multiple for first fractional Pontryagin class

Question

Consider the Whitehead tower of the classifying space of the special orthogonal group $BSO(n)$.

$$\cdots\to BString(n)\to BSpin(n)\to BSO(n)$$

There is an obstruction to lifting classifying maps into $BSpin(n)$ to $BString(n)$ that lives in $H^4(BSpin(n),\mathbb{Z})$. This obstruction is the so called "fractional first Pontryagin class" and is typically denoted by $\frac{1}{2}p_1$. The justification for naming is that if $\pi:BSpin(n)\to BSO(n)$ is the map coming from the tower and $p_1\in H^4(BSO(n),\mathbb{Z})$ is the classical first Pontryagin class then

$$\pi^\ast(p_1)=2(\frac{1}{2}p_1).$$ This result seems to be folklore and is stated for example in Section 4.4 of this paper. I am hoping to get a justification for the 2 in the above formula. I know from the construction of the Whitehead tower that $H^4(BSpin(n),\mathbb{Z})$ is infinite cyclic and that $\frac{1}{2}p_1$ is a generator. From this it follows that $\pi^\ast(p_1)=n(\frac{1}{2}p_1)$ and that it is just a matter of finding the correct multiple.

My guess is that the correct approach is to just compute things for a particular (hopefully simple) example and then the multiple will just fall out of the computation, but I don't know what the right example is or how to compute the relevant classes. Any suggestions or references would be greatly appreciated.

Answer 1

Denoting a generator of $H^4(BSpin(n); \mathbb{Z})$ by $\frac{1}{2}p_1$ isn't a sensible thing to do until you know that $\pi^*p_1$ is twice a generator. For example, in the paper they denote a generator of $H^4(BSpin(d); \mathbb{Z})$ by $\omega$, and then state that $\pi^*p_1 = 2\omega$ which motivates the notation $\omega = \frac{1}{2}p_1$. I believe your question is really, given that $\pi^*p_1 = n\omega$, why is $n = 2$?

As kamills explains in their answer, this follows from the Serre spectral sequence applied to $B\mathbb{Z}_2 \to BSpin \xrightarrow{\pi} BSO$. Recall that $Spin$ is $2$-connected with $\pi_3(Spin) \cong \mathbb{Z}$, so $BSpin$ is $3$-connected with $\pi_4(BSpin) \cong \mathbb{Z}$, and hence $H^4(BSpin; \mathbb{Z}) \cong \mathbb{Z}$. Since $E^{3,1}_{\infty} = E^{2,2}_{\infty} = E^{1,3}_{\infty} = 0$, there is a short exact sequence $0 \to E^{4,0}_{\infty} \to H^4(BSpin;\mathbb{Z}) \to E^{0,4}_{\infty} \to 0$. Unwinding the definitions (in particular, see Theorem $5.9$ of McCleary's User's Guide to Spectral Sequences and the text that precedes it), this gives a short exact sequence

$$0 \to \pi^*H^4(BSO; \mathbb{Z}) \hookrightarrow H^4(BSpin; \mathbb{Z}) \to i^*H^4(BSpin; \mathbb{Z}) \to 0$$

where $i$ is the inclusion of a fiber. Note that $H^4(BSO; \mathbb{Z})$ is generated by $p_1$, so $\pi^*H^4(BSO; \mathbb{Z})$ is generated by $\pi^*p_1$. The spectral sequence calculation shows that $i^*H^4(BSpin; \mathbb{Z}) \cong E_{\infty}^{0,4} \cong \mathbb{Z}_2$, so $\pi^*p_1 = 2\omega$ where $\omega$ is a generator of $H^4(BSpin; \mathbb{Z}) \cong \mathbb{Z}$.

The fact that $p_1$ is divisible by $2$ for a spin bundle $E$ can also be explained by the relation $p_1(E) \equiv w_2(E)^2 \bmod 2$. However, this only shows that $\pi^*p_1 =n\omega$ for some even $n$, and by replacing $\omega$ with $-\omega$ if necessary, we can suppose $n$ is non-negative. It follows that $p_1(E)$ is divisible by $n$ for every spin bundle $E$, so to show $n = 2$, it is enough to find one example of a spin bundle $E$ where $p_1(E)$ is only divisible by $2$. One such example is the tangent bundle of $\mathbb{HP}^2$ since $p_1(T\mathbb{HP}^2) = 2u$ where $u$ is generator for $H^4(\mathbb{HP}^2; \mathbb{Z})$.

Answer 2

One can show that $H^4(BSpin)$ is infinite cyclic since the map $BSpin(3) = \mathbb{H}P^\infty \to BSpin$ is highly connected (so it induces an isomorphism on $H^4$).

Note that given the double covering $Spin(n) \to SO(n)$ we have a fibration $B\mathbb{Z}/2 \to BSpin(n) \to BSO(n)$. To make notation simpler we can send $n$ to infinity (we're at least treating $n \geq 4$ and we're going to be looking for $H^4$, and we know that the stable/unstable classifying spaces have isomorphic cohomology in degrees less than $n$), and note that $B\mathbb{Z}/2 \simeq \mathbb{R}P^\infty$, so we have a fibration $\mathbb{R}P^\infty \to BSpin \to BSO$.

Using the Serre spectral sequence for this fibration is another way to see that the generator $y$ of $H^4(BSpin)$ satisfies $2y = p_1$, which justifies the notation $\frac{1}{2}p_1$ for this generator.

The Serre spectral sequence provides a bit more information than this, however; namely, the map $\pi^*$ is the composition $H^4(BSO;\mathbb{Z}) = E_2^{4,0} \xrightarrow{e} E_\infty^{4,0} \to H^4(BSpin;\mathbb{Z})$, where $e$ is the edge map (and in this case it's in fact an isomorphism basically just because there are no nonzero differentials into or out of the $(4,0)$ position) and the second map is multiplication by $2$ (follows from convergence, and the same fact that told us that $2y = p_1$ above). Hence $\pi^*$ must be multiplication by $2$.

This is a bit of a rough sketch since I'm too lazy to draw out all the spectral sequence diagrams/introduce all notation. But perhaps this will get you started and I can provide more details if necessary.