Cohomology calculation for maps to the 2-sphere.

Question

Let $Y^3$ be a closed 3-manifold and $f\colon Y\to \operatorname{SO}(3)$, $g\colon Y\to S^2$ be smooth maps.

Define $g'\colon Y\to S^2$ be the following composition: \begin{equation}Y\xrightarrow{f\times g}SO(3)\times S^2 \xrightarrow{\pi} S^2, \end{equation} where $\pi\colon \operatorname{SO(3)}\times S^2\to S^2$ is the map $(A,x)\mapsto Ax$ for all $x\in S^2\subset \mathbb{R}^3$.

By composing embedding $S^2\hookrightarrow \mathbb{C}\mathbb{P}^\infty$, $g$ and $g'$ represent the cohomology class in $H^2(Y;\mathbb{Z})=[Y,\mathbb{C}\mathbb{P}^\infty]$. (Here, $[A,B]$ denotes the homotopy classes of maps from $A$ to $B$.)

The following is claimed in the Ozsvath-Szabo's famous annals paper.

\begin{equation} [g']=[g]+f^*(w)\in H^2(Y;\mathbb{Z}) \end{equation} where $w\in H^2(\operatorname{SO}(3);\mathbb{Z})=\mathbb{Z}/2$ is the generator.

What is the full detail of this claim?

Thanks in advance.

Answer 1

First, recall that the bijection between $H^2(Y)$ and $[Y,\mathbb{C}P^\infty]$ is obtained by pulling back the canonical generator of $H^2(\mathbb{C}P^\infty)$ and that the inclusion $S^2\rightarrow \mathbb{C}P^\infty$ pulls back the generator of $H^2(\mathbb{C}P^\infty)$ to the generator of $H^2(S^2)$. Hence, $[g]$ is really just shorthand for $g^\ast (z)$ where $z\in H^2(S^2)$ is a generator.

Likewise, $[g']$ is really just shorthand for $(f\times g)^\ast \pi^\ast(z)$

(I'm being a bit sloppy with the notion of "the" generator, but as long as you are consistent, the choice doesn't matter.)

So, the claim is that $(f\times g)^\ast \pi^\ast(z) = g^\ast(z) + f^\ast(w)$ where $z\in H^2(S^2)$ generates and $w\in H^2(SO(3))$ generates.

Claim 1: $\pi^\ast z = w + z \in H^2(SO(3)\times S^2)\cong H^2(SO(3))\oplus H^2(S^2)$

Proof: We can write $\pi^\ast z = aw+bz$ where $a\in \mathbb{Z}/2$ and $b\in\mathbb{Z}$. To determine $a$, notice the map $s:SO(3)\rightarrow S^2$ given by $A\mapsto An$ where $n$ is the north pole factors through $SO(3)\times S^2$. It follows that $s^\ast(z) = aw$. Further, the map $s$ is a fiber bundle with fiber $S^1$ as it comes from the natural group action $S^1\rightarrow SO(3)\rightarrow SO(3)/S^1 \cong S^2$.

If $a = 0$, then the map $s:SO(3)\rightarrow S^2$ is $0$ on $H^2$, hence by this MSE answer the map $s$ has a lift $\tilde{s}:SO(3)\rightarrow S^3$ to the Hopf fibration $p:S^3\rightarrow S^2$. This means the bundle $S^1\rightarrow SO(3)\rightarrow S^2$ is the pull back of the Hopf bundle by the identity map, which implies the bundle $S^1\rightarrow SO(3)\rightarrow S^2$ is isomorphic to the Hopf bundle. In particular $SO(3)$ and $S^3$ must then be diffeomorphic, an obvious contradiction. Hence, $a\neq 0$.

To determine $b$, follow a similar but easier process. The identity map $S^2\rightarrow S^2$ factors through $SO(3)\times S^2\rightarrow S^2$, and so $b = 1$.

Claim 2: $(f\times g)^\ast(w+z) = f^\ast w + g^\ast z$.

Proof: Almost by definition. We have $f = \pi_1\circ (f\times g)$ and $g = \pi_2\circ(f\times g)$ where $\pi_1$ and $\pi_2$ are the two projections onto $SO(3)$ and $S^2$. So, $(f\times g)^\ast w = (f\times g)^\ast(\pi_1^\ast w) = f^\ast w$ and likewise for $(f\times g)^\ast z$.

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Finally, putting Claim 1 and Claim 2 together gives \begin{align*} [g'] &= (f\times g)^\ast(\pi^\ast(z)) \\ &= (f\times g)^\ast(w+z)\\ &= f^\ast w + g^\ast z\\ &= f^\ast w + [g] \end{align*} as claimed.