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Chris
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I asked this question on math stack exchange here, but I wonder if it would be better received here.

Let $\pi:P\rightarrow M$ and $\pi':P'\rightarrow M$ be principal $G$ and $H$ bundles respectively, and $f:G\rightarrow H$ a Lie group homomorphism. Recall that an $f$-reduction of principal bundles is the principal bundle $P$ together with a smooth bundle homomorphism $F:P\rightarrow P'$ that is $f$ equivariant. In other words we have the following two properties: $$\pi'\circ F=\pi$$ and: $$F(p\cdot g)=F(p)\cdot f(g)$$ for all $p\in P$, and $g\in G$.

My question is then this: let $U\subset M$ be an open neighborhood of $x\in M$, and $(U,\phi)$ be a principal bundle chart for $P'$, that is an $H$ equivariant diffeomorphism: $$\phi:\pi'^{-1}(U)\longrightarrow U\times H$$ Does there then exist a principal bundle chart $(U,\psi)$ for $P$ that is compatible with the $f$ reduction, in the sense that: $$\phi\circ F=(\text{Id}_U\times f)\circ \psi$$ In the case where $f$ is a Lie group isomorphism I believe this trivial. If $f$ is an embedding, then $F$ is a $G$ reduction of $H'$, and so $P$ is a principal subbundle of $P'$. In this case, let: $$\phi(p')=(\pi'(p'),h(p'))$$ and: $$\psi(p)=(\pi(p),g(p))$$ for some smooth equivariant maps $g:P_U\rightarrow G$ and $h:P_U\rightarrow H$, then our requirement reduces to: $$\phi(F(p))=(\pi'\circ F(p),h(F(p)))=(\pi(p),f(g(p)))=(\text{Id}_U\times f)\circ \psi(p)$$ since $F$ is a bundle homomorphism we have that: $$h(F(p))=f(g(p))$$ Which I believe is always the case, due to the equivariance of $F$ and $h$, but I can't rigorously demonstrate why. If I am mistaken please let me know.

The case I am more interested is when $f$ is a surjective Lie group homomorphism. In particular, this implies that $G/\ker f\cong H$, as Lie groups, and that $f$ is a smooth submersion. In this case, I strongly suspect that it is not true, as I suspect it would imply that global sections of $f$ exist, which is false in general.

My motivation for this problem, is to specifically apply it when $M$ is an oriented Riemannian spin manifold, and $P=\text{Spin}(M)$, and $P'=SO(M)$, i.e. $P$ is a spin structure associated to the bundle of oriented orthonormal frames.

If such charts existed, I would be easily able to prove that for every local section of $SO(M)$, there exist precisely two local sections of $\text{Spin}(M)$ that map to original local section of $SO(M)$ under $F$.

Any help would be greatly appreciated, as I have not been able to find anything regarding questions such as this in the literature.

Edit: At least in the case of $\text{Spin}(M)$ and $SO(M)$, I believe the existence of such charts to be true actually. This is because th existence of such sections of $\text{Spin}(M)$ implies the existence of such charts.

I asked this question on math stack exchange here, but I wonder if it would be better received here.

Let $\pi:P\rightarrow M$ and $\pi':P'\rightarrow M$ be principal $G$ and $H$ bundles respectively, and $f:G\rightarrow H$ a Lie group homomorphism. Recall that an $f$-reduction of principal bundles is the principal bundle $P$ together with a smooth bundle homomorphism $F:P\rightarrow P'$ that is $f$ equivariant. In other words we have the following two properties: $$\pi'\circ F=\pi$$ and: $$F(p\cdot g)=F(p)\cdot f(g)$$ for all $p\in P$, and $g\in G$.

