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.


  • $\begingroup$ I don't understand your final sentence (before the "Thanks"). Can you state a bit more precisely what you mean here? $\endgroup$
    – Paul Levy
    Commented 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
    Commented 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
    Commented May 24, 2017 at 19:07
  • $\begingroup$ yes.....correct $\endgroup$
    – Z.A.Z.Z
    Commented 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
    Commented May 24, 2017 at 21:56

1 Answer 1


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.

  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented 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
    Commented 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
    Commented May 25, 2017 at 13:26

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