections of $E$ form a locally-free $C^infty(M)$ module of dimension equal to the rank of $E$. I don't have a good reference to hand, so hopefully you'll be able to track one down.
$endgroup$
– Tyrone
Jan 14 at 10:19
1
$begingroup$
@Hempelicious You have $mathbb{Q}otimes_mathbb{Z}mathbb{Z}congmathbb{Q}$ and $Hom_mathbb{Z}(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})$. Therefore $Hom_mathbb{Z}(mathbb{Q},mathbb{Q}otimes_mathbb{Z}mathbb{Z})cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})cong Hom(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}$, so it holds. Or have I made a mistake?
$endgroup$
– Tyrone
Jan 14 at 12:59
1
$begingroup$
@Tyrone: No, the mistake is mine! I meant $A=S=mathbb{Q}$ and $R=B=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 14 at 13:45
$begingroup$
I don't think this is true. Take $A=B=mathbb{Q}$ and $R=S=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 13 at 23:47
$begingroup$
I don't think this is true. Take $A=B=mathbb{Q}$ and $R=S=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 13 at 23:47
$begingroup$
You are right that the generalization does not hold. But there is something different about the case stated. Fiber wise, we have an isomorphism of vector spaces.
$endgroup$
– CL.
Jan 14 at 7:48
$begingroup$
You are right that the generalization does not hold. But there is something different about the case stated. Fiber wise, we have an isomorphism of vector spaces.
$endgroup$
– CL.
Jan 14 at 7:48
$begingroup$
I believe that the map $nu$ defined by $nu(fotimes c)(a)=f(a)otimes c$ is an isomorphism $nu:Hom_R(A,B)otimes_RCcong Hom_R(A,Botimes_RC)$ when either $A$ or $C$ is a finitely generated projective $R$-module. (I don't see the problem with Hempelicious's example above). Then the statement should hold since the sections of $E$ form a locally-free $C^infty(M)$ module of dimension equal to the rank of $E$. I don't have a good reference to hand, so hopefully you'll be able to track one down.
$endgroup$
– Tyrone
Jan 14 at 10:19
$begingroup$
I believe that the map $nu$ defined by $nu(fotimes c)(a)=f(a)otimes c$ is an isomorphism $nu:Hom_R(A,B)otimes_RCcong Hom_R(A,Botimes_RC)$ when either $A$ or $C$ is a finitely generated projective $R$-module. (I don't see the problem with Hempelicious's example above). Then the statement should hold since the sections of $E$ form a locally-free $C^infty(M)$ module of dimension equal to the rank of $E$. I don't have a good reference to hand, so hopefully you'll be able to track one down.
$endgroup$
– Tyrone
Jan 14 at 10:19
1
1
$begingroup$
@Hempelicious You have $mathbb{Q}otimes_mathbb{Z}mathbb{Z}congmathbb{Q}$ and $Hom_mathbb{Z}(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})$. Therefore $Hom_mathbb{Z}(mathbb{Q},mathbb{Q}otimes_mathbb{Z}mathbb{Z})cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})cong Hom(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}$, so it holds. Or have I made a mistake?
$endgroup$
– Tyrone
Jan 14 at 12:59
$begingroup$
@Hempelicious You have $mathbb{Q}otimes_mathbb{Z}mathbb{Z}congmathbb{Q}$ and $Hom_mathbb{Z}(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})$. Therefore $Hom_mathbb{Z}(mathbb{Q},mathbb{Q}otimes_mathbb{Z}mathbb{Z})cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})cong Hom(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}$, so it holds. Or have I made a mistake?
