What is the significance of monodromy matrix in the context of differential equations? I have seen some papers(1,2,3 etc) in CFT which use the monodromy method to compute conformal blocks at large central charge. Can someone discuss (or give some basic references) what is actually this monodromy specially in the context of CFT.

  • $\begingroup$ The use of monodromy for calculating semiclassical conformal blocks is perhaps due to Zamolodchikov's 1987 paper on recursion relations for conformal blocks (although the paper itself doesn't explain how it arises). I am not sure that the actual derivation of the differential equation and monodromy is written in comprehensible form anywhere in the literature:) I thought I saw it somewhere, but couldn't find now. I can try to recover the derivation a bit later, if you wish $\endgroup$ – Peter Kravchuk Feb 22 '16 at 7:55
  • $\begingroup$ @PeterKravchuk Thanks! Please let me know if you can recover it. :) $\endgroup$ – Physics Moron Feb 22 '16 at 12:09
  • $\begingroup$ okay, here's the derivation. A bit longer then I hoped. Let me know if it is comprehensible. $\endgroup$ – Peter Kravchuk Feb 24 '16 at 4:46

I'm going to explain how the monodromy approach to computation of semi-classical conformal block arises.

The goal is to compute the conformal block corresponding to the exchange of operator $\mathcal{O}$ in the four-point function $$ \langle V_1(z_1)V_2(z_2)V_3(z_3)V_4(z_4)\rangle,\qquad (1) $$ in $V_1V_2-V_3V_4$ channel. Here note that since we are talking about something determined completely by the conformal algebra, we can actually consider only the holomorphic problem. So here $V_i$ are the formal operators characterized by conformal weight $h_i$.

Now, some reparametrization of the problem is convenient, $$ h_i=\frac{1}{b^2}\delta_i,\quad h_\nu=\frac{1}{b^2}\delta_\nu, \quad \delta_i=\frac{1-\lambda_i^2}{4},\quad \delta_\nu=\frac{1-\nu^2}{4},\\ c=1+6(b+b^{-1})^2, $$ where $h_\nu$ is the dimension of exchanged $\mathcal{O}$.

We now consider the 5-point function $$ \langle V_{(1,2)}(z)V_1(z_1)V_2(z_2)V_3(z_3)V_4(z_4)\rangle, $$ where $V_{(1,2)}$ is a degenerate Virasoro field (I'm using the notation of Di Francesco, I believe). Of course, this field need not exist in the theory, we are just formally considering it here, since it reflects the properties of the conformal algebra. This correlator satisfies a differential equation due to degeneracy of $V_{(1,2)}$, $$ \left[\frac{1}{b^2}\frac{\partial^2}{\partial z^2}+\sum_{i=1}^4\left(\frac{h_i}{(z-z_i)^2}+\frac{1}{z-z_i}\frac{\partial}{\partial z_i}\right)\right]\langle V_{(1,2)}(z)V_1(z_1)V_2(z_2)V_3(z_3)V_4(z_4)\rangle=0, $$ and any conformal block of this correlator satisfies the same equation (because this equation is a property which follows from conformal algebra alone; you can check this explicitly by inserting projection operators in the correlator, and using the fact that they commute with all Virasoro generators).

We then consider a specific conformal block, the one given by the picture where you fuse $V_1$ and $V_2$ to obtain $\mathcal{O}$, then $\mathcal{O}$ fuses with $V_{(1,2)}$ to become some $\mathcal{O}'$ and then $\mathcal{O}'$ becomes $V_3$ and $V_4$. You should think of this as the conformal block for $(1)$ where the intermediate $\mathcal{O}$ interacts with $V_{(1,2)}$.

We consider the limit of large central charge $c$, corresponding to $b\to 0$, and simultaneously we take $h_i$,$h_\nu$ to be large, keeping $\delta_i,\delta_\nu$ fixed. The physical assumption is that since the scaling dimension of $V_{(1,2)}$ remains finite in this limit, the 5-point conformal block is given by the formula $$ \langle V_{(1,2)}(z)V_1(z_1)V_2(z_2)V_3(z_3)V_4(z_4)\rangle_{CB}=\psi(z|z_i)e^{\frac{1}{b^2}f_{\nu,\delta_i}(z_i)}, $$ where $\psi(z|z_i)$ is the ``wavefunction'' of the light $V_{(1,2)}$ in the background of semiclassical 4-point conformal block $e^{\frac{1}{b^2}f_{\nu,\delta_i}(z_i)}$. I am not sure if there is a proof of this statement in the literature. What we do know is that it can be tested and it has been tested.

