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8h
comment When is the metric completion of a Riemannian manifold a manifold with boundary?
@Vertex: If you have the time, can you post the answer? I'd be interested in knowing what it is.
20h
comment When is the metric completion of a Riemannian manifold a manifold with boundary?
@PhillipAndreae: You're right (on both counts)! So, I also have issues with my examples: as you say, the path metric on the graph of $\sin(1/x)$ doesn't coincide with the restriction of the standard metric on $\mathbb{R}^2$. (Likewise on $X$ with center point deleted - the path metric gives the remaining components infinite distance.) Sorry about that. Nonetheless, I still suspect that the completion of a Riemannian manifold as a metric space can fail to be (homeomorphic to) a manifold (with or without boundary).
20h
comment When is the metric completion of a Riemannian manifold a manifold with boundary?
@user1952009: Based on Vertex's wording, I think we are to assume the metric completion of a Riemannian manifold is simply the usual metric space completion. That is, form the Riemannian manifold $(M,g)$, form the metric space $(M,d)$ with $d$ defined as the infimum among paths connecting points. Then, take the usual metric space completion. (One typical approach is using Cauchy sequences.) Said another way, $\hat{M}$ is the completion of $M$ if there is a map $\phi:M\rightarrow \hat{M}$ with dense image with the property that $d_M(x,y) = d_{\hat{M}}(\phi(x), \phi(y)).$
22h
comment When is the metric completion of a Riemannian manifold a manifold with boundary?
Even worse, if you delete the center point from the letter "X", (as well as the 4 end points), you have a nice smooth manifold whose metric completion is not even a manifold with boundary. If I had to, I'd guess the question is very hard to answer in general. (Also, the graph of $\sin(1/x)$ for $x > 0$ is a perfectly good Riemannian manifold whose metric completion is the graph together with $\{0\}\times [-1,1]$, so fails to be a manifold terribly.)
1d
comment Are there proper Ad invariant sets on simple lie algebras?
Stupid observation: $\mathcal{C} = \{0\}$, it satisfies $1.$ and $2.$, so you really do want something like $3.$ Also, $\mathfrak{g}\setminus\{0\}$ satisfies $1.$ and $2.$
1d
comment vector bundles of $\mathbb{P}^2$
Smooth vector bundles over $\mathbb{P}^2$ have been completely classified (I assume here you mean $\mathbb{C}P^2$). In particular, they are determined by their characteristic classes (The second and fourth Stiefel-Whitney class, as well as the first Pontryagin class and Euler class). On the other hand, I have no idea if holomorphic vector bundles are classified or not.
1d
comment For which Lie groups $G$ can one write $g$ as the exponential $\exp X$ of some $X \in {\frak g}$ for every element $g \in G$?
It's true for all compact groups (like $SU(2)$) and also $\mathbb{R}^n$ and products $G\times \mathbb{R}^n$ where $G$ is compact, but I don't know if it's true or not for anything else. Maybe someone else can chime in? (Later, I can supply a handwavy proof for compact groups.)
Feb
2
comment What does the notation “*” mean?
Could it be the pullback of a fiber bundle/covering space?
Feb
2
comment To prove that a generator-candidate is sufficient to find all elements in $SO(3)$
Another issue with your approach: If a subset of a space has the same volume as the space, it need not be the full space - consider removing a point. Unfortunately, I don't know how to prove what you're asking for...
Feb
2
comment Homotopy equivalence between $X=\{0\}\cup \{\frac{1}{n},n\in \mathbb{N}\}$ and a discrete space
Hint: 1. Show that if $Y$ is discrete and $f:Y\rightarrow Y$ is any function, then any homotopy of $F$ is constant in time. 2. As you noticed, if $f:X\rightarrow Y$ is continuous, then $f(X)$ is a finite subset of $Y$. Now, if $g:Y\rightarrow X$ then $f\circ g$ has finite image. Now apply 1... ( Feel free to write up your own answer. And note that this proof works for any infinite $Y$, but does not work if $Y$ is finite.)
Jan
31
awarded  Enlightened
Jan
31
awarded  Nice Answer
Jan
30
comment There is a theorem analogous to the Brouwer fixed-point for the 2-dimensional sphere?
There are also maps with precisely one fixed point - take a vector field on $\mathbb{R}^2$ which always points left, but whose length decays to $0$ as $|(x,y)|\rightarrow\infty$, then use stereographic projection to make a corresponding vector field on $S^2$. For short enough time, the flow only has one fixed point (the point at infinity).
Jan
30
awarded  Enlightened
Jan
30
awarded  Nice Answer
Jan
29
comment Quadratic form equals zero
I don't think that $\gamma'(t)^T A \gamma'(t) = 0$ should hold in general (though I don't have time to cook up a counter example now). The fact that $\gamma(t)^T A \gamma(t) = 0$ for all $t$ implies, via the product rule, that $\gamma'(t)^T A \gamma(t) + \gamma(t)^T A \gamma'(t) = 0$, though.
Jan
29
comment Proof of “removing a point” strategy
It's probably worth pointing out the that the $\implies$ direction did not use Hausdorffness, and is the direction typically needed in terms of applying this statement to show two spaces are not homeomorphic.
Jan
28
comment Properties of minimal $\mathbb{Z}$-actions on infinite compact spaces
What's a minimal action again?
Jan
26
comment Punctured complex projective space
I'm not sure I understand that circle action. Note that having no fixed points is weaker than having a free action.
Jan
25
comment Is there a rational surjection $\Bbb N\to\Bbb Q$?
Here's an alternative argument (which still needs some fleshing out): Write your rational function as $p(x) + \frac{q(x)}{r(x)}$ where $p,q,r$ are polynomials with $\deg q < \deg r$. Then your function is asymptotic to $p$. But $p$ is bounded above or below on $\mathbb{N}$, and there are only finitely many elements of $\mathbb{N}$ for which your rational function is more than $1$ away from $p(x)$, so your rational function must also be bounded above or below.