$endgroup$
– Tyrone
Jan 14 at 12:59
1
1
$begingroup$
@Tyrone: No, the mistake is mine! I meant $A=S=mathbb{Q}$ and $R=B=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 14 at 13:45
$begingroup$
@Tyrone: No, the mistake is mine! I meant $A=S=mathbb{Q}$ and $R=B=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 14 at 13:45
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Well, the book doesn't just state this fact; it cites Theorem 16.13 from earlier in the book as justification. That theorem says that the operation of tensoring and Homming sections of vector bundles over the ring $Omega^0(M)$ corresponds to the tensor and Hom operations on the vector bundles themselves. In particular, there are natural isomorphisms $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^2(M) otimes_{Omega^0(M)} Omega^0(xi))cong Omega^0(operatorname{Hom}(xi,wedge^2T^*Motimes xi))$$ and $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^0(xi)) otimes _{Omega^0(M)} Omega^2(M)cong Omega^0(operatorname{Hom}(xi,xi)otimeswedge^2T^*M).$$
The desired isomorphism then follows from the isomorphism of vector bundles $operatorname{Hom}(xi,wedge^2T^*Motimes xi)cong operatorname{Hom}(xi,xi)otimeswedge^2T^*M$, which on fibers is just the natural isomorphism of vector spaces $operatorname{Hom}(V,Wotimes V)cong operatorname{Hom}(V,V)otimes W$.
answered Jan 15 at 5:23
Eric WofseyEric Wofsey
186k14214341
$endgroup$
add a comment |
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$begingroup$
Well, the book doesn't just state this fact; it cites Theorem 16.13 from earlier in the book as justification. That theorem says that the operation of tensoring and Homming sections of vector bundles over the ring $Omega^0(M)$ corresponds to the tensor and Hom operations on the vector bundles themselves. In particular, there are natural isomorphisms $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^2(M) otimes_{Omega^0(M)} Omega^0(xi))cong Omega^0(operatorname{Hom}(xi,wedge^2T^*Motimes xi))$$ and $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^0(xi)) otimes _{Omega^0(M)} Omega^2(M)cong Omega^0(operatorname{Hom}(xi,xi)otimeswedge^2T^*M).$$
The desired isomorphism then follows from the isomorphism of vector bundles $operatorname{Hom}(xi,wedge^2T^*Motimes xi)cong operatorname{Hom}(xi,xi)otimeswedge^2T^*M$, which on fibers is just the natural isomorphism of vector spaces $operatorname{Hom}(V,Wotimes V)cong operatorname{Hom}(V,V)otimes W$.
answered Jan 15 at 5:23
Eric WofseyEric Wofsey
186k14214341
$endgroup$
add a comment |
$begingroup$
Well, the book doesn't just state this fact; it cites Theorem 16.13 from earlier in the book as justification. That theorem says that the operation of tensoring and Homming sections of vector bundles over the ring $Omega^0(M)$ corresponds to the tensor and Hom operations on the vector bundles themselves. In particular, there are natural isomorphisms $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^2(M) otimes_{Omega^0(M)} Omega^0(xi))cong Omega^0(operatorname{Hom}(xi,wedge^2T^*Motimes xi))$$ and $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^0(xi)) otimes _{Omega^0(M)} Omega^2(M)cong Omega^0(operatorname{Hom}(xi,xi)otimeswedge^2T^*M).$$
The desired isomorphism then follows from the isomorphism of vector bundles $operatorname{Hom}(xi,wedge^2T^*Motimes xi)cong operatorname{Hom}(xi,xi)otimeswedge^2T^*M$, which on fibers is just the natural isomorphism of vector spaces $operatorname{Hom}(V,Wotimes V)cong operatorname{Hom}(V,V)otimes W$.
answered Jan 15 at 5:23
Eric WofseyEric Wofsey
186k14214341
$endgroup$
add a comment |
$begingroup$
Well, the book doesn't just state this fact; it cites Theorem 16.13 from earlier in the book as justification. That theorem says that the operation of tensoring and Homming sections of vector bundles over the ring $Omega^0(M)$ corresponds to the tensor and Hom operations on the vector bundles themselves. In particular, there are natural isomorphisms $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^2(M) otimes_{Omega^0(M)} Omega^0(xi))cong Omega^0(operatorname{Hom}(xi,wedge^2T^*Motimes xi))$$ and $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^0(xi)) otimes _{Omega^0(M)} Omega^2(M)cong Omega^0(operatorname{Hom}(xi,xi)otimeswedge^2T^*M).$$
The desired isomorphism then follows from the isomorphism of vector bundles $operatorname{Hom}(xi,wedge^2T^*Motimes xi)cong operatorname{Hom}(xi,xi)otimeswedge^2T^*M$, which on fibers is just the natural isomorphism of vector spaces $operatorname{Hom}(V,Wotimes V)cong operatorname{Hom}(V,V)otimes W$.
