Parameterized p-subset of a manifold M

[spaghetti3451]

Homework Statement

A parameterized $p$-subset $(U,F)$ of a manifold $M^{n}$ is ''irregular'' at $u_0$ if rank $F<p$ at $u_0$.

Show that if $\alpha^{p}$ is a form at such a $u_0$ then $F^{*}\alpha^{p}=0$.

Homework Equations

The Attempt at a Solution

A parameterized $p$-subset of a manifold $M^{n}$ is a pair $(U,F)$ consisting of a region $U$ in $\mathbb{R}^{p}$ and a differentiable map $F:U(u)\rightarrow M^{n}(x)$.

If $\alpha^{p}$ is a form at a $u_0$ where rank $F<p$, then

$\displaystyle{F^{*}\alpha^{p}=(F^{*}\alpha^{p})\bigg[\frac{{\bf{\partial}}}{{\bf{\partial}} u^{i_{1}}}, \dots , \frac{{\bf{\partial}}}{{\bf{\partial}} u^{i_{p}}}\bigg]du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$

$\displaystyle{F^{*}\alpha^{p}=\alpha^{p}\bigg[F_{*}\frac{{\bf{\partial}}}{{\bf{\partial}} u^{i_{1}}}, \dots , F_{*}\frac{{\bf{\partial}}}{{\bf{\partial}} u^{i_{p}}}\bigg]du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$

$\displaystyle{F^{*}\alpha^{p}=\alpha^{p}\bigg[\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}}\frac{{\bf{\partial}}}{{\bf{\partial}} y^{j_{p}}}, \dots , \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}\frac{{\bf{\partial}}}{{\bf{\partial}} y^{j_{p}}}\bigg]du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$

$\displaystyle{F^{*}\alpha^{p}=\alpha_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$

Next, I need to use the fact that the matrix of partial derivatives has rank less than $p$. Can you suggest a trick to show that $F^{*}\alpha^{p}$ is $0$?

 

[Orodruin]

What is the determinant of a square matrix with rank lower than its size?

Alternatively you can try acting with the form on p vectors and use the fact that at least one of the pushforwards is linearly dependent on the others.

 

[spaghetti3451]

What is the determinant of a square matrix with rank lower than its size?

The determinant is surely zero, but in this case, we have $\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}$, which is clearly not the determinant of the matrix of partial derivatives.

 

[Orodruin]

The determinant is surely zero, but in this case, we have $\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}$, which is clearly not the determinant of the matrix of partial derivatives.

You are contracting it with something antisymmetric and hence ... (fill in the dots)

On the other hand, the alternative approach may be more aesthetic ...

 

[spaghetti3451]

I get it.

Thank you so very much!

 

[spaghetti3451]

Oh, just one thing.

I have not really used the fact that the rank of $F$ is less than $p$.

 

[Orodruin]

Oh, just one thing.

I have not really used the fact that the rank of $F$ is less than $p$.

This would be much easier to address if you told me what you did.

 

[spaghetti3451]

$\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}$ is symmetric but $\displaystyle{du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$ is antisymmetric. This seems to imply that the product of the two is $0$, regardless of the rank of $F$.

Oh wait, $\alpha_{j_{1}\dots j_{p}}$ is also antisymmetric, so that the total product is symmetric and not necessarily $0$.

So, I am not really sure how to approach the problem.

 

[Orodruin]

$\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}$ is symmetric but $\displaystyle{du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$ is antisymmetric. This seems to imply that the product of the two is $0$, regardless of the rank of $F$.

Oh wait, $\alpha_{j_{1}\dots j_{p}}$ is also antisymmetric, so that the total product is symmetric and not necessarily $0$.

So, I am not really sure how to approach the problem.

No, this is incorrect. The matrix of partial derivatives is not symmetric.

 

[Orodruin]

You still have both my suggestions left. Either use that the determinant is zero or that the vector pushforwards are linearly dependent.

 

[spaghetti3451]

No, this is incorrect. The matrix of partial derivatives is not symmetric.

Well, the matrix of partial derivatives is not necessarily symmetric - I agree.

