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What is the dimension of Borel subgroup of $Sp(2n,\mathbb C) $ ?

I know that if I choose the bilinear form nicely then Borel subgroup of $Sp(2n,\mathbb C)$ is just the intersection $Sp(2n,\mathbb C) \cap T(2n,\mathbb C)$, where $T(2n,\mathbb C)$ is the set of upper triangular invertible matrices. In this scenario, my calculation shows that the dimension should be $n^2 + \dfrac{n(n+1)}{2}$. Is it correct ?

Saikat
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    Welcome Saikat, to MO. The answer seems to be $n^2+n$. In any case, this question is not appropriate for MO since MO is for research mathematicians asking for not well known references, or proofs not set down in print, or sometimes, some technical questions. For general questions like yours, math stackexchange may be more suitable. –  May 02 '19 at 11:28
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    @Saikat: This is a pretty well known computation, given by the rank + the number of positive roots, and applies uniformly to all simple algebraic groups over an algebaically closed field of arbitrary characteristic $p \geq 0$. So Venkatarama is correct. –  May 03 '19 at 16:51

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