Mixing
Transitive action of $H^2(M;Bbb Z)$ on $Spin^c$ structures over $M$
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
I’ve a problem understanding why the action of the second cohomology group (integer coefficients) of an oriented smooth manifold $M$ is free and transitive on the set of $Spin^c$. I’m following these notes on the nLab (Prop D.43) but they dont prove that the action is transitive.
For every two $Spin^c$-structure on $M$, say $P,P’$ the author constructs a chern class $delta$ (Distinguishing chern class) and then claim that $$P’cong Ptimes_K U(Bbb L_delta)$$ where $K=ker picolon Spin^c(n)to SO(n)$ and $U(Bbb L_delta)$ is the $S^1$-bundle over $M$ associated to the chern class $delta$.
how can I start building a map between $P’$ and $Ptimes_K U(Bbb L_delta)$ for example?
I'm aware that this question is somehow technical, but i can’t find any resource spelling out the details of transitiveness and freenes of such action.
algebraic-topology differential-topology principal-bundles spin-geometry
asked Jul 18 at 18:37
Luigi M
1,5081821
add a comment |Â
up vote
4
down vote
favorite
I’ve a problem understanding why the action of the second cohomology group (integer coefficients) of an oriented smooth manifold $M$ is free and transitive on the set of $Spin^c$. I’m following these notes on the nLab (Prop D.43) but they dont prove that the action is transitive.
For every two $Spin^c$-structure on $M$, say $P,P’$ the author constructs a chern class $delta$ (Distinguishing chern class) and then claim that $$P’cong Ptimes_K U(Bbb L_delta)$$ where $K=ker picolon Spin^c(n)to SO(n)$ and $U(Bbb L_delta)$ is the $S^1$-bundle over $M$ associated to the chern class $delta$.
how can I start building a map between $P’$ and $Ptimes_K U(Bbb L_delta)$ for example?
I'm aware that this question is somehow technical, but i can’t find any resource spelling out the details of transitiveness and freenes of such action.
algebraic-topology differential-topology principal-bundles spin-geometry
asked Jul 18 at 18:37
Luigi M
1,5081821
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I’ve a problem understanding why the action of the second cohomology group (integer coefficients) of an oriented smooth manifold $M$ is free and transitive on the set of $Spin^c$. I’m following these notes on the nLab (Prop D.43) but they dont prove that the action is transitive.
For every two $Spin^c$-structure on $M$, say $P,P’$ the author constructs a chern class $delta$ (Distinguishing chern class) and then claim that $$P’cong Ptimes_K U(Bbb L_delta)$$ where $K=ker picolon Spin^c(n)to SO(n)$ and $U(Bbb L_delta)$ is the $S^1$-bundle over $M$ associated to the chern class $delta$.
how can I start building a map between $P’$ and $Ptimes_K U(Bbb L_delta)$ for example?
I'm aware that this question is somehow technical, but i can’t find any resource spelling out the details of transitiveness and freenes of such action.
algebraic-topology differential-topology principal-bundles spin-geometry
asked Jul 18 at 18:37
Luigi M
1,5081821
I’ve a problem understanding why the action of the second cohomology group (integer coefficients) of an oriented smooth manifold $M$ is free and transitive on the set of $Spin^c$. I’m following these notes on the nLab (Prop D.43) but they dont prove that the action is transitive.
For every two $Spin^c$-structure on $M$, say $P,P’$ the author constructs a chern class $delta$ (Distinguishing chern class) and then claim that $$P’cong Ptimes_K U(Bbb L_delta)$$ where $K=ker picolon Spin^c(n)to SO(n)$ and $U(Bbb L_delta)$ is the $S^1$-bundle over $M$ associated to the chern class $delta$.
how can I start building a map between $P’$ and $Ptimes_K U(Bbb L_delta)$ for example?
I'm aware that this question is somehow technical, but i can’t find any resource spelling out the details of transitiveness and freenes of such action.
algebraic-topology differential-topology principal-bundles spin-geometry
asked Jul 18 at 18:37
Luigi M
1,5081821
edited Jul 18 at 19:44
asked Jul 18 at 18:37
Luigi M
1,5081821
asked Jul 18 at 18:37
Luigi M
1,5081821
asked Jul 18 at 18:37
Luigi M
1,5081821
1,5081821
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
You consider the fiber sequence $BbbCP^infty to BtextSpin^c(n) to BSO(n).$ Things are easiest if you choose a model of this fibration in which $BbbCP^infty$ is a group, and this is a principal $G$-bundle.
