# Homework Help: Parameterized p-subset of a manifold M

1. Oct 16, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

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 $F^{*}\alpha^{p}=0$.

2. Relevant equations

3. 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 $0$?

Last edited: Oct 17, 2016
2. Oct 17, 2016

### Orodruin

Staff Emeritus
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.

3. Oct 17, 2016

### spaghetti3451

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.

4. Oct 17, 2016

### Orodruin

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

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

5. Oct 17, 2016

### spaghetti3451

I get it.

Thank you so very much!

6. Oct 17, 2016

### spaghetti3451

Oh, just one thing.

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

7. Oct 17, 2016

### Orodruin

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

8. Oct 17, 2016

### 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.

9. Oct 17, 2016

### Orodruin

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

10. Oct 17, 2016

### Orodruin

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

11. Oct 17, 2016

### spaghetti3451

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?

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.

12. Oct 17, 2016

### Orodruin

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

13. Oct 17, 2016

### spaghetti3451

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

14. Oct 17, 2016

### Orodruin

Staff Emeritus
And therefore ...

15. Oct 17, 2016

### spaghetti3451

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.

16. Oct 17, 2016

### Orodruin

Staff Emeritus
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?

17. Oct 17, 2016

### 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.

18. Oct 17, 2016

### Orodruin

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

19. Oct 17, 2016

### spaghetti3451

So, do you mean the following:

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

20. Oct 17, 2016

### Orodruin

Staff Emeritus
Yes. Now what does this tell you?

21. Oct 17, 2016

### spaghetti3451

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 $0$.

22. Oct 17, 2016

### Orodruin

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

23. Oct 17, 2016

Thank you.