Is there an elementary description of the affinization of the algebraic cotangent bundle of $CP^n$? I know that it can be described as some sort coadjoint orbits, but I am interested in a translation of that to concrete equations. Is it correct that the cotangent bundle can be obtained from the affinization by a blowup at a point? If the general equations are too complicated, I would be interested in some low dimensional examples such as n=2,3...
Probably matrices of rank 1 with fixed trace gives the answer.
The condition rank not more than 1 is given by algebraic equations on minors.
The condition trace = C - will delete zero and nilpotent from this set.
Any such matrix M= column*row, the map M-> row, gives the "projection" to P^n.
E.g. in 2x2 it is very simple we get condition det(M)=0, Tr(M)=1. Tr(M) = 1 can be resolved explicitly taking matrix of the form [a b; c 1-a] and we get equation a-a^2-bc = 0 - quadratic equation, can be rewritten as (a-1/2)^2+bc=1/4 - so we recognize hyperboloid which is known to be affine version of T^*P^1.
This corresponds to coadjoint orbit description - rank = 1 Trace = C - is clearly a coadjoint orbit.
Never checked the details but I think this is should be true
The punctured cotangent bundle (excluding the zero section) of the complex projective space of dimension $n$ can be identified with the space:
$(P,Q) | P^2=P, tr(P)=1, PQ+QP = Q, tr(Q) = 0, Q\ne 0$,
Where $P$ and $Q$ are $(n+1)$ dimensional Hermitian matrices. Please see Furutani and Tanaka's article(Proposition 2.3.).
Sketch of the proof: The space of the matrices $P$ is isomorphic as a homogeneous space to $CP^n$, since the action of $U(n+1)$ on it is transitive and the isotropy group of any point is $U(n)$. The conditions on the matrix $Q$ are obtained by taking the derivatives of the conditions on $P$, and the identification of $T^*CP^n$ and $TCP^n$ vuia the Fubini-Study metric.