A brief point on principal bundles: given a principal $G$-bundle $P to B$, there is an associated bundle of groups, $textAut(P)$; there is an isomorphism of groups $textAut(P)_x cong G$ (an automorphism of the $G$-set $P_x$ is given by right-multiplication by some element of $G$). This bundle is not necessarily trivial, but if $P$ had transition functions $rho_alpha beta: U_alpha beta to G$ acting on elements of $G$ by left-multiplication, then the transition functions for $textAut(P)$ act on elements by conjugation by $rho_alpha beta$ instead of left multiplication. In particular, if $G$ is abelian, this bundle is trivial, and sections of $textAut(P)$ are just maps $B to G$.
Now, given two sections $sigma_1$ and $sigma_2$ of $P$, there is an automorphism $g$ of $P$ determined by sending $sigma_1(x)g(x) = sigma_2(x)$. Because both $sigma_1$ and $sigma_2$ are continuous sections, as is the inversion operation in $G$, so too is $g: B to textAut(P)$ continuous. We see that $Gamma(P)$ is acted on simply transitively by $Gamma(textAut(P))$ - provided $Gamma(P)$ is nonempty. When $G$ is abelian, $Gamma(textAut(P)) = textMap(M, G)$, as above.
Suppose you have an oriented vector bundle $E$ (coded by a map $f: X to BSO(n)$), $textspin^c$ structures on $E$ are the same thing as lifts to $tilde f: X to BtextSpin^c(n)$. This is the same as the space of sections of a principal $BbbCP^infty$-bundle written $textSpin^c(E)$, the bundle whose fiber above $x$ is the space of $textspin^c$ structures on $E_x$ (this is the pullback to $X$ of the fiber sequence in the first line of this answer). What was written above is that picking an element of $Gamma(textSpin^c(E))$ gives a canonical bijection with $textMap(X, BbbCP^infty)$. (Of course, this is assuming there exists a $textspin^c$ structure on $E$!)
Passing to homotopy classes, we see that the space of $textspin^c$ structures on $E$ up to isomorphism is acted on simply transitively by $[X, BbbCP^infty] = [X, K(Bbb Z,2)] = H^2(X;Bbb Z)$, assuming there was a $textspin^c$ structure to begin with.
Another argument (essentially the same as this) runs through obstruction theory, which determines whether or not lifts exist and how many they are; the point being that $H^n(X;pi_n F)$ is zero only for $n = 2$, where we get $H^2(X;Bbb Z)$. The existence of a lift is determined by a class in $H^3(X;Bbb Z)$, which turns out to be $beta w_2(E)$, where $beta$ is the Bockstein map.
answered Jul 18 at 20:51
Mike Miller
33.9k364128
1
To be clear, this assumes the existence of a lift in the first place: you can't act on the empty space.
– Mike Miller
Jul 18 at 21:59
I guess my problem was "Two sections of a principal bundle differ by multiplication by a map to $G$" I didn't know this result
– Luigi M
Jul 19 at 0:41
1
@LuigiM I should edit my answer: that is not true as phrased. That is true here because $G$ is abelian (many arguments; it is $BG$ for $G$ a commutative group, you can explicitly write $BbbCP^infty$ as the projectivization of the field $Bbb C(x)$, and it inherits a group structure from that...). In general you differ by a section of $textAut(P)$.
– Mike Miller
Jul 19 at 0:53
1
I have edited and rephrased the answer in more bundle-theoretic language.
– Mike Miller
Jul 19 at 1:23
Thanks a lot! That's an impressive answer. I'll meditate on that a little bit before accepting it:)
– Luigi M
Jul 19 at 2:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You consider the fiber sequence $BbbCP^infty to BtextSpin^c(n) to BSO(n).$ Things are easiest if you choose a model of this fibration in which $BbbCP^infty$ is a group, and this is a principal $G$-bundle.
