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I really appreciate it when you take time to answer my questions. Thanks!


Apr
8
comment What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different?
So I guess the point here would be, you cannot glue Hermitian metric holomorphically, but you can do so smoothly. This means that the isomorphism doesn't work for the case of holomorphic sections, but should work for the case of smooth sections
Apr
8
comment What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different?
Hm I see. I'm not sure in the smooth case if they are really different then. (Tend to say they should be the same) But what you wrote down for $H^0$ is definitely holomorphic sections, and so take the complex structure into account.
Apr
8
comment Weird definition of Kodaira-Spencer map (What's a relative Kähler differential on a manifold?)
@user40276, not really - the thing is people generally learn about locally ringed space when they do scheme theory, and rarely looked at smooth manifolds this way. I am not 100% certain, but let's say I'm 90% certain, that for your question it should work in both smooth category and holomorphic category.
Apr
8
comment What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different?
and that depends on the metric on vector space. The norm on finite dimensional vector space is essentially unique, but when you glue them over different charts, it's not. So if you can put a metric on your bundle - you have the isomorphism. This is the case for real vector bundles (over say, compact manifolds), but not the case for complex vector bundles (over complex manifolds, because you don't have partition of unity)
Apr
8
comment What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different?
The fiberwise isomorphism fails because there's no canonical way to identify $V$ with $V^*$.
Apr
8
comment Weird definition of Kodaira-Spencer map (What's a relative Kähler differential on a manifold?)
I just want to mention that your definition of relative differentials would also work for manifolds, as long as you regard your (smooth) manifold as a locally ringed space with the sheaf of smooth functions. Of course, that's equivalent to what Eric wrote down.
Apr
3
comment Can every continuous function on the closed unit disc be approximated uniformly by polynomials?
You are correct. Uniform approximation by polynomials (which are holomorphic) on a compact set, means that the limit must be holomorphic too.
Mar
19
comment Prove: If A is invertible, then adj(A) is invertible and $[adj(A)]^{-1}=\frac{1}{det(A)}A=adj(A^{-1})$
$$adj(A^{-1}) = \frac{1}{det(A^{-1})} A^{-1}$$ The left side gives you an expression of $A^{-1}$. Can you proceed then?
Mar
18
awarded  Yearling
Mar
10
comment Contractibility of $ S^2 $
@KomanR, exactly :)
Mar
10
comment Contractibility of $ S^2 $
@KomanR, precisely.
Mar
10
comment Contractibility of $ S^2 $
Because there's nowhere your south pole can be pulled along continuously.
Feb
28
comment How prove this $arg((1+ia)(2+ia)(3+ia)\cdots(n+ia))=\arctan{\frac{a}{1}}+\arctan{\frac{a}{2}}+\cdots+\arctan{\frac{a}{n}}$
You already saw that it's not even true for $n=1$..
Feb
21
comment If $x,y,z\in(0;1)$, prove that $(x+1)(y+1)(z+1)\ge \sqrt{8(x+y)(y+z)(z+x)}$.
$(x+1)(y+1) \ge 2(x+y) \Leftrightarrow (1-x)(1-y) \ge 0$.
Feb
17
comment E: $y^2+y=x^3$ an elliptic curve over $F_{2}$. How to prove the number of $E(F_{2^n})$ = $2^n+1$ if n is odd, …
@AdamStaples, I think the zeta function solution is exactly the one you posted - the proof of the theorem has Hasse-Weil L function lurking behind. On the other hand, you can also proceed by Jacobi sum. The case where $n$ is even is already done by Jyrki Lahtonen above, and the case where $n$ involves a Jacobi sum for the cubic characters.
Feb
12
comment class field theory via schemes?
mathoverflow.net/questions/73054/…
Feb
12
comment Why are de Rham cohomology and Cech cohomology of the constant sheaf the same
You are right..
Feb
6
comment How prove $\sum_{cyc}(f(x^3)+f(xyz)-f(x^2y)-f(x^2z))\ge 0$
Actually, using $1/(1-x) = 1+x+x^2+...$ + schur should work already.
Feb
1
answered Is there a general pattern behind the decimal expansion of $\frac{1}{7}$ being $.14+.0028+.000056+.00000112+…=.\overline{142857}$?
Feb
1
comment How prove $\sum_{cyc}(f(x^3)+f(xyz)-f(x^2y)-f(x^2z))\ge 0$
See artofproblemsolving.com/Forum/viewtopic.php?t=22364?ml=1 post #4. Your problem is a special case of a general form of Popoviciu's inequality, since $\frac{1}{1-e^x}$ is a convex function in your range. Using Vasc's inequality in that post, you get automatically that $$f(x^3)+f(y^3)+f(z^3) + \frac{3}{2} f(xyz) \ge \frac{3}{4} (f(x^2y) + \cdots + f(zx^2))$$ This is weaker than what you are proposing, but since your $f$ is special it's quite possible to wiggle the proof of that inequaity (which is for convex functions) for this case.