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 $$ X^a:=(X_1,...,X_{k-1},X_a,X_{k+1},...,X_{p+1}) \in \mathfrak{X}^{p+1} $$ $$ X^b:=(X_1,...,X_{k-1},X_b,X_{k+1},...,X_{p+1}) \in \mathfrak{X}^{p+1}$$ $$ X':=(X_1,...,X_{k-1},g X_a+h X_b,X_{k+1},...,X_{p+1}) \in \mathfrak{X}^{p+1} $$ $$ X^{(k)}:=(X_1,...,X_{k-1},X_{k+1},...,X_{p+1}) \in \mathfrak{X}^{p} $$