A brief point on principal bundles: given a principal $G$-bundle $P to B$, there is an associated bundle of groups, $textAut(P)$; there is an isomorphism of groups $textAut(P)_x cong G$ (an automorphism of the $G$-set $P_x$ is given by right-multiplication by some element of $G$). This bundle is not necessarily trivial, but if $P$ had transition functions $rho_alpha beta: U_alpha beta to G$ acting on elements of $G$ by left-multiplication, then the transition functions for $textAut(P)$ act on elements by conjugation by $rho_alpha beta$ instead of left multiplication. In particular, if $G$ is abelian, this bundle is trivial, and sections of $textAut(P)$ are just maps $B to G$.
Now, given two sections $sigma_1$ and $sigma_2$ of $P$, there is an automorphism $g$ of $P$ determined by sending $sigma_1(x)g(x) = sigma_2(x)$. Because both $sigma_1$ and $sigma_2$ are continuous sections, as is the inversion operation in $G$, so too is $g: B to textAut(P)$ continuous. We see that $Gamma(P)$ is acted on simply transitively by $Gamma(textAut(P))$ - provided $Gamma(P)$ is nonempty. When $G$ is abelian, $Gamma(textAut(P)) = textMap(M, G)$, as above.
Suppose you have an oriented vector bundle $E$ (coded by a map $f: X to BSO(n)$), $textspin^c$ structures on $E$ are the same thing as lifts to $tilde f: X to BtextSpin^c(n)$. This is the same as the space of sections of a principal $BbbCP^infty$-bundle written $textSpin^c(E)$, the bundle whose fiber above $x$ is the space of $textspin^c$ structures on $E_x$ (this is the pullback to $X$ of the fiber sequence in the first line of this answer). What was written above is that picking an element of $Gamma(textSpin^c(E))$ gives a canonical bijection with $textMap(X, BbbCP^infty)$. (Of course, this is assuming there exists a $textspin^c$ structure on $E$!)
Passing to homotopy classes, we see that the space of $textspin^c$ structures on $E$ up to isomorphism is acted on simply transitively by $[X, BbbCP^infty] = [X, K(Bbb Z,2)] = H^2(X;Bbb Z)$, assuming there was a $textspin^c$ structure to begin with.
Another argument (essentially the same as this) runs through obstruction theory, which determines whether or not lifts exist and how many they are; the point being that $H^n(X;pi_n F)$ is zero only for $n = 2$, where we get $H^2(X;Bbb Z)$. The existence of a lift is determined by a class in $H^3(X;Bbb Z)$, which turns out to be $beta w_2(E)$, where $beta$ is the Bockstein map.
answered Jul 18 at 20:51
Mike Miller
33.9k364128
1
To be clear, this assumes the existence of a lift in the first place: you can't act on the empty space.
– Mike Miller
Jul 18 at 21:59
I guess my problem was "Two sections of a principal bundle differ by multiplication by a map to $G$" I didn't know this result
– Luigi M
Jul 19 at 0:41
1
@LuigiM I should edit my answer: that is not true as phrased. That is true here because $G$ is abelian (many arguments; it is $BG$ for $G$ a commutative group, you can explicitly write $BbbCP^infty$ as the projectivization of the field $Bbb C(x)$, and it inherits a group structure from that...). In general you differ by a section of $textAut(P)$.
– Mike Miller
Jul 19 at 0:53
1
I have edited and rephrased the answer in more bundle-theoretic language.
– Mike Miller
Jul 19 at 1:23
Thanks a lot! That's an impressive answer. I'll meditate on that a little bit before accepting it:)
– Luigi M
Jul 19 at 2:37
add a comment |Â
up vote
4
down vote
accepted
You consider the fiber sequence $BbbCP^infty to BtextSpin^c(n) to BSO(n).$ Things are easiest if you choose a model of this fibration in which $BbbCP^infty$ is a group, and this is a principal $G$-bundle.
A brief point on principal bundles: given a principal $G$-bundle $P to B$, there is an associated bundle of groups, $textAut(P)$; there is an isomorphism of groups $textAut(P)_x cong G$ (an automorphism of the $G$-set $P_x$ is given by right-multiplication by some element of $G$). This bundle is not necessarily trivial, but if $P$ had transition functions $rho_alpha beta: U_alpha beta to G$ acting on elements of $G$ by left-multiplication, then the transition functions for $textAut(P)$ act on elements by conjugation by $rho_alpha beta$ instead of left multiplication. In particular, if $G$ is abelian, this bundle is trivial, and sections of $textAut(P)$ are just maps $B to G$.