Plugging this ansatz into the differential equation for the 5-point conformal block, we find a wave equation $$ \frac{\partial^2\psi}{\partial z^2}+\psi\sum_{i=1}^4\left(\frac{\delta_i}{(z-z_i)^2}+\frac{c_i}{z-z_i}\right)=0, $$ where $$ c_i=\frac{\partial f_{\nu,\delta_i}}{\partial z_i} $$ are the so-called accessory parameters. (There are four of them, but only one is independent. This is because the semiclassical 4-point conformal block is conformally covariant, and this induces relations between $c_i$. E.g. $\sum_{i=1}^4 c_i=0$ because of the translational invariance of $f_{\nu,\delta_i}$).

Now we are in position to understand where does the monodromy problem come from. You can see that we obtained a second-order ODE for $\psi$, which has two linearly independent solutions, whereas $\psi$ should determine, it seems, just one 5-point conformal block. So what is the interpretation of the two solutions? As it turns out, there are actually two conformal blocks we are determining. Indeed, when we were specifying the 5-point conformal block, the operator $\mathcal{O}$ turned into an operator $\mathcal{O}'$ after the interaction with $V_{(1,2)}$. What is this operator? Well, since $V_{(1,2)}$ is a degenerate field, it can only have a non-zero three point function $\langle \mathcal{O}V_{(1,2)}\mathcal{O}'\rangle$ if the scaling dimension of $\mathcal{O}'$ is $h'=h_\nu\pm \frac{\nu}{2}$.

So we are in fact computing two 5-point blocks, and that is why there are two solutions for $\psi$. Now, how can we identify the two specific linear combinations corresponding to the blocks we are looking for? In computing the 5-point block, we replace $V_1(z_1)V_2(z_2)$ with $\mathcal{O}(z_1)$ and its descendants, and then, for $|z-z_1|>|z_1-z_2|$, we replace $\mathcal{O}(z_1)V_{(1,2)}(z)$ by $\mathcal{O}'(z_1)$. Thinking about the scaling dimensions, one can find that the 5-point conformal block must have an expansion for $|z_2-z_1|<|z-z_1|<|z_3-z_1|,|z_4-z_1|$ of the form $$ \sum_n a_n (z-z_1)^{h_{\mathcal{O}}+h_{(1,2)}-h_{\mathcal{O}'}} $$. It then follows that the monodromy of the 5-point block when $z$ goes counterclockwise around $z_1$ and $z_2$ (we need to go around $z_2$ since this expansion only works for $|z_2-z_1|<|z-z_1|$) is $$ \exp\left\{2\pi i\left(h_{\mathcal{O}}+h_{(1,2)}-h_{\mathcal{O}'}\right)\right\}=-\exp(\pm i\pi\nu), $$ where the $\pm$ corresponds to the choice of conformal block as in $h'=h_\nu\pm \frac{\nu}{2}$.

We thus see that the basis of solutions corresponding to the two possible 5-point conformal blocks diagonalizes the monodromy around the points $z_1$ and $z_2$, and the monodromies have to be very specific. In this basis, the monodromy matrix for the contour $\gamma_{12}$ going around $z_1,z_2$ counterclockwise is then $$ M(\gamma_{12})=\begin{pmatrix} -e^{i\pi\nu} & 0\\ 0 & -e^{-i\pi\nu} \end{pmatrix}. $$ An basis-invariant way to characterize this monodromy is to say $$ \mathrm{tr} M(\gamma_{12})=-2\cos\pi\nu, $$ and this uniquely determines the eigenvalues (because the monodromy matrix has to be unimodular in this case, but I can't remember why).

Now, its a good time to step back and recap. Instead of computing the semiclassical 4-point conformal block directly, we considered how it enters the calculation of a 5-point conformal block with a degenerate field. We found that the 5-point block is then determined by a second order ODE. The coefficients in this ODE are determined by the semiclassical 4-point conformal block. Consistency then requires this ODE to have a specific monodromy property; this constrains the coefficients of the ODE and thus the semiclassical 4-point conformal block.

  • $\begingroup$ There is actually a cool story how monodromy problem can be mapped into a 1D Schrödinger equation in a periodic potential. This can be found in Zamolodchikov's 1987 paper on recursion relations. $\endgroup$ – Peter Kravchuk Feb 24 '16 at 4:48

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