Path Integral as Functor

[Anonymous No.8566711]

How do we realize the path integral as a (symmetric,monoidal) functor [math]Z:\operatorname{nCob} \to \operatorname{Vect} [/math] ?

[Anonymous No.8566762]

>>8566711
Here is what I got so far.

Suppose we have a field [math]\phi :\Sigma \to M[/math] from some n-dim manifold to our target space. Governed by a simple action [math]S\left[ \phi \right] = \int\limits_\Sigma {{{\left| {\operatorname{d} \phi } \right|}^2}} [/math].


We get a functional measure [math]\left[ {D\phi } \right][/math] on the space of maps [math]\left\{ {\phi :\Sigma \to M} \right\}[/math] via some generalization of the Wiener measure.


If sigma is bounded we define measures [math]\mu \left[ \phi \right][/math] on the space of maps [math]\left\{ {\phi :\partial \Sigma \to M} \right\}[/math].
There give a "conditional path integral measure" obeying the rule [math]\int {\left[ {D\phi |\mu } \right] \cdot } F\left( {{{\left. \phi \right|}_{\partial \Sigma }}} \right) = \int {\operatorname{d} \mu \left[ \phi \right]} \cdot F\left( \phi \right)[/math].


Using this we define numbers [math]{U_\Sigma }\left[ \mu \right] \equiv \int {\left[ {D\phi |\mu } \right] \cdot } \exp \left( { - S\left[ \phi \right]} \right)[/math]. These can be manipulated to obey a variant of gluing axioms.

View sigma as cobordism [math]\Sigma :\partial \Sigma \to \emptyset [/math].

So functorialy we have a map [math]Z\left( \Sigma \right):Z\left( {\partial \Sigma } \right) \to Z\left( \emptyset \right)[/math] where [math]Z\left( \emptyset \right) = \mathbb{C}[/math] by definition.

Let [math]{\left\{ {{\mu _i}} \right\}_{i \in I}}[/math] be a basis for the space measures [math]\mu \left[ \phi \right][/math].

Then we get our equivalence by letting [math]Z\left( {\partial \Sigma } \right) \cong \operatorname{span} {\left\{ {{\mu _i}} \right\}_{i \in I}}[/math] and [math]Z\left( \Sigma \right)\left( \mu \right) = {U_\Sigma }\left[ \mu \right][/math].


How do I show this is symmetric and monoidal?

[Anonymous No.8566810]

can you put this in the form of a frog meme?

[Anonymous No.8566881]

>>8566711
https://ncatlab.org/nlab/show/FQFT

[Anonymous No.8566910]

>>8566881
>https://ncatlab.org/nlab/show/FQFT
That does not show what I want to show.

[Anonymous No.8568246]

>>8566711
It is trivial.