@MikeMiller My computer got busted, sorry. If $g \in G$ acts on $M$ by reversing orientation, then look at the orbit $G_x$. $x$ and $gx$ has opposite local orientation. But then look at the image of the orbit via the quotient. $M/G$ is orientable, so lift a loop from there. Lift an open cover of the loop to get an open cover for the path connecting $x$ and $gx$. Coherence of orientation in $M$ is now totally void.
@hippa The way you mesure it is you take the ground distance between the starting line and the nearest point of the javelin from this line, or the point where it's set in the ground.
@Balarka: OK, that's fine. I would have phrased it like this: there is an orientation on $M$ pulled back by the orientation on $M/G$. The induced maps of $\pi g$ and $\pi$ on $H_n(M,M-x)$ are the same because they're locally around the exact same map. So we see $g$ sends the pullback orientation at $H_n(M,M-x)$ to the pullback orientation at $H_n(M,M-gx)$.
Too much trouble on my phone, Mike. It was an intuitive discussion of diff forms, and he seems to miss a big point about integrability, thinking I'm being overly pedantic, when he's just being wrong.
@TedShifrin Ah. If we can set something fairly certain, I will plan a trip out there. I will be there tomorrow, but I don't want to make a trip out there on Friday if we are not actually going to have a chance to meet.
@Ted: I saw the thing you're referring to. I didn't read it carefully but what a crummy tone. He posted an incorrect comment on another question I saw. This is my only interaction with him
What a horrible situation: the game of Simona Halep is postponed because of the rain, and this night I need to wake up at 04:45. I really wanted to watch the game...
@Balarka: The fact that noncompact manifolds have $H_n(M) = 0$ is not really that hard, at least for smooth (PL probably too?) manifolds. It's classical, and combinatorial, that such a thing deformation retracts onto an $(n-1)$-dimensional subcomplex, sometimes called a core.
In the first of your multivarible calculus lectures on youtube you say that the tangent is the best linear approximation of a function at a point, what exactly does "best" mean here? I mean what's the criteria that lets you compare 2 different approximation and decide which is better? @Ted
I believe you should have introduced the little $o$-notation when you were talking about the relative error, i.e., that the error tends to $0$ faster than all the other errors, etc.
Best way to confuse students. -_-
Hmm, @MikeMiller, now that I think about it, your deformation retract claim makes sense.
It works for all the examples I can think of off the top of my head
@PVAL: There is only one PL/smooth embedding $B^k \to M$ up to isotopy when $M$ is connscted. Do you know if this is true for continuous embeddings and continuous isotopy with no local flatness assumptions?
@MikeMiller Nah that's not true. The Alexander Horned sphere bounds a 3 ball in S^3 and there's no way to isotope that to a standard 3-ball since the complement isn't even a manifold with boundary (or is not simply-connected or whatever you want to use). There are analogues to this in higher dimensions and codimensions as well, i.e. the cone on a non-trivial knot cannot be isotopic to a slice disk for that knot.
@MikeMiller I do not know the answer if you don't want to consider ambient isotopies. It should be sufficient for the ball to be locally flat at a point on the boundary (just by isotoping the other parts of the embedding into smaller and smaller balls), and I do not know of any topological embedding which isn't locally flat at some point.
@PVAL: There are wild knots that don't have a tams arc so we can take the cone on one of those. Actually this is precisely why I care: it would be nice to know that any wild knot is still trivial in $S^4$.
(In the sense of being continuously isotopic to the unknot.)
Greetings folks. It's been a long time since I was last here. It's also been a long time since I've thought about proofs, so I could use a little kick start if somebody could spare a moment to help...
I want to prove that, given the sets $A=\{n\in\mathbb{Z}\mid 30 \text{ divides } n\}$ and $B=\{n\in\mathbb{Z}\mid 10 \text{ divides } n \wedge 6 \text{ divides } n\}$, that $A=B$. I have one direction already (proving that $x\in A\Rightarrow x\in B$.. but I'm not sure how to start the other direction.
For the first part, I just assumed that $x\in A\Rightarrow x=30a,\ a\in \mathbb{Z}$ and went from there...
