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How to construct a Cartan subalgebra for ${\frak so}_{2n}(\mathbb C)$?

In here: http://ncatlab.org/nlab/show/Cartan+subalgebra it said:

For a compact Lie group $G$ with Lie algebra $\frak g$, a Cartan subalgebra of $\frak g$ is a sun-Lie algebra $\frak t \hookrightarrow \frak g$ that is the Lie algebra of a maximal torus $T \hookrightarrow \ G$.

So in our case, ${\frak g}={\frak so}_{2n}(\mathbb C)$ and $G=SO_{2n}(\mathbb C)$. And I know that

$$T=\begin{bmatrix} R_1 & & 0\\ & \ddots & \\ 0& & R_n \end{bmatrix}$$ where $T$ is an $2n \times 2n$ matrix with $R_i \in SO_2$

If the above argument is right, than the corresponding Lie algebra $\frak t$ for $T$ is a Cartan subalgebra of ${\frak so}_{2n}(\mathbb C)$? Can someone show me how to pass $T$ (Lie group) to $\frak t$ (Lie algebra).

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  • $\begingroup$ This is expkained in most textbooks treating lie groups; Lee's book on smooth manifolds being one of them, an a particularly good one, in fact. $\endgroup$ – Mariano Suárez-Álvarez Sep 24 '15 at 19:03
  • $\begingroup$ So $\frak t$ is equal to the tangent space of $T$ at $1 $ (identity matrix)? $\endgroup$ – SamC Sep 24 '15 at 19:10
  • $\begingroup$ Well, that is precisely the definition of the Lie algebra of a Lie group! $\endgroup$ – Mariano Suárez-Álvarez Sep 24 '15 at 19:10

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