Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am reading Floer's Instanton-Invariant paper, and am stuck on a sentence. To set the stage:

Consider a closed connected oriented 3-manifold $M$ and the nonabelian group $SU_2$. Denote the equivalence classes of representations $\mathcal{R}(M)=Hom(\pi_1M,SU_2)/\text{ad}(SU_2)$.

Now given a Heegaard splitting $M=M_+\cup_SM_-$, one can consider $\mathcal{R}(M)$ as the intersection of $\mathcal{R}(M_+)$ and $\mathcal{R}(M_-)$ in $\mathcal{R}(S)$. Indeed, Seifert van-Kampen's theorem gives $\pi_1(M)\cong\pi_1(M_+)\ast_{\pi_1(S)}\pi_1(M_-)$ and then the statement follows by the universal property of amalgamated free products.

The resulting intersection number (ignoring the trivial representation) can be shown to be independent of the particular Heegaard splitting.
[The "result" refers to the integer-valued Casson invariant, which assigns a sign to each intersection $a\in\mathcal{R}$].

How is this done?

share|improve this question
    
I don't understand what you mean by "the resulting intersection." –  Qiaochu Yuan Oct 7 '12 at 7:21
1  
Presumably what they're getting at is that $\mathcal R(M)$ is an idea that is independent of any particular presentation of $\pi_1 M$. –  Ryan Budney Oct 7 '12 at 7:47

1 Answer 1

up vote 1 down vote accepted

(For completeness)

This is explained/proved in the main reference of the invariant: Casson's Invariant for Oriented Homology 3-Spheres (by Akbulut and McCarthy).

share|improve this answer

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.