Right, so for one the volume form eats an orthonormal basis of $T_pM$ and takes the value $1$, because that's the volume of the box. I think in general the value of the form on a set of basis vectors of $T_p M$ is the determinant of the Gram-Schmidt matrix with which you orthonormalize that basis. I'll have to write something out to prove this.
For one, I need orientability to have a top dimensional form. But after that I have two choices of orientation: a positive orientation gives that the top dimensional form takes value $1$ on the orthonormal basis, and $-1$ if it's the other orientation.
So I need to choose an orientation to have a well defined unique volume form. Might as well choose a positive orientation, because $-1$ requires two characters to write, while $1$ requires one.
@BalarkaSen Well, it's "oriented Riemannian manifold", not positively oriented. But yes, I want a positively oriented basis.
@BalarkaSen OK, so that's all I wanted from you for now. Indeed, all you needed to do to define the volume form $\omega$ is to define $\omega_p(e_1, \dots, e_n) = 1$, where $e_i$ is an oriented orthonormal basis for $T_p M$.
Because $\Lambda^n T_pM$ is 1-dimensional, this specifies it on every vector. (To use your language: If you write out the matrix $\{v_1, \dots, v_n\}$ with respect to the above basis, this spits out the determinant of that matrix.)
One then checks that this is a smoothly varying differential form (this just comes from picking a set $E_n$ of locally defined, smoothly varying vector fields, that form an oriented orthonormal basis; that this is possible is more or less Gram-Schmidt)
And that it didn't depend on the choice of oriented orthonormal basis (can you tell me why that's true right now?)
And that this is the same thing as your differential form (This is either clear or an exercise)
Tell me when you're done with this.
@Balarka Maybe you could look at this later so I could say the other things I wanted to say? I am a little short on time this morning.
Thanks... The other thing was to go back earlier when you were defining the 2-form on a surface in $\Bbb R^3$. You wrote down its coordinate expression in an evocative way: $n_3 dx_1 \wedge dx_2 + \dots$.
The "differential forms only" way of doing this is to say that one considers the vector field $N$ along the surface, its unit oriented normal vector. If $\omega$ is the volume form on $\Bbb R^3$, then the volume form on the surface is $\iota_N \omega$, defined by $\iota_N \omega(v,w) = \omega(N,v,w)$. This is more or less the same thing as what you said, but stated in the language of differential forms.
$\iota_N \omega$ is called the "interior product".
1) See why this, also, gives the volume form on the surface. 2) Generalize this construction to submanifolds of higher codimension. You will need to work locally here, so this might be less inspiring.
3) In $\Bbb R^3$, I've now told you about two dualities. vector fields are equated with 2-forms by $V \iff \iota_V \omega$. (See, in general, why this gives an equality $\mathfrak X(M) = \Omega^{n-1}(M)$ on manifolds with a volume form; as $C^\infty(M)$-modules if you like.) We also can get there on a Riemannian manifold via identifying vector fields with 1-forms and then by identifying 1-forms with (n-1)-forms (Hodge duality).
See why these two dualities are the same thing: why $\iota_V \omega_g = \ast_gX^\sharp$ ($\sharp$ is the map sending 1-forms to vector fields; or possibly that's denoted $\flat$, I always forget which).
@Hippalectryon related to what I previously showed you, the math world I entered for some years is a pretty tough one. I mean saying that you're self-educated or other stuff like that won't impress anyone, no one cares about that, you can be simply told not to publish anything. I mean that being self-educated makes you just far more responsible than the people with all kind of degrees, and a self-educated one really has to come up with mavellous stuff if the case of publishing things.
4) See why an orientation on a manifold $M$ is equivalent to choosing a nonvanishing volume form, up to multiplication with a positive smooth function, and the existence of such a volume form is the same thing as being orientable. Use this definition of orientation to see why the boundary of a manifold is orientable. Give it a canonical orientation.
@user1618033 True, but that's also understandable. You need some kind of safeguard to avoid being flooded by "bad" submissions. And self educated or not, once you've already made one recognized publication it's much easier I believe.
5) Let $M$ be an oriented manifold, and $N$ a codimension 1 hypersurface. Prove that $N$ is orientable if and only if it has a nonvanishing normal vector field. (Interpret normal appropriately.) What's special about boundaries?
@Hippalectryon Yeah, I know. I just wanted to emphasize that being self-educated is never an excuse or a reason for the other to be nicer with you (when trying to publish anything). It just makes you more responsible, and the expectations are higher.
