3,970 reputation
820
bio website
location Prague, Czech Republic
age 28
visits member for 3 years, 5 months
seen 21 hours ago

I study Mathematics at the Charles University in Prague and have a degree in Theoretical Physics from the same school. I am quite fluent in Computer Science and several programming languages as well.

I work on topological aspects of analysis (or perhaps analytical aspects of topology?) such as K-theory, index theory and whatnot. But I also enjoy studying differential topology, groups, number theory and basically anything else I come across. And the more I learn the more I believe in the old cliche that "There is only one mathematics".

People I enjoy reading/watching the most: Serre, Atiyah, Milnor, Bott, A[a-z] (still looking for this guy).


Jan
5
answered Mirror point with respect to Riemann circle (Möbius transformation)
Jan
5
comment How a group represents the passage of time?
@wonghang: precisely. Another often occurring group is integers that occurs in discrete dynamics. You obtain it by simply taking any map $f$ on a given space and iterating it and its inverse. Although to be fair, most dynamics are not reversible, maps not invertible, so one uses natural numbers instead and moves from groups to monoids or semigroups. But dynamics coming from physics usually are reversible (with the notable exception of heat equation).
Jan
4
comment Uniform continuity of $f(x)=x+\frac{\sin x}{x}$ in $(0,\infty)$
@GinKin, your argument is wrong. You can't calculate a limit of $\infty - \infty$ by simply saying it's $\infty$. It's undefined and you need to use some other way. Taylor series for $\sin(x)$ is $x - x^3 / 6 + \cdots$ so that $\sin(x)/x$ = $1 - x^2 / 6 + \cdots$ has $0$ derivative at $0$.
Jan
4
comment Open/Closed Subsets of Metric Space
I'll add that $U$ is not closed, since $f \equiv 5$ is a limit of functions in $U$. For $V$, consider the function $f \equiv 0$ and argue that in all of its neighborhoods there are functions with negative measure.
Jan
4
comment Doubt on differential
Glad to hear that, you're welcome.
Jan
4
comment Doubt on differential
I think your confusion is only notational. The expression $df(v)$ defines a function on $U$ by $[df(v)](p) =[df(p)]v(p)$. All of these objects (vector fields, form fields, functions) are local, so by applying one object to the other what we really mean is that we do it at every point $p \in U$.
Jan
4
comment Doubt on differential
What do you mean by "vector field doesn't stay in $U$"? Where does it go? I am also not clear about your confusion in the last paragraph. You know how to apply $v$ pointwise, so what's the problem with applying it globally? Same way $v(p)$ corresponds to derivation at $p$, $v$ globally corresponds to a derivation $X \in Der(U)$ in the algebra of derivations on $U$ that take smooth functions to smooth functions and obey Leibniz's rule.
Jan
4
answered How a group represents the passage of time?
Jan
4
answered Disprove that the group ring $\mathbb{Z}G(2)$ of $G(2)$ with coefficients $\mathbb{Z}$ is a principal ideal ring
Jan
4
answered The Hairy ball theorem and Möbius transformations
Jan
4
comment irrep of a non unit element in the finite group
I like this argument but not this particular hint. Regular representation has $\rho(g) \neq E$ by the very definition, so there is nothing to conclude. The real work is in reducing into irreducibles.
Jan
4
answered irrep of a non unit element in the finite group
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