This is a quite specific question continuing the problems I have with computing the expectation value of intersecting Wilson loops I laid out here. Using the tools from the answer there, I quite quickly arrive at the following expression for the local factor associated to a vertex, where two Wilson loops with reps $\alpha_1$ and $\alpha_2$ meet, and where the four surrounding regions have reps $\beta_1$ to $\beta_4$:

$$ G(\alpha_1,\alpha_2,\beta_{1,2,3,4})_{\mu\nu}^{\sigma\rho} := \epsilon_\mu^{ijk}(\alpha_1,\beta_1,\beta_4)\epsilon_\nu^{lmn}(\alpha_2,\beta_1,\beta_2){\epsilon^*}^\sigma_{ijk}(\alpha_1,\beta_2,\beta_3){\epsilon^*}^\rho_{lmn}(\alpha_2,\beta_3,\beta_4)$$

The Greek indices are the indices incurred from decomposing tensor products as $a^i \otimes b^j \otimes b^k = \epsilon_\mu^{ijk}e^\mu$, leading to integral results like

$$\int \alpha_i(V_b)^i_{i'} \beta_c(V_b)^j_{j'} \beta_{c'}(V_b)^k_{k'} \mathrm{d} V_b = {\epsilon^*}^\mu_{i'j'k'}\epsilon^{ijk}_\mu$$ (see previous answer). Since the $\epsilon^*$ that has the $\mu$ this is summed with lives on the opposite end of the (part of) the Wilson line, the Greek indices must necessarily remain open at the vertices. I am totally fine with this being the result of the computation, but I am still puzzled why the relation to the 6j symbol is so casually tossed about.

Let me first remark that the above equation is already suspiciously similar to the very first equation in the definition of $6j$ symbols, but the free indices are irritating me. If $\epsilon_\mu^{ijk}(\alpha_l,\beta_m,\beta_n)$ is the $3jm$ symbol (with $i,j,k$ playing the role of the $m$ and the reps corresponding to the $j$), what is the additional index $\mu$ doing here? If it is not the $3jm$ symbol (which I am currently thinking), then why would the $G$ defined above be the $6j$ symbol (and why has it free indices)? (If these are neither $3jm$ nor $6j$ symbols, then why do Witten, Ramgoolam, Moore, etc. insist they are?)

Note that the $6j$ symbol cannot arise after summing the Greek indices, since the $G$s the second index belongs to are, in general, at other vertices, and so have not exactly the same 6 reps as arguments.

Furthermore, the $3jm$ symbols are, if I understand them correctly, essentially the Clebsch-Gordan coefficients for expanding a tensor product of two irreducible reps in a third, and the $\epsilon$ above expand the tensor product of three irreducible reps in all possible fourths (which are then summed over in form of the Greek indices).

Something does not add up here, and I heavily suspect it is only in my understanding of the symbols, so I would really appreciate someone clearing up my confusion.


Ok, I think I have found something, but I am still a far shot from solving this riddle, and it requires to think more carefully about the coefficients $\epsilon^{ijk}_\mu$:

Let $\alpha,\beta,\gamma$ be reps with basis elements $a^i,b^j,c^k$as before. Then, we can decompose the tensor product stepwise instead of at once as:

$$ a^i \otimes b^j \otimes c^k = \sum_{\rho \subset \alpha \otimes \beta} C(\alpha,\beta,\rho)^{ij}_\zeta e(\rho)^\zeta \otimes c^k = \sum_{\rho \subset \alpha \otimes \beta} \sum_{\sigma \subset \rho \otimes \gamma} C(\alpha,\beta,\rho)^{ij}_\zeta C(\rho,\gamma,\sigma)^{\zeta k}_\mu e(\sigma)^\mu$$

(I apologize for the abundance of symbols, but it really becomes clearer what's happening that way.)

Here, the $C(j_1,j_2,j_3)$ are now manifestly Clebsch-Gordan coefficients for $j_1,j_2$ in $j_3$, and thus essentially $3jm$ symbols, and the notation $\rho \subset \alpha \otimes \beta$ means that the irreducible rep $\rho$ occurs as a subrep in $\alpha \otimes \beta$. Since Clebsch-Gordan coefficients for reps not appearing in a given tensor product are zero, we can drop the constraint on the sums and sum over all irreducible reps. Thus, $\epsilon^{ijk}_\mu = \sum_\rho C(\alpha,\beta,\rho)^{ij}_\zeta C(\rho,\gamma,\sigma)^{\zeta k}_\mu$.

Now, in the integral result above, by the Peter-Weyl theorem (see also previous answer), the $\epsilon$ are only summed over the $\mu$ belonging to trivial subreps of $\alpha \otimes \beta \otimes \gamma$, i.e $\sigma = 0$, if we denote the trivial rep by $0$ in analogy to $j = 0$ in the spin case. Therefore, we have that the result of the integral is

$$I := \int \alpha(g)^i_{i'}\beta(g)^j_{j'}\gamma(g)^k_{k'} = \left(\sum_\rho C(\alpha,\beta,\rho)^{ij}_\zeta C(\rho,\gamma,0)^{\zeta k}_\mu C^*(\alpha,\beta,\rho)_{i'j'}^\eta C^*(\rho,\gamma,0)_{\eta k'}^\mu\right) $$

But the trivial irreducible rep has only one dimension, so the sum over the $\mu$ is just the multiplicity $n(\rho,\gamma,0)$ of $0$ in $\rho \otimes \gamma = \bigoplus_\sigma n(\rho,\gamma,\sigma)\sigma$, i.e.

