more smooth manifolds questions (differential forms edition)

[MathNugget]

Given: M is a smooth manifold, $dim(M)> 1$, and $\alpha \in \Omega^1(M)$, yet $\alpha \wedge \beta =0, \forall \beta \in \Omega^1(M)$. $\Omega^1{M}$ are the differential 1-forms.

Question: $\alpha$ must be 0?

I've decided that I want it to be (it would be weird otherwise).

Given that $\Omega^1{M}$ is a vector space (or so I think), I guess we can build a matrix of independent differential 1-forms, A. $\alpha A=0$ , because there's products of 1 forms happening.

And now when we return through $A^{-1}$, $0 A^{-1}=0$. Am I even close?

 

[MathNugget]

Maybe it was $\alpha \wedge A$, now that I think of it. And here I generalize vector- matrix product for 1-forms...

 

[blamocur]

I guess we can build a matrix of independent differential 1-forms

What does this mean? I can see how you can build a matrix representing exterior products in $\mathbb R^2$ (it looks rather simple), but for manifolds your best bet would be to localize and prove that $\alpha=0$ for every local map.

 

[MathNugget]

What does this mean?

I don't know. I thought this would work for $\mathbb{R}^3$, so I said to myself...maybe it is true in an arbitrary vector space, and won't even need to use that this is differential geometry...

but for manifolds your best bet would be to localize and prove that $\alpha=0$ for every local map.

I'll see what I can do about this. I'll take a few hours to look into what this means

 

[blamocur]

so I said to myself...maybe it is true in an arbitrary vector space,

You are right. Not sure why I thought of 2d only. Maybe because you mentioned a matrix, which would make sense in 2D, but not sure about higher dimensions.

and won't even need to use that this is differential geometry...

But you are asking about differential forms on manifolds, which only makes sense in the context of differential geometry. I still think that reducing this to local neighborhoods/maps, which effectively means reducing it to cartesian space $\mathbb R^n$, is the way to go.
In $\mathbb R^n$ your proof would depend on the definition of the exterior, a.k.a. "wedge", product, and lemmas and theorems you can use. I think it can be proven using somewhat abstract definition from Wikipedia.

 

[MathNugget]

So, we choose a basis of $\Omega^1(M)$, write $\alpha, \beta$ in that basis, then do the wedge product?
$\alpha \wedge \beta=\sum_{0< i< j< n} (\alpha_i\beta_j-\alpha_j\beta_i)(e_i\wedge e_j)$, where $e_i$ are the elements of the basis?

 

[blamocur]

So, we choose a basis of $\Omega^1(M)$, write $\alpha, \beta$ in that basis, then do the wedge product?
$\alpha \wedge \beta=\sum_{0< i< j< n} (\alpha_i\beta_j-\alpha_j\beta_i)(e_i\wedge e_j)$, where $e_i$ are the elements of the basis?

Except that $\Omega^1(M)$ might not have any basis, but a small enough neighborhood would.

 

[MathNugget]

Except that $\Omega^1(M)$ might not have any basis, but a small enough neighborhood would.

Does that happen because $\Omega^1(M)$ might be infinite dimensional?

 

[blamocur]

Does that happen because $\Omega^1(M)$ might be infinite dimensional?

No, but because $M$ is a manifold. Try finding vector basis for $\mathbb S^2$

 

[MathNugget]

No, but because $M$ is a manifold. Try finding vector basis for $\mathbb S^2$

True, but not very obvious as to how I'd prove it. It's pretty clear that for the 2 ways I know to locally map it to open sets of $\mathbb{R}^2$, the 2 vector basis doesn't go in same place for all inverse maps.
If the atlas had a single map though, it would be certain that the manifold has a basis. Maybe only manifolds that are globally homeomorphic to open sets of $\mathbb{R}^n$ have a basis?

 

[MathNugget]

So, we choose a basis of $\Omega^1(M)$, write $\alpha, \beta$ in that basis, then do the wedge product?
$\alpha \wedge \beta=\sum_{0< i< j< n} (\alpha_i\beta_j-\alpha_j\beta_i)(e_i\wedge e_j)$, where $e_i$ are the elements of the basis?

returning to this...do I pick $\beta_1=1$ and the rest 0, then $\beta_2=1$ and the rest 0, and so on?
I'll think about it tomorrow, where these calculations get me. Thank you for help.

