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1d

comment 
cycle class as Chern class
This is GrothendieckRiemannRoch. 
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/00255726/35/3/A02 
2d

answered  Is the orthogonal complement of an admissible subcategory admissible itself? 
Dec 8 
comment 
Properties of finite quotients of quasiprojective varieties
@Anton: For finite group acting on a quasiprojective variety in char 0  yes. But not only in this case. 
Dec 8 
comment 
Properties of finite quotients of quasiprojective varieties
$X/G$ is the coarse moduli space of $[X/G]$ when it is a categorical quotient. For more details see MumfordFogartyKirwan. 
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  Pushforward of locally free sheaves 
Nov 11 
awarded  Pundit 
Nov 10 
comment 
Analogue of BorelBottWeil 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 BorelBottWeil 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! 