$$ I = \sum_\rho n(\rho,\gamma,0)C(\alpha,\beta,\rho)^{ij}_\zeta C(\rho,\gamma,0)^{\zeta k} C^*(\alpha,\beta,\rho)_{i'j'}^\eta C^*(\rho,\gamma,0)_{\eta k'}$$

This gets rid of the annoying $\mu$, would lead to the $G$ from the beginning of the question to be comprised of the sum over a product of 8 $3jm$ symbols (the $C(\alpha,\beta,\gamma)$), of which 4 each are summed over their $m$ indices, yielding the product of two $6j$ symbols summed over one of their $j$s (the $\rho$). Also, the index structure in $e^{ijk}e^*_{ijk}$ in $G$ would translate to an index structure of the $C$ exactly matching that of the $3jm$ in a $6j$ symbol.

But before I work that out, can anybody tell me if this is the right track or if I butchered something along the way (I am not comfortable enough with my skills yet to fully trust my reasoning when it leads me to an answer whose shape I already know)? Or should I perhaps take this to the mathematicians, since the answer seems to be purely group theoretic so far?


1 Answer 1


Alright, let's go on a thrilling tour through the theory of representations. Notation in the following is the same as in the OP, except that we call the trivial representation $\boldsymbol{1}$, as is canon.

We start from my expression in the question

$$ I := \int \alpha(g)^i_{i'}\beta(g)^j_{j'}\gamma(g)^k_{k'} = \sum_\rho \sum_{\mu = 0}^{n(\rho,\gamma,\boldsymbol{1})} C(\alpha,\beta,\rho)^{ij}_\zeta C(\rho,\gamma,\boldsymbol{1})^{\zeta k}_\mu C^*(\alpha,\beta,\rho)_{i'j'}^\eta C^*(\rho,\gamma,\boldsymbol{1})_{\eta k'}^\mu $$

Now, the first thing we must find out is: How often does $\boldsymbol{1}$ appear in $\rho \otimes \gamma$?

The answer is depressingly simple. By Schur's lemma, every morphism of vector spaces $V_\rho \rightarrow V_\gamma $ that commutes with the group action is either an isomorphism of the zero map. Let us denote the set of morphism that commute with the group action by $\mathrm{Hom}_G(V_\rho,V_\gamma)$. Now, $n(\rho,\gamma,\boldsymbol{1})$ is the dimension of $\mathrm{Hom}_G(V_{\boldsymbol{1}},V_\rho \otimes V_\gamma)$. For finite vector spaces, $V_\rho \otimes V_\gamma \cong \mathrm{Hom}(V_\rho,V_\gamma^*)$, where the star denotes the dual space on which we have the dual representation $\gamma^*$.Thus, we seek the dimension of $\mathrm{Hom}_G(V_{\boldsymbol{1}},V_\rho \otimes V_\gamma) \cong \mathrm{Hom}_G(V_{\boldsymbol{1}},\mathrm{Hom}(V_\rho,V_\gamma^*)) \cong \mathrm{Hom}_G(V_\rho,V_\gamma^*)$. But Schur's lemma tells us that this space is only non-zero when $V_\rho$ and $V_\gamma^*$ are isomorphic! (And it is easy to see that even then, it is only one-dimensional.) So, we have found a precise criterion for the occurence of the trivial subrep: $\rho = \gamma^*$. Thus, the sum over $\mu$ and $\rho$ is killed, and all that is left is

$$I = C(\alpha,\beta,\gamma^*)^{ij}_\zeta C(\gamma^*,\gamma,\boldsymbol{1})^{\zeta k}C^*(\alpha,\beta,\gamma^*)^\eta_{i'j'}C^*(\rho,\gamma,\boldsymbol{1})_{\eta k'}$$

Now, we must think a bit more carefully than I did before about the factors at a vertex: The three representations associated with the $\epsilon$ factor are the $\alpha_1$ of the Wilson line and the $\beta_1,\beta_4$ of the regions adjacent to it. But for one of the two regions, the Wilson line runs against its natural orientation, so it will be w.l.o.g. a factor $\beta_4(g^{-1})$ in the integral. Now, since we consider unitary representations, $\beta_4(g^{-1}) = \beta_4^*(g)$, and the factor associated to that part of the Wilson line at the vertex is

$$\epsilon(\alpha_1,\beta_1,\beta_4^*)^{ij}_k := C(\alpha_1,\beta_1,\beta_4)^{ij\zeta}C(\beta_4,\beta_4^*,\boldsymbol{1})_{\zeta k}$$

which is the $3jm$-symbol for the reps $\alpha_1,\beta_1,\beta_4$.

Thinking carefully about the four lines meeting at the vertex and the relative orientations of the regions and the lines, one finally finds the total factor at one vertex to be

$$ G(\alpha_1,\alpha_2,\beta_{1,2,3,4}) = {\epsilon(\alpha_1,\beta_1,\beta^*_4)^{ij}}_k{\epsilon(\alpha_2,\beta_2,\beta_1^*)^{lm}}_j {\epsilon^*(\alpha_1,\beta_2,\beta^*_3)_{im}}^n {\epsilon^*(\alpha_2,\beta_3,\beta^*_4)_{ln}}^k $$

which has exactly the right structure of being the product of 4 $3jm$ symbols summed over their $m$ to be a $6j$ symbol. Thus, as is unsurprising, Witten is correct in asserting that we get the $6j$ symbol at a vertex, though I still consider this to be highly non-obvious.


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