# What does reduction of structure group of principal bundle say?

Let $G$ be a Lie group and $\pi:P\rightarrow M$ be a principal $G$ bundle.

The notion of reduction of structure group is standard but I will recall here in case some one needs it.

Let $f:P(M,G)\rightarrow P'(M',G')$ be a morphism of principal bundles such that $f:P\rightarrow P'$ is an imbedding and $f:G\rightarrow G'$ is a monomorphism. If $M=M'$ abd the induced map $f:M\rightarrow M'$ is identity map, we call $P(M,G)$ to be reduced bundle for $P'(M,G')$ .

Given a principal bundle $P’(M’,G’)$ and a Lie subgroup $G$ of $G’$, we say the structure group $G’$ is reduced to $G$ if there is a reduced bundle $P(M,G)$.

Reduction of structure group says some thing interesting about manifolds involving it. For example,

• A manifold admits an almost-complex structure if the frame bundle on the manifold, whose fibers are $GL(2n,\mathbb{R})$, can be reduced to the group $GL(n,\mathbb{C})\subset GL(2n,\mathbb{R})$.
• A manifold is orientable if and only if its frame bundle can be reduced to the special orthogonal group, $SO(n,\mathbb{R})\subset GL(n,\mathbb{R})$.

I am interested in knowing similar results about reduction of structure group. Please add references (if possible, a sketch of the proof) for results you quote here. One result in one answer please.

• Results that are not very standard are most welcome. If you use some result in your paper that has something to do with reduction of structure group, please consider adding that as an answer. – Praphulla Koushik Aug 26 '18 at 4:44
• Your understanding of proof of any of the above three results I have mentioned in your own words is also most welcome :) – Praphulla Koushik Aug 26 '18 at 7:23
• @Arun I would call that a lift of the structure group, but I think if you said "reduction of structure group to Spin(n)" it would probably cause minimal confusion. – Mike Miller Aug 26 '18 at 17:24
• You have it backwards: $P'=P \times_G G'$ is automatically determined from $P$, so it is actually $P$ that is the reduction of structure group of $P'$. – Ben McKay Aug 26 '18 at 18:56
• Given any group homomorphism $\phi\colon H\to G$ and a principal $G$-bundle $P\to M$ (or even a $G$-fibre bundle), one can ask for the existence of a principal $H$-bundle $Q\to M$ such that $P \simeq Q\times_H G$, where $H$ acts on $G$ via $\phi$. This is equivalent to asking if the Čech cocycle arising from the transition functions for $P$ is in the image of $H^1(M,H) \to H^1(M,G)$. There are a large number of other equivalent formalisations of this. For the fibre bundle case, the version using transition functions is the one that makes immediate sense. – David Roberts Aug 26 '18 at 23:56

Let $G$ be a topological group and $M$ be a smooth manifold. Then, a reduction of the structure group of the frame bundle from $\mathrm{GL}_n(\mathbb R)$ to $\mathrm{GL}_n(\mathbb R)\times G$ is equivalent data to a principal $G$-bundle $Q\to M$.
Here's a proof sketch: if $H$ and $H'$ are groups, an $(H\times H')$-torsor is the same thing as a product of an $H$-torsor and an $H'$-torsor. So the fiber of the principal $(\mathrm{GL}_n(\mathbb R)\times G)$-bundle we obtained from the frame bundle at some $x\in M$ is a product of a $\mathrm{GL}_n(\mathbb R)$-torsor $P_x$ and a $G$-torsor $Q_x$, and $P_x$ is the $\mathrm{GL}_n(\mathbb R)$ of bases of $T_xM$. Since $Q_x$ varies smoothly, it defines a principal $G$-bundle $Q\to M$.
• I am not familiar with this definition of reduction of structure group where we are not considering injective morphism of groups... Are you saying there is some natural map $GL_n(\mathbb{R})\times G\rightarrow GL_n(\mathbb{R})$ with which we are thinking of structure group reduction?? I am not able to think of any natural map... I see there is a map $GL_n(\mathbb{R})\rightarrow GL_n(\mathbb{R})\times G$ that sends $A$ to $(A,1)$... – Praphulla Koushik Aug 26 '18 at 16:52
• @PraphullaKoushik right, the definition of "reduction of the structure group" that I learned is not quite the one on Wikipedia; it allows for noninjective maps. In particular, the natural map $\mathrm{GL}_n(\mathbb R)\times G\to\mathrm{GL}_n(\mathbb R)$ is projection onto the first factor. – Arun Debray Aug 26 '18 at 16:58
• Why cannot $Q_x$ somehow "entangle" with $P_x$ along the way? Say, if $\mathrm{GL}_n(\mathbb R)$ contains a copy of $G$, in principle it might secretly "swap" with the "external" $G$... – მამუკა ჯიბლაძე Aug 27 '18 at 4:22
• @მამუკაჯიბლაძე I think in that case you won't get back the frame bundle when you take the quotient, but rather some other $\mathrm{GL}_n(\mathbb R)$-bundle. – Arun Debray Aug 27 '18 at 13:33