# affinization of T^*CP^n

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...

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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.

PS

This corresponds to coadjoint orbit description - rank = 1 Trace = C - is clearly a coadjoint orbit.

PSPS

Never checked the details but I think this is should be true

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Alexander -- I think this works but you have to consider projections (i.e. maps $f$ such that $f\circ f=Id$), not just any linear maps; the trace than will automatically be 1. –  algori Aug 7 '12 at 22:38
this is not the "affinization", this is the smoothing. the affinization is given by taking things of trace=0 and rank at most 1 and T^*CP^n will be given by blowing up at the 0-matrix. –  Daniel Pomerleano Aug 8 '12 at 0:36
@algori Rank 1, trace 1 forces $f^2=f$ (which is what you meant). –  David Speyer Aug 8 '12 at 11:58
David -- yes, of course, for rank 1 linear maps the trace is 1 iff the map is idempotent. Daniel -- I interpreted "affinization" as "an affine variety isomorphic to"; if this is not what is meant here, then please disregard my remark. –  algori Aug 8 '12 at 14:52
Affinization usually means Spec(\Gamma(O_X))... –  Daniel Pomerleano Aug 8 '12 at 21:53
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.