Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

160 submissions , 132 unreviewed
4,164 questions , 1,543 unanswered
5,015 answers , 21,272 comments
1,470 users with positive rep
586 active unimported users
More ...

  $S^3$ partition function from the 1-category of $S^1$ in Chern Simons theory

+ 6 like - 0 dislike
82 views

I am trying to learn about some categorical aspects of topological quantum field theories. For concreteness, I am considering Chern Simons theory with gauge group $G$ in three dimensions. As I understand, I can describe the theory as a functor $F$ on a cobordism category where there are $0$, $1$ and $2$ dimensional objects (manifolds) and morphisms between objects. Let me denote some $\mathbb C$-linear $n$-categories as $C_n$, such that the functor $F$ leads to following assignments: $$F(S^3) = C_{-1} =: Z_{S^3}\,, \qquad F(S^2) = C_0 =: \mathcal H_{S^2}\,, \qquad F(S^1) = C_1\,, \qquad F(\{\mathrm{pt}\}) = C_2\,.$$ where $S^n$ is the $n$-dimensional sphere, $\mathcal H_{S^2}$ is a vector space (with an inner product), and $Z_{S^3}$ is a complex number, this question does not concern the 2-category $F(\{\mathrm{pt}\})$. The 1-category $F(S^1)$ is a category of some representations of the affine Lie group $\widehat G$.

Question: How do I compute the number $Z_{S^3}$ given only the 1-category $F(S^1)$?

As a clarification of what I am asking let me explain how I would proceed for the simpler problem of computing $Z_{S^3}$ given the 0-category $F(S^2)$. Given $\mathcal H_{S^2}$, I can construct $S^3$ as $S^3 = \bar D^3 \sqcup_{S^2} \bar D^3$ where I am gluing two closed three dimensional discs along their $S^2$ boundaries and my understanding is that (I would like to know if this is incorrect) $F$ assigns to $\bar D^3$ an element of $F(\partial \bar D^3) = F(S^2) = \mathcal H_{S^2}$, i.e. $F(\bar D^3)$ is a vector and the result of the gluing is that $F(S^3) = \langle F(\bar D^3), F(\bar D^3) \rangle$ where $\langle -, - \rangle$ is an inner product (which can be written as a path integral over $S^3$).

I would like to know what is the analogous procedure if I start from one dimension bellow. If the answer is standard or trivial I will be happy with a reference, and if it helps anyone willing to write an answer, I am more familiar with the physics side of the subject than the math side. Thanks in advance for any help.

Edit: (some attempts) I can try to construct $F(S^2)$ from the knowledge of $F(S^1)$ by gluing two $\bar D^2$'s along their boundary $S^1$. $F(\bar D^2)$ should be an object of the 1-category $F(S^1)$, so $F(\bar D^2)$ is a $\widehat G$-module, therefore the image of two disjoint $\bar D^2$'s (with one "incoming" and one "outgoing" boundary) under $F$ should be $$F\left(\bar D^2 \sqcup \left(\bar D^2\right)^*\right) = F(\bar D^2) \otimes F(\bar D^2)^*\,.$$ Finally from the physical point of view the following seems reasonable: $$F(S^2) = F\left(\bar D^2 \sqcup_{S^1} \left(\bar D^2\right)^*\right) = \left(F(\bar D^2) \otimes F(\bar D^2)^*\right)^{\widehat G}\,. \tag{1}$$ Then I can proceed along the line described earlier for the simpler situation of having the data of $F(S^2)$.

This kind of answers my original question. For the sake of being able to do computation I have two followup questions:

Followup question 1: How exactly do I decide which element of $F(S^2)$ to assign to $F(\bar D^3)$ (without using a path integral definition of a wave functional)? Is it part of the axioms or is there a derivation?

Followup question 2: In (1), if I choose a basis $\{e_i\}$ for $F(\bar D^2)$ (and a dual basis $\{e_i^*\}$ for $F(\bar D^2)^*$), then $F(S^2)$ seems to be the one dimensional space $\mathrm{Span}\left(\sum_i e_i \otimes e_i^*\right)$, if $F(\bar D^2)$ is infinite dimensional then is there a problem with normalization for the vectors in $F(S^2)$?

Perhaps I am making some silly mistakes, sorry for the lengthy post!

This post imported from StackExchange MathOverflow at 2017-01-11 10:39 (UTC), posted by SE-user Nafiz Ishtiaque Ornob
asked Jan 8, 2017 in Theoretical Physics by nio (60 points) [ no revision ]
retagged Jan 11, 2017
I suspect that to compute $Z(S^3)$, you'll need additional structure on the 1-category attached to $S^1$. In your decategorified example with $\mathcal H_{S^2}$, you need more than a Hilbert space structure to know what $F(\overline D^3)$ is: it's determined by the unit or counit of the Frobenius algebra structure on $\mathcal H_{S^2}$. Similarly, the $1$-category $Z(S^1)$ attached to a 3D TQFT $Z$ should come with additional data.

This post imported from StackExchange MathOverflow at 2017-01-11 10:39 (UTC), posted by SE-user Arun Debray
@ArunDebray You're right, $\mathcal H_{S^2}$ comes with two maps $\eta: \mathbb C \to \mathcal H_{S^2}$ and $\varepsilon: \mathcal H_{S^2} \to \mathbb C$, and $Z(S^3)$ is formally given by $\varepsilon \circ \eta (1)$ (using a morphism $\emptyset \to S^2 \to \emptyset$). I will update my post once I understand the $S^1$ 1-category better. Thank you for the comment.

This post imported from StackExchange MathOverflow at 2017-01-11 10:39 (UTC), posted by SE-user Nafiz Ishtiaque Ornob

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysics$\varnothing$verflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...