Is there an elementary description of the affinization of the algebraic cotangent bundle of $\mathbb CP^n$? 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 $n=2,3\dotsc$.
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asked Aug 7, 2012 at 7:40
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2 Answers
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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
answered Aug 7, 2012 at 8:12
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$\begingroup$ 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. $\endgroup$– algoriCommented Aug 7, 2012 at 22:38
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1$\begingroup$ 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. $\endgroup$ Commented Aug 8, 2012 at 0:36
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2$\begingroup$ @algori Rank 1, trace 1 forces $f^2=f$ (which is what you meant). $\endgroup$ Commented Aug 8, 2012 at 11:58
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$\begingroup$ 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. $\endgroup$– algoriCommented Aug 8, 2012 at 14:52
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1$\begingroup$ Affinization usually means Spec(\Gamma(O_X))... $\endgroup$ Commented Aug 8, 2012 at 21:53
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The punctured cotangent bundle (excluding the zero section) of the complex projective space of dimension $n$ can be identified with the space:
$$\{(P,Q) \mid P^2=P, tr(P)=1, PQ+QP = Q, \operatorname{tr}(Q) = 0, Q\ne 0\},$$
where $P$ and $Q$ are $(n+1)$ dimensional Hermitian matrices. Please see Furutani and Tanaka's article A Kähler structure on the punctured cotangent bundle of complex and quaternion projective spaces and its application to a geometric quantization I (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 $TCP^n$ via the Fubini–Study metric.
