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Given the $\mathfrak{u}(N)$ algebra with generators $L^a$ and commutation relations $ [L^a,L^b] = \sum_c f^{a,b}_{c} L^c $ , the WZW currents of $U(N)_k$ $$ J(z) = \sum_{n \in \mathbb{Z}} J^a_n z^{-n-1} $$ satisfy the Kac-Moody commutation relations of their mode expansion $$ [J^a_n, J^b_m] = \sum_c f^{a,b}_c J^c_{n+m} + k \cdot n \delta_{n+m,0} \mathrm{tr}( L^a L^b ) $$.

These currents generate the algebra of operators of the WZW theory.

[EDIT: THIS IS NOT QUITE TRUE. See the below explanation.]

The algebra of operators of WZW theory (say for $U(N)_k$) is generated of two kinds of operators.

  1. The currents $J^a(z)$.
  2. The 'primary fields' $\Phi^\rho$. Mathematically, $\Phi^\rho(0)$ is the intertwiner $M_{\mathrm{Id}} \to M_{\rho}$. The vacuum module $M_{\mathrm{Id}}$ has highest-weight vector $|\mathrm{Id}\rangle$. $M_{\rho}$ is generated by a highest weight vector $|\rho\rangle$.

The currents $J$ preserve the module $M_\rho \to M_\rho$. In general, $\Phi^\rho(z)$ is also an intertwining operator from $M_{\mathrm{Id}} \to M_{\rho}$. It is related to $\Phi^\rho(z)$ by conjugation with the evolution operator $e^{z \cdot H}$, where the Hamiltonian $H$ is a function of the currents $J$.

In particular, the primary fields $\Phi^\rho$ satisfy the OPEs $$J^a(y)\Phi^\rho(z) = - \frac{\rho(L^a)\Phi^\rho(z)}{y-z} + \text{reg.}$$ where $\rho$ is an `integrable' representation of the algebra and $\Phi^\rho(z)$ live in the representation space of the algebra.

So in principle, it should be possible to write all primary fields purely in terms of WZW currents. Or it should be possible to write correlations in terms of current correlations. Has this been done before?

EDIT: In lieu of Nikita's response, I think a better question would be can nonzero correlations of vertex operators be written in terms of the current algebra? Although by 'NOTE 2' below, maybe even this isn't necessary.

In the case of products of vertex operators (say for a single boson, which corresponds to a $U(1)_1$ WZW model), vertex correlations with some neutrality condition $\sum_i {\alpha_i} = 0$, the correlator $\langle e^{i \alpha_1 \phi(x_1)} \cdots e^{i \alpha_n (x_n)} \rangle$ can be written in terms of the currents $\epsilon_{\mu\nu} \partial^\nu \phi$ since the argument of the exponential can be written as

$$\sum_i \alpha_i \phi(x_i) = - \alpha_1 \int_{x_1}^{x_2} dx^\mu \partial_\mu \phi(x) - (\alpha_1 + \alpha_2) \int_{x_2}^{x_3} dx^\mu \partial_\mu \phi(x) - \cdots - (\alpha_1 + \alpha_2 + \cdots + \alpha_{n-1}) \int_{x_{n-1}}^{x_n} dx^\mu \partial_\mu \phi(x).$$

As such, these products of vertex operators are written as a 'non-local' integral of currents away from the insertion points. However, note that these representations are not 'universal' in the sense that there may be many ways to write these integrals.

For $U(1)_k$, primary fields correspond to restricting $\alpha_i \in \frac{1}{\sqrt{k}} \mathbb{Z}$, so such a presciption does indeed exist. The question is now to generalize the above to $U(N)_k$.

NOTE: Regarding $U(1)_k$, a somewhat more aesthetic way to write the correlations are in terms of $k$ independent boson fields $\{\phi^{(1)} \cdots \phi^{(k)}\}$, $e^{i \frac{n}{\sqrt{k}} \phi(x)} \mapsto e^{i \frac{n}{k} \sum_{i=1}^{k} \phi^{(i)}(x)}$, which trades a nasty factor of $1/\sqrt{k}$ for a factor $1/k$. This is also nice because the current of $U(1)_k$ can be written as $J = \sum_{i=1}^{k} \epsilon_{\mu \nu} \partial^\nu \phi^{(i)}$, the diagonal algebra of $k$ boson currents. So, the notion of $k$ independent layers of the currents should somehow play a role in such constructions.

NOTE 2: Note that if we impose $\phi(z) \xrightarrow{z \to \infty} 0$, we'd have $\phi(z) = \int_{\infty}^{z} dx^{\mu} \partial_\mu \phi(x)$, so even the operators $e^{\alpha \phi(z)}$ by themselves have such a representation.

