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What is the outer automorphism of $SL_{2n}$ and how do you get the Symplectic group as a fixed locus of this automorphism ? I know that the automorphism reverses the Dynkin diagram which is a line in this case. I heard that it is $X \mapsto (X^T)^{-1}$ but the fixed points in this case is the orthogonal group. Then how do you get the Symplectic group out of this ? If at all you get the Symplectic group, then what could be the Symplectic form associated to this automorphism ?

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One way to describe this automorphism is as $X\mapsto f^{-1}(X^T)^{-1}f$, where $f=\operatorname{antidiag}(1,\ldots,1,-1,\ldots,-1)$. The matrix $f$ is exactly the Gram matrix of the form to be preserved by the elements of the symplectic group.

Note that the automorphisms are uniquely determined by their action on the generators, and in case of Chevalley groups one has to check if a bijection preserves a very few relations, see Section 12.2 of "Simple groups of Lie type" by R. W. Carter. So it is often easier to look at the generators and relations rather than the matrices.

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    $\begingroup$ There is only one outer automorphism of $SL_{2n}$. So the outer automorphism in the question $X \mapsto (X^T)^{-1}$ and the automorphism you are given are equivalent somehow. But the for one you get the orthogonal group and for the other you get the symplectic group which are not isomorphic to each other. Why so ? $\endgroup$
    – icmes imrf
    May 16, 2017 at 1:46
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    $\begingroup$ The usual wording "one outer automorphism" means that the quotient group $\operatorname{Aut}(G)/\operatorname{Inn}(G)$ is $C_2$, but in fact any composition of an inner automorphism and an outer automorphism is outer. This is exactly the case, the two automorphisms in question differ by composition with $X\mapsto f^{-1}Xf$. $\endgroup$ May 16, 2017 at 8:40
  • $\begingroup$ I want to see concretely why this automorphism is coming from a dynkin automorphism ? In particular is should map $\alpha_i \mapsto \alpha_{2n-i}$. But, I don't think this automorphism does that. $\endgroup$
    – icmes imrf
    May 16, 2017 at 22:16
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    $\begingroup$ Take an elementary generator $t_{ij}(\xi)=e+\xi e_{ij}$ in $SL_{2n}$ and apply this automorphism — first permute columns into the reverse order, then transpose, then permute rows. The thing here is that it also changes the sign of $\xi$, that is how you can get the long roots of $C_\ell$ (the corresponding generators are $t_{i,2n-i+1}(\xi)$), and can't have such elements in $D_\ell$. $\endgroup$ May 16, 2017 at 22:53

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