2
$\begingroup$

I am studying TQFT and have a question on one standard property. A remark in Wikipedia (see the link above) says:

If for a closed manifold $M$ we view $Z(M)$ as a numerical invariant, then for a manifold with boundary we should think of $Z(M) ∈ Z(∂M)$ as a "relative" invariant. Let $f : Σ × I → Σ × I$ be an orientation preserving diffeomorphism, and identify opposite ends of $Σ × I$ by $f$. This gives a manifold $Σ_f$ and our axioms imply $$ Z(\Sigma_f)=\text{Trace}\ \Sigma(f) $$ where $Σ(f)$ is the induced automorphism of $Z(Σ)$.

How can one see this formula? A standard way to cut the manifold is to cut it "at the middle" to obtain two manifolds with boundary, each of which is isomorphic to $Σ × I$, but this does not help me get the formula.

$\endgroup$

1 Answer 1

3
$\begingroup$

If you cut $\Sigma _f$ "in the middle", as you say, you obtain indeed two manifolds $M_1,M_2$ diffeomorphic to $\Sigma \times I$. For $M_1$ the two embeddings of $\Sigma $ in the boundary are standard, hence the element $Z(M_1)$ of $\mathrm{Hom(Z(\Sigma ),Z(\Sigma ))}$ is the identity. For $M_2$ the two embeddings differ by $f$, hence $Z(M_2)\in\mathrm{Hom(Z(\Sigma ),Z(\Sigma ))}$ is $\Sigma (f)$. Now the associativity axiom tell you that $Z(\Sigma_f )=\mathrm{Tr}(\mathrm{Id}\circ \Sigma (f))=\mathrm{Tr}(\Sigma (f))$.

$\endgroup$
4
  • $\begingroup$ Thanks for the answer, but I don't understand the last sentence. Where did you use the associativity axiom and how did you get the trace? $\endgroup$
    – user2013
    Feb 24, 2014 at 12:34
  • $\begingroup$ The associativity axiom tells you that $Z(\Sigma _f)=\langle Z(M_1),Z(M_2)\rangle$, where $\langle \ ,\ \rangle$ is the natural pairing between $Z(\partial M_1)=\mathrm{Hom}(Z(\Sigma ),Z(\Sigma ))$ and $Z(\partial M_2)=Z(\partial M_1)^*$. Now the natural pairing between $\mathrm{Hom}(V,V)$ and $\mathrm{Hom}(V,V)^*\cong\mathrm{Hom}(V,V)$ is given by $(u,v)\mapsto \mathrm{Tr}(uv)$ . $\endgroup$
    – abx
    Feb 24, 2014 at 12:59
  • $\begingroup$ Can I ask why your paring is the natural one? $\endgroup$
    – user2013
    Feb 24, 2014 at 13:53
  • $\begingroup$ This is part of the axioms. I suggest Atiyah's beautiful The geometry and Phisics of Knots, which is much better than the Wikipedia article. $\endgroup$
    – abx
    Feb 24, 2014 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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