A question about the existence of spin maps

Question

Let $M, N$ be two smooth manifolds, not necessarily spin. My question is the following:

How can we construct a non-constant spin map $f:M\to N$ of degree zero?

Here spin map means that $f$ preserves the second Stiefel-Whitney classes, i.e., $f^*w_2(TN)=w_2(TM)$.

More generally, are there some useful methods to construct spin maps between two given manifolds which are not necessarily spin? Most of the time, the existence of a spin map is just a hypothesis of some theorem in advance, however, I have no idea about extra concrete examples of spin maps when the manifolds are not spin.

I am sorry that I am not sure this question is quite naive or very advanced for Mathoverflow. Could you please give me some details? Thanks in advance.

Answer 1

• If $M$ and $N$ are spin, then every map between them is a spin map. In particular, there exist spin maps $M \to N$ of degree zero.
• If $M$ is spin and $N$ is not spin, then $f : M \to N$ is a spin map if and only if $f^*w_2(TN) = 0$. So nullhomotopic maps are spin maps of degree zero for example.
• If $M$ is not spin and $N$ is spin, then there are no spin maps $f : M \to N$ (the condition to be satisfied is $f^*0 = w_2(TM) \neq 0$).
• If $M$ and $N$ are both not spin, there may or may not be spin maps between them, and even if there are, there may be none of degree zero.

The following types of maps are always spin maps:

• covering maps,
• the projection of a principal bundle,
• the natural degree one map $X\# Y \to Y$ where $X$ is spin.

However, given $M$ and $N$ both not spin, I know of no general theorem which will give conditions for the existence of spin maps $M \to N$. What follows is a collection of examples and particular cases when $M$ and $N$ are both not spin.

Let $M = \mathbb{CP}^2$ and $N = (S^2\times S^2)/\mathbb{Z}_2$ where $\mathbb{Z}_2$ acts diagonally by the antipodal map. Both $M$ and $N$ are non-spin, but there are no spin maps between them. To see this, note that any map $f : \mathbb{CP}^2 \to (S^2\times S^2)/\mathbb{Z}_2$ lifts through the universal covering $\pi : S^2\times S^2 \to (S^2\times S^2)/\mathbb{Z}_2$ to a map $\tilde{f} : \mathbb{CP}^2 \to S^2\times S^2$ since $\mathbb{CP}^2$ is simply connected. As $f = \pi\circ\tilde{f}$ and $S^2\times S^2$ is spin, we see that $f^*w_2(TN) = \tilde{f}^*\pi^*w_2(TN) = \tilde{f}^*w_2(T(S^2\times S^2)) = \tilde{f}^*0 = 0 \neq w_2(TM)$.

If $M = N = \mathbb{CP}^2$, then a map $f : M \to N$ is a spin map if and only if it has odd degree. In particular, there are no spin maps of degree zero. For an example with $M \neq N$, let $M = \mathbb{CP}^2\#(S^2\times S^2)$ and $N = \mathbb{CP}^2$, then the natural degree one map $M \to N$ is a spin map. One can also show there are no spin maps of degree zero here either. More generally, we have the following:

Let $M$ be a closed, connected, oriented four-manifold. Then there is a spin map $f: M \to \mathbb{CP}^2$. Moreover, there is a spin map $f : M \to \mathbb{CP}^2$ of degree zero if and only if $w_2(TM)$ admits an integral lift $c$ with $c^2 = 0$.

Proof: Note that $[M, \mathbb{CP}^2] \cong [M, \mathbb{CP}^{\infty}] \cong H^2(M; \mathbb{Z})$. Under this correspondence, $[f] \mapsto f^*\alpha$ where $\alpha$ is a generator of $H^2(\mathbb{CP}^2; \mathbb{Z})$ . Since $\alpha$ reduces mod $2$ to $w_2(T\mathbb{CP}^2)$, we see that $f^*\alpha$ reduces mod $2$ to $f^*w_2(T\mathbb{CP}^2)$.

Note that $M$ is spin$^c$, so there is $c \in H^2(M; \mathbb{Z})$ whose mod $2$ reduction is $w_2(TM)$. Choosing an $f$ such that $f^*\alpha = c$, we see that $f^*w_2(T\mathbb{CP}^2) = w_2(TM)$, so $f$ is a spin map.

If $\beta$ denotes an oriented generator of $H^4(M; \mathbb{Z})$, then $(f^*\alpha)^2 = f^*\alpha^2 = (\deg f)\beta$, so $f$ has degree zero if and only if $(f^*\alpha)^2 = 0$. If $f$ is a spin map, then $c = f^*\alpha$ is an integral lift of $w_2(TM)$ and if $f$ has degree zero, then $c^2 = 0$. Conversely, if $c$ is an integral lift of $w_2(TM)$ with $c^2 = 0$, then the corresponding map (unique up to homotopy) $f : M \to \mathbb{CP}^2$ is a spin map of degree zero. $\square$

So, for example, $M = (S^2\times S^2)/\mathbb{Z}_2$ admits a spin map $M \to \mathbb{CP}^2$ of degree zero because $w_2(TM) \neq 0$ has integral lift the unique non-zero element $c$ of $H^2(M; \mathbb{Z}) \cong \mathbb{Z}_2$ which necessarily satisfies $c^2 = 0$. An example where $c$ is not torsion is $M = \mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ and $c = (1, 1)$ under the natural isomorphism $H^2(M; \mathbb{Z}) \cong \mathbb{Z}\oplus\mathbb{Z}$.

A corollary of the highlighted statement above is that if there is a spin map $f : M \to \mathbb{CP}^2$ of degree zero, then $w_2(TM)^2 = 0$ (as is the case for $M = (S^2\times S^2)/\mathbb{Z}_2$ and $M = \mathbb{CP}^2\#\overline{\mathbb{CP}^2}$).

One can use the same arguments to prove the following generalisation:

Let $M$ be a closed, connected, oriented $4n$-manifold. Then there is a spin map $f : M \to \mathbb{CP}^{2n}$ if and only if $M$ is spin$^c$. Moreover, there is a spin map $f : M \to \mathbb{CP}^{2n}$ of degree zero if and only if $w_2(TM)$ admits an integral lift $c$ with $c^{2n} = 0$.