What's that mean? You can take the determinant of n vectors. There's only a slight problem getting a number as output of your manifold is not oriented.
@MikeMiller I'm confused because do Carmo explains orientation in terms of chart transition functions, not in terms of orientation of the tangent spaces.
So without going through all of the proofs in Lee that connect the two, I don't see how one is supposed to know that "parallel transport preserves orientation" even makes sense.
@MikeMiller Ah, because another basis $\{\partial/\partial y_i\}$ will be related by the transition function (chain rule) and this is orientation preserving in the vector space sense?
Pick an orientation on one tangent space and parallel transport it to every other tangent space. If that's well-defined, it gives you a consistent orientation on every tangent space. It seems you've proved that parallel transport preserves orientation.
Take determinant with respect to one basis. Change bases (by a matrix A, say). Then the new determinant is the original times det A. Or maybe the inverse of that, I forget.
In particular if you can't say "Take determinant with respect to an orthonnormal basis" unless you can throw the adjective oriented in there - your determinant is not well-defined.
@RudytheReindeer Basically, I'm trying to show that normality implies: given closed $A$, open $U\supset A$, then there is a closed $B$ such that $A\subset B\subset U$
And that question came up in my construction of $B$
I think I found a construction that does not depend on that, but my first try seemingly did
$$ T = \{ \varnothing, \{a\}, \{b\}, \{a,b\}, \{b,c\}, \{a,b,c\} =X\}$$ with $U = \{a\}$ and $V=\{b,c\}$. Then the closure of $U$ is $\{b,c\}$ so they are not disjoint.
oops, I didn't check if this space is connected.... let's see.
@0celo7 You implicitly require $U \cup V$ to be the whole space? That's impossible in a connected space. But if you don't require that you could to something like $U=[-1,0)$ and $V=({1\over 2}, 1]$ in $[-1,1]$ with the standard topology.
@Ted: The determinant defines a homomorphism $\Lambda^n TM \to \Bbb R$, if it's well-defined. A nowhere zero section would be the inverse image of 1, aka "the unique element with determinant 1."
Consider the silly case of the the standard basis in $\Bbb R^n$. Change basis to the same with first vector $e_1/2$. I think the determinant of the old atandard basis in this new basis is 2.
In fact if $A$ is the new basis and $V$ is the old vectors, I claim the new basis is $A^{-1}V$ or $VA^{-1}$ or something like that.
But in @0celo's case, we have an orientable manifold, and he's got a parallel frame along the curve. So we just can use that frame field to normalize. I guess I didn't make that clear.
Including the 4-skew line question, probably something about Morse theory, combinatorics and computing how many homogeneous polynomials of degree $k$ in $n$ variables there are, etc.
Yes, yes. I do think there's lots to say about connections. At least that's the easiest route between representation theory and differential geometry I know. :)
@0celo7 I remember taking a music class this past quarter. everyone in the class kept saying "oh mozart is my fav, oh beethoven is my fav." and finally, one of my homies in the back was just like "yeah, i listen to 2 Chainz"
@TheGreatDuck Yeah! I felt amazed when I started studying abstract algebra. I felt like those kinds of basic operations should be explained more deeply early on
@StanShunpike Throwing out 5 of my favorite songs: Till I Collapse (Em), Renegade (Jay Z, Em), John (Wayne, Ross), Birthday Song (2 Chainz, Kanye), Black Skinhead (Kanye)
@0celo7 I was telling @StanShunpike that my approach towards teaching would be to teach as many of the atomic operations as possible. And then I rattled off a bunch of different operations. The idea is that while some of the, might define each other, they can be used to define all other functions... Or at least, all functions that everyday ordinary people learn.
@StanShunpike thanks. I admit they might not be "atomic" per se, but it is probably the best mentality for a simple analysis. If one needs to add another one it could be modulo which is the remainder when a is divided by b.
@TheGreatDuck Building in an understanding of basic operations is crucial to learning how to think mathematically
@TheGreatDuck Sometimes when i've been tutoring, I get the impression the people I am teaching simply lack a basic sense of these operations and as a result their equation solving becomes mechanical and they don't really realize these operations are....exactly that operations. it is just sort of a mindless solving process they go through
@StanShunpike exactly! Plus, if one can express all other operations in terms of lesser "atom-like" operations, then the other operations become meaningless. To use a computer science term... They are "sugar syntax".
@StanShunpike i hate statistics and it relies on integration heavily, but cs definitely needs to be taught more. It's concepts are useful beyond programming.
I never even considered that statements in math returned true or false until programming
And even then it wasnt until i actually posted a q&a question implementing an equality conditional
someone mentioned the iverson bracket "sugar syntax".
needless to say, it was quite... Intriguing.
The tv show i am watching right now is so weird
It is an episode of star trek where they land on a modern day rome...
O.o
i feel like its the beginning of a bad bar joke
"three gladiators and captain kirk walk into a bar..."
Essentially, the one using extensions picks two ($n$-fold and $m$-fold) extensions, fixes it so that the modules are projective over the ground ring (when working over a Hopf algebra which is projective over the ground ring), and applies the Kunneth theorem so that the tensor product (in a suitable order) becomes an $n + m$-fold extension.
The one appealing directly to projective resolutions seems easier.
@AndrewThompson Yeah, I don't think I have ever seen the one using such extensions. I am not even sure I have ever seen a proper proof that $Ext^n$ can be described using those, and it never really seems relevant
