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could any one tell me why not all automorphisms of a root system are elements of weyl group? For example in $A_n, n>2$ the automorphism $\alpha\mapsto -\alpha$ is not in the weyl group. I do not understand why? it is also a reflection and must be an element of the weyl group. thank you. well $A_n$ means root systm of rank $n$.

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The automorphism group of the root systems $Aut \Phi$ is the semi-direct sum of the Weyl group with the group of automorphisms of the Dynkin diagram. In the case of $A_n$ there exists a non-trivial automorphism of the diagram which flips over the two tails. It is not the one which sends $\alpha$ to $-\alpha$ as you claim. In the case of $D_4$ this group is isomorphic to $S_3$.

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could you tell me what is semi direct sum? I have not reached upto dynkin diaram, could it be possible to explain in some other way? –  El Angel Exterminador Oct 1 '12 at 11:00
    
The OP does not claim that $\alpha \mapsto −\alpha$ is the automorphism outside of $W(A_n)$ one sees in the Dynkin diagram. It still is in $Aut(A_n) \setminus W(A_n)$ for $n \ge 2$, it's just the flip times some element (in fact, the longest element $w_0$) of the Weyl group. That is, $-id = flip \times w_0$. –  Torsten Schoeneberg Mar 5 at 15:18

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