# $\pi_0$ in the long exact sequence of a fibration and quaternionic projective space

I am doing a past paper for an introductory course in algebraic topology. The question is

Calculate the homology of the quaternionic projective space. What can you say about its homotopy groups?

I figured that we have a CW decomposition with one cell in every $4k$-th dimension. All boundary maps are 0, so that $$H^{4k}(\mathbb HP^n,\mathbb Z)=\mathbb Z\\H^i(\mathbb HP^n,\mathbb Z)=0,\quad i\neq 4k$$

In particular, if the fundamental group is Abelian, it must be trivial.

For the homotopy groups, there is a fibration $$\mathrm{Sp}(1)\to S^{4n+3}\to\mathbb HP^n$$ so that $\mathbb HP^n$ has higher homotopy groups isomorphic to those of $S^{4n+3}$.

But the long exact sequence of the fibration ends in $$\cdots\to\pi_1S^{4n+3}\to\pi_1\mathbb HP^n\to\pi_0\mathrm{Sp}(1)\to\pi_0S^{4n+3}\to\pi_0\mathbb HP^n\to 0$$

But $\mathrm{Sp}(1)$ has two path-components, so this seems to suggest that $\pi_1\mathbb HP^n=\mathbb Z_2$, which contradicts the fact that $H_1(\mathbb HP^n,\mathbb Z)=0$. Where have I gone wrong?

## 1 Answer

Sp(1) doesn't have two path components. It's just the unit quaternions, which are $S^3$.

• This also means that the higher homotopy of $\mathbb{H}P^n$ will be somewhat more complicated. – Miha Habič Jul 7 '13 at 20:26
• Oh, you're right. I mixed up Sp(1) with Spin(1). A simple dimension count should have shown this. Sorry for the confusion and thank you for your answer. The long exact sequence then allows me to calculate the homotopy groups (properly this time). – Earthliŋ Jul 7 '13 at 20:28