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In a nutshell, the derivation of the Stiefel-Whitney and Chern classes for a (let's say) closed space $X$ is as follows: For $k = {\mathbb{R}}$ or $\mathbb{C}$, any $k$-line bundle $\xi \to X$ is the pullback $f^*\gamma$ of the tautological bundle $\gamma \to k\mathbb{P}^\infty$; and since $k{\mathbb{P}}^\infty$ is a $K(\pi, n)$, the pullback $f^*[k\mathbb{P}^\infty]$ classifies $\xi$ up to homotopy. For $k = \mathbb{R}$, we have $\mathbb{RP}^\infty = K(\mathbb{Z}_2, 1)$; and for $k = \mathbb{C}$, we have $\mathbb{CP}^\infty = K(\mathbb{Z}, 2)$. Thus in the real case, we have a characteristic class $w_1(\xi)\in H^1(X, \mathbb{Z}_2)$; and in the complex case, we have a characteristic class $c_1(\xi)\in H^2(X, \mathbb{Z})$. For a general bundle $\xi \to X$, we can use the splitting principle to define classes $w(\xi)\in H^*(X, \mathbb{Z}_2)$ and $c(\xi)\in H^{2*}(X, \mathbb{Z})$.

This construction seems to rely on the apparent coincidence that $\mathbb{RP}^\infty$ and $\mathbb{CP}^\infty$ are both Eilenberg-MacLane spaces. I therefore have a question:

We can perform the analogous construction for quarternionic line bundles (unless I'm mistaken) and similarly show that they're all pullbacks of the tautological line bundle $\gamma\to \mathbb{HP}^\infty$. Unfortunately, $\mathbb{HP}^\infty$ is not a $K(\pi, n)$. Is there an analogue of the Stiefel-Whitney or Chern class that classifies quaternionic bundles; and if so, how is it derived?

My apologies in advance for the vagueness of the question. (Something like, "Go read [book]" is a totally valid response.)

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  • $\begingroup$ As far as book recommendations, I think the canonical reference is Milnor and Stasheff. $\endgroup$
    – Qiaochu Yuan
    Jun 19, 2014 at 6:54
  • $\begingroup$ I've actually gone through Milnor and Stasheff. I liked it, but it seemed to present a lot of ad-hoc constructions rather than going through, for example, a full treatment of classifying spaces. $\endgroup$
    – anomaly
    Jun 19, 2014 at 7:01

1 Answer 1

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The formalism you want is that of principal bundles. For $K = \mathbb{R}, \mathbb{C}, \mathbb{H}$, classifying $n$-dimensional $K$-vector bundles is equivalent to classifying principal $\text{GL}_n(K)$-bundles. For any topological group $G$, there is a classifying space $BG$ for classifying principal $G$-bundles, and by the Yoneda lemma a characteristic class of principal $G$-bundles is precisely a cohomology class in the cohomology of $BG$.

For the case of quaternionic line bundles the relevant topological group is $G = \text{GL}_1(\mathbb{H})$, which up to homotopy can be replaced with $G = \text{Sp}(1)$. Hence the classifying space of quaternionic line bundles is $B\text{Sp}(1)$, and you want to compute the cohomology of this classifying space. The answer is that

$$H^{\bullet}(B \text{Sp}(1), \mathbb{Z}) \cong \mathbb{Z}[p_1]$$

is a polynomial ring on the universal first Pontryagin class $p_1 \in H^4(B \text{Sp}(1), \mathbb{Z})$. Similarly the integral cohomology of $B \text{Sp}(n)$, which gives integral characteristic classes for $n$-dimensional quaternionic vector bundles, is a polynomial ring on classes $p_k \in H^{4k}, k \le n$.

I don't really understand your second question.

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  • $\begingroup$ That's wonderful, thank you very much. As for the second question...yeah, it wasn't really well-formed. I think I'll just stick to the first one. $\endgroup$
    – anomaly
    Jun 19, 2014 at 6:31
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    $\begingroup$ I'll also warn you that it might be tempting to hypothesize that $B \text{Sp}(1)$ is a $K(\mathbb{Z}, 4)$. In fact it has the wrong cohomology - the cohomology of $K(\mathbb{Z}, 4)$ is more complicated, which reflects the existence of more complicated cohomology operations once you get this high - as well as the wrong homotopy; $\text{Sp}(1) \cong S^3$, so the homotopy groups of $B \text{Sp}(1)$ are those of $S^3$ shifted. $\endgroup$
    – Qiaochu Yuan
    Jun 19, 2014 at 6:34
  • $\begingroup$ Ah, because cohomology operations are represented by elements of some $H^*(K(\pi, n), \pi')$ by the Yoneda lemma. Certainly the integral homotopy groups of $S^3$ are a bit complicated. This is terrific--- thank you again for explaining things so clearly. $\endgroup$
    – anomaly
    Jun 19, 2014 at 6:42
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    $\begingroup$ No problem. To relate this construction back to the construction using the splitting principle, from the perspective of classifying spaces the point is that the operation of taking the direct sum of $n$ line bundles is represented (let me consider only the real case for now) by a map $B \text{O}(1) \times ... \times B \text{O}(1) \to B \text{O}(n)$, and that the pullback in cohomology over $\mathbb{F}_2$ along this map is 1) injective, 2) by symmetry, lands in the $S_n$-invariant part of the cohomology of the product, and 3) is in fact isomorphic to this $S_n$-invariant part. $\endgroup$
    – Qiaochu Yuan
    Jun 19, 2014 at 6:47
  • $\begingroup$ (And a similar story for the Chern and Pontryagin classes over $\mathbb{Z}$.) $\endgroup$
    – Qiaochu Yuan
    Jun 19, 2014 at 6:48

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