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What is symmetric variety ? Let G be a semisimple simply connected algebraic group over the complex numbers. Let $\sigma$ : G $\to$ G be an automorphism of order 2 and H be the sub group of G of the element fixed by by $\sigma$. Then prove that G/H is a symmetric variety.

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  • $\begingroup$ This seems to be an exercise. From where? According to a standard definition, there is nothing to be done to solve your problem. Presumably, then, your definition is different... $\endgroup$ Sep 20 '15 at 8:35
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By definition, a symmetric variety is a quotient $G/H$ of a reductive algebraic group $G$ by the subgroup $H$ fixed by some involution of $G$. So there is nothing to show. Symmetric varieties are a subclass of spherical varieties. To give an example, consider the quotient $GL_n/O_n$, which is a symmetric variey and thus spherical. Indeed, let $GL_n$ act by conjuguation on the non-degenerate quadratic form $q(x_1, \cdots , x_n) = \sum_i x_i^2$. The orbit is isomorphic to the quotient $GL_n/O_n$. The involution $σ$ is defined by $\sigma(g) = (g^{-1})^t$.

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  • $\begingroup$ Are number of involution single or finite or infinite ? $\endgroup$
    – Amit
    Sep 20 '15 at 9:14
  • $\begingroup$ A single involution, but arbitrary. $\endgroup$ Sep 21 '15 at 12:53

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