answered Jan 15 at 5:23
Eric WofseyEric Wofsey
186k14214341
$endgroup$
Well, the book doesn't just state this fact; it cites Theorem 16.13 from earlier in the book as justification. That theorem says that the operation of tensoring and Homming sections of vector bundles over the ring $Omega^0(M)$ corresponds to the tensor and Hom operations on the vector bundles themselves. In particular, there are natural isomorphisms $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^2(M) otimes_{Omega^0(M)} Omega^0(xi))cong Omega^0(operatorname{Hom}(xi,wedge^2T^*Motimes xi))$$ and $$operatorname{Hom}_{Omega^0(M)}(Omega^0(xi), Omega^0(xi)) otimes _{Omega^0(M)} Omega^2(M)cong Omega^0(operatorname{Hom}(xi,xi)otimeswedge^2T^*M).$$
The desired isomorphism then follows from the isomorphism of vector bundles $operatorname{Hom}(xi,wedge^2T^*Motimes xi)cong operatorname{Hom}(xi,xi)otimeswedge^2T^*M$, which on fibers is just the natural isomorphism of vector spaces $operatorname{Hom}(V,Wotimes V)cong operatorname{Hom}(V,V)otimes W$.
answered Jan 15 at 5:23
Eric WofseyEric Wofsey
186k14214341
answered Jan 15 at 5:23
Eric WofseyEric Wofsey
186k14214341
answered Jan 15 at 5:23
Eric WofseyEric Wofsey
186k14214341
answered Jan 15 at 5:23
Eric WofseyEric Wofsey
186k14214341
186k14214341
add a comment |
add a comment |
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$begingroup$
I don't think this is true. Take $A=B=mathbb{Q}$ and $R=S=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 13 at 23:47
$begingroup$
You are right that the generalization does not hold. But there is something different about the case stated. Fiber wise, we have an isomorphism of vector spaces.
$endgroup$
– CL.
Jan 14 at 7:48
$begingroup$
I believe that the map $nu$ defined by $nu(fotimes c)(a)=f(a)otimes c$ is an isomorphism $nu:Hom_R(A,B)otimes_RCcong Hom_R(A,Botimes_RC)$ when either $A$ or $C$ is a finitely generated projective $R$-module. (I don't see the problem with Hempelicious's example above). Then the statement should hold since the sections of $E$ form a locally-free $C^infty(M)$ module of dimension equal to the rank of $E$. I don't have a good reference to hand, so hopefully you'll be able to track one down.
$endgroup$
– Tyrone
Jan 14 at 10:19
1
$begingroup$
@Hempelicious You have $mathbb{Q}otimes_mathbb{Z}mathbb{Z}congmathbb{Q}$ and $Hom_mathbb{Z}(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})$. Therefore $Hom_mathbb{Z}(mathbb{Q},mathbb{Q}otimes_mathbb{Z}mathbb{Z})cong Hom_mathbb{Z}(mathbb{Q},mathbb{Q})cong Hom(mathbb{Q},mathbb{Q})otimes_mathbb{Z}mathbb{Z}$, so it holds. Or have I made a mistake?
$endgroup$
– Tyrone
Jan 14 at 12:59
1
$begingroup$
@Tyrone: No, the mistake is mine! I meant $A=S=mathbb{Q}$ and $R=B=mathbb{Z}$.
$endgroup$
– Hempelicious
Jan 14 at 13:45