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Apr
21
comment Rational function with absolute value $1$ on unit circle
Great, thank you very much. I guess this stronger identity theorem is implicit in his "counting of poles/zeros" of a rational function, which for him are always finite determined by the degrees of the polynomials in the numerator/denominator. $M (z) - 1$ is rational and has infinitely many zeros (at least on the unit circle), so it has to be the zero function. (I totally forgot about $\frac{a−b}{1 - \overline a b}$, thank you! Now there are no more loose ends.)
Apr
21
comment Rational function with absolute value $1$ on unit circle
The book is Ahlfors Complex analysis (I have the second edition). The Lagrange interpolation says that for any $n$ values, you can find a polynomial P of degree $< n$ such that P takes the specified values. If one could show that the points in the unit circle are "generic enough", one might argue that there couldn't be any other rational function, satisfying the uncountably many "constraints" $M (z) = 1$ for $\lvert z \rvert = 1$... (But I don't know if I'm on the right track...)
Apr
21
comment Rational function with absolute value $1$ on unit circle
The previous section has as exercise "verify the Cauchy-Riemann equations for $z^2$ and $z^3$". In other words, almost no "tools", only the definition of a holomorphic/rational function. There is a theorem "If all zeros of a polynomial lie in a half plane, then all zeros of the derivative lie in the same half plane", but I don't think it's relevant. The reader also learned how to develop partial fractions (and construct the Lagrange interpolation polynomial by proving $\frac{P (z)}{Q (z)} = \sum_{k=1}^n \frac{P (a_k)}{Q' (a_k)(z-a_k)}$, where $\deg P < n$ and $a_i$'s are the roots of $Q$.)
Apr
21
comment Rational function with absolute value $1$ on unit circle
@DanielFischer I don't quite understand the "identity theorem" you cite. We have that $M (z) = R (z) . \overline{R (1/\overline z)}$ satisfies $M (z) = 1$ for $\lvert z \rvert = 1$. But the set of such $z$ is not open, so the usual identity theorem does not apply. Also, at the point of the book, the identity theorem hasn't been mentioned, so I was wondering if there is an easier argument. (The previous exercise asks to construct the Lagrange interpolation polynomial.)
Feb
11
comment What is the flaw of this proof (largest integer)?
@TripeHound 13 fingers on each hand or on both hands? Wait... does God have two hands?
Jun
1
comment Is this a known formula? $\det (id, \partial_x, \partial_y)^t (f, g, h)$
@BCLC I found it while calculating automorphisms of rank-two vector bundles on a non-commutative deformation of a surface. This expression comes from rewriting three expressions containing Poisson structures as this determinant. The expressions with Poisson structures don't simplify to anything nice, so I am wondering if maybe this determinant expression is something nice I just don't know.
May
17
comment A proof for Atiyah-Macdonald Exercise I.21.iii
@TakumiMurayama That is where my confusion lies. Maybe it is a shorthand for the set of all the elements in the contractions of the prime ideals containing $\mathfrak b$ (or a shorthand for the ideal generated by all these elements). The first equality in the last chain of equalities says $r (E^c) = r (E)^c$, which I only know to hold when $E$ is an ideal. I'm asking if there is a way to make sense of this "proof", because if the answer is yes, it is shorter than the other proof I found. But if the answer is no, then I will simply ignore it.
Feb
5
comment First examples for topology of non-Hausdorff spaces
@EricWofsey What is an "inductive Mayer-Vietoris argument"?
Feb
5
comment First examples for topology of non-Hausdorff spaces
@PedroTamaroff The only broad thing about the question is the last paragraph, which is optional (should I say "bonus question"?).
Jul
22
comment Connection in fibre bundle from discontinuous group action
@Bombyxmori I think it is reasonable, considering that I "act in every direction". The formal argument I know invokes Poincaré duality to conclude: if a group $\Gamma$ acts (faithfully) on a contractible manifold $X$ and $\mathrm{cd} \, \Gamma = \dim X$, then $X / \Gamma$ is compact. ($\mathrm{cd}$ being the "cohomological dimension".)
Jun
12
comment Number of Zeros of a Section vs Integral First Chern Class
Thanks. Any hint as to why this is true?
Jun
12
comment Number of Zeros of a Section vs Integral First Chern Class
"complex line bundles are actually classified by their first Chern class" Is this true for bundles over any base space?
Mar
19
comment Can one exchange fibre and base space in a fibre bundle?
@geodude My question was about "general fibre bundles", I'm not really interested in direct products. Thank you for the link.
Jan
23
comment All almost complex structures on a manifold
@JimBelk $\mathcal O(k)$ is the complex line bundle (real rank 2 bundle) of Chern class $k$ over $\mathbb P^1\cong S^2$ (see the notation on Wikipedia). I loosely expect the space to be finite-dimensional, because the space of complex structures on $\mathcal O(k)$, $k$ neg., is finite-dimensional (this one has to calculate). Even if it isn't finite-dimensional, though, one should be able to construct at least a one-parameter family of almost complex structures for the theorem to be of any practical value.
Jan
22
comment Intuitively what is it if making a modification of a torus?
As far as I can tell, your modification is not an equivalence relation.
Dec
13
comment Derived functors and coboundary operator
@ZhenLin I'm not exactly sure what your point is, but as far as I understand, you can calculate group cohomology either from an injective resolution (which is difficult to find) or from a projective resolution with the Hom functor applied to the whole resolution.
Dec
13
comment Derived functors and coboundary operator
@ZhenLin In principle, yes. I added more details in the question body.
Dec
12
comment Derived functors and coboundary operator
@ZhenLin Edited. Is this more clear?
Sep
7
comment Cohomology of $\mathcal O(k)$
@atricolf The total space of the bundle $\mathcal O(k)$.
Jul
7
comment Cell decomposition for connected sum
@DanielRust Thanks. Is the idea to use another Mayer-Vietoris sequence for the "punctured" manifolds, or is there another way to calculate their homology?