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 Yearling
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Jan
31
comment Under what conditions the quotient space of a manifold is a manifold?
@AlexM. - I'm not sure I understand your question correctly, but isn't it enough to observe that every closed embedded submanifold is proper? Regardless, I was saying that if $M/E$ inherits a smooth structure from $M$, then $E$ is a closed submanifold of $M\times M$ and the first projection $E\to M$ is a submersion.
Dec
14
awarded  Yearling
Nov
27
awarded  Popular Question
Nov
15
awarded  Nice Question
Sep
15
comment Why is $[0, 1] \cap \mathbb{Q}$ not compact in $\mathbb{Q}$?
I would like to point out that your question title talks about one space being compact in another. There's no such thing. A space is compact or is not compact, and any "surrounding space" is irrelevant. (This is different from closedness for example, where you always ask about one space being closed in another.)
Sep
15
comment Coordinate map from cotangent space is smooth
I guess you want to show that certain (maybe only locally defined) maps $T^*M\to\mathbb{R}$ are smooth (namely the coordinate projections). In particular, you need a smooth structure on $T^*M$. Which definition of the smooth structure on $T^*M$ are you using?
Sep
15
comment Coordinate map from cotangent space is smooth
Could you explain your sentence "the coordinate map $\pi_i$ of $T_xM$ is nothing but the composition of the chart and coordinate map in $\mathbb{R}^n$"?
Sep
15
comment The identity element of the Tensor Product of Vector Bundles
Let us continue this discussion in chat.
Sep
15
comment The identity element of the Tensor Product of Vector Bundles
To show it's a diffeomorphism, you check smoothness of the map and its inverse (which you both know explicitly -- you should probably check that they are indeed inverses). I'm not sure what you mean by class representative.
Sep
15
comment The identity element of the Tensor Product of Vector Bundles
(Indeed, a canonical choice of basis in the fibers is nothing but a trivialization of the bundle.)
Sep
15
comment The identity element of the Tensor Product of Vector Bundles
Yes, you're right, you can always find an isomorphism between a fiber and $\mathbb{R}^n$, but there is no standard choice of such an isomorphism (i.e. the fiber does not have a canonical choice of basis). It is the naturality of the isomorphism $\mathbb{R}\otimes F\cong F$ that makes things work. I'm saying things a little vaguely now, and one can make them precise, but it won't fit nicely in a comment.
Sep
15
comment Is this backslash a typo here in showing a function?
Yes, I also use the backslash, but I often see the minus-sign. The TeX command for the backslash is actually explicitly called \setminus, so it seems like a good choice.
Sep
15
comment Is this backslash a typo here in showing a function?
en.wikipedia.org/wiki/… It is sometimes written as $\mathbb{R}^n - \{0\}$.
Sep
15
answered The identity element of the Tensor Product of Vector Bundles
Sep
15
comment Calculating torus surface area
Is it obvious that this should work? (In other words, is Pappus' centroid theorem obvious?)
Dec
14
awarded  Yearling
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 $(w(t),W(t))$.