3
$\begingroup$

Consider the affine Kac-Moody algebra $\mathfrak g=\widehat{\mathfrak{sl}}_r(\mathbb C((t)))$ and consider the two involutions $$a(t)\rightarrow \sigma(a(t))=-\,^ta(-t),$$ and when $r$ is even $$a(t)\rightarrow \tau(a(t))=-J_r\,^ta(-t)J_r^{-1},$$ where $$J_r=\begin{pmatrix} 0& D_{r/2} \\ -D_{r/2}& 0\end{pmatrix},$$ where $D_{r/2}$ is the anti-diagonal matrix of size $r/2$ with all entries equal $1$.

What is the normalized $2-$cocycle defining the twisted Kac-Moody algebra $\mathcal L(\mathfrak{sl}_r,\sigma)$ and $\mathcal L(\mathfrak{sl}_r,\tau)$? In other words, what is the canonical central element of these algebras w.r.t. that of $\mathfrak{g}$?

The expectation is that they are the same for $\sigma$ case, and it is half that of $\mathfrak{g}$ in the othercase.

Thanks

$\endgroup$
7
  • $\begingroup$ I don't understand your final sentence (before the "Thanks"). Can you state a bit more precisely what you mean here? $\endgroup$
    – Paul Levy
    May 24, 2017 at 18:35
  • $\begingroup$ If $c$ is the central element of $\mathfrak g$, then the canonical central element for $\mathcal L(\sigma)$ is again $c$ and for $\mathcal L(\tau)$ is $c/2$. I expect this from diffrent geometric approch. $\endgroup$
    – Z.A.Z.Z
    May 24, 2017 at 19:06
  • $\begingroup$ I assume you mean $c/2$ and not $1/2c$ in the latter case then... $\endgroup$
    – Paul Levy
    May 24, 2017 at 19:07
  • $\begingroup$ yes.....correct $\endgroup$
    – Z.A.Z.Z
    May 24, 2017 at 19:08
  • $\begingroup$ I'm not familiar with the way you are writing things here - in the formulation in Kac's book (a good place to start), you pick a periodic automorphism of the finite-dimensional Lie algebra and associate a Kac-Moody Lie algebra to that - for any outer automorphism of $\mathfrak{sl}_n$ ($n>2$) we obtain the same twisted affine type Kac-Moody algebra, which won't have much to do with $\mathfrak{sl}_n({\mathbb C}((t)))$. (I don't know what you mean by $\widehat{\mathfrak{sl}_n}$.) I can only assume that by $\sigma$ and $\tau$ you want to take involutions of $\mathfrak{sl}_n$ with fixed points... $\endgroup$
    – Paul Levy
    May 24, 2017 at 21:56

1 Answer 1

2
$\begingroup$

Right - I think I see now... you consider the two twisted loop algebras as subalgebras of $\mathfrak{sl}_n({\mathbb C}((t)))$, in fact as fixed points for the involutions $\sigma$, $\tau$ on this larger algebra.

Let $\alpha_0,\ldots ,\alpha_{2r-1}$ be the simple positive roots of $\tilde{A}_{2r-1}$ with $\alpha_0$ the affine node. The canonical central element $c$ is the sum of these simple co-roots. In the case of $\tau$, we can assume after conjugating that $e_{\pm\alpha_i}\mapsto e_{\pm\alpha_{2r-i}}$ where $1\leq i\leq 2r-1$, and then we are forced to choose $e_{\pm\alpha_0}\mapsto e_{\pm\alpha_0}$. Let $\beta_0,\ldots ,\beta_r$ be the simple co-roots in $\tilde{A}_{2r-1}^{(2)}$. We will use the involution $\sigma$ to identify the corresponding simple co-roots with certain elements of the untwisted affine Lie algebra. For $1\leq i\leq r-1$ we can identify $e_{\pm\beta_i}$ with $e_{\pm\alpha_i}+e_{\pm\alpha_{2r-i}}$; $e_{\pm\beta_r}$ identifies with $e_{\pm\alpha_r}$. (Note that these are all fixed by $\tau$.) For the affine root we identify $e_{\beta_0}$ with something like $[e_{\alpha_0},e_{\alpha_1}-e_{\alpha_{2r-1}}]$, and similarly for $e_{-\beta_0}$. (We need to obtain an element of the $(-1)$ eigenspace for $\tau$.) I don't think this is quite right but it does tell us what the $\beta_i^\vee$ are in terms $\alpha_0^\vee,\ldots ,\alpha_{2r-1}^\vee$: we have $$\beta_0^\vee = 2\alpha_0^\vee+\alpha_1^\vee+\alpha_{2r-1}^\vee,\; \beta_1^\vee=\alpha_1^\vee+\alpha_{2r-1}^\vee,\ldots ,\; \beta_{r-1}^\vee=\alpha_{r-1}^\vee+\alpha_{r+1}^\vee,\; \beta_r^\vee=\alpha_r^\vee.$$ Now we look at Kac's tables (or work it out directly) to see that the canonical central element is $\beta_0^\vee+\beta_1^\vee+2(\beta_2^\vee+\ldots +\beta_r^\vee)=2\sum_0^r \alpha_i^\vee$. So I obtain $2c$.

In the case of $\sigma$, we think of the affine roots $\beta_0,\ldots ,\beta_{r-1}$ as forming the type $D_r$ subsystem which gives us $\mathfrak{so}_{2r}\subset\mathfrak{sl}_{2r}$, and make $\beta_r$ behave as the affine node. After conjugating we can assume that $\sigma(e_{\pm\alpha_i})=-e_{\pm\gamma(\alpha_i)}$. Then $e_{\beta_{r-1}}$ identifies with $e_{\alpha_1}-e_{\alpha_{2r-1}}$ and similarly $e_{\beta_i}$ for $1\leq i\leq r-2$ identifies with $e_{\alpha_{r-i}}-e_{\alpha_{r+i}}$. To complete the type $D_r$ subsystem we identify $e_{\beta_0}$ with $e_{\alpha_{r-1}+\alpha_r}-e_{\alpha_r+\alpha_{r+1}}$. It only remains to observe that $e_{\alpha_0}\mapsto -e_{\alpha_0}$ and so the last root element is $e_{\beta_r}=e_{\alpha_0}$. Again this isn't quite right but it will suffice for determining co-roots: $$\beta_0^\vee = \alpha_{r-1}^\vee+2\alpha_r^\vee+\alpha_{r+1}^\vee,\; \beta_1^\vee=\alpha_{r-1}^\vee+\alpha_{r+1}^\vee,\; \ldots ,\beta_{r-1}^\vee=\alpha_1^\vee+\alpha_{2r-1}^\vee, \beta_r^\vee=\alpha_0^\vee.$$ Once again we have $\beta_0^\vee+\beta_1^\vee+2(\beta_2^\vee+\ldots +\beta_r^\vee) = 2\sum_0^r\alpha_i^\vee=2c$.

So in both cases I obtain $2c$, but it is entirely possible that I have misunderstood your question and/or dualized somewhere (or got something wrong). But it seems to me that the central element you obtain in the twisted case will be something like the index times $c$.

EDIT: I see you are saying (as one might expect) that $\widehat{\mathfrak{sl}}_r$ is the untwisted affine Lie algebra. But then I really don't understand why you are taking the loop algebra over that. I must have a different edition of Kac's book to you - I don't have a Remark 8.6 and I only have a Thm. 8.5.

$\endgroup$
8
  • $\begingroup$ Thank you for your answer. It is just a metter of notation, I usually denote the affine Lie Algebra by $\widehat{\mathfrak{sl}}_r(K)$ where $K=\mathbb C((t))$. I assume that $e_{\alpha_i}$ are the Chevalley generators... also what is $\gamma$? $\endgroup$
    – Z.A.Z.Z
    May 25, 2017 at 12:21
  • $\begingroup$ Whoops - that got left in there from an earlier edit. There was also a mistake in that sentence, which I think I have fixed now. $\endgroup$
    – Paul Levy
    May 25, 2017 at 12:30
  • $\begingroup$ Remark 8.5: sais and I quote: "... the isomorphism class of $\mathcal L(\mathfrak{g},\sigma,m)$ depends only on the connected component of $Aut(\mathfrak{g})$ containing $\sigma$. .... " which is a consequence of Proposition 8.5 (if obviously it is the same as that in your version). By the way I have the 3rd version. $\endgroup$
    – Z.A.Z.Z
    May 25, 2017 at 12:57
  • $\begingroup$ Right, well there you go - these are both outer automorphisms so they will give isomorphic twisted loop algebras. (There are only two connected components of ${\rm Aut}(\mathfrak{sl}_n)$.) $\endgroup$
    – Paul Levy
    May 25, 2017 at 13:02
  • $\begingroup$ But, as I mentioned in the last comment above, by Lemma 8.5, $\tau$ and $\sigma$ give isomorphic twisted Kac-Moody algebras iff $$\tau=f(t)\sigma f(-t)^{-1},$$ for some $f(t)\in Aut(\mathcal L(\mathfrak{sl}_r))$, this is not clear for me, I don't see such $f(t)$. $\endgroup$
    – Z.A.Z.Z
    May 25, 2017 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.