Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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...

share|improve this question
add comment

2 Answers 2

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

share|improve this answer
    
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
1  
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
1  
@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
1  
Affinization usually means Spec(\Gamma(O_X))... –  Daniel Pomerleano Aug 8 '12 at 21:53
show 3 more comments

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.