What happened to the Vacuum Hypothesis in TQFT? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:41:51Z http://mathoverflow.net/feeds/question/17383 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17383/what-happened-to-the-vacuum-hypothesis-in-tqft What happened to the Vacuum Hypothesis in TQFT? Charlie Frohman 2010-03-07T15:13:24Z 2010-03-07T17:39:59Z <p>I remember that in the beginning, there was an axiom for $(n+1)$-dimensional TQFT that said that the state space $V(\Sigma)$ assigned to an $n$-dimensional oriented manifold is spanned by the invariants of all $n+1$-dimensional oriented manifolds $M$ with $\partial M=\Sigma$. If we call the invariant of $M$, $Z(M)\in V(\Sigma)$, this just says that $V(\Sigma)$ is spanned by all $Z(M)$. Maybe it was in Segal's Swansea notes, maybe it was in an early version of Atiyah's axioms. It doesn't seem to have made it into "The Geometry and Physics of Knots"</p> <p>For instance if you read "Topological Quantum Field Theories Derived from the Kauffman Bracket" by Blanchet, Habegger, Masbaum and Vogel, the vacuum hypothesis is implicit in their constructions.</p> <p>There is a theorem that the category of Frobenius algebras is equivalent to the category of $1+1$-dimensional TQFT's. For instance, let $A=\mathbb{C}[x]/(x^3)$, with Frobenius map $\epsilon(1)=\epsilon(x)=0$ and $\epsilon(x^2)=-1$. This is the choice that Khovanov made to construct his $sl_3$-invariant of links. At this point, no one would deny that this gives rise to a TQFT.</p> <p>The state space associated to a circle is just $A$. The $2$-manifolds with boundary the circle are classified by genus. Using TQFT to compute them, I get that a disk has invariant $1$, a surface of genus one with one boundary component has invariant $3x^2$, and any other surface has invariant $0$. The invariants don't span $A$.</p> <p>The problem as I learned from Chris French is that $A$ is not semisimple. In fact, $A$ being spanned by the invariants of surfaces with one boundary component is equivalent to the semisimplicity of $A$.</p> <p>Here is my question. At what point, and why was the vacuum hypothesis abandoned?</p> http://mathoverflow.net/questions/17383/what-happened-to-the-vacuum-hypothesis-in-tqft/17400#17400 Answer by Scott Morrison for What happened to the Vacuum Hypothesis in TQFT? Scott Morrison 2010-03-07T17:39:59Z 2010-03-07T17:39:59Z <p>At least in your example about <a href="http://arxiv.org/abs/math.QA/0304375" rel="nofollow">Khovanov's TQFT</a> for $\mathfrak{sl}_3$ link homology, part of the answer is that that theory also allows 'singular surfaces', locally modelled on a book with 3 pages. Now you can take a "punctured torus with a disk stuck in its throat" (see <a href="http://arxiv.org/pdf/math.GT/0612754" rel="nofollow">my paper</a>, for example), as a singular 2-manifold with boundary that circle, and you'll see that this gives the missing element of $A$.</p> <p>I'm not sure, though, what the general story is here.</p>