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Mar
20
reviewed Close Looking for Open Source Math Software with Poor Documentation
Mar
20
answered Showing when a permutation matrix is diagonizable over $\mathbb R$ and over $\mathbb C$
Mar
20
comment Fibered product of symplectic fiber bundles
Why would $P$ carry a symplectic form? There is an obvious dimensional restriction, as $P$ need not be even dimensional.
Mar
19
answered Morphism of vector bundles covering maps of the bases
Mar
19
revised Lie group. How is manifold defined?
added 738 characters in body
Mar
19
answered Lie group. How is manifold defined?
Mar
18
comment Does the elementary knot move really preserve the orientation?
You can get from the triangle on the left to that on the right by a rotation $\in SO(3)$, which representes the pinnacle of orientation preservation. So I don't think there is a problem. Maybe you mean the fact that those triangles don't induce the same orientation on the plane they are drawn on, but an orientation on a vector space ($\Bbb R^3$ here) doesn't induce one on its subspaces (for instance the two dimensional plane that carries both triangles), so that would also be a moot point. Do you have something else in mind?
Mar
17
comment Covering space BSO_n-> BO_n
Is it really just the covering space stuff that you are interested in? You should be able to work it out by hand if you are familiar with the constructions of both $BSO_n$ and $BO_n$. At every step of the construction, you get a two-fold covering. To be frank, I have never worked the details out myself, so there might be a little disagreable fiddling around with the topology, but conceptually things should be clear. Although, come to think of it, since everything is cellular, topology shouldn't be much of a problem.
Mar
17
comment Covering space BSO_n-> BO_n
What information do you want? You are probably aware that the covering map can be described in term of infinite grassmannians, $BSO_n$ is the infinite grassmannian of oriented $n$-planes in $\Bbb R^\infty$, and it projects onto the infinite grassmannian of $n$-planes, i.e. onto $BO_n$. This is a double cover.
Mar
17
comment Weak topology generated by the collection of functions from $X$ to itself that contains the identity function contains any topology
@Jose27 Right, I misread the question. Thanks!
Mar
17
comment Weak topology generated by the collection of functions from $X$ to itself that contains the identity function contains any topology
It doesn't make sense to include the identity function in the collection $F$ : $F$ must consist of set maps $f_i:X\to Y_i$ where each $Y_i$ is already endowed with a topology. Then the weak (or initial) topology on $Y$ is the coarsest topology rendering all maps $f_i$ continuous.
Mar
16
comment A kind of uniqueness for the double of a manifold
I'll just point out that one should impose connectedness, as there are uninteresting (disconnected) counter examples in $2$ dimensions. Take for instance $M=D^2\sqcup D^2\sqcup\Sigma^2$ and $N=S^2\sqcup (T^2\setminus D^2) \sqcup (T^2\setminus D^2)$. Then the boundaries are two circles for both $M$ and $N$, and the doubles are both diffeomorphic to the disjoint union of two $2$-spheres and two genus $2$ surfaces $\Sigma^2$.
Mar
16
comment Equilateral triangle in a circle
You should write down your geometric solution. I for one am perfectly happy with an algebraic argument, it is enough to convince me the proposition is true, but I'd like to read your geometric solution too.
Mar
16
revised Equilateral triangle in a circle
added 137 characters in body
Mar
16
answered Equilateral triangle in a circle
Mar
15
answered Is the set $A = \left\lbrace \frac{2^k}{3^l}: k,l\in\mathbb{Z}\right\rbrace$ dense?
Mar
14
comment Homomorphism of $\langle a,b \mid a^5b^{-3}, b^3(ab)^{-2} \rangle$ and $A_5$
Any relation to this question math.stackexchange.com/questions/1190045/… ?
Mar
14
revised Understanding an example from Hatcher - cellular homology
added 717 characters in body
Mar
14
answered Understanding an example from Hatcher - cellular homology
Mar
14
comment Convolution of functions defined on manifold
I doubt you can define convolution without some group structure on the manifold.