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?



Is a 4-dimensional submanifold of a spin manifold always spin?
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.


