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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
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 $Z(\Sigma_f )=\mathrm{Tr}(\mathrm{Id}\circ \Sigma (f))=\mathrm{Tr}(\Sigma (f))$.

Improve this answer
$\endgroup$
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
    Commented 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
    Commented Feb 24, 2014 at 12:59
  • $\begingroup$ Can I ask why your paring is the natural one? $\endgroup$
    – user2013
    Commented 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
    Commented Feb 24, 2014 at 14:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .