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The dimensionality of the indefinite orthogonal group $O(p, q)$ equals $n(n-1)/2$, where $n := p + q$. Is it similarly true that the dimensionality of all the indefinite Lie groups with signature $(p, q)$ depends only on $n := p + q$? For example, is the dimensionality of $U(p, q)$ equal to $(p + q)^2$?

This statement seems intuitively obvious, because we should be able to convert any $U(p, q)$ matrix to a $U(p + q)$ matrix simply by judiciously inserting some factors of $i$, but I cannot prove it.

tparker
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    Yes, this should be true. It's enough to show that the dimensions of the complexifications of their Lie algebras match, and that's because they'll in fact be isomorphic (e.g the complexification of $\mathfrak{so}(p, q)$ is $\mathfrak{so}(p + q, \mathbb{C})$). That is, the various Lie algebras will all be real forms of the same complex Lie algebra. – Qiaochu Yuan Oct 21 '15 at 04:09
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    While the statement is true (either via looking at complexifications or by direct considerations), it is absolutely not possible to convert a matrix from $U(p,q)$ to a matrix in $U(p+q)$ by inserting factors of $i$. For a matrix in $U(p+q)$ any entry has absolute value $\leq 1$, while matrices in $U(p,q)$ can have arbitrarily large entries. This only works on the level of the Lie algebras $\mathfrak{u}(p,q)$ and $\mathfrak{u}(p+q)$. – Andreas Cap Oct 24 '15 at 14:56

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