# Regular Values (Introduction to Smooth Manifolds)

1. Mar 30, 2013

### BrainHurts

1. The problem statement, all variables and given/known data
Consider the map $\Phi$ : ℝ4 $\rightarrow$ ℝ2

defined by $\Phi$ (x,y,s,t)=(x2+y, yx2+y2+s2+t2+y)

show that (0,1) is a regular value of $\Phi$ and that the level set $\Phi^{-1}$ is diffeomorphic to S2 (unit sphere)

2. Relevant equations

3. The attempt at a solution

So I have the Jacobian

D$\Phi$ = [2x 2y 0 0; 2x 2y+1 2s 2t]

the reduced row echelon form of D$\Phi$ is a rank 2 matrix which implies that F is a smooth submersion.

So I guess I'm a little confused on the definition: Let M and N be smooth manifolds

If F:M→N is smooth and let p $\in$ M, then p is a regular point if D$\Phi$ at p is onto.

D$\Phi$ is a rank 2 matrix which implies that D$\Phi$ is onto right?

Another question that I have is that the pre-image of (0,1) under $\Phi$ is a subset of ℝ4 any hints on the diffeomorphism part?

2. Mar 30, 2013

### micromass

Staff Emeritus
I don't get the same Jacobian

But $D\Phi$ is not a matrix at all. You should evaluate it in different points. You should ask for which $(x,y,s,t)$ that $D\Phi_{(x,y,s,t)}$ is surjective. This is not true for all $(x,y,s,t)$. For example $(0,0,0,0)$.

3. Mar 30, 2013

### BrainHurts

Shoot!
$\Phi$(x,y,s,t) = (x2+y, x2+y2+s2+t2+y)

D$\Phi$ = [2x 1 0 0; 2x 2y+1 2s 2t]

How is D$\Phi$ not a matrix? I'm confused on that?

I'm going to switch notation up a bit

If we have a smooth map F: M → N and let p $\in$ M, then p is a regular point if

TpF is onto. In this case TpF is the total derivative which we used D$\Phi$.

TpF : TpM → TF(p)N

However I do see your point about D$\Phi$(0,0,0,0), we would have a rank 1 matrix in that case.

Last edited: Mar 30, 2013
4. Mar 30, 2013

### micromass

Staff Emeritus
I still don't agree.

It's not a matrix in the sense that it depends on specific values of $(x,y,s,t)$. If you're given specific values, then it is a matrix.

Can you classify all points such that $D\Phi_{(x,y,s,t)}$ is surjective?

5. Mar 30, 2013

### BrainHurts

sorry my equation was wrong

6. Mar 30, 2013

### micromass

Staff Emeritus

So, can you now figure out for which $(x,y,s,t)$ that your Jacobian has rank 2?

7. Mar 30, 2013

### BrainHurts

First sorry about that reply thing, I realized that once I made the edit, thought about re-editing so that I made my mistake, but then that's just too many re-edits, in any case I won't make that mistake again!

so the point (0,1), the pre-image of (0,1) under $\Phi$ is a subset of ℝ4

so we will look at the specific case:

(x2+y,x2+y2+s2+t2+y) = (0,1)

this means that

x2+y = 0

and that x2+y2+s2+t2+y = 1

this reduces the second component to

y2+s2+t2 = 1 (which is nice sine S2 : y2+s2+t2)

$\Phi^{-1}$(0,1) = {(x,y,s,t)|(x2+y=0 and y2+s2+t2 = 1}

so for all (a,b,c,d) $\in$ $\Phi^{-1}$(0,1)

we have D$\Phi$(a,-a2,c,d) is a rank 2 matrix

I think that's good now right?

8. Mar 30, 2013

### micromass

Staff Emeritus
If it's clear to you why $D\Phi(a,-a^2,c,d)$ has rank 2, then it's good.

9. Mar 30, 2013

### BrainHurts

for all p $\in$ $\Phi^{-1}$(0,1), p is a regular point, and it follows that (0,1) is a regular value. Thanks so much! That really helped a lot. I think I have the second part.