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Let $G=GL(n,\mathbb{C})$ and $B$ be a Borel in $G$.

The variety $G/B$ is known to be a moduli of flags (i.e., a variety of Borels) with a bijection between a class of Borels and points of $G/B$.

The cotangent bundle of $G/B$ has a description
$$ T^*(G/B)= G\times_{B} (\mathfrak{g}/\mathfrak{b})^* = G\times_B \mathfrak{n} $$

$$ = \{ (\mathfrak{b'}, \langle n, -\rangle) \in G/B\times \mathfrak{g}^* : n\in \mathfrak{b}' \mbox{ and } n \mbox{ is nilpotent } \}. (\star) $$

  1. Besides the upper triangular matrices in $G$, what other Borels are in $G$? If this question is too hard to answer, how about just for the case $n=2$ or $3$?

  2. Concretely, say for $n=2$, can you give an example of this description $ (\star)$?

  3. How is $G$ acting on $T^*(G/B)$? I thought I saw in several papers that $$ g.[h,n] = [hg^{-1},gn] \in G\times_{B}\mathfrak{n}, $$ but in some other references, I thought I saw $$ g.[h,n] = [hg^{-1},gng^{-1}] \in G\times_{B}\mathfrak{n}. $$ Which one is correct?

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For 1), all Borels are conjugate, and all conjugates of $B$ are Borels. –  M Turgeon Jul 8 '12 at 22:56
    
Thank you M Turgeon for this reply. So if $B$ and $B'$ are two Borels in $GL(n,\mathbb{C})$, there exists some $g\in G$ so that $gBg^{-1}=B'$? –  math-visitor Jul 8 '12 at 23:00
    
Yes, exactly. And vice-versa. –  M Turgeon Jul 8 '12 at 23:01

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