But the product of partial derivatives $\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}$ is surely symmetric, is it not?

You still have both my suggestions left. Either use that the determinant is zero or that the vector pushforwards are linearly dependent.

Well, the determinant of the matrix of partial derivatives is $0$, since the matrix has rank less than its number of dimensions. Also, the product $\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}$ is being contracted on both sides, on the left with the antisymmetric $\alpha_{j_{1}\dots j_{p}}$ and on the right with the antisymmetric $du^{i_{1}}\wedge \dots \wedge du^{i_{p}}$.

I know that the determinant of the matrix of partial derivatives is given by $\displaystyle{\epsilon_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}$, where the epsilon tensor is antisymmetric, but I am not sure how this relates to the above.

 

[Orodruin]

How many linearly independent p-forms exist in a p-dimensional space?

 

[spaghetti3451]

In a $p$-dimensional space, there is $1$ linearly independent $p$-form.

 

[Orodruin]

In a $p$-dimensional space, there is $1$ linearly independent $p$-form.

And therefore ...

 

[spaghetti3451]

And therefore ...

I give up!

I understand that this means that $\displaystyle{F^{*}\alpha^{p}=\alpha_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}=\alpha_{1\dots p}\frac{\partial y^{1}}{\partial u^{1}} \dots \frac{\partial y^{p}}{\partial u^{p}}du^{1}\wedge \dots \wedge du^{p}}$,

but I can't carry this idea any further.

 

[Orodruin]

I give up!

I understand that this means that $\displaystyle{F^{*}\alpha^{p}=\alpha_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}=\alpha_{1\dots p}\frac{\partial y^{1}}{\partial u^{1}} \dots \frac{\partial y^{p}}{\partial u^{p}}du^{1}\wedge \dots \wedge du^{p}}$,

but I can't carry this idea any further.

No it does not mean that. You missed several (most) terms of the sum. Can you write down a particular p-form using the permutation symbol?

 

[spaghetti3451]

$\displaystyle{{\alpha}_{{\mu}_{\sigma(1)}\dots{\mu}_{\sigma(p)}}}=\text{sgn}(\sigma)\alpha_{\mu_{1}\dots\mu_{p}}$,

where the signature $\text{sgn}(\sigma)$ of a permutation $\sigma=\sigma(1)\dots\sigma(p)$ is defined to be $+1$ if $\sigma$ is even and $−1$ if $\sigma$ is odd.

 

[Orodruin]

No, I am asking if you can write down a p-form using $\epsilon_{i_1\ldots i_p}$.

 

[spaghetti3451]

So, do you mean the following:

$\alpha_{j_{1}\dots j_{p}}=\epsilon_{j_{1}\dots j_{p}}\alpha_{1\dots p}$?

 

[Orodruin]

So, do you mean the following:

$\alpha_{j_{1}\dots j_{p}}=\epsilon_{j_{1}\dots j_{p}}\alpha_{1\dots p}$?

Yes. Now what does this tell you?

 

[spaghetti3451]

Yes. Now what does this tell you?

It tells me that

$\displaystyle{F^{*}\alpha^{p}=\alpha_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$

$\displaystyle{F^{*}\alpha^{p}=\epsilon_{j_{1}\dots j_{p}}\alpha_{1\dots p}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$.

Now, $\displaystyle{\epsilon_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}$ is the determinant of the matrix of partial derivatives, so that $F^{*}\alpha^{p}$ is $0$.

 

[Orodruin]

It tells me that

$\displaystyle{F^{*}\alpha^{p}=\alpha_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$

$\displaystyle{F^{*}\alpha^{p}=\epsilon_{j_{1}\dots j_{p}}\alpha_{1\dots p}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}$.

Now, $\displaystyle{\epsilon_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}}$ is the determinant of the matrix of partial derivatives, so that $F^{*}\alpha^{p}$ is $0$.

Technically it is $\epsilon_{i_1 \ldots i_p}$ multiplied by the determinant, but it is still equal to zero since the determinant is.

 

[spaghetti3451]

Thank you.