Now, given two sections $sigma_1$ and $sigma_2$ of $P$, there is an automorphism $g$ of $P$ determined by sending $sigma_1(x)g(x) = sigma_2(x)$. Because both $sigma_1$ and $sigma_2$ are continuous sections, as is the inversion operation in $G$, so too is $g: B to textAut(P)$ continuous. We see that $Gamma(P)$ is acted on simply transitively by $Gamma(textAut(P))$ - provided $Gamma(P)$ is nonempty. When $G$ is abelian, $Gamma(textAut(P)) = textMap(M, G)$, as above.
Suppose you have an oriented vector bundle $E$ (coded by a map $f: X to BSO(n)$), $textspin^c$ structures on $E$ are the same thing as lifts to $tilde f: X to BtextSpin^c(n)$. This is the same as the space of sections of a principal $BbbCP^infty$-bundle written $textSpin^c(E)$, the bundle whose fiber above $x$ is the space of $textspin^c$ structures on $E_x$ (this is the pullback to $X$ of the fiber sequence in the first line of this answer). What was written above is that picking an element of $Gamma(textSpin^c(E))$ gives a canonical bijection with $textMap(X, BbbCP^infty)$. (Of course, this is assuming there exists a $textspin^c$ structure on $E$!)
Passing to homotopy classes, we see that the space of $textspin^c$ structures on $E$ up to isomorphism is acted on simply transitively by $[X, BbbCP^infty] = [X, K(Bbb Z,2)] = H^2(X;Bbb Z)$, assuming there was a $textspin^c$ structure to begin with.
Another argument (essentially the same as this) runs through obstruction theory, which determines whether or not lifts exist and how many they are; the point being that $H^n(X;pi_n F)$ is zero only for $n = 2$, where we get $H^2(X;Bbb Z)$. The existence of a lift is determined by a class in $H^3(X;Bbb Z)$, which turns out to be $beta w_2(E)$, where $beta$ is the Bockstein map.
answered Jul 18 at 20:51
Mike Miller
33.9k364128
1
To be clear, this assumes the existence of a lift in the first place: you can't act on the empty space.
– Mike Miller
Jul 18 at 21:59
I guess my problem was "Two sections of a principal bundle differ by multiplication by a map to $G$" I didn't know this result
– Luigi M
Jul 19 at 0:41
1
@LuigiM I should edit my answer: that is not true as phrased. That is true here because $G$ is abelian (many arguments; it is $BG$ for $G$ a commutative group, you can explicitly write $BbbCP^infty$ as the projectivization of the field $Bbb C(x)$, and it inherits a group structure from that...). In general you differ by a section of $textAut(P)$.
– Mike Miller
Jul 19 at 0:53
1
I have edited and rephrased the answer in more bundle-theoretic language.
– Mike Miller
Jul 19 at 1:23
Thanks a lot! That's an impressive answer. I'll meditate on that a little bit before accepting it:)
– Luigi M
Jul 19 at 2:37
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You consider the fiber sequence $BbbCP^infty to BtextSpin^c(n) to BSO(n).$ Things are easiest if you choose a model of this fibration in which $BbbCP^infty$ is a group, and this is a principal $G$-bundle.
A brief point on principal bundles: given a principal $G$-bundle $P to B$, there is an associated bundle of groups, $textAut(P)$; there is an isomorphism of groups $textAut(P)_x cong G$ (an automorphism of the $G$-set $P_x$ is given by right-multiplication by some element of $G$). This bundle is not necessarily trivial, but if $P$ had transition functions $rho_alpha beta: U_alpha beta to G$ acting on elements of $G$ by left-multiplication, then the transition functions for $textAut(P)$ act on elements by conjugation by $rho_alpha beta$ instead of left multiplication. In particular, if $G$ is abelian, this bundle is trivial, and sections of $textAut(P)$ are just maps $B to G$.
Now, given two sections $sigma_1$ and $sigma_2$ of $P$, there is an automorphism $g$ of $P$ determined by sending $sigma_1(x)g(x) = sigma_2(x)$. Because both $sigma_1$ and $sigma_2$ are continuous sections, as is the inversion operation in $G$, so too is $g: B to textAut(P)$ continuous. We see that $Gamma(P)$ is acted on simply transitively by $Gamma(textAut(P))$ - provided $Gamma(P)$ is nonempty. When $G$ is abelian, $Gamma(textAut(P)) = textMap(M, G)$, as above.
