Talk:Aufgaben:Problem 10
I don't think the proof for asymmitry was sufficient, it was only shown that we can pull out a minus sign from every sum, but the sum are disordered after the permutation and have be put back together. to show that we get back \(-d\omega\) and not something different
Carl (talk) 14:01, 18 June 2015 (CEST)
I do not see a problem, what do you mean by disordered sum, where? He starts with dw(..k+1,k..) and gets back -dw(..k,k+1..), doesn't he? And this generalises to every permutation.
Mario (talk) 18:35, 30 June 2015 (CEST)
-> View histroy
Carl (talk) 21:34, 27 July 2015 (CEST)
Typos in a)?
I'm not too confident with this subject so I won't change nothing but in the prop from page 77 the summation index should run up to n instead of p. Further I think we should continue to label the V_j: ... \(V_j(x^{j_i})\) ... Thanks for the nice solution
Mario (talk) 15:55, 30 June 2015 (CEST)
I think your right on both, although you probably mean: \(V_i(x^{j_i})\) --> yepp, thats what I meant.
Carl (talk) 16:24, 30 June 2015 (CEST)
Another point. Unused as I am, it took me quite a while to get the meaning of what you mean by V(gX) in your warning, (or X1(gX2) respectively for the commutator). Maybe one could make a explanationary note that (as in other cases where you did explicitly note it) here X is not a vectorfield anymore but a function X(f) as the argument was omitted. Am I right? Well I just see that it is actually written in the task itself when defining [,]..
Mario (talk) 18:35, 30 June 2015 (CEST)
I didn't actually write the solution, but feel free to add explanation if you think they will help.
Carl (talk) 21:09, 30 June 2015 (CEST)
I don't think, I've understood this topic very well, so probably it's a petty dumb question, but maybe anyone can help me nevertheless:
I'm not quite sure, wheter there's an error right in the first equation.
\[V(f)(z) = \sum \limits_{j=1}^n V(x^j)(z) \cdot \frac{\partial}{\partial x^j }f(z)\]
As far as I know, the "\(x^j\)" are supposed to be functions on \(\mathbb{R}^n\), since later it is written:
"...the vector field \(V_k\) has already been evaluated at the smooth function \(x^{j_k}\) ."
But then I don't understand, why we have n such functions \(x^, ... x^n\), they can't form a basis to \(C^inf\), because this is an infinite dimensional vectorspace. And in addition I don't see how we can take the derivative with respect to a function in the second part:
\[\frac{\partial}{\partial x^j }\]
So, as you can see I'm quite confused about what all those things in this equation really mean. Maybe someone can help me. (Or it is indeed a typo, but actully I doubt that...)
Jo (talk) 10:30, 29 July 2015 (CEST)
This is the way I understand it: the smooth function \(x^j\) is the coordinate map given by \(x\mapsto x^j\) so \(x^j\) can be thought of as the coordinate of \(x\) but also as a smooth function on \(U\) that is evaluated at \(x\) to give you the \(j\)-th coordinate. I agree that this is confusing but I think it is still formally correct. These function don't form a basis, but the derivatives with respect to the coordinates form a "basis" (probably not the right word) of all the smooth vector fields on \(U\), as the equation states (proof \(\rightarrow\) lecture notes).
Carl (talk) 12:04, 29 July 2015 (CEST)
I don't see a way to shorten the multilinearity apart from doing the addition and multiplication at once like to show that $$d\omega(X_1,...,X_{k-1},g X_a+h X_b,X_{k+1},...,X_{p+1})=g\cdot d\omega(X_1,...,X_{k-1},X_a,X_{k+1},...,X_{p+1})+h\cdot d\omega(X_1,...,X_{k-1},X_b,X_{k+1},...,X_{p+1})$$
or defining some typographic abbreviations i.e $$ X^a:=(X_1,...,X_{k-1},X_a,X_{k+1},...,X_{p+1}) \in \mathfrak{X}^{p+1} $$ $$ X^b:=(X_1,...,X_{k-1},X_b,X_{k+1},...,X_{p+1}) \in \mathfrak{X}^{p+1}$$ $$ X':=(X_1,...,X_{k-1},g X_a+h X_b,X_{k+1},...,X_{p+1}) \in \mathfrak{X}^{p+1} $$ $$ X^{(k)}:=(X_1,...,X_{k-1},X_{k+1},...,X_{p+1}) \in \mathfrak{X}^{p} $$
--Brynerm (talk) 17:03, 3 August 2015 (CEST)
What I meant is to basically do cravens solution. Such that you first prove the asymmetry and then only need to prove multilinearity for the first element, as you can permute any element to the front and then back, also as you said one can show additivity and homogeneity at the same time. This sort of abbreviation might also be a good idea.
Carl (talk) 17:18, 3 August 2015 (CEST)
Nice idea doing the antisymmetry first. That's indeed a bit less messy
