Timeline for WZW primary correlations in terms of current algebra?
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| Oct 20, 2023 at 14:32 | comment | added | Joe | Also, by zero mode, I mean the zero mode of the Laplacian on the whole plane, not the constant mode in the space of states of the Hilbert space. These are different. One can certainly have field configurations on the circle that don't average to zero. Tossing away these modes is a different statement than imposing a vanishing condition at infinity. | |
| Oct 20, 2023 at 14:29 | comment | added | Joe | 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. | |
| Oct 19, 2023 at 0:40 | comment | added | Nikita Grygoryev | 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. | |
| Oct 18, 2023 at 21:52 | comment | added | Joe | 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. | |
| Oct 18, 2023 at 21:07 | comment | added | Nikita Grygoryev | 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. | |
| Oct 17, 2023 at 18:35 | comment | added | Joe | 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? | |
| Oct 16, 2023 at 23:57 | history | edited | Nikita Grygoryev | CC BY-SA 4.0 |
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| S Oct 16, 2023 at 21:49 | review | First answers | |||
| Oct 17, 2023 at 5:03 | |||||
| S Oct 16, 2023 at 21:49 | history | answered | Nikita Grygoryev | CC BY-SA 4.0 |