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

In her article Higher string topology operations, Godin in particular construct for each surface with $n$ incoming and $m \geq 1$ outgoing boundary circles an operation $H_\ast(BMod(S);det^{\otimes d}) \otimes H_\ast(LM)^{\otimes n} \to H_\ast(LM)^{\otimes m}$, where $Mod(S)$ is the mapping class group $\pi_0(Diff^+(S;\partial S))$.

As an example she gives claims that the generator of $H_1(BMod(Cylinder))$ (the twisting is trivial here) gives the BV-operator, where of course the generator of $H_1(BMod(Cylinder))$ corresponds to the generator of $Mod(Cylinder)$ corresponding to the Dehn twist around one of the boundary components. My first question is: how does it follow from Godin's construction that this generator acts as the BV-operator?

In general, the mapping class group of a surface with boundary is generated by a finite number of Dehn twists (e.g. A primer on mapping class groups v.4.02, page 131). My second question is: Do all of these have an action in string topology similar to a BV-operator?

Finally, in many cases (e.g. the pair of pants) $BMod(S)$ is an H-space, being the classifying space of an abelian group (thanks to Chris Schommer-Pries for pointing out a mistake here originally). This means there is an induced product in homology, and also a shifted product on twisted homology if the twistings is trivial (section 4.5 of Godin's article tells us this is the case is at most one boundary component is completely free). My third question is: how does this product interact with the string topology operations?

share|improve this question
    
@ skupers: You said: "BMod(S) is an H-space, being the classifying space of a group...". This is false, perhaps you meant something else? A topological group G is an H-space, but its classifying space is not generally going to be an H-space. What could the product be? If G=A is abelian then you get a map $$BA \times BA \to BA$$ making BA topological group. You don't have this for general groups G. –  Chris Schommer-Pries Aug 6 '10 at 18:54
    
Yes, apparently I did mean abelian group. Thanks for pointing that out, I'll edit it. At least in the case of the pair of pants, we know that the mapping class group is abelian: it is $\mathbb{Z}^3$, though. –  skupers Aug 6 '10 at 20:51
    
@skupers : The mapping class group of a compact orientable surface with boundary is abelian only for the sphere, the disc, the annulus, and the pair of pants, which doesn't seem to me like "many cases". –  Andy Putman Aug 6 '10 at 22:04
2  
One way to think of (some of) these loops in $BMod(S)$ is as loops in $M_{g,n}$ encircling a point in the boundary of Deligne-Mumford space (a stable nodal curve). Non-triviality of the resulting operations on loopspace homology would be an obstruction to extending the string topology TCFT to D-M space. This would be of interest, for instance, in the picture of $H_{-\ast}(LM)$ as symplectic cohomology $SH^\ast(T^\ast M)$. –  Tim Perutz Aug 7 '10 at 1:01

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.