4,137 reputation
1023
bio website shapetales.wordpress.com
location Prague, Czech Republic
age 29
visits member for 4 years
seen Nov 24 at 20:34

Math PhD. student and functional programming enthusiast.


Jun
6
comment Why does differentiating a polynomial reduce its degree by $1$?
This is exactly Mitchell's answer above. Why post the same thing again?
Jun
3
comment Decomposition of cohomology group on $S^{n}$
Can you at least please fix the typos? The question is almost unreadable as it stands.
Jun
2
comment Poincare dual of unit circle
@PeterM: I see, I didn't get that part. I think you misunderstood the definition of Poincare dual then, let me edit it into my answer.
Feb
6
comment What are necessary and sufficient conditions for the product of spheres to be paralellizable?
But otherwise this is a very nice and probably classical question that hopefully some local expert will answer soon. To add my gut feeling -- you can't expect parallelizability in general, tangent bundles of spheres are quite complicated and there's no a priori reason for their direct sums to be trivial besides the trivial reason of adding the normal bundle.
Feb
6
comment What are necessary and sufficient conditions for the product of spheres to be paralellizable?
I don't think clutching functions are helpful here, at least when used naively. That is because the contractible neighborhoods for $S^i \times S^k$ will be products of hemispheres, so there's four of them in total with mutual intersections being homotopic to either $S^{i-1}$, $S^{k-1}$ or $S^{i-1} \times S^{k-1}$. Correspondingly, there will be multiple clutching functions depending on which overlap we're talking about -- the one you mention is for passing from southern to northern hemispheres in both spheres simultaneously.
Jan
16
comment existence of double covering
$GL(n, {\bf C})$ has the homotopy type of $U(n)$ which has the same fundamental group as $U(1)$ (this isomorphism is induced by $\det: U(n) \to U(1)$ and backwards by the inclusion $U(1) \to U(n)$ as scalars).
Jan
15
comment Dolbeault cohomology and analytic regularity
Indeed, nobody can stop you from doing that. But in that case you should change the question's title and body since it has nothing to do with complexes in general and Dolbeault cohomology in particular.
Jan
15
comment Dolbeault cohomology and analytic regularity
I don't really understand the question, even with the edit. When you have a complex $M^{\bullet}$, all the spaces are fixed ($C^1$, say) beforehand. You can't just decide that some of them will be $C^2$ when it suits you because you need to quotient. That doesn't make any sense.
Jan
14
comment How do I maximize $|t-e^z|$, for $z\in D$, the unit disk?
There are obviously other solutions even for $y_0 = 0$ because also $\phi = \pi$ works and moreover normal to the curve at every point intersects real axis somewhere. Of course, one needs to check which of the these three candidate points is the true maximizer of the distance.
Jan
13
comment Easy question about $C_c^\infty(0,T)$ and $C_c^\infty((0,T);X)$
@janmarqz: it's standard definition for the space of smooth functions defined on the interval $(0, T)$.
Jan
13
comment Easy question about $C_c^\infty(0,T)$ and $C_c^\infty((0,T);X)$
@janmarqz: $g$ doesn't it anything. It's a generalization of $f$ that takes values in general Banach space $X$ rather than in $\bf R$. Otherwise it's the same.
Jan
13
comment Properties of $\pi_n$ from a category theoretical point of view
Aren't the structures one is looking for essentially $n$-groupoids? E.g. crossed module is a $2$-group and generally I'd expect $n$-type to be captured by $n$-groupoid that should be an $n$-truncation of $\infty$-groupoid, which should reflect the full homotopy type. But your last paragraph suggests that the situation is not very well understood. Does the problem lie in finding nice algebraic definition for $n$-groupoids?
Jan
10
comment — Cartan matrix for an exotic type of Lie algebra --
@Idear: are you sure that this is the same matrix? At first sight it seems to be quite a different object just accidentally having a same name (I'm sure Cartan has looked at more than one matrix during his life...). Also, I think you are doing neither modular representation theory nor working with associative algebras, right?
Jan
10
comment Fundamental group of CW-Complex only depends on 2-Skeleton
math.cornell.edu/~hatcher/AT/AT.pdf Corollary 4.12 in the Hatcher's book should do the trick.
Jan
10
comment Fundamental group of CW-Complex only depends on 2-Skeleton
The intuition seems correct to me. Of course, it's sweeping few pages of technical details under the rug, as you mention yourself :)
Jan
10
comment Motivation for introducing algebraic topology?
@KCd: good points but your comment is quite imprecise. All ${\bf R}^n$ have homotopy type of a point, so that their homology and actually all algebraic topological properties coincide. What's different is e.g. the homology of ${\bf R}^n \setminus {\rm pt}$ which has the homotopy type of a sphere. Another way would be using compactly supported cohomology but these invariants are of course not preserved by homotopy.
Jan
10
comment Motivation for introducing algebraic topology?
Even more than that, maps between Riemann surfaces (and even branched covers) were understood almost a century before algebraic topology appeared. I think the invariance of domain might be a better example (although I'm not sure it doesn't have a simple proof without using AT).
Jan
10
comment Motivation for introducing algebraic topology?
Very good answer, I agree with most of what you have said. But your example $S^2 \to T^2$ is actually pretty simple. Just write $S^2$ as union of hemispheres. Image of the equator must be contractible, so that image of one of the hemispheres is then a disk, while the image of the other one contains two great circles (generators of homology of $T^2$) and wouldn't be contractible.
Jan
9
comment Orthogonal basis for infinite-dimensional vector spaces
@Ketty: the standard definition of basis for inner product spaces is not the same as the one for linear algebra, (see e.g. here en.wikipedia.org/wiki/Inner_product_space#Orthonormal_sequences). I'm voting to close because of your uncooperativeness and refusal to make yourself and your question clear even after much discussion with your peers.
Jan
9
comment Calculating the lie algebra of $SO(2,1)$
The equation here is then $A = - \eta A^t \eta$. It's quite simple to solve but as a baby step you can try $SO(1,1)$ with $\eta = {\rm diag}(1, -1)$ first.