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?
  • 13
    $\begingroup$ 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

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.

  • 7
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.