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Let $G=GL(n, \mathbb{C})$ and $M=\{Q \in GL(n, \mathbb{C}) | Q^t = Q \}$. Suppose $G$ acts on $M$ through the following: $\forall g \in G, \forall m \in M$, $g\cdot m = gmg^t$. Question: Is the action transitive or not?

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up vote 7 down vote accepted

Every symmetric matrix $Q\in M$ is the matrix of a quadratic form of maximum rank over $\mathbb{C}^n$. The action of $G$ represents base change for the quadratic form represented by $Q$. Every quadratic form has orthonormal bases. Thus the identity matrix is in every orbit, so yes, the action is transitive.

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Ok thanks. What happens if $\mathbb{C}$ is changed to $\mathbb{R}$? Is the action still transitive? – Eric Apr 24 '12 at 13:18
Have you heard of the signature of a real (or complex hermitian) quadratic form? There are a total of $n+1$ orbits in the real (or complex hermitian) case for non degenerate quadratic forms, one for each signature. You'll probably find stuff on wikipedia under signature or Sylvester's law of inertia (direct translation from french, not sure if correct name in English.) – Olivier Bégassat Apr 24 '12 at 13:25
Over $\mathbb{R}$ non degenerate quadratic forms are not all equivalent (you can check that $I_n$ and $-I_n$ are in different orbits because the former defines a positive quadratic form while the latter is negative). Two invertible symmetric matrices are in the same orbit if and only if they have the same number of positive (and negative) eigenvalues, we call this number the signature of the quadratic form. – Joel Cohen Apr 24 '12 at 13:27
Thanks a lot! Unfortunately, I am not aware of Slyvester's law of inertia... – Eric Apr 24 '12 at 13:45

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