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Let $M^d$ be a $d$-dimensional orientable spin manifold, and $N^4$ is a closed $4$-dimensional orientable submanifold of $M^d$.

  1. Is $N^4$ always spin?
  2. If $d=5$, is $N^4$ always spin?
  3. If $N^4$ is a boundary in $M^d$, is $N^4$ always spin?
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    $\begingroup$ Any $4$-dimensional manifold is a submanifold of $S^9$ so without some restrictions on $M^d$ you cannot say much. $\endgroup$ – Liviu Nicolaescu Apr 30 '17 at 22:40
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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.

  1. 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.

  2. 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.

  3. 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.

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    $\begingroup$ It might be worth pointing out that there are already counterexamples in codimension 2: $\mathbb{CP}^2\subset\mathbb{CP^3}$ and $\mathbb{CP}^3$ is spin while $\mathbb{CP}^2$ is not. $\endgroup$ – Robert Bryant May 1 '17 at 0:56
  • $\begingroup$ @RobertBryant: Nice example. Here $\nu = \mathcal{O}_{\mathbb{CP}^3}(1)|_{\mathbb{CP}^2} = \mathcal{O}_{\mathbb{CP}^2}(1)$ and $w_2(\mathcal{O}_{\mathbb{CP}^2}(1)) \neq 0$. $\endgroup$ – Michael Albanese May 1 '17 at 0:59
  • $\begingroup$ I think in 2 you are tacitly using that $N$ or equivalently $\nu$ is orientable (at least I understand "spin" as $w_1(TN)=w_2(TN)=0$). But this is not automatic. There is a nontrivial real line bundle over the Klein bottle $N'$ whose total space is orientable. Consider the double of the unit disk bundle. This is an orientable 3-manifold $M'$, which is automatically spin. Now, take a cross product of all this with $S^2$ to get a counterexample. $\endgroup$ – Sebastian Goette May 1 '17 at 10:32
  • $\begingroup$ @SebastianGoette: Yes I was. In the question it is assumed that $N$ is orientable, but I should have made it clear that orientability is needed when I made the more general statement. I will make the necessary edit now. $\endgroup$ – Michael Albanese May 1 '17 at 13:51

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