4,139 reputation
822
bio website shapetales.wordpress.com
location Prague, Czech Republic
age 28
visits member for 3 years, 11 months
seen 9 hours ago

Math PhD. student and functional programming enthusiast.


Jan
4
comment Are any of those quotient rings isomorphic to other well known rings?
Yes, (1) seems to be something like $C(\beta {\bf R} \setminus {\bf R})$. Essentially, any function with non-trivial or non-existent limit corresponds to a new point in the compactification.
Jan
4
revised “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology
Removed the wrong remark about the box topology.
Jan
4
comment “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology
@Stefan: 1) you are correct, it seems your $B(x,r)$ generate a topology that lies strictly between uniform and box topology. I'll remove the note. 2) any open set containing $z$ must also contain $B(z, \epsilon)$ for some $\epsilon$ (that's the definition of basis).
Jan
4
comment “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology
@Daniel: done. And thanks for the discussion. I stared at this question for quite some time until I realized where the problem is; ..it's so natural to take openness for granted :)
Jan
4
revised “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology
Added explanation of unopenness
Jan
4
comment “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology
@Daniel: ah, I see what you mean by the radius, fine. Still, consider the OP's $B(x, 1)$ with $x = (1, 1/2, \dots)$. $(1, 1, \dots) \in B(x, 1)$ but there is no open subset of $B(x, 1)$ containing it.
Jan
4
comment “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology
@Daniel: are you saying that uniform topology is finer than the box topology? And also that $g$ is not continuous?
Jan
3
answered “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology
Jan
3
awarded  Good Question
Jan
3
comment $H^1$ of a constant sheaf
@Georges: indeed, thank you. But OP didn't mention irreducibility in their reasoning at all, therefore my comment.
Jan
3
comment $H^1$ of a constant sheaf
The argument you offer doesn't really use properties of either $H^1$ or $k(X)$. Are you sure you haven't just proved that $H^1(X, A)$ vanishes for any constant sheaf $A$ whatsoever (an absurdity to be sure)?
Jan
3
comment $H^1$ of a constant sheaf
The nerve certainly need not be a simplex, are you sure Serre said that?
Jan
1
comment How prove this limit $\lim_{n\to\infty}a_{n}=0$
Let $f(x) = x$, $a_n = 1$, $b_n = 1$, $c_n = 0$. Then the conditions on the sequences are satisfied but $\lim_{n\to\infty} a_n = 1$.
Jan
1
comment When is the converse of Casorati–Weierstrass false?
I'd say the converse is the trivial direction. If the function has a removable singularity at a point, it's obviously bounded near it, so the image can't be dense. Similarly, if the function has a pole then its modulus must be bounded away from zero and the image will now miss the nbhd of 0. I suspect this is why the theorem is not stated as an equivalence.
Jan
1
comment Why does $\operatorname{rank}(\mathbf{X})$ equal $\operatorname{rank}(\mathbf{X^TX})$? What is $\dim(\mathbf{X^TX})$?
OP mentions all of this in the question, why repeat it? On the other hand, I have no idea what the real question is supposed to be.
Jan
1
comment Computing complex integrals
Are you familiar with the residue theorem? It lets you relate the integral over the real line to the residues of $f$ provided you can show the integral over the half circle vanishes in the limit of $r \to \infty$.
Dec
24
comment Are these functions isomorphisms of the first binary structure with the second? - Fraleigh p.34 3.13, 3.15
It was not magic :) If you work on your own and try to prove e.g. injectivity, you are quickly lead to consider values at $x=0$. After that, any function that is not equal to $0$ at $x=0$ works. Thankfully, B.S. posted full details, check his answer.
Dec
20
comment What is the importance of conformal vector fields on Riemannian manifolds?
Flow is just a family of mappings, one for each $t$ in some interval. So it preserves something when all of those mappings do. I do not understand your second question. What do you mean by physical? I'll give you one example, perhaps it will help. In complex plane any holomorphic map is conformal. So a conformal flow here is just a smooth family of holomorphic maps, e.g. $z + te^z$.
Dec
19
comment Inverse image of a PID is a PID
@rschwieb: I had something bit more non-trivial in mind. For example, considering $S = {\bf C}$, which $R$ mapping onto it are PIDs? And similarly for other choices of $S$ or even whole classes of $S$.
Dec
19
comment Multidimensional complex integral of a holomorphic function with no poles
I deleted my answer because as Daniel pointed out (thank you), the proof by the Stokes' theorem theorem only works when $2n - 1 = n$, i.e. when $n=1$ because we'd like to integrate an $n$-form over the boundary of an $2n$-dimensional domain. The reason we need an $n$-form is that there are $n$ independent conditions of holomorphy in $n$ dimensions that we need to use. So it's impossible to even set up a reasonable integral when $n > 1$.