# Isomorphism as Lie algebra induced by an isomorphism between flag varieties

This may be a dumb question.
I want to prove a following lemma.

Let $$V,V'$$ be vector spaces of dimension $$r$$ over $$\mathbb{C}$$.
Assume there is an isomorphism $$\Phi$$ between the flag varieties, $$\Phi:\operatorname{Fl}(V)\rightarrow \operatorname{Fl}(V')$$.
Then, $$\Phi$$ induces an isomorphism as Lie algera, $$\Psi: \operatorname{End}_{0}(V)\rightarrow \operatorname{End}_{0}(V')$$.
( $$\operatorname{End}_{0}(V)$$ is the Lie algebra consisting of trace-zero endomorphisms. )

I don't know how to define $$\Psi$$ explicitly and why it is an isomorphism as Lie algebra.
Let $$T_{Fl(V)}$$ be the tangent bundle; then $$H^0(Fl(V), T_{Fl(V)}) \cong End_0(V).$$ Therefore, $$\Psi$$ can be defined as the differential of $$\Phi$$. To be more precise, the differential of $$\Phi$$ defines an isomorphism $$T_{Fl(V)} \stackrel{d\Phi}\to \Phi^*T_{Fl(V')},$$ hence the induced morphism on global sections gives $$End_0(V) \cong H^0(Fl(V), T_{Fl(V)}) \stackrel{d\Phi}\to H^0(Fl(V),\Phi^*T_{Fl(V')}) \cong H^0(Fl(V'),T_{Fl(V')}) \cong End_0(V').$$
• Thanks. This is very helpful for me. But why $H^0(\operatorname{Fl}(V),\operatorname{T}_{Fl(V)}) \cong \operatorname{End}_{0}(V)$? If it is not easy to explain, could you tell me some reference?
• It is a standard fact that $H^0(G/P,T) = \mathfrak{g}$ for any semisimple algebraic group $G$ and its parabolic subgroup $P$; this should be in any textbook on algebraic groups. In the case $G = \mathrm{SL}(V)$ and $P = B$ a Borel subgroup, this result reduces to what you want. Oct 2, 2020 at 17:46