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location Prague, Czech Republic
age 28
visits member for 3 years, 5 months
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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
19
answered Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$
Dec
18
answered How many degrees of freedom does a metric have on a psuedo-Riemannian manifold?
Dec
18
comment $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$
@Idear: did you read the answer? It's written explicitly in the second paragraph.
Dec
18
comment $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$
Ah, all right, I didn't understand that you are leaving the general case to the OP :)
Dec
18
comment $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$
Incidentally, there is a some information on $\pi_n(S^3)$ (which is isomorphic to $\pi_n(S^2)$) in this old question of mine: math.stackexchange.com/questions/50377/homotopy-groups-of-s2
Dec
18
comment $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$
Is there really an extension problem here? $SU(2)$ is simply connected, $U(1)$ has free abelian homotopy group and $SO(3)$ is a quotient of $SU(2)$. Where could the problem possibly arise?
Dec
18
answered $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$
Dec
18
comment Fixed point of a shrinking map. Proof.
If you check out the proof in the first link from the first comment, you'll see that the sought for fixed-point can be obtained as a limit of the sequence $f^n(x)$ for any $x \in X$. In other words, limit of $f^n(X)$ is a single point. So your $A$ is actually a one-point set. It seems very hard (if not impossible) to prove anything about $A$ (like your $A \subset f(A)$ without using that information.
Dec
18
comment Fixed point of a shrinking map. Proof.
In (5) you say 'by continuity'. But of what? Notice that on the LHS you have $f^n$ which depends on $n$, not a single function.
Dec
18
comment Using quaternions to represent an affine transformation?
I know I read about some parts of the above on wikipedia and in Hatcher's book on Vector Bundles and K-theory (there are many other fantastic insights in that book). But I got much of my intuition from Bott and Tu's Differential Forms in Algebraic Topology. That one is a gem.
Dec
18
comment Using quaternions to represent an affine transformation?
hm, not sure where I have the idea from. One source is definitely reading about K-theory. To classify rank $k$ vector bundles on $S^n$, you can consider maps from the equator (which is $S^{n-1}$) to $GL(k)$. This tells you how to glue the north and south portions of the bundle (each of which is a trivial bundle because disk is contractible). So you need to find the homotopy type of $GL(n)$ and it turns out it's $O(n)$ resp. $U(n)$ for reals resp. complexes (by Gramm-Schmidt). So in the end one is studying homotopy groups of classical groups.
Dec
18
comment Using quaternions to represent an affine transformation?
thank you for your kind words. Although I'm not as fluent as I'd like to be, I'm beginning to see the light here and there :) By the way, one piece of analysis was missing: we always stay in the positive-determinant component of $GL(2)$. That's why it's enough to work with $SO(2)$. Otherwise there would be a further ambiguity of adding a reflection.
Dec
18
comment Using quaternions to represent an affine transformation?
As $GL(2)$ has the homotopy type of $O(2)$, this is the only ambiguity we get. The remaining degrees of freedom (scaling and skewing) can always be divided in half uniquely.
Dec
18
comment Using quaternions to represent an affine transformation?
Nice answer. As for when the method works: basically what you're doing is picking a point on the group of transformations, connecting it to the origin using the geodesic and picking a point in the middle. So the method works when the geodesic is unique (this happens almost always). Your statement that it works for rigid rotations is not quite correct. If you pick rotation by $\pi$ then there are two solution $\pi/2$ and $3\pi/2$. Your equation $(c + I_2)$ also degenerates in this case since $c=-I_2$. So there is inherent ambiguity of left vs. right.
Dec
18
comment Using quaternions to represent an affine transformation?
Quaternions are not used to represent any 3d transformations. They are used to represent rotations. The group of rotations is $SO(3)$ and unit quaternions with quaternion multiplication form a double cover of that group. That's the sole reason they are useful. If you are doing anything other than 3d rotations (in particular arbitrary 2d transformations) chances are quaternions won't help you one bit.
Dec
18
comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots
@Goos: we can agree to disagree. It largely depends on context and audience when you should conflate two problems into one (for example when teaching limits for the first time, you should definitely not conflate the two). I for one prefer to be explicit and don't consider making myself clear a pain. Rather I find it painful if someone doesn't take the effort in presentation and assume everybody thinks the same way they do and knows the same facts as well. If you want to continue the discussion let's move it into chat, I think we spammed this thread enough :)
Dec
18
comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots
@Goos: if it's a standard usage then be it (I don't think it is) but I still need not like it. Same thing with that multiple of $2$. It's really two distinct questions, one for integrality and one for divisibility by $2$. Conflating them into one question doesn't really do anyone a service. You save about three words at the expense of possibly confusing anyone who doesn't follow exact same conventions you do.
Dec
18
comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots
@Goos: there is nothing really interesting in this question as I said. One can asks dozen similar questions. But the accidental connection to Legendre polynomial pointed out in Ragib's answer is neat, so that's the only interesting thing going on here.
Dec
18
comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots
@DanielV: thank you Captain Obvious. Since the question talks about negativity of roots it obviously assumes they are real (since there is no notion of negativity on complex roots, there being no order on complex numbers). If this is not so then the question's title is misleading and should state that OP wants to know whether the roots are real.
Dec
18
comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots
@Goos: fair enough. But I don't think the interesting part of the question is in proving the reality of the roots.