Regular Values (Introduction to Smooth Manifolds)

[BrainHurts]

Homework Statