Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What generalizations of Seiberg-Witten theory to 4-manifolds with boundary do exist?

I would be especially interested in theories which "behave good" under gluing along the boundary (comparable to TQFTs).

share|improve this question
    
I added the TQFT tag. –  Chris Schommer-Pries Mar 16 '10 at 3:15
add comment

1 Answer 1

up vote 15 down vote accepted

Hi Fabian! Kronheimer and Mrowka's book Monopoles and three-manifolds lays out comprehensively the construction of a Seiberg-Witten TQFT, called monopole Floer homology. It is conjectured to be isomorphic to the Heegaard Floer homology TQFT of Ozsváth-Szabó. There are also beautiful constructions due to Froyshov and to Manolescu which do not apply in quite so much generality.

The structure of monopole Floer homology is as follows. The TQFT is a functor on the cobordism category $COB_{3+1}$ whose objects are connected, smooth, oriented 3-manifolds. In fact, the TQFT consists of a trio of functors, denoted $\widehat{HM}_{\bullet}$, $\overline{HM_\bullet}$ and $\check{HM}_{\bullet}$ (the last of these is such a sophisticated invariant that you have to download a special LaTeX package just to typeset it properly). These are $\mathbb{Z}[U]$-modules; there's a story about gradings that's too long to be worth summarising here. There are natural transformations which, for any connected 3-manifold $Y$, define the maps in a long exact sequence $$ \cdots\to \widehat{HM} _{\bullet}(Y) \to \overline{HM_\bullet}(Y)\to \check{HM}_{\bullet}(Y) \to \widehat{HM}_{\bullet}(Y) \to \cdots $$ Why this structure? Well, the theory is based on the Chern-Simons-Dirac functional $CSD$ on a global Coulomb gauge slice through a space of (connection, spinor) pairs. $CSD$ is a $U(1)$-equivariant functional, and $\check{HM}_{\bullet}$ is, philosophically, its $U(1)$-equivariant semi-infinite Morse homology. $\overline{HM}_\bullet$ is the part coming from the restriction of $CSD$ to the $U(1)$-fixed-points, and $\widehat{HM}_\bullet$ is the equivariant homology relative to the fixed point set.

Now here's a subtlety for the TQFT enthusiasts out there to get your teeth into (axiomatize, explain...)! The invariant of a closed 4-manifold $X$ in any of the three theories is... zero. The famous SW invariant of a 4-manifold with $b_+>0$ comes about via a secondary operation, not part of the TQFT itself. Delete two balls from $X$ to get a cobordism from $S^3$ to itself. When $b_+(X)>0$, there are generically no reducible SW monopoles on this cobordism, and this implies that the TQFT-map $\widehat{HM}_\bullet(S^3) \to \widehat{HM}_\bullet(S^3)$ lifts canonically to a map $\widehat{HM}_\bullet(S^3) \to \check{HM}_\bullet(S^3)$; it is this lift that carries the SW invariant.

share|improve this answer
    
If what you say is true, then these can't be TQFTs is the sense of the Atiyah-Segal axioms (i.e. as functors from $COB_{3+1}$ to Vect). It follows from the axioms that for any such TQFT and any 3-manifold X the value of the TQFT on the 4-manifold $X \times S^1$ is the dimension of the vector space associated to X. If this is zero for all X, then the TQFT is the zero TQFT =><=. My recollection is that what you get is similar to an Atiyah-Segal TQFT, but not quite defined on all bordisms. –  Chris Schommer-Pries Mar 16 '10 at 3:12
    
Ahhh, this is why you insisted on connected 3-manifolds! Now I see. –  Chris Schommer-Pries Mar 16 '10 at 3:13
    
Thank you for this extensive answer. I heard about those three approaches, but was not sure which was the most relevant, and if there was something I missed. I see more clearly now. –  J. Fabian Meier Mar 16 '10 at 15:46
    
@Chris. Indeed - since we're not allowed disconnected 3-manifolds, we can't call this a monoidal functor, and we don't get all the usual consequences of the TQFT axioms. For instance, HM-bar is always an infinite rank Z-module. –  Tim Perutz Mar 16 '10 at 19:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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