Feel free to ping me with responses even if I'm not in here. I don't think this is more than what you should be able to do within the span of a few hours to a day.
and one advantage of the above notation is that it lets me drop back down into a coordinate-based representation for computational purposes easily, while still retaining the geometry
@Hippalectryon I've been waiting for an important article to be published - which presents a new very cool idea ... (all was fine so far - waiting for the very last step - the final decision)
You could, like, use the Hodge dual to identify them with wedge products of vector fields. But by the time you're wedging vector fields we've thrown out babies and bathwaters and lots more.
the other thing that gets weird for the purposes of applications is that, if you're doing 4D, that usually means you're really doing something in 3+1 spacetime
in which case the notion of 'geometry' becomes far less transparent
e.g. the hodge dual of $dx_1\wedge dx_2$ would be $dx_3\wedge dt$ (up to a choice of sign w/e)
By the way, to finish what I was saying above: in the formality of Riemannian metric I described, the parametrized surface $S$ in $\Bbb R^3$ gets a natural Riemannian metric: $[E, F; F, G]$, using the notation already introduced. This is apparently called the "first fundamental form" on $S$. Not really deep, just a cute context we're putting our first half of the discussion in.
The "first fundamental form" is old-fashioned terminology for the Riemannian metric. :)
It was Gauss's remarkable discovery (his theorema egregium) that it was what actually carried the notion of 'curvature', not the ambient embedding, that leads us to think about the modern idaes of Riemannian geometry.
I want to show the embedding $W^{1,p}(0,1) \subset C^0 [0,1]$.
So we pick a $u \in W^{1,p}(0,1)$ and want to show that $u \in C^0 [0,1]$.
Let $x_n \to x$. We want to show that $u(x_n) \to u(x)$.
Since $u \in W^{1,p}(0,1)$ we have that $u \in L_p$ and $u' \in L_p$.
And you told me to use t...
then its classical derivative exists a.e. and is equal to $u'$ a.e.; this is distribution derivative also; hence in the sense of distributions, $(u-\overline{u})' = 0
Therefore, $u = \overline{u} + c$ for some constant $c$ a.e. and the latter is absolutely continuous
So it is uniformly continous hence extends to boundary continuously
@Mambo Do we deduce that its classical derivative exists a.e, since each integrable function is differentiable? Or doesn't this hold? And why a.e. and not everywhere?
We suppose that a function $f$ on $(a,b)$ can be extended to $[a,b]$ continuously . Then the limits $\lim_{x \to a} f(x)$ and $\lim_{x \to b} f(x)$ exist. But how can show that f is uniformly continuous? In this case we cannot use the definition. Right? @Mambo
According to this post the problem with MathJax in the sidebar should already have been solved. But I still sometimes see problems like this: i.stack.imgur.com/NtEZK.png
A translation of the comment I added to the question I just posted:
"This question is inspired by my comment to a recent question, and my unwillingness to carry out the resulting algebra." = "I was bored enough to come up with this question, but too lazy to figure it out myself."
user174558
I forgot. How is your PhD coming along @Semiclassical?
the only annoying thing was figuring out how to get tex to work, which in practice is "I write everything as $latex math [some technical garbage that says my tex should have the right font and color]$
@Vrouvrou The you only have the inequality, not the equality. So it's not interesting for the $\sup$.
@Vrouvrou It's like saying : we're trying to find $\sup (h(a),a\in A)$. We know that $\forall a\in A,h(a)\le M$. Furthermore for a particular $a_0\in a$ (so, a special case), $h(a_0)=M$. Thus $\sup (h(a),a\in A)=M$.
@MikeMiller Sorry! I fell ill within an hour after the conversation we had. But I did do some thinking. I have solutions to a couple of the things you asked, and have thoughts about the rest. Not sure if I should write down the solutions now or wait until I have thought more and have all the solutions.
AFAIK its the only algebra book which develops multilinear algebra carefully. I'm sure some Diff. geo. books do as well, but I haven't read any of those.
Hmm. How in-depth does it get on geometric algebras? I'm starting to find a lot of connections between some groups in my work and noticing how much everything acts like Rotors and Blades.
I want to show the embedding $W^{1,p}(0,1) \subset C^0 [0,1]$.
So we pick a $u \in W^{1,p}(0,1)$ and want to show that $u \in C^0 [0,1]$.
Let $x_n \to x$. We want to show that $u(x_n) \to u(x)$.
Since $u \in W^{1,p}(0,1)$ we have that $u \in L_p$ and $u' \in L_p$.
And you told me to use t...