Suppose you have an oriented vector bundle $E$ (coded by a map $f: X to BSO(n)$), $textspin^c$ structures on $E$ are the same thing as lifts to $tilde f: X to BtextSpin^c(n)$. This is the same as the space of sections of a principal $BbbCP^infty$-bundle written $textSpin^c(E)$, the bundle whose fiber above $x$ is the space of $textspin^c$ structures on $E_x$ (this is the pullback to $X$ of the fiber sequence in the first line of this answer). What was written above is that picking an element of $Gamma(textSpin^c(E))$ gives a canonical bijection with $textMap(X, BbbCP^infty)$. (Of course, this is assuming there exists a $textspin^c$ structure on $E$!)
Passing to homotopy classes, we see that the space of $textspin^c$ structures on $E$ up to isomorphism is acted on simply transitively by $[X, BbbCP^infty] = [X, K(Bbb Z,2)] = H^2(X;Bbb Z)$, assuming there was a $textspin^c$ structure to begin with.
Another argument (essentially the same as this) runs through obstruction theory, which determines whether or not lifts exist and how many they are; the point being that $H^n(X;pi_n F)$ is zero only for $n = 2$, where we get $H^2(X;Bbb Z)$. The existence of a lift is determined by a class in $H^3(X;Bbb Z)$, which turns out to be $beta w_2(E)$, where $beta$ is the Bockstein map.
answered Jul 18 at 20:51
Mike Miller
33.9k364128
You consider the fiber sequence $BbbCP^infty to BtextSpin^c(n) to BSO(n).$ Things are easiest if you choose a model of this fibration in which $BbbCP^infty$ is a group, and this is a principal $G$-bundle.
A brief point on principal bundles: given a principal $G$-bundle $P to B$, there is an associated bundle of groups, $textAut(P)$; there is an isomorphism of groups $textAut(P)_x cong G$ (an automorphism of the $G$-set $P_x$ is given by right-multiplication by some element of $G$). This bundle is not necessarily trivial, but if $P$ had transition functions $rho_alpha beta: U_alpha beta to G$ acting on elements of $G$ by left-multiplication, then the transition functions for $textAut(P)$ act on elements by conjugation by $rho_alpha beta$ instead of left multiplication. In particular, if $G$ is abelian, this bundle is trivial, and sections of $textAut(P)$ are just maps $B to G$.
Now, given two sections $sigma_1$ and $sigma_2$ of $P$, there is an automorphism $g$ of $P$ determined by sending $sigma_1(x)g(x) = sigma_2(x)$. Because both $sigma_1$ and $sigma_2$ are continuous sections, as is the inversion operation in $G$, so too is $g: B to textAut(P)$ continuous. We see that $Gamma(P)$ is acted on simply transitively by $Gamma(textAut(P))$ - provided $Gamma(P)$ is nonempty. When $G$ is abelian, $Gamma(textAut(P)) = textMap(M, G)$, as above.
Suppose you have an oriented vector bundle $E$ (coded by a map $f: X to BSO(n)$), $textspin^c$ structures on $E$ are the same thing as lifts to $tilde f: X to BtextSpin^c(n)$. This is the same as the space of sections of a principal $BbbCP^infty$-bundle written $textSpin^c(E)$, the bundle whose fiber above $x$ is the space of $textspin^c$ structures on $E_x$ (this is the pullback to $X$ of the fiber sequence in the first line of this answer). What was written above is that picking an element of $Gamma(textSpin^c(E))$ gives a canonical bijection with $textMap(X, BbbCP^infty)$. (Of course, this is assuming there exists a $textspin^c$ structure on $E$!)
Passing to homotopy classes, we see that the space of $textspin^c$ structures on $E$ up to isomorphism is acted on simply transitively by $[X, BbbCP^infty] = [X, K(Bbb Z,2)] = H^2(X;Bbb Z)$, assuming there was a $textspin^c$ structure to begin with.
