twisted affine algebras

Let $g$ a finite-dimensional complex simple Lie algebra and $\sigma$ a finite order Dynikin diagram automorphism of $g$.

Consider $\tilde g$ as the loop algebra associated to $g$, and $\tilde g^\sigma$ as the twisted affine Lie algebra (associated to $g$) in the spirit of the theory developed in the book ''Infinite dimensional Lie algebra" by Kac, or ''Lie algebras of Finite and Affine Type'' by Roger Carter.

It is easy to see that given an action of $\tilde g$ we have an action of $\tilde g^\sigma$ which is given by the restriction of the action of $\tilde g$. On the opposite way, my question is: Given an action of $\tilde g^\sigma$ on a module $M$, is it possible to extend this action to $\tilde g$?

Moreover, if we restrict to the context of universal highest-weight modules for $\tilde g^\sigma$ can we produce a highest-weight module for $\tilde g$ by extending the action of $\tilde g^\sigma$?

What should be a reference for this subject concerning about these extensions?

Thanks,

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I added the [lie-algebras] tag –  David Roberts Sep 6 '11 at 23:42
@David: Great, thanks! –  Chris Sep 6 '11 at 23:44
One may try to take an induced or coinduced module of $M$, and, using that the loop algebra and its twisted subalgebra are "close enough", to try to find there a submodule or a quotient which may be "close" to $M$. This is a sheer speculation, I haven't tried to work it out. –  Pasha Zusmanovich Oct 21 '11 at 16:29
@Pasha: Do you have a bet for what the term "close" means? –  Chris Oct 22 '11 at 1:22
@Chris: Sorry, no. This is a sheer speculation, like I said. –  Pasha Zusmanovich Oct 22 '11 at 6:44

To add to what Carnahan has posted. There is a difference between the representation theories of untwisted and twisted affine algebras. But they seem to be unified through vertex algebra theory.

Vertex operator algebras themselves are "untwisted" but admit twisted modules. Haisheng Li has some results that make this statement more concrete, but essentially, while there are twisted VOA modules, there are no twisted VOAs.

So though Affine VOAs are associated with untwisted affine algebras, they do include representations of the corresponding twisted affine.

You may want to take a look at some of Haisheng Li's work in this area. For example:

Li, Haisheng A new construction of vertex algebras and quasi-modules for vertex algebras. Adv. Math. 202 (2006), no. 1, 232–286. 17B69

Take a look at section 7, specifically, Proposition 7.4

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Thanks, I just opened the paper on arxiv, it seems very helpful. –  Chris Sep 7 '11 at 15:50

I don't have Kac's book next to me right now, but I was under the impression that representations of the affine algebra do not give rise to representations of the twisted algebra in any straightforward way. In particular, the twisted algebra is not the subalgebra of fixed points of an algebra automorphism, and in fact does not admit a nonzero algebra map to the affine algebra. The fractional powers of $z$ in the twisted algebra seem to form the main obstruction here.

In conformal field theory, modules of the affine algebra live on ordinary points of a Riemann surface, while modules of the twisted algebra live on orbifold points of order $|\sigma|$. This at least suggests that they are substantially different animals.

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@S. Carnahan: The first two lines in your post seem very weird to me, since $\tilde g^\sigma$ is a Lie subalgebra of $\tilde g$, and, then, by restriction of an action of $\tilde g$ we get naturally an action of $\tilde g^\sigma$. Is not it? –  Chris Sep 7 '11 at 15:56
The category of $\tilde{\mathfrak g}^\sigma$-representations is a module over the $\tilde{\mathfrak g}$-representations. More concisely, $\tilde{\mathfrak g}^\sigma$-reps is a ($\tilde{\mathfrak g}$-reps)-module. –  André Henriques Sep 29 '11 at 21:55

Affine Lie algebras are not exactly loop algebras, they are central extensions of loop algebras. If I understand correctly, commutation relations for twisted and untwisted affine Lie algebras are different.

For untwisted Lie algebra we have $\hat{\mathfrak{g}}$$=(\mathbb{C} [t,t^-1]\otimes\mathfrak{g})\oplus \mathbb{C}c\oplus\mathbb{C}d$

Commutation relations are $[t^m\otimes X,t^n\otimes Y]=t^{m+n}\otimes [X,Y]+m\delta_{m+n,0}(X,Y)c$.

For twisted affine Lie algebra with order $r$ automorphism $\sigma$ we have $\mathbb{Z}/r\mathbb{Z}$ grading on $\mathfrak{g}$ and

$\hat{\mathfrak{g}}^{\sigma}$ =$(\sum_{n\in\mathbb{Z}}t^{n}\otimes\mathfrak{g}_{n\; \mathrm{mod}\; r})\oplus \mathbb{C}c\oplus \mathbb{C}d$.

Commutation relations in this case are $[t^m\otimes X,t^n\otimes Y]=t^{m+n}\otimes [X,Y]+m\delta_{m+n,0}\frac{(X,Y)}{r}c$.

(This is from the book by Minoru Wakimoto Infinite-dimensional Lie algebras)

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There is no issues when you are taking brackets. Moreover, informally talking, if you do work on finite-dimensional representations, then you do not need care about $c$ and $d$. –  Chris Sep 30 '11 at 13:05