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 ?
2026-03-30 23:14:29.1774912469
Outer automorphism of SLn
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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.