Let $S^m$ be a standard sphere of dimension $m=n+4k$, and let $M$ be any closed Riemaninan manifold of dimension $n$ with nonnegative curvature operator.
My question: Is there always a smooth spin map $f:S^m\to M$ of nonzero $\widehat{A}$-degree?
Could you please give me some details? Thanks in advance.
For now, let $M$ be any closed oriented $n$-dimensional manifold, and $f : S^m \to M$ a smooth map.
First suppose $k \geq 1$.
Since $m = n + 4k > 2$, we have $H^2(S^m; \mathbb{Z}_2) = 0$, and hence $f^*w_2(TM) = 0 = w_2(TS^m)$, so every map $f : S^m \to M$ is a spin map.
If $\omega$ is a volume form on $M$ with $\displaystyle\int_M\omega = 1$, then the $\hat{A}$-degree of $f$ is $\displaystyle\int_{S^m}f^*\omega\wedge\hat{A}_k(TS^m) = 0$ since $\hat{A}_k(TS^m)$ is a polynomial in the Pontryagin classes of $TS^m$ which vanish as $TS^m$ is stably trivial.
Now suppose $k = 0$, so $m = n$.
If $m \neq 2$, then the same argument applies to show that every map $f : S^m \to M$ is spin. If $m = 2$, then $M$ is a closed oriented surface and hence spin, so $f^*w_2(TM) = f^*0 = 0 = w_2(TS^2)$; again, we see that every map is a spin map.
When $k = 0$, the $\hat{A}$-degree of $f$ is precisely the usual degree of $f$. There exists a non-zero degree map $f : S^m \to M$ if and only if $\widetilde{M}$, the universal cover of $M$, is a rational homology sphere, see here. There are examples of such manifolds with non-negative curvature operator, such as the Wu manifold $SU(3)/SO(3)$.
