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visits member for 2 years, 11 months
seen Nov 18 at 4:57

Oct
23
comment If the absolute value of a function is continuous, is the function continuous?
I assumed the domain was the integers because the way I see it, a number $x$ has to be an integer for the statement '$x$ is odd' to make sense. I guess that's a matter of convention, though. The wording 'if $x$ is an odd integer', plus an indication of what the domain is, would make the answer clearer.
Oct
5
comment If the absolute value of a function is continuous, is the function continuous?
This function is continuous. (I assume its domain is the integers?)
Jul
2
awarded  Curious
Jun
16
comment Natural connection on tautological bundle over real Grassmannian
Thanks, that makes a lot of sense. Let me summarize how parallel transport works in my own words. Let $W(t)$ be a path in $G(k,V)$ and $w\in W(0)$ (so that the pair $(w,W(0))$ is in the tautological bundle). There is a unique path $w(t)$ in $V$ such that $w(t)\in W(t)$ and $w'(t) \perp W(t)$ for all $t$. Parallel transport of $(w,W(0))$ along $W(t)$ is $(w(t),W(t))$.
Jun
16
accepted Natural connection on tautological bundle over real Grassmannian
Jun
12
comment Prove that $|x - y| \geq |x| - |y|$
As a general strategy for finding where your mistake is, plug in some easy numbers.
Jun
12
answered Prove that $|x - y| \geq |x| - |y|$
Jun
12
asked Natural connection on tautological bundle over real Grassmannian
Apr
10
comment explicit cocycle representing Stiefel-Whitney class in Milnor and Stasheff
In problem 6-C you only have an embedding of $G_n(\mathbb{R}^m)$ into $G_{n+1}(\mathbb{R}^{m+1})$, and which sends $r$-cells to $r$-cells. However, I think that not all $r$-cells have to be in the image of this embedding (as can be seen by counting $r$-cells). In particular, the induced map in cohomology $H^r(G_{n+1}(\mathbb{R}^{m+1})) \to H^r(G_{n}(\mathbb{R}^{m}))$ is not injective. Therefore, you cannot deduce the Stiefel-Whitney class in $H^r(G_{n+1}(\mathbb{R}^{m+1}))$ from the one in $H^r(G_{n}(\mathbb{R}^{m}))$. Or at least, I don't see how.
Mar
13
comment Under what conditions the quotient space of a manifold is a manifold?
This is in fact equivalent to $M/E$ having a smooth structure compatible with the quotient topology such that $M\to M/E$ is a submersion, according to math.illinois.edu/~ruiloja/Math519/notes.pdf, theorem 8.3 on page 63. (if by submanifold you mean embedded submanifold)
Mar
2
reviewed No Action Needed Under what conditions are trigonometric integrals over a period zero?
Dec
14
awarded  Yearling
Oct
28
awarded  Self-Learner
May
22
accepted Every principal $G$-bundle over a surface is trivial if $G$ is compact and simply connected: reference?
May
16
comment Every principal $G$-bundle over a surface is trivial if $G$ is compact and simply connected: reference?
Thanks for the answer. Any reference I can take a look at?
May
15
answered Deduce dice configuration knowing 2 adjacent faces
May
14
comment The number of words that can be made by permuting the letters of _MATHEMATICS_ is
Example: the number of words that can be made by permuting the letters of BEER is 12. The possibilities are EEBR, EERB, EBER, EREB, EBRE, ERBE, BERE, REBE, BREE, RBEE, BEER, REEB.
May
14
awarded  Caucus
Apr
25
comment Every principal $G$-bundle over a surface is trivial if $G$ is compact and simply connected: reference?
@EricO.Korman - Thank you, that helps (though I have to think about it a little bit).
Apr
23
comment Every principal $G$-bundle over a surface is trivial if $G$ is compact and simply connected: reference?
I am familiar with the fundamental group and with singular and de Rham (co)homology. I can define the higher homotopy groups, but I cannot state any relevant "big theorems" by heart. I think I know what classifying spaces are about, meaning that in particular I more or less know the correspondence between homotopy classes of maps $\Sigma\to BG$ and $G$-bundles over $\Sigma$. I have never really worked with higher homotopy groups. (And I'm aware it may be hard to give a good answer to someone lacking this knowledge.)