@TedShifrin Apparently there's a neat definition of orientation that also works fine in dimension 0; Mike told me like a week ago but I'm only starting to get it now
@TedShifrin so given a vector space V, GLn(R) acts on the set of ordered bases of V, and also on the set {-1, 1}; you take the amalgamated product and call it the set of orientations on V
I learned a cool proof that subgroups of free groups are free earlier (actually months ago, but I forgot about it until I studied it again for an upcoming exam today)
No, it's an approach you probably won't enjoy as much as the covering spaces business :P
By Bass-Serre theory a group acting freely on a tree is free, and a subgroup of a free group acts freely on the Cayley graph of the latter (which is a tree since it's the Cayley graph of a free group)
@TedShifrin oh, I thought it was you who insisted that the determinant of the empty matrix is 1; but I can't find the relevant message in the chat history
anyway the 0x0 matrices form the trivial ring where everything is a unit
It's the same issue, I suppose, with trying to define $\Lambda^0(V)$ when $V$ has dimension $0$. No matter how you say it, it's a matter of definition/convention.
@TedShifrin I don't think there should be any issue defining $\Lambda^0(V)$... it should just be $\Lambda^\ast(V)$ and that satisfies some universal property; even $\Lambda^0(V)$ has a universal property; anyway they are both the zero algebra?
@Ted I wrote like 4 proofs yesterday. For proving whether some numbers were irrational. I thought the cube roots would be hard, but it was pretty similar to the square ones.
@Lucas: If it's the thing you pinged me for yesterday, I did. I'm sorry that one of the most important vector fields has an unfortunate picture, but get over it.
@TedShifrin oh I thought about something today and I don't know if this is nonsense: a manifold being orientatable is equivalent to having a non-vanishing top-form because the top-form sort of assigns an outward-pointing normal (if you imbed your manifold locally to the euclidean space of one dimension higher)
Oh @Ted, remember one more thing: you said that a good way to parametrize a 2D curve is to set $y = tx$, because $t$ is the slope; your teacher said that and you never forgot. But I suppose this is always true, right? I mean, "if a subset $S \subset R^2$ is parametrizable by some function $\vec{f}(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}$ then $y(t) = tx(t)$" or something like that using derivatives.
@TedShifrin sure, I know the proof, but I don't think it gives me that much intuition as to why the theorem is true. I'm not trying to talk about some philosophical nonsense like what is "is", I just want some sort of intuition
@Eric: The purchasing power is very inadequate in places like CA, especially where I live, however. I have a good friend who works as a cab driver and can't even afford rent.
Certain cities and states have raised it, @CaptainAmerica, but our generous rich legislators of the you-know-which-party only care about the rich and entitled.
@AkivaWeinberger I like how these diagrams somehow triggering very strange scenes in last night dream, leading me to a possibly better understanding of what is motion
Exposing to diagrams of hyperbolic spaces and games does really help on intuition on curved spacetime, because locally your view changes depending on where you are on the manifold, similar to how in these hyperbolic animation, the centre piece always look flat
