Reputation
636
Top tag
Next privilege 1,000 Rep.
Create tags
Badges
3 9
Impact
~10k people reached

  • 0 posts edited
  • 0 helpful flags
  • 1 vote cast
Apr
20
awarded  Popular Question
Mar
18
comment Prove the Inverse of a Nonconstant Harmonic Function is Unbounded
I think that $u^{-1}(c)$ here means the preimage: $u^{-1}(c)=\{z\in \mathbb{C}\, |\, u(z)=c\}$
Feb
23
comment The complement $\mathbb{P}_k^2 \setminus C$ of a plane curve $C$ is affine.
Yes that's what I mean. I also have posted a full answer below to be more clear
Feb
23
answered The complement $\mathbb{P}_k^2 \setminus C$ of a plane curve $C$ is affine.
Feb
23
comment The complement $\mathbb{P}_k^2 \setminus C$ of a plane curve $C$ is affine.
Well, you know that $\mathbb{P}^n_k\setminus H \cong \mathbb{A}^n_k$ and then you have a closed subset of affine space that is by definition affine.
Feb
23
comment The complement $\mathbb{P}_k^2 \setminus C$ of a plane curve $C$ is affine.
Using the Veronese embedding corresponding to the degree of the curve you can reduce to the situation of $\mathbb{P}^n$ minus an hyperplane.
Feb
16
answered Relating the cardinality to surjectivity
Feb
16
comment Can a product of a Stein manifold and a compact manifold be again Stein?
You're welcome!
Feb
16
comment Can a product of a Stein manifold and a compact manifold be again Stein?
I think that the answer should be no, as a closed submanifold of a Stein manifold is Stein, and $Y$ is a closed submanifold of $ X \times Y$.
Feb
16
awarded  Supporter
Dec
20
awarded  Caucus
Dec
15
comment Flexes of cubic curve
@evinda: I wrote an answer that hopefully explains enough. Sorry that it took me so long and tell me if you would like further clarifications.
Dec
15
answered Flexes of cubic curve
Dec
9
comment Flexes of cubic curve
sorry I have to go now, I will write a more extensive explanation later. Anyway you can try to have a look at wikipedia: en.wikipedia.org/wiki/Projective_space
Dec
9
comment Flexes of cubic curve
No, by definition $\mathbb{P}^2(\mathbb{C})=(\mathbb{C}^2\setminus \{0\}) / \mathbb{C}^*$.
Dec
9
comment Flexes of cubic curve
What is your definition of nonsingular? In this case you can just check that the partial derivatives of $F$ cannot vanish all together in a point of the curve.
Dec
9
comment Flexes of cubic curve
First you could try to prove that the curve is nonsingular, and then compute the interesection points of the curve with the Hessian. In this case the computations should be pretty straighforward.
Dec
4
answered Vanishing of global sections with very negative twists.
Nov
11
awarded  Yearling
Nov
11
comment Showing $A_5$ is simple
Yes, usually by $A_5/H$ you mean the set of all cosets of $H$, and if $H$ is normal you can give this the structure of a group and call it the quotient group (or also the factor group I guess, even if I never heard that name).