3,955 reputation
820
bio website
location Prague, Czech Republic
age 28
visits member for 3 years, 5 months
seen 2 days 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).


Feb
5
comment Question about a functional equation
Finally, the open segments thesis is clear. If $B(t) = 0$ for some $t$ and, as we have by $(2)$, $\partial_t B = -1$ then $t$ is an isolated zero of $B$ and therefore the set of all $t$ where $B$ and $B^2$ is dependent consists only of isolated points. Therefore its complement is open and dense.
Feb
5
comment Question about a functional equation
Well, $(2)$ is clear, if $B(t, \cdot) = 0$, then almost all the terms die and you are left with $(\partial_t B + 1) = 0$. But $(3)$ is puzzling. It is true that from it it follows that $B(t,T) = B(t)$ is independent of $T$. I think what is meant here is that there exists $t$ such that $B(t) = 0$. Existence of such a $t$ probably follows from the stuff above which I don't really understand..
Feb
4
comment Union of two self-intersecting planes is not a surface
Regarding 4 connected components, it's correct. But to prove this, you'd need something on the order of Jordan Curve theorem (to show that number of components outside a simple curve in $\mathbb R^2$ is at most two).
Feb
4
comment Union of two self-intersecting planes is not a surface
Regarding "but V is any arbitrary open set.", it's not arbitrary. Since $V$ is homeomorphic to $S$, it must be connected and simply connected, i.e. it is a ball. Minus a point, it is an annulus. So $\pi_1(V \setminus \{f^{-1}(p)\}) = \mathbb Z$. But $\pi_1(X)$ is something else (try to show this).
Jan
31
answered Is the inverse of a linear transformation linear as well?
Jan
29
comment Embedded surface in $\mathbb{R}^3$
Because you can work locally and pick a neighbourhood of $\sigma(\bar x$) small enough, so that it looks like (flat) circle, and the tubular neighbourhood like a cylinder. In this cylinder, everything should be regular, but by definition of $\bar x$, there are singular points arbitrarily close (in the normal direction) to $\sigma(\bar x)$.
Jan
29
comment Solving to get free falling coordinate as function of arbitrary coordinate
@Aftnix: I think all that's going on here is plugging into the equation the general Taylor expansion (up to second order terms) and comparing coefficients. Do you need an explanation how to do this?
Jan
28
comment Gauss-Bonnet theorem for spheres that almost look like a torus
I think it's also worth emphasizing that the first part is actually quite negligible, since it's enough to cut out arbitrarily small area of negative curvature. It is the second part (i.e. bending the torus inwards) that accounts for the difference.
Jan
28
comment Embedded surface in $\mathbb{R}^3$
sorry, I don't have time right now, that's why I'm only posting comments. Suppose for contradiction that no such $\epsilon$ exists. Produce a sequence of points $(u,v)$ that map under $\tau_{\delta}$ to a singular point. Now, you can pick a convergent subsequence in $\bar V$ and this will under $\sigma|_{\bar V}$ map to a singular point of $\sigma(\bar V)$, a contradiction. Now the general case follows by restricting to an open subset $V$ of $\bar V$.
Jan
28
comment Embedded surface in $\mathbb{R}^3$
Yes, that's correct. See the picture here: en.wikipedia.org/wiki/Tubular_neighborhood . Now, try to think of conditions that guarantee that self-intersections don't occur. The (relative) compactness, (i.e. boundedness) is essential here.
Jan
28
comment Embedded surface in $\mathbb{R}^3$
Do you understand why $\epsilon$ can't be arbitrary in general (i.e. must be small for certain surfaces)? Understanding this point will take you half-way to the answer.
Jan
26
comment Calculating the limit $\lim \limits_{n \to \infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\frac{1}{n^2}) + 17}$
Do you know how to compute the limit for $n^{1/n}$?
Jan
26
comment Differential of the exponential map on the sphere
Correct. Glad I could help.
Jan
26
answered Differential of the exponential map on the sphere
Jan
25
answered Computation of Laplace-Beltrami operator in a conformally equivalent metric
Jan
25
comment Is Goedel term (in incomleteness theorem) both true and unproveable?
@Suzan: in logic you have two sides: syntactic and semantic. Syntactic is what you can prove formally, while semantic is what is really true (in some model). Ideal theory for a given model should be both sound (you can only prove syntactically what is true semantically) and complete (you can prove syntactically everything that is true semantically). So by adding $\neg G$ you obtain a theory that is not sound anymore (relative to the model for $T$ you were working with previously).
Jan
25
answered Is Goedel term (in incomleteness theorem) both true and unproveable?
Jan
25
comment Is Goedel term (in incomleteness theorem) both true and unproveable?
@Berci: but it is also important to note that extending by $\not \phi$ produces a theory that is not sound since we already know that $\phi$ is true (semantically).
Jan
24
comment $Sp(V)$ acts transitively on $V^*-\{0\}$ where $\Omega$ here is symplectic 2 form
@rschwieb: all of the stuff mentioned in the question is standard and natural. Still, since $\Omega$ is non-degenerate, we can translate all questions about forms to equivalent questions about vectors by $v \mapsto \Omega(v, \cdot)$ and its inverse, so it's a bit puzzling that the author insists on working with $V^*$.
Jan
24
answered normalize subgroup