How to find the dimension of the group $O_{p,q}(\mathbb R)= \{g \in GL_n(\mathbb R): g^TI_{p,q} \ g = I_{p,q}\}$, where $I_{p,q}= diag(1,..., 1,-1,...,-1)$ and $p+q=n$?


closed as off-topic by Derek Holt, Watson, Bobson Dugnutt, vrugtehagel, colormegone Mar 10 '16 at 23:16

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  • $\begingroup$ Do you know about the Lie algebra of a Lie group (or an algebraic group) ? $\endgroup$ – Captain Lama Mar 10 '16 at 7:11
  • $\begingroup$ Thanks for the answer, I wanted to use only lie groups theory.. and I found a theorem (Regular Value Theorem) talks about that $\endgroup$ – Ronald Mar 10 '16 at 23:36

If you know about Lie algebras, then you can check that the Lie algebra of this group is $\{ a\in M_n(\mathbb{R}) | a^T I_{p,q} + I_{p,q}a = 0\}$, which is just $I_{p,q} \mathfrak{so}_n(\mathbb{R})$, so it has the same dimension as $\mathfrak{so}_n(\mathbb{R})$, ie $n(n-1)/2$.

If not, there is a little trick : if you tensor with $\mathbb{C}$, then $O_{p,q}(\mathbb{R})\otimes \mathbb{C} = O_{p,q}(\mathbb{C})$, but since all non-degenerated quadratic forms over $\mathbb{C}$ are isometric, $O_{p,q}(\mathbb{C}) \simeq O_n(\mathbb{C})$.

So $O_{p,q}(\mathbb{R})\otimes \mathbb{C} = O_n(\mathbb{R})\otimes \mathbb{C}$, and since tensoring with $\mathbb{C}$ doubles the real dimension in both cases, you see that $O_{p,q}(\mathbb{R})$ has the same dimension as $O_n(\mathbb{R})$ (meaning $n(n-1)/2$).


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