Let $M^d$ be a $d$-dimensional orientable spin manifold, and $N^4$ is a closed $4$-dimensional orientable submanifold of $M^d$.
- Is $N^4$ always spin?
- If $d=5$, is $N^4$ always spin?
- If $N^4$ is a boundary in $M^d$, is $N^4$ always spin?
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Sign up to join this communityLet $M^d$ be a $d$-dimensional orientable spin manifold, and $N^4$ is a closed $4$-dimensional orientable submanifold of $M^d$.
Let $i$ denote an immersion $N \to M$. There is an exact sequence of vector bundles on $N$ given by
$$0 \to TN \to i^*TM \to \nu \to 0$$
where $\nu$ is the normal bundle. As total Stiefel-Whitney classes are multiplicative in short exact sequences (alternatively, $i^*TM \cong TN\oplus\nu$ smoothly), it follows that
\begin{align*} i^*w_1(M) &= w_1(N) + w_1(\nu)\\ i^*w_2(M) &= w_2(N) + w_1(N)w_1(\nu) + w_2(\nu). \end{align*}
If $M$ and $N$ are orientable, then we see that $w_1(\nu) = 0$ (i.e. $\nu$ is an orientable bundle) and hence $i^*w_2(M) = w_2(N) + w_2(\nu)$. If $M$ is spin, then we see that $N$ is spin if and only if $w_2(\nu) = 0$. More generally, if two of $M$, $N$, $\nu$ are spin, then so is the third.
No. There are examples of $N$ non-spin which embed in a spin manifold $M$ with $w_2(\nu) \neq 0$. In fact, we don't need any of the above to see that the answer is no. Note that any manifold $N$ embeds in $M = \mathbb{R}^d$ (or $S^d$ if you want something closed) for $d$ large enough by the Whitney embedding theorem, regardless of whether or not $N$ is spin.
Yes. In general, if $\dim M = \dim N + 1$, then $N$ has codimension one so $w_2(\nu) = 0$ and hence $w_2(N) = 0$. If $N$ is also orientable, we see that $N$ is spin.
No. The non-spin four-manifold $N = \mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ is orientedly null-cobordant, so there is a compact five-dimensional manifold with boundary $X$ with $\partial X = \mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. Again by the Whitney embedding theorem, $X$ embeds into $M = S^d$ for $d$ large enough. Even though $N$ is a boundary in $M$, $N$ is not spin.