# How hard is it to endow a $\textit{Spin}^{c}$ structure on four-dimensional manifolds?

I am in a certain math conference and we came across Seiberg-Witten equations. Since I am really novice in the field, I asked if all "reasonable" four manifolds carry a $\textit{spin}^{c}$ structure. I was under the impression that $\textit{Spin}$ structure is something rather rigid, and consider the fact we had uncountably many smooth structure in dimension 4, it seems quite unlikely that a $\textit{spin}^{c}$ structure would exist for all smooth four manifolds.

Not long afterwards a colleague told me this is settled; a source claimed that $\textit{spin}^{c}$ structure exists for all smooth manifolds of dimension less than or equal to 4. I was a bit surprised and decided to double check. It was not clear to me why it would work for all four dimensional manifolds, and if it works - why stop at 4? (this statement is obviously wrong for trivial reasons, as $\textit{spin}^{c}$ structure only exists on oriented manifolds).

I think I now want to ask about this question seriously. I read the wikipedia article carefully, but after tracking back the source, I do not see the book (Kirby calculus and 4 manifolds) proved any statement like that. On other hand, I think I saw on the Seiberg-Witten invariant page that $\textit{spin}^{c}$ structure exists for all smooth, compact oriented four manifolds. This is a reasonable statement to believe, but how to prove it? The wikipedia article on $\textit{spin}^{c}$ structure claimed that:

"A $\textit{spin}^{c}$ structure exists if the manifold is orientable and ....in other words, the third integral Stiefel-Whitney class vanishes)"

The proof is in the section "details". Now, I have trouble believing the all smooth, compact oriented four manifolds have $w_{3}(M)=0$. Since an axiomatic point of view is obviously not helpful, I tried to review the corresponding chapter in Hatcher, which says(bottom of page 75):

"..For each cell the obstruction to extending lies in $\pi_{2}(SO(n))$. This group happens to be $0$ for all $n$, so the section automatically extends over $B^{3}$. "

According to this, for all reasonable manifolds(not just dimension 4); $w_{3}(M)=0$. If I am not mistaken this fact $(\pi_{2})(G)=0$ if $G$ is a semisimple Lie group) is proved by Bott. However, this definition (Whitney's original definition) seems to be subtlely different from the modern definition(see the next page in Hatcher). So I want to ask if my reasoning process works through. I thought I do "know" characteristic classes, but obviously I only knew them at a superficial level that I cannot prove this fact myself without referring anything.

• Four times as hard as it is to do for one manifold? (I'm joking, but I do think writing four-manifold or four-dimensional manifold is much clearer. See object–verbal noun compounds as well as the Wikipedia page on 4-manifolds.) – Zev Chonoles Aug 3 '13 at 7:10
• The point is the hyphen in these words, although "four dimensional manifolds" is at least an improvement on "four manifolds" (it would only be ambiguous if there were something called a "dimensional manifold", which is not the case as far as I know. – Zev Chonoles Aug 3 '13 at 7:26
• @ZevChonoles: Thanks for the advice; I guess I am having an Ive League rank personal editor... – Bombyx mori Aug 4 '13 at 6:03
• Spin is quite strong condition, but spin$^\mathbb{C}$ isn't all that strong at all—as one well-known noncommutative geometer calls it, it's "orientability $+ \epsilon$", and virtually any concrete orientable manifold one might ever deal with will be spin$^\mathbb{C}$. Indeed, whilst being spin$^\mathbb{C}$ does fail to be automatic for orientable manifolds of dimension greater than $4$, the actual counterexamples are apparently rather recherché. Moreover, pretty much all additional structures applied to rientable manifolds—symplectic, almost-complex, Kähler, spin—imply spin$^\mathbb{C}$. – Branimir Ćaćić Aug 4 '13 at 8:50
• As for a more low-tech proof, perhaps the one at the end of these notes (after Friedrich's book on Dirac operators) might do the trick? mathematik.uni-regensburg.de/ginoux/spincstruct.pdf – Branimir Ćaćić Aug 4 '13 at 8:59

## 2 Answers

After consulting the usual reference (John Morgan's book), it seems this is proved in the Chapter introducing the $\textit{Spin}^{c}$ structure. However his proof involves $\mathbb{Z}/2^{k}\mathbb{Z}$ homology classes and is not very readable from my point of view. I suspect an independent proof by myself needs to be constructed. This seems standard enough that probably too low for mathoverflow.

• I was browsing Friedrich's book earlier today. I think Branimir's comment is worthwhile. If you haven't looked at that yet it's probably worth a few minutes. – James S. Cook Aug 27 '14 at 5:27

Every orientable manifold of dimension less than or equal to three is spin and therefore spin$${}^c$$.

Every smooth orientable four-manifold is spin$${}^c$$. Another reference for this fact is this note by Teichner and Vogt. The benefit of this proof is that it also works for non-compact manifolds.

In dimensions five and above, there are orientable manifolds which are not spin$${}^c$$. For example, the Wu manifold $$SU(3)/SO(3)$$ is an orientable five-manifold which is not spin$${}^c$$. It follows that $$S^n\times(SU(3)/SO(3))$$ is not spin$${}^c$$ for every $$n$$.