# Fibre sequences involving $BO(2)$, $BSO(3)$ and $BSU(2)$

In an answer to another question, it is remarked that there are fibre sequences $$\mathbf{RP}^2 \to BO(2) \to BSO(3),$$ $$\mathbf{RP}^\infty \to BSU(2) \to BSO(3).$$

The second one is explained by another fact stated in this answer, namely: if we have an exact sequence $$1 \to H \to G \to G/H \to 1$$, then it induces a fibre sequence $$BH \to BG \to B(G/H).$$ I don't know how to obtain the first one though.

Does someone have a reference for either of these facts?

You can directly verify that a short exact sequence of groups induces a fiber sequence of classifying spaces by using the long exact sequence in homotopy to compute say the homotopy groups of the fiber, and noting that classifying spaces are characterized by their homotopy groups.

Or you can write down models of classifying spaces and check. Let $$EG$$ be a contractible space on which $$G$$ acts freely, etc. Then the map $$BG \to B(G/H)$$ is modeled by $$BG \simeq EG \times_G E(G/H) \simeq BH \times_{G/H} E(G/H) \to * \times_{G/H} E(G/H) \simeq B(G/H),$$ whose homotopy fiber is $$BH$$.

On the other hand, the sequence $$\mathbb{R}P^2 \to BO(2) \to BSO(3)$$ is induced by the group homomorphism $$O(2) \hookrightarrow SO(3)$$ sending $$A$$ to $$A \oplus \det A$$. The fiber of $$BO(2) \to BSO(3)$$ is $$SO(3)/O(2)$$, which can be identified with $$\mathbb{R}P^2$$. ($$SO(3)$$ acts transitively on lines in $$\mathbb{R}^3$$, and the stabilizer of a line is $$O(2)$$.)

• I'm not sure about your very first paragraph; how do you mean that classifying spaces are characterised by their homotopy groups? The first answer I linked to gives two classifying spaces (of different groups though) having the same homotopy groups, while not being homotopy equivalent, so I think I'm missing a subtlety in that statement. Commented Nov 18, 2020 at 8:54
• I was imagining that the groups $G$ and $H$ are discrete, in which case $BG$ is a $K(G,1)$. But the answer you linked to would not be a counterexample anyway since $O(2) \not\cong SO(2) \times \mathbb{Z}/2$ as groups.
– JHF
Commented Nov 18, 2020 at 16:02
• The groups are indeed not isomorphic of course. But since really there is (up to equivalence) only one classifying space for a group, saying a classifying space is characterised by its homotopy group is a bit meaningless. Anyway, the rest of your answer answered my question. Thanks! Commented Nov 18, 2020 at 16:12
• I guess the point was that if you could show that the fiber $F$ of $BG \to B(G/H)$ has homotopy groups concentrated in degree 1, then $F$ is determined up to equivalence: it is a classifying space for $\pi_1 F$.
– JHF
Commented Nov 18, 2020 at 16:18
• Ah, I see what you meant. Commented Nov 18, 2020 at 16:20