more smooth manifolds questions (differential forms edition)

[MathNugget]

some computations: for n vectors in basis, I get n equations, and out of [imath]n^2[/imath] sum terms, n+1 are 0...
[imath]\sum_{1\leq i < j} \alpha_i(e_i \wedge e_j) - \sum_{j< k \leq n}\alpha_k(e_j \wedge e_k)=0[/imath] ,[imath]\forall j =\overline{1,n}[/imath].
I guess the 2nd sum needed k instead of i, for clarity. I also see why everyone likes the Einstein summation notation, but I simply don't see what I am doing anymore when using that.

So I guess this is wrong?

 

[MathNugget]

I asked my question because I believe a good first step is to prove the statement in op for [imath]\mathbb R^n[/imath]. Do you see that?

Oh, so trying to do the exercise for [imath]M=\mathbb{R}^n?[/imath]

 

[blamocur]

Oh, so trying to do the exercise for [imath]M=\mathbb{R}^n?[/imath]

Yes.

 

[MathNugget]

some computations: for n vectors in basis, I get n equations, and out of [imath]n^2[/imath] sum terms, n+1 are 0...
[imath]\sum_{1\leq i < j} \alpha_i(e_i \wedge e_j) - \sum_{j< k \leq n}\alpha_k(e_j \wedge e_k)=0[/imath] ,[imath]\forall j =\overline{1,n}[/imath].
I guess the 2nd sum needed k instead of i, for clarity. I also see why everyone likes the Einstein summation notation, but I simply don't see what I am doing anymore when using that.

so...this? make [imath]\beta=(1, 0, ... 0)[/imath], then [imath]\beta=(0, 1, 0, ..., 0)[/imath] and so on?

 

[blamocur]

so...this? make β=(1,0,...0)\beta=(1, 0, ... 0)β=(1,0,...0), then β=(0,1,0,...,0)\beta=(0, 1, 0, ..., 0)β=(0,1,0,...,0) and so on?

I don't think it is a good idea asking questions about every step along the way. A much better -- and more efficient -- approach would be to try solving the problem, then posting the whole solution. This way we can have a much more meaningful conversation.

 

[MathNugget]

For [imath]\alpha, \beta \in \Omega^1(\mathbb{R}^n)[/imath], [imath]\alpha=\sum_{i=1}^n \alpha_i e_i[/imath], similar notation for beta.
[imath]\alpha \wedge \beta=\sum_{1 \leq i < j \leq n} (\alpha_i\beta_j-\alpha_j\beta_i)(e_i\wedge e_j)[/imath]
Since it's for all [imath]\beta[/imath], I choose [imath]\beta=e_1[/imath]
[imath]\alpha \wedge e_1=\sum_{1 < i < n}\alpha_i(e_i \wedge e_1)[/imath]
Since [imath]\alpha \wedge \beta \in \Omega^2(\mathbb{R}^n)[/imath], [imath]\{e_i \wedge e_j \mid 1 \leq i < j \leq n \}[/imath] is a basis, and a linear combination of them cannot be 0 unless they're all multiplied by 0...
So, if [imath]\alpha \wedge e_i =0,[/imath] [imath]\alpha_i=0, \forall i=\overline{2,n}[/imath].
Then I just pick [imath]\beta=e_2[/imath] to show [imath]\alpha_1=0[/imath].

 

[MathNugget]

I think I wrote the same thing before, that's why I asked what I am supposed to do.
But here it is again.

 

[blamocur]

I think I wrote the same thing before,

Which post?

 

[blamocur]

For [imath]\alpha, \beta \in \Omega^1(\mathbb{R}^n)[/imath], [imath]\alpha=\sum_{i=1}^n \alpha_i e_i[/imath], similar notation for beta.
[imath]\alpha \wedge \beta=\sum_{1 \leq i < j \leq n} (\alpha_i\beta_j-\alpha_j\beta_i)(e_i\wedge e_j)[/imath]
Since it's for all [imath]\beta[/imath], I choose [imath]\beta=e_1[/imath]
[imath]\alpha \wedge e_1=\sum_{1 < i < n}\alpha_i(e_i \wedge e_1)[/imath]
Since [imath]\alpha \wedge \beta \in \Omega^2(\mathbb{R}^n)[/imath], [imath]\{e_i \wedge e_j \mid 1 \leq i < j \leq n \}[/imath] is a basis, and a linear combination of them cannot be 0 unless they're all multiplied by 0...
So, if [imath]\alpha \wedge e_i =0,[/imath] [imath]\alpha_i=0, \forall i=\overline{2,n}[/imath].
Then I just pick [imath]\beta=e_2[/imath] to show [imath]\alpha_1=0[/imath].

I agree with this proof for [imath]\mathbb R^n[/imath]. In the case of an arbitrary differential manifold it can be used to show that for any [imath]p\in M[/imath] there exists a neighborhood [imath]U: p\in U \subset M[/imath] such that [imath]\alpha=0[/imath] in [imath]U[/imath].

 

[MathNugget]

Which post?

True. The attempt at 13# wasn't too close, now that I think of it. Sorry.

I agree with this proof for [imath]\mathbb R^n[/imath]. In the case of an arbitrary differential manifold it can be used to show that for any [imath]p\in M[/imath] there exists a neighborhood [imath]U: p\in U \subset M[/imath] such that [imath]\alpha=0[/imath] in [imath]U[/imath].

So I take a map [imath](U, \phi)[/imath], and through [imath]\phi^{-1}[/imath] I move the basis [imath]e_i[/imath] into a basis on U? internet seems to say a homeomorphism maps basis to basis.

Thank you very much for help!!

 

[blamocur]

True. The attempt at 13# wasn't too close, now that I think of it. Sorry.


So I take a map [imath](U, \phi)[/imath], and through [imath]\phi^{-1}[/imath] I move the basis [imath]e_i[/imath] into a basis on U? internet seems to say a homeomorphism maps basis to basis.

Thank you very much for help!!

You link points to a question about topological bases, not vector spaces. But if you revisit the definition of differential manifolds you'll see that all differential structures are defined for points' neighborhoods.

Glad we've worked this out In the future feel free to post your self-contained solutions -- they are easier to discuss than references to previous posts. And by all means don't be afraid of making mistakes in your posts: addressing and fixing them is, in my experience, one of the most effective ways to learn new stuff.

 

[MathNugget]

You link points to a question about topological bases, not vector spaces. But if you revisit the definition of differential manifolds you'll see that all differential structures are defined for points' neighborhoods.

Well then...I think an isomorphism of vector spaces sends basis to basis, and a homeomorphism is an isomorphism. Just to make sure I can have a basis on an open set homeomorphic to an open set in [imath]\mathbb{R}^n[/imath]

Glad we've worked this out In the future feel free to post your self-contained solutions -- they are easier to discuss than references to previous posts. And by all means don't be afraid of making mistakes in your posts: addressing and fixing them is, in my experience, one of the most effective ways to learn new stuff.

Alright, I'll keep that in mind. I expected the process of solving this exercise to be completely illogical, so I thought all attempts would be too naive. In hindsight, the exercise appears reasonable though.
Thanks for help!!