3,960 reputation
820
bio website
location Prague, Czech Republic
age 28
visits member for 3 years, 5 months
seen 50 mins 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
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 $n > 1$.
Dec
19
comment Inverse image of a PID is a PID
What lead you to conjecture this? There's hardly any map to PIDs that would have the property you mention. In fact, it would be fun to post another question: find some conditions on a ring so that your statement holds :)
Dec
19
comment How can I get the two identities
math.stackexchange.com/questions/81203/… here is proved your first identity.
Dec
19
comment How can I get the two identities
It might help you if you used consistent notation. What you write as $D$ and as $\nabla_0$ is the same thing, the covariant derivative. Then it's quite obvious that $\Delta_0$ is actually a trace of the Hessian.
Dec
19
comment Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$
@Tim: okay, I should officially go to sleep :D For some reason I thought the problem asked for equality now. Duh...
Dec
19
comment Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$
@Tim: $3^n + 5^n$ is always even whereas $n^2 - 1$ would be odd.
Dec
19
comment Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$
@Tim: even $n$ can't be a solution.
Dec
19
comment Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$
@Tim, Ian: thank you both, I forgot that $n$ must be prime, although it's obvious in retrospect -- for general $n$ one needs to use $\phi(n)$ to satisfy the equation at every prime dividing $n$. Perhaps there's some way to leverage that here although I do not see it yet.
Dec
19
revised Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$
added explanation