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  • $\begingroup$ To reiterate Nikita's point, if you write down what you mean by "in terms of WZW currents" you see it's incorrect. What I think you're saying is that every integrable representation $M$ of the affine VOA $V(\mathfrak{g})$ is generated by a single vector $m\in M$, which is false. Since if $M$ is generated by a single vector there's a surjection $V(\mathfrak{g})\twoheadrightarrow M$, i.e. $M$ must be a quotient of $V(\mathfrak{g})$. So it's easy to create counterexamples: $M=V(\mathfrak{g})\oplus V(\mathfrak{g})$, or indeed most smooth representations of $\hat{\mathfrak{g}}$ (i.e. reps of V(g)). $\endgroup$
    – Pulcinella
    Commented Oct 25, 2023 at 1:04
  • $\begingroup$ I understand now that simple currents keep one in same VOA module (essentially by definition). As above, the better question is about correlators of primary fields. I think my question translates as follows. Take primaries $\Phi_1(z_1), \cdots, \Phi_n(z_n)$ corresponding to punctures $z_i$ and modules $M_i$. How does one write the projection to the vacuum sector of the operator $\Phi_1(z_1) \cdots \Phi_n(z_n)$ in terms of the VOA? This is only nonzero if $M_1 \otimes \cdots \otimes M_n$ decomposes with an identity representation. $\endgroup$
    – Joe
    Commented Oct 25, 2023 at 20:04
  • $\begingroup$ In the boson example above, the $e^{i \alpha \phi(0)}$ correspond to different vacua $| \alpha \rangle$. More generally, they act as intertwiners taking the $| \beta \rangle$-module $M_\beta$ to the $| \alpha + \beta \rangle$-module $M_{\alpha+\beta}$ for all $\beta$. In this case, with $\sum_i \alpha_i = 0$, the operator $e^{i \alpha_n \phi(z_n)} \cdots e^{i \alpha_1 \phi(z_1)}$ maps modules $M_0 \to M_{\alpha_1} \to M_{\alpha_1 + \alpha_2} \to \cdots \to M_{\alpha_1 + \cdots \alpha_n} = M_0$. And indeed we can write this in terms of currents like above. $\endgroup$
    – Joe
    Commented Oct 25, 2023 at 20:12

1 Answer 1

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It is not in general true that given a set of currents you can express all primary fields in terms of some functions of currents. If you think of this in terms of representation theory of corresponding VOA (which in this case is in 1-to-1 correspondence with representations of affine Kac-Moody Lie algebra), differential polynomials in currents only give you "vacuum" module - parabolic with highest weights $h_i=0$ (it is the space of VOA itself). Primary operators give you VOA modules (in this case integrable) and there's no universal way to write them as currents.
In some cases you can get around it, for example for $k=1$ you have a bosonisation that allows you to write primary fields as $V_\alpha=:e^{i(\alpha,\phi)}:$, but I'm not aware of what one would do in general case (maybe something with Wakimoto construction).

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    $\begingroup$ Hi, thanks for the comment. I edited my post to talk about how products of the above primary fields can be written in terms of integrals of the currents over the whole space. The integrals will involve currents that are away from the insertion points, so I'd expect them to be 'non-local' in some sense. Although I agree that such expressions aren't universal in the sense that there may be many ways to write these integrals. Is the statement that the current algebra generates the whole operator algebra even true in general? $\endgroup$
    – Joe
    Commented Oct 17, 2023 at 18:35
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    $\begingroup$ I'm not sure what might be the meaning of the representation you provide. It's interesting that you can write them exactly in the cases where you get non-zero correlators. But as for whether the currents generate the whole algebra I would say it's not the case even for free Boson, since you can't write operator $:e^{\alpha \phi(z)}:$ in terms of $\partial \phi(z)$ simply because you can't shift zero mode. In general it should be even more problematic, since you don't get any sort of nice formula for vertex operators. $\endgroup$
    – Nikita Grygoryev
    Commented Oct 18, 2023 at 21:07
  • $\begingroup$ Even $e^{\alpha \phi(z)}$ has such a representation. Typically, one quotients out the zero mode, which corresponds to imposing a vanishing condition $\phi(z) \xrightarrow[]{z \to \infty} 0$. After doing so, we'd have $\phi(z) = \int_{\infty}^{z} dx^\mu \partial_\mu \phi(x)$. The point is that since the $J$'s generate the whole algebra of operators of the theory, all operators should have some expansion in the $J$'s. Although it may be an infinite sum of products of operators, as is what happens expanding the exponential in the boson case above. $\endgroup$
    – Joe
    Commented Oct 18, 2023 at 21:52
  • $\begingroup$ I would be very cautious about throwing away both zero mode an it's shift. I think if one where to write just $$V_\alpha(z):=:e^{\sum_{n\neq 0} \frac{a_n}{-n}z^{-n}}:$$ they would get something like $$V_\alpha(z)V_\beta(w)=(1-\frac{z}{w})^{\alpha\beta}:V_\alpha(z)V_\beta(w):$$ which is not a behavior you expect from operators in a Lorentz invariant theory. $\endgroup$
    – Nikita Grygoryev
    Commented Oct 19, 2023 at 0:40
  • $\begingroup$ Right, I guess in the operator language it can be more subtle. In the sense of computing those correlations in the path integral and interpreting determinants of Laplacians, I believe it's okay toss away the zero mode. This is standard in analyzing 'analytic torsion' of compact manifolds, which are regulated Laplacian determinants with the zero mode thrown out. $\endgroup$
    – Joe
    Commented Oct 20, 2023 at 14:29

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