A simple example of a $G$-principal bundle that cannot be lifted to an $F$-principal bundle along an epimorphism $F\to G$.

Question

I am trying to understand spin structures, in particular how they may fail to exist. To start with, I would like to see a most simple example of a $G$-principal bundle that cannot be lifted to an $F$-principal bundle along a given epimorphism $F\to G$.

Since it is written that all compact, orientable manifolds of dimension 3 or less are spin, I hope that there is an homomorphism $F\to G$ for which an example where liftings fail to exist is easier to construct and to understand than for $\operatorname{Spin}(n)\to SO_n$.

To be precise, I am asking for 3 things:

1. a principal $G$-bundle $E\to B$,
2. an epimomorphism of topological groups $F\to G$,
3. an argument that shows that there exist no principal $F$-bundle $D\to B$ such that the given principal $G$-bundle $E\to B$ could be viewed as its quotient by the action of the kernel of $F\to G$.

I hope I understand correctly that this would be "analogous" to the absence of spin structures. If I am mistaken, I'll appreciate an explanation why not.

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Update.

I have a conjecture that the $\mathbf{Z}/2\mathbf{Z}$-principal bundle $S^2\to\mathbf{R}P^2$ with the epimorphism $\mathbf{Z}/4\mathbf{Z}\to\mathbf{Z}/2\mathbf{Z}$ could be an example I am looking for.

Answer 1

Indeed, your $\mathbb{Z}_4\rightarrow \mathbb{Z}_2$ map is a perfect example. Because there are two $\mathbb{Z}_2$s in play (the quotient, and the kernel), I'll use $K$ for the kernel and $Q$ for the quotient.

What you want to see is:

There is no principal bundle $\mathbb{Z}_4\rightarrow P\rightarrow \mathbb{R}P^2$ for which a) $P/K \cong S^2$ and
b) $Q$ acts on $P/K$ via the antipodal map.

But, in fact, more is true: asking for condition a) is already too much.

Proposition: There is no principal bundle $\mathbb{Z}_4\rightarrow P\rightarrow \mathbb{R}P^2$ for which condition a) above holds.

To see this, let's assume for a contradiction that we have such a principal bundle $\mathbb{Z}_4\rightarrow P\rightarrow \mathbb{R}P^2$ for which the subaction by $K$ on $P$ has quotient $S^2$.

Now, $P$ must be a $4$-fold covering of $\mathbb{R}P^2$, so there are a few possibilities.

1. $P$ is a disjoint union of $4$ copies of $\mathbb{R}P^2$
2. $P$ is a disjoint union of $2$ copies of $\mathbb{R}P^2$ and one copy of $S^2$.
3. $P$ is a disjoint union of $2$ copies of $S^2$.

We will rule these out one at a time.

In case 1, because the quotient $P/\mathbb{Z}_4$ must be connected, it follows that $\mathbb{Z_4}$ permutes the copies of $\mathbb{R}P^2$ around. But then $P/K$ is a disjoint union of two copies of $\mathbb{R}P^2$, so is not $S^2$. Thus, case 1 cannot occur.

In case 2, because each element of $\mathbb{Z}_4$ acts by a diffeomorphism on $P$, while $S^2$ and $\mathbb{R}P^2$ are not diffeomorphic, it follows that no element of $\mathbb{Z}_4$ maps a point in $S^2$ to a point in either copy of $\mathbb{R}P^2$. In particular, the quotient $P/\mathbb{Z}_4$ is necessarily disconnected, so case 2 cannot occur either.

Thus, we are left with case 3. We first claim that the element $1\in \mathbb{Z}_4 = \{0,1,2,3\}$ must interchange the two copies of $S^2$. Indeed, because $1$ acts as a diffeomorphism on $P$ it either preserves both copies of $S^2$ or it interchanges them. But $1$ generates $\mathbb{Z}_4$, so if it preserves both copies, then all of $\mathbb{Z}_4$ preserves both copies, which then gives a disconnected quotient $P/\mathbb{Z}_4$, a contradiction.

Now, because $1$ interchanges the two copies of $S^2$, it now follows that $2 = 1 + 1$ preserves the two copies of $S^2$. But $2$ generates $K$, so now we reach the final contradiction that $P/K$ is disconnected, so is not diffeomorphic to $S^2$.