 

[blamocur]

True, but not very obvious as to how I'd prove it.

How to prove what? That there is no basis on $\mathbb S^2$ for tangent vectors? You don't really have to, but you can read on this here: https://en.wikipedia.org/wiki/Hairy_ball_theorem

Maybe only manifolds that are globally homeomorphic to open sets of Rn\mathbb{R}^nRn have a basis?

You can find a basis on a torus.

 

[MathNugget]

returning to this...do I pick $\beta_1=1$ and the rest 0, then $\beta_2=1$ and the rest 0, and so on?
I'll think about it tomorrow, where these calculations get me. Thank you for help.

some computations: for n vectors in basis, I get n equations, and out of $n^2$ sum terms, n+1 are 0...
$\sum_{1\leq i < j} \alpha_i(e_i \wedge e_j) - \sum_{j< k \leq n}\alpha_k(e_j \wedge e_k)=0$ ,$\forall j =\overline{1,n}$.
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.

 

[blamocur]

some computations: for n vectors in basis, I get n equations, and out of $n^2$ sum terms, n+1 are 0...
$\sum_{1\leq i < j} \alpha_i(e_i \wedge e_j) - \sum_{j< k \leq n}\alpha_k(e_j \wedge e_k)=0$ ,$\forall j =\overline{1,n}$.
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.

What is the basis of $\bigwedge^2 (\mathbb R)$ ?

 

[MathNugget]

What is the basis of $\bigwedge^2 (\mathbb R)$ ?

I'll follow wikipedia's steps to find that out...
$e_1=\{1\}$ would be a basis for $\mathbb{R}$. It is very weird though, as repeating the process I see on wikipedia leads me to believe it doesn't have a basis, as $e_1 \wedge e_1 =0$, and there's a single 'vector' in the basis. I have found the answer was actually {0}, also on same wikipedia page.
Sounds like you want me to find $dim(\Omega^1(M))$, relative to $dim(M)$?

 

[blamocur]

I'll follow wikipedia's steps to find that out...
$e_1=\{1\}$ would be a basis for $\mathbb{R}$. It is very weird though, as repeating the process I see on wikipedia leads me to believe it doesn't have a basis, as $e_1 \wedge e_1 =0$, and there's a single 'vector' in the basis. I have found the answer was actually {0}, also on same wikipedia page.
Sounds like you want me to find $dim(\Omega^1(M))$, relative to $dim(M)$?

You are right, my mistake: I meant to ask what is the basis of $\bigwedge^2 \mathbb R^{\mathbf n}$.

 

[MathNugget]

You are right, my mistake: I meant to ask what is the basis of $\bigwedge^2 \mathbb R^{\mathbf n}$.

Well, it appears to be $\{e_i \wedge e_j \mid 1 \leq i < j \leq n \}$.
I suppose $\Omega^1{M}\subset \bigwedge^1(M)$, so they use same basis? and then $\alpha \wedge \beta$ would be an element of $\bigwedge^2(M)$, so it has same basis?
Did I made a mistake earlier, and that's why I do these calculations?

 

[blamocur]

Well, it appears to be {ei∧ej∣1≤i<j≤n}\{e_i \wedge e_j \mid 1 \leq i < j \leq n \}{ei∧ej∣1≤i<j≤n}.

Agree with this.

I suppose Ω1M⊂⋀1(M)\Omega^1{M}\subset \bigwedge^1(M)Ω1M⊂⋀1(M), so they use same basis? and then α∧β\alpha \wedge \betaα∧β would be an element of ⋀2(M)\bigwedge^2(M)⋀2(M), so it has same basis?

As I mentioned earlier there is no basis on a manifold in the general case.

Did I made a mistake earlier, and that's why I do these calculations?

I don't understand this part.

 

[MathNugget]

I don't understand this part.

Why did I calculate this:

What is the basis of $\bigwedge^2 (\mathbb R)$ ?

 

[blamocur]

Why did I calculate this:

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