Another argument (essentially the same as this) runs through obstruction theory, which determines whether or not lifts exist and how many they are; the point being that $H^n(X;pi_n F)$ is zero only for $n = 2$, where we get $H^2(X;Bbb Z)$. The existence of a lift is determined by a class in $H^3(X;Bbb Z)$, which turns out to be $beta w_2(E)$, where $beta$ is the Bockstein map.
answered Jul 18 at 20:51
Mike Miller
33.9k364128
edited Jul 19 at 1:23
answered Jul 18 at 20:51
Mike Miller
33.9k364128
answered Jul 18 at 20:51
Mike Miller
33.9k364128
answered Jul 18 at 20:51
Mike Miller
33.9k364128
33.9k364128
1
To be clear, this assumes the existence of a lift in the first place: you can't act on the empty space.
– Mike Miller
Jul 18 at 21:59
I guess my problem was "Two sections of a principal bundle differ by multiplication by a map to $G$" I didn't know this result
– Luigi M
Jul 19 at 0:41
1
@LuigiM I should edit my answer: that is not true as phrased. That is true here because $G$ is abelian (many arguments; it is $BG$ for $G$ a commutative group, you can explicitly write $BbbCP^infty$ as the projectivization of the field $Bbb C(x)$, and it inherits a group structure from that...). In general you differ by a section of $textAut(P)$.
– Mike Miller
Jul 19 at 0:53
1
I have edited and rephrased the answer in more bundle-theoretic language.
– Mike Miller
Jul 19 at 1:23
Thanks a lot! That's an impressive answer. I'll meditate on that a little bit before accepting it:)
– Luigi M
Jul 19 at 2:37
add a comment |Â
1
To be clear, this assumes the existence of a lift in the first place: you can't act on the empty space.
– Mike Miller
Jul 18 at 21:59
I guess my problem was "Two sections of a principal bundle differ by multiplication by a map to $G$" I didn't know this result
– Luigi M
Jul 19 at 0:41
1
@LuigiM I should edit my answer: that is not true as phrased. That is true here because $G$ is abelian (many arguments; it is $BG$ for $G$ a commutative group, you can explicitly write $BbbCP^infty$ as the projectivization of the field $Bbb C(x)$, and it inherits a group structure from that...). In general you differ by a section of $textAut(P)$.
– Mike Miller
Jul 19 at 0:53
1
I have edited and rephrased the answer in more bundle-theoretic language.
– Mike Miller
Jul 19 at 1:23
Thanks a lot! That's an impressive answer. I'll meditate on that a little bit before accepting it:)
– Luigi M
Jul 19 at 2:37
1
1
To be clear, this assumes the existence of a lift in the first place: you can't act on the empty space.
– Mike Miller
Jul 18 at 21:59
To be clear, this assumes the existence of a lift in the first place: you can't act on the empty space.
– Mike Miller
Jul 18 at 21:59
I guess my problem was "Two sections of a principal bundle differ by multiplication by a map to $G$" I didn't know this result
– Luigi M
Jul 19 at 0:41
I guess my problem was "Two sections of a principal bundle differ by multiplication by a map to $G$" I didn't know this result
– Luigi M
Jul 19 at 0:41
1
1
@LuigiM I should edit my answer: that is not true as phrased. That is true here because $G$ is abelian (many arguments; it is $BG$ for $G$ a commutative group, you can explicitly write $BbbCP^infty$ as the projectivization of the field $Bbb C(x)$, and it inherits a group structure from that...). In general you differ by a section of $textAut(P)$.
– Mike Miller
Jul 19 at 0:53
@LuigiM I should edit my answer: that is not true as phrased. That is true here because $G$ is abelian (many arguments; it is $BG$ for $G$ a commutative group, you can explicitly write $BbbCP^infty$ as the projectivization of the field $Bbb C(x)$, and it inherits a group structure from that...). In general you differ by a section of $textAut(P)$.
– Mike Miller
Jul 19 at 0:53
1
1
I have edited and rephrased the answer in more bundle-theoretic language.
– Mike Miller
Jul 19 at 1:23
I have edited and rephrased the answer in more bundle-theoretic language.
– Mike Miller
Jul 19 at 1:23
Thanks a lot! That's an impressive answer. I'll meditate on that a little bit before accepting it:)
– Luigi M
Jul 19 at 2:37
Thanks a lot! That's an impressive answer. I'll meditate on that a little bit before accepting it:)
– Luigi M
Jul 19 at 2:37
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2855874%2ftransitive-action-of-h2m-bbb-z-on-spinc-structures-over-m%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password