Suppose $G$ is a Lie group (or I guess a linear algebraic group), $P \subset G$ a Lie subgroup with Lie algebras $\mathfrak{g}$ and $\mathfrak{p}$ respectively. In Chriss and Ginzburg's book "Representation theory and complex geometry" they have the following result (Lemma 1.4.9):
There is a natural $G$-equivariant isomorphism $$ T^*(G/P) \cong G \times_P \mathfrak{p}^\perp $$ where $\mathfrak{p}^\perp$ is the annihilator of $\mathfrak{p}$ in $\mathfrak{g}^*$ and $P$ acts on $\mathfrak{p}^\perp$ by the coadjoint action.
They give a proof however I don't understand it at all. I will will reproduce it for convenience below (with some annotations that are my own!). I was hoping someone could help me fill in the details. I am actually interested in the algebraic version of the result but help understanding the proof in either the differential category or the algebraic one would be very useful!
Proof. Let $e = 1 \cdot P/P \in G/P$ be the base point. We have $T_e(G/P) = \mathfrak{g}/\mathfrak{p}$ and $T_e^*(G/P) = (\mathfrak{g}/\mathfrak{p})^* = \mathfrak{p}^\perp \subset \mathfrak{g}^*$. (NOTE: ok up to here I understand, but the rest I have no idea). It follows that, for any $g \in G$ $$T_{g \cdot e}^*(G/P) = g\mathfrak{p}^\perp g^{-1}. $$ Question 1: So I understand that somehow I am meant to be able to transfer the tangent space at $e$ to the point $g \cdot e$ using the $G$ action but I'm unsure how to write this down.
This shows that the vector bundles $T^*(G/P)$ and $ G \times_P \mathfrak{p}^\perp $ have the same fibers at each point of $G/P$, hence are equal as sets (NOTE: ok I am happy with this if I can understand the above). To prove that they are isomorphic as manifolds, one can refine the argument as follows.
Consider the trivial bundle $\mathfrak{g}_{G/P} = G/P \times \mathfrak{g}$ on $G/P$ with fiber $\mathfrak{g}$. The infinitesimal $\mathfrak{g}$-action on $G/P$ gives rise to a vector bundle morphism $\mathfrak{g}_{G/P} \longrightarrow T(G/P)$.
Question 2: I understand that each element of the Lie algebra $X \in \mathfrak{g}$ gives rise to a vector field $\xi_X$, is the image of $(gP,X)$ under this map simply $\xi_X(gP)$?
It is clear that the kernel of this morphism is the sub bundle $E \subset \mathfrak{g}_{G/P}$ whose fiber at a point $x \in G/P$ is the isotropy Lie algebra $\mathfrak{p}_x \subset \mathfrak{g}$ at $x$. This gives an isomorphism $T(G/P) \cong \mathfrak{g}_{G/P}/E$.
Question 3: I don't know what is meant by the isotropy Lie algebra at the point but if I am correct in question 2 I would guess this is simply the tautological thing - i.e. the elements of the lie algebra that act infinitesimally by zero?
Further, the description of the fibers of $E$ gives an isomorphism $E \cong G \times_P (\mathfrak{g}/\mathfrak{p})$.
Question 4: I don't understand this at all
Hence, $T(G/P) \cong G \times_P (\mathfrak{g}/\mathfrak{p})$,
Question 5: Didn't we just say that $T(G/P)$ was a quotient by this bundle?
and the result follows by taking duals on each side. (NOTE: That I am happy with).