My question is then this: let $U\subset M$ be an open neighborhood of $x\in M$, and $(U,\phi)$ be a principal bundle chart for $P'$, that is an $H$ equivariant diffeomorphism: $$\phi:\pi'^{-1}(U)\longrightarrow U\times H$$ Does there then exist a principal bundle chart $(U,\psi)$ for $P$ that is compatible with the $f$ reduction, in the sense that: $$\phi\circ F=(\text{Id}_U\times f)\circ \psi$$ In the case where $f$ is a Lie group isomorphism I believe this trivial. If $f$ is an embedding, then $F$ is a $G$ reduction of $H'$, and so $P$ is a principal subbundle of $P'$. In this case, let: $$\phi(p')=(\pi'(p'),h(p'))$$ and: $$\psi(p)=(\pi(p),g(p))$$ for some smooth equivariant maps $g:P_U\rightarrow G$ and $h:P_U\rightarrow H$, then our requirement reduces to: $$\phi(F(p))=(\pi'\circ F(p),h(F(p)))=(\pi(p),f(g(p)))=(\text{Id}_U\times f)\circ \psi(p)$$ since $F$ is a bundle homomorphism we have that: $$h(F(p))=f(g(p))$$ Which I believe is always the case, due to the equivariance of $F$ and $h$, but I can't rigorously demonstrate why. If I am mistaken please let me know.

The case I am more interested is when $f$ is a surjective Lie group homomorphism. In particular, this implies that $G/\ker f\cong H$, as Lie groups, and that $f$ is a smooth submersion. In this case, I strongly suspect that it is not true, as I suspect it would imply that global sections of $f$ exist, which is false in general.

My motivation for this problem, is to specifically apply it when $M$ is an oriented Riemannian spin manifold, and $P=\text{Spin}(M)$, and $P'=SO(M)$, i.e. $P$ is a spin structure associated to the bundle of oriented orthonormal frames.

If such charts existed, I would be easily able to prove that for every local section of $SO(M)$, there exist precisely two local sections of $\text{Spin}(M)$ that map to original local section of $SO(M)$ under $F$.

Any help would be greatly appreciated, as I have not been able to find anything regarding questions such as this in the literature.

I asked this question on math stack exchange here, but I wonder if it would be better received here.

Let $\pi:P\rightarrow M$ and $\pi':P'\rightarrow M$ be principal $G$ and $H$ bundles respectively, and $f:G\rightarrow H$ a Lie group homomorphism. Recall that an $f$-reduction of principal bundles is the principal bundle $P$ together with a smooth bundle homomorphism $F:P\rightarrow P'$ that is $f$ equivariant. In other words we have the following two properties: $$\pi'\circ F=\pi$$ and: $$F(p\cdot g)=F(p)\cdot f(g)$$ for all $p\in P$, and $g\in G$.

My question is then this: let $U\subset M$ be an open neighborhood of $x\in M$, and $(U,\phi)$ be a principal bundle chart for $P'$, that is an $H$ equivariant diffeomorphism: $$\phi:\pi'^{-1}(U)\longrightarrow U\times H$$ Does there then exist a principal bundle chart $(U,\psi)$ for $P$ that is compatible with the $f$ reduction, in the sense that: $$\phi\circ F=(\text{Id}_U\times f)\circ \psi$$ In the case where $f$ is a Lie group isomorphism I believe this trivial. If $f$ is an embedding, then $F$ is a $G$ reduction of $H'$, and so $P$ is a principal subbundle of $P'$. In this case, let: $$\phi(p')=(\pi'(p'),h(p'))$$ and: $$\psi(p)=(\pi(p),g(p))$$ for some smooth equivariant maps $g:P_U\rightarrow G$ and $h:P_U\rightarrow H$, then our requirement reduces to: $$\phi(F(p))=(\pi'\circ F(p),h(F(p)))=(\pi(p),f(g(p)))=(\text{Id}_U\times f)\circ \psi(p)$$ since $F$ is a bundle homomorphism we have that: $$h(F(p))=f(g(p))$$ Which I believe is always the case, due to the equivariance of $F$ and $h$, but I can't rigorously demonstrate why. If I am mistaken please let me know.

The case I am more interested is when $f$ is a surjective Lie group homomorphism. In particular, this implies that $G/\ker f\cong H$, as Lie groups, and that $f$ is a smooth submersion. In this case, I strongly suspect that it is not true, as I suspect it would imply that global sections of $f$ exist, which is false in general.

My motivation for this problem, is to specifically apply it when $M$ is an oriented Riemannian spin manifold, and $P=\text{Spin}(M)$, and $P'=SO(M)$, i.e. $P$ is a spin structure associated to the bundle of oriented orthonormal frames.

If such charts existed, I would be easily able to prove that for every local section of $SO(M)$, there exist precisely two local sections of $\text{Spin}(M)$ that map to original local section of $SO(M)$ under $F$.

Any help would be greatly appreciated, as I have not been able to find anything regarding questions such as this in the literature.

Edit: At least in the case of $\text{Spin}(M)$ and $SO(M)$, I believe the existence of such charts to be true actually. This is because th existence of such sections of $\text{Spin}(M)$ implies the existence of such charts.

Source Link
Chris
  • 391
  • 1
  • 11

Existence (or non existence) of principal bundle charts compatible with an $f$-reduction

I asked this question on math stack exchange here, but I wonder if it would be better received here.

Let $\pi:P\rightarrow M$ and $\pi':P'\rightarrow M$ be principal $G$ and $H$ bundles respectively, and $f:G\rightarrow H$ a Lie group homomorphism. Recall that an $f$-reduction of principal bundles is the principal bundle $P$ together with a smooth bundle homomorphism $F:P\rightarrow P'$ that is $f$ equivariant. In other words we have the following two properties: $$\pi'\circ F=\pi$$ and: $$F(p\cdot g)=F(p)\cdot f(g)$$ for all $p\in P$, and $g\in G$.

My question is then this: let $U\subset M$ be an open neighborhood of $x\in M$, and $(U,\phi)$ be a principal bundle chart for $P'$, that is an $H$ equivariant diffeomorphism: $$\phi:\pi'^{-1}(U)\longrightarrow U\times H$$ Does there then exist a principal bundle chart $(U,\psi)$ for $P$ that is compatible with the $f$ reduction, in the sense that: $$\phi\circ F=(\text{Id}_U\times f)\circ \psi$$ In the case where $f$ is a Lie group isomorphism I believe this trivial. If $f$ is an embedding, then $F$ is a $G$ reduction of $H'$, and so $P$ is a principal subbundle of $P'$. In this case, let: $$\phi(p')=(\pi'(p'),h(p'))$$ and: $$\psi(p)=(\pi(p),g(p))$$ for some smooth equivariant maps $g:P_U\rightarrow G$ and $h:P_U\rightarrow H$, then our requirement reduces to: $$\phi(F(p))=(\pi'\circ F(p),h(F(p)))=(\pi(p),f(g(p)))=(\text{Id}_U\times f)\circ \psi(p)$$ since $F$ is a bundle homomorphism we have that: $$h(F(p))=f(g(p))$$ Which I believe is always the case, due to the equivariance of $F$ and $h$, but I can't rigorously demonstrate why. If I am mistaken please let me know.

The case I am more interested is when $f$ is a surjective Lie group homomorphism. In particular, this implies that $G/\ker f\cong H$, as Lie groups, and that $f$ is a smooth submersion. In this case, I strongly suspect that it is not true, as I suspect it would imply that global sections of $f$ exist, which is false in general.

My motivation for this problem, is to specifically apply it when $M$ is an oriented Riemannian spin manifold, and $P=\text{Spin}(M)$, and $P'=SO(M)$, i.e. $P$ is a spin structure associated to the bundle of oriented orthonormal frames.

If such charts existed, I would be easily able to prove that for every local section of $SO(M)$, there exist precisely two local sections of $\text{Spin}(M)$ that map to original local section of $SO(M)$ under $F$.

Any help would be greatly appreciated, as I have not been able to find anything regarding questions such as this in the literature.