# 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

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.
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 $$n$$.