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1d
comment cycle class as Chern class
This is Grothendieck--Riemann--Roch.
2d
comment cycle class as Chern class
One always have $[Z] = ch_p(O_Z)$ (the $p$-the coefficient of the Chern character), and in fact you don't need to resolve the singularities. Expressing it in terms of Chern classes and taking into account that $ch_i(O_Z) = c_i(O_Z) = 0$ for $0 < i < p$, one can prove the formula you want, but without the "funny factors"!
2d
comment Is the orthogonal complement of an admissible subcategory admissible itself?
iopscience.iop.org/0025-5726/35/3/A02
2d
answered Is the orthogonal complement of an admissible subcategory admissible itself?
Dec
8
comment Properties of finite quotients of quasi-projective varieties
@Anton: For finite group acting on a quasi-projective variety in char 0 --- yes. But not only in this case.
Dec
8
comment Properties of finite quotients of quasi-projective varieties
$X/G$ is the coarse moduli space of $[X/G]$ when it is a categorical quotient. For more details see Mumford--Fogarty--Kirwan.
Nov
24
comment When a proper morphism of schemes is a closed imbedding?
There is a notion of a closed subfunctor (introduced by Grothendieck), see e.g. [FGA]. Of course $X \to Y$ is a closed embedding if and only if $Mor(-,X)$ is a closed subfunctor in $Mor(-,Y)$.
Nov
23
comment Determinant of the oriented adjacency matrix of a tree
Expanding darij grinberg's comment --- it looks like the answer is $(-1)^n$ where $n$ is the number of "$v$-antioriented edges", i.e. the number of edges orientation of which should be switched to make all edges oriented from $v$.
Nov
20
awarded  Enlightened
Nov
20
awarded  Nice Answer
Nov
19
answered Push-forward of locally free sheaves
Nov
11
awarded  Pundit
Nov
10
comment Analogue of Borel--Bott--Weil for General Equivariant Vector Bundles
It follows easily from the equivalence of the category of equivariant bundles and the category of representations of Borel subgroup, since Borel is solvable.
Nov
10
comment Analogue of Borel--Bott--Weil for General Equivariant Vector Bundles
On flag variety any equivariant vector bundle is an iterated extension of line bundles. So, its Euler characteristic can be computed by applying BBW to the factors and summing up.
Nov
7
answered When does a cubic surface pass through five lines?
Nov
4
comment What are the higher homotopy groups of a K3 suface?
I guess you could say that $V_3$ should be the kernel of the multiplication map $S^2H^2(X,Q) \to H^4(X,Q)$, so its dimension is $b_2(b_2+1)/2 - 1 = 252$. Of course this is equivalent to your computation, but does not require choosing a basis and a bit simpler.
Nov
3
comment What are the higher homotopy groups of a K3 suface?
But the higher homotopy gorups of an elliptic curve is easy to find, and this is definitely a better analogy for a K3 surface.
Oct
28
awarded  Enlightened
Oct
21
comment Why is it so hard to compute $\pi_n(S^n)$?
Probably, to check nontriviality of the Hopf bundle one can compute its Euler class?
Oct
14
comment The linear projection of projective spaces
@abx: of course the sign was wrong, now it is corrected, thanks!