Why is the plaquette operator in the string-net model a projection operator?

In the string-net model, the plaquette operator is defined as $B_P = \sum_{s}a_s B_{P}^{s}$, where $s$ runs over the string types $\{0,1,2,\dots,n\}$. It is claimed on page 19 of http://arxiv.org/abs/cond-mat/0404617 that for the parameter choice $a_s = \frac{d_s}{\sum_k d_k^2}$, one can use the relation $B_P^{s_1}B_{P}^{s_2}= \sum_{k}\delta_{k^{*}s_1s_2}B_P^k$ to show that $B_P$ is a projection operator. where $\delta_{k^{*}s_1s_2}$ equals $1$ if the string type $\{k^*,s_1,s_2\}$ is allowed, and equals $0$ otherwise. I'm having trouble proving this statement.

To show $B_P$ is a projection operator, one needs to show that $B_P^2 = B_P$. \begin{align} B_P^2 &= \sum_{s_1,s_2}a_{s_1}a_{s_2}B_P^{s_1}B_P^{s_2} \\ &= \sum_{s_1,s_2,k}\delta_{k^{*}s_1s_2} a_{s_1}a_{s_2}B_P^k \\ &= \sum_{s_1,s_2,k}\delta_{k^{*}s_1s_2} \frac{d_{s_1}d_{s_2}}{(\sum_l d_l^2)^2}B_P^k\\ ?&= \sum_{s}\frac{d_s}{\sum_l d_l^2}B_P^s \\ &= \sum_s a_s B_P^s = B_P \end{align} The crucial step is the one with the question mark in the front, which I don't know how to proceed.

• Did you try to prove this using the "fattened lattice" picture (Appendix C of the paper)? – Norbert Schuch Aug 26 '15 at 23:16
• I did. The relation $B_P^{s_1}B_P^{s_2} = \sum_k \delta_{k^* s_1s_2}B_P^k$ comes from putting $s_2$-string and then $s_1$-string into the fattened lattice and then applying $F$ moves. – Zitao Wang Aug 26 '15 at 23:28

The relevant part of the sum as

$\sum_{k^*,s_1,s_2}\delta_{k^*,s_1s_2}d_{s_1}d_{s_2}B^k_P$

Let me assume that the fusion category has no multiplicities, so $N_{ab}^c=0,1$, which I think Levin and Wen also assumed. We can write the sum as

$\sum_{k^*,s_1}d_{s_1}B^k_P\sum_{s_2\in s_1\times \bar{k}}d_{s_2}$

This is because if $\delta_{k^*,s_1s_2}=1$, it means $N_{s_1s_2}^k=1$, which implies $N_{s_1 \bar{k}}^{s_2}=1$. We then use the definition of quantum dimensions $d_a d_b=\sum_c N_{ab}^c d_c$, to write

$\sum_{s_2\in s_1\times \bar{k}}d_{s_2}=d_{s_1}d_{\bar{k}}=d_{s_1}d_k$

So we get $\sum_{k^*,s_1,s_2}\delta_{k^*,s_1s_2}d_{s_1}d_{s_2}B^k_P=(\sum_{s_1,k}d_{s_1}^2)d_k B_P^k$.

This is basically what you need to kill that question mark :-)

• Thanks Meng :) I missed the part $\sum_{s_2}\delta_{k^* s_1s_2}d_{s_2} = \sum_{s_2}N^{s_2}_{s_1\bar{k}}d_{s_2}$, which is why I cannot proceed. – Zitao Wang Aug 27 '15 at 2:29