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Mar 4, 2013 at 21:35 comment added Paul Levy In type $A_n$ we know that every outer automorphism is $\gamma\circ\sigma$ for some inner $\sigma$, where $\gamma$ is a graph involution. So it isn't possible for an outer automorphism of $L$ to restrict to an element of the Weyl group on $\Delta$. Another way of thinking of it is that an outer automorphism can only act as $-w$ on a Cartan subalgebra (and there's no way $-w$ can equal $w'$ for $w,w'\in W$).
S Mar 3, 2013 at 14:41 comment added sunny ... to $\Delta$ lies inside the Weyl group (i.e. it is not reflection on $D$), and which brings the Kac diagram of $f$ to Kac diagram of $h$? Such $\phi$ will be inner.
S Mar 3, 2013 at 14:41 comment added sunny ... are related by a rotation on $D$, then as you say $f$ and $h$ are conjugate by inner automorphism, because the rotation on $D$ is the restriction of an inner $L$-automorphism to the Cartan. But I do not see an easy proof for the converse (though I also think the converse is true): Suppose that the Kac diagrams of $f$ and $h$ are related by a reflection on $D$ and not a rotation on $D$. We look for $\phi$ so that $f = \phi h \phi^{-1}$. Indeed the obvious $\phi$ given by reflection on $D$ is an outer $L$-automorphism. But how do we know there is no other choice of $\phi$ whose restriction
S Mar 3, 2013 at 14:41 comment added sunny (sorry the "add comment" button does not work so I write my response here) @ Paul Levy: Thank you very much for your comments. I assume you use Cartan subalgebra and the resulting action on the roots $\Delta$. But I do not see one direction of the proof. Take $A_n$ for example, and consider the affine Dynkin diagram $D$ which is a $(n+1)$-gon. Choose a Cartan subalgebra and a simple system so that the vertices of $D$ represent the simple and lowest roots. Let $f, h \in F(L)$ be represented by non-negative integers on the vertices of the $(n+1)$-gon (i.e. Kac diagrams). If the two Kac diagrams
Mar 2, 2013 at 22:20 history answered Paul Levy CC BY-SA 3.0