3,955 reputation
820
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location Prague, Czech Republic
age 28
visits member for 3 years, 5 months
seen 7 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).


Dec
18
comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots
Minor remark: the fact that the roots must be negative is obvious since the coefficients are all positive. A second minor remark: almost all polynomials have distinct roots and so does probably yours as well. It might be very hard to prove it but unless something special is going on (like the binomial formula for $(1+x)^n$) it's really not surprising, or even interesting. You could ask similar question for a polynomial having coefficients given by any function of binomial numbers whatsoever (except the trivial one I just mentioned).
Dec
18
revised Aplanar covering of $S^1 \vee S^1$?
Found all aplanar finite degree coverings
Dec
18
answered Aplanar covering of $S^1 \vee S^1$?
Dec
18
answered Are these functions isomorphisms of the first binary structure with the second? - Fraleigh p.34 3.13, 3.15
Dec
18
comment Are these functions isomorphisms of the first binary structure with the second? - Fraleigh p.34 3.13, 3.15
You cannot suspect it, you just have to figure it out. Either find a proof or counterexample. A map can fail to be an isomorphism because of only three reason: it's not injective or not surjective or not homomorphism. So check each property in turn.
Dec
17
comment Aplanar covering of $S^1 \vee S^1$?
Universal covering is a tree, so certainly planar. But I'm not sure about finite degree coverings. They seem intricate.
Dec
17
comment Irreducible and regular representation
@James: there are many ways to decompose representations into the irreducibles. It's hard to help you without knowing what you know and what you're supposed to use. E.g., are you required to use pure linear algebra to block-diagonalize the matrices? Or can you use standard character theory or other tools from rep. theory?
Dec
14
comment Cohomology groups
@Daniel, thanks for pointing this out. I was implicitly assuming finite generation.
Dec
13
answered How to identify surfaces of revolution
Dec
13
comment How to identify surfaces of revolution
I agree with you that physicists like to exploit any symmetry to simplify the problem (in this case to work in cylindrical coordinates given by the axis of rotation). But usually the process can't be reversed. You can't expect any symmetry in general surfaces, even in those having pretty inertia tensors.
Dec
13
comment How to identify surfaces of revolution
@bubba: okay, doesn't matter whose idea it is :) Consider for example the deformed annuloid (one that can be actually embedded in a plane) such that it's inner boundary is an ellipse-like curve and its outer boundary is another ellipse-like curve such that the surface is thinner where its farther away from origin. It's really not hard to arrange the matters such that the inertia tensor coincides with that of some circle. It's not easy to find these surfaces explicitly but simply from dimensional counting it's obvious that inertia tensor can't help you here.
Dec
13
comment How to identify surfaces of revolution
@Darcy: I'm not sure how your comment helps. It's true that when the surface is a surface of revolution then it will have an eigenvalue of multiplicity at least two. On the other hand, this will be true of many surfaces that are not made by revolution as well. You can't hope to capture this information by a mere $3 \times 3$ matrix. The map $S \mapsto I_S$ from surfaces to their inertia tensors is hugely degenerate.
Dec
13
answered Cohomology groups
Dec
12
answered Homological problem help
Dec
12
comment Homological problem help
Indeed. So, for $g = 1$ these groups coincide. And for the OP's convenience, here's a link to your statement on abelianization known as Hurewicz theorem (which I'm sure you know): en.wikipedia.org/wiki/Hurewicz_theorem
Dec
12
comment Riemann tensor symmetries
Riemann tensor can be equivalently viewed as curvature 2-form $\Omega$ with values in a Lie algebra $\mathfrak g$ of group $G = SO(n)$. The antisymmetry in one pair comes from being a 2-form, the antisymmetry in the other pair comes from the antisymmetry of ${\mathfrak so}(n)$. This holds even when the connection has torsion. But it won't neccesarily hold for non-metric connections (such as symplectic connections). en.wikipedia.org/wiki/Curvature_form
Dec
12
awarded  Popular Question
Dec
12
comment How prove this $f=C$ if $4f(x,y)=f(x-1,y)+f(x,y-1)+f(x+1,y)+f(x,y+1)$
The first link of @lhf is not only related, it completely solves the question. In short: since $f(x,y)$ is average of its neighbors, there can be no non-trivial extrema. Because $f$ is bounded, there is a sequence of $(x,y)$ approaching its suppremum and using harmonic property again one eventually shows that $f$ must be equal to that suppremum everywhere by first observing that this holds along that sequence.
Dec
12
answered Top homology of an oriented, compact, connected smooth manifold with boundary
Dec
12
comment Map of constant rank 1
My two cents: the Jacobian of the function defines a line bundle over ${\bf R}^2$. This in turn defines a unit vector field. The claim is then that $f$ maps all of its integral curves diffeomorphically to the $y$-axis. While very plausible, I can't quite prove it. Perhaps someone else can finish this argument.