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Jan
25
comment the top chern class of the holomorphic tangent bundle is the euler class
@user125763: Not a big problem, I did this in my diplom thesis and this is (almost) a straight copy and paste from the source. Hope it is of any use to you.
Jan
25
revised the top chern class of the holomorphic tangent bundle is the euler class
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Jan
25
comment the top chern class of the holomorphic tangent bundle is the euler class
Dear @user125763: I have proved this once for nonsingular complex projective varieties using a couple of big hammers: Borel-Serre identity, Hirzebruch-Riemann-Roch and Hodge Decomposition. If you want, I can show you how that works. I deal with varieties much more than with manifolds, but in the above case the two notions overlap. Concerning complex conjugation, I unfortunately do not know. This sounds like a basic fact though, if it is true. Have you checked standard literature?
Jan
23
answered the top chern class of the holomorphic tangent bundle is the euler class
Jan
22
comment Let $f: U \rightarrow W$ be a morphism of affine algebraic sets and $f': k[W] \rightarrow k[U]$ be the k-algebra morphism of coordinate rings.
Oh, that's true.
Jan
22
answered Let $f: U \rightarrow W$ be a morphism of affine algebraic sets and $f': k[W] \rightarrow k[U]$ be the k-algebra morphism of coordinate rings.
Jan
5
answered If $xa=xb$ then $a=b$
Dec
24
revised Maximal solvable subgroup not Borel
edited tags
Dec
21
awarded  Constituent
Dec
11
answered quotient is a projective variety
Dec
8
awarded  Caucus
Dec
6
revised Invariants of $O(2) \times O(2)$ under simultaneous conjugation
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Dec
5
answered Invariants of $O(2) \times O(2)$ under simultaneous conjugation
Dec
4
revised action of $GL_3$ on $P^2$
added 128 characters in body
Dec
4
answered action of $GL_3$ on $P^2$
Dec
4
revised action of $GL_3$ on $P^2$
edited tags
Dec
4
comment Order of product of abelian group
I think you have only shown that the order of $c=ab$ divides $\operatorname{lcm}(m,n)$. This does not prove that they are equal.
Dec
4
comment Order of product of abelian group
No, this is not true. from $b^n=a^{-m}$ you can not conclude $b^{n-1}=a^{-(m-1)}$ because $a^{-(m-1)}=a^{-m+1}=b^{n+1}\ne b^{n-1}$ in general. I have a feeling the order should be the least common multiple of $m$ and $n$.
Dec
3
comment How does Molien series describe polynomial invariants?
@monomorphic: You got it. Unfortunately, I am afraid I have never seen a proof of Molien's theorem and you might best be served asking this as another question, I am sure someone here can help you with that.
Dec
3
comment How does Molien series describe polynomial invariants?
@monomorphic: Alright =)