# ways to see whether the Pontryagin class of a quaternionic line bundle over a CW-complex is zero

the first pontryagin class of a quaternionic line bundle over a CW-complex is zero if and only if the quaternionic line bundle is trivial or not?

Let $\xi^\mathbb{H}$ be a given quaternionic line bundle. Is there any method to see whether $p_1(\xi^{\mathbb{H}})=0$? Could you give references?

Quaternionic line bundles are classified by maps to $B GL_1(\mathbb{H}) \cong BSp(1)$. This has cohomology a polynomial algebra

$$H^{*}(BSp(1)), \mathbb{Z}) \cong \mathbb{Z}[p_1]$$

on the Pontryagin class $p_1 \in H^4$.

For complex line bundle and chern class, I obtained that the first chern class is zero if and only if the complex line bundle is trivial. This is obtained by $$c_1(\xi^\mathbb{C})=e((\xi^\mathbb{C})_\mathbb{R})=o_2((\xi^\mathbb{C})_\mathbb{R}).$$ By page 140, 143, 158, Charachateristic class, Milnor and Stasheff, we obtain that the first chern class is zero if and only if the complex line bundle is trivial.

But for quaternionic line bundle $\xi^{\mathbb{H}}$, page 174, Charachateristic class, Milnor and Stasheff, $$p_1(\xi^\mathbb{H})=p_1((\xi^\mathbb{H})_\mathbb{R})=-c_{2}((\xi^\mathbb{H})_\mathbb{R}\otimes\mathbb{C}).$$ I do not know how to do.

• For your edited question, again, as with your question about complex line bundles, it depends on what you know about your quaternionic line bundle. – Qiaochu Yuan Feb 4 '15 at 3:28

No, this is not true. Quaternionic line bundles are classified by maps to $B GL_1(\mathbb{H}) \cong BSp(1)$. This has cohomology a polynomial algebra
$$H^{\bullet}(BSp(1)), \mathbb{Z}) \cong \mathbb{Z}[p_1]$$
on the Pontryagin class $p_1 \in H^4$, so there are no more integral characteristic classes, but a quaternionic line bundle is not determined by its Pontryagin class, and in particular the map $BSU(2) \to B^4 \mathbb{Z}$ coming from $p_1$ is not a homotopy equivalence.
To see this it suffices to recall that $\pi_i(BG) \cong \pi_{i-1}(G)$, so that in particular
$$\pi_5(BSp(1)) \cong \pi_4(Sp(1)) \cong \pi_4(S^3) \cong \mathbb{Z}_2.$$
Hence there is a nontrivial quaternionic line bundle over $S^5$, but such a quaternionic line bundle must have trivial Pontryagin class. In fact since $S^3$ has infinitely many nonzero homotopy groups there are nontrivial quaternionic line bundles over spheres of arbitrarily large dimension.