Timeline for Gauss-Milgram formula for fermionic topological order?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jul 22, 2015 at 0:52 | comment | added | Xiao-Gang Wen | There are two sets: $S^{TC},T^{TC}$ ($N$ by $N$ matrix) and $S,T$ ($N/2$ by $N/2$ matrix). The ground state degeneracy is $N/2$. In our paper, we mainly used $S^{TC},T^{TC}$ . | |
| Jul 21, 2015 at 2:58 | comment | added | Yingfei Gu | Following the above comment, the fermionic version of Gauss-Milgram in my mind was something derived from the restricted modularity(no matter what it is). Now you suggest to consider an embedding to bosonic TO, and I saw your paper with T. Lan conjectures the $c_-$ mod 1/2 is independent of embedding, I am very interested and excited about this point. Is there any hint or heuristic argument for that? | |
| Jul 21, 2015 at 2:52 | comment | added | Yingfei Gu | And $\Theta=0$ seems to suggest $S^\dagger TS=0$ for fermionic TO, so I am curious about the physical meaning of it. I was naively guessing there should be some restricted modularity for the fermionic TO, which preserve the spin structure of the torus, e.g. a subgroup of $\mathbb{SL}_2(\mathbb{Z})$ maybe generated by $T^2$ an S. Does your result $S^\dagger TS=0$ something along this line? | |
| Jul 19, 2015 at 4:06 | history | edited | Xiao-Gang Wen | CC BY-SA 3.0 |
deleted 1 character in body
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| Jul 19, 2015 at 1:18 | history | answered | Xiao-Gang Wen | CC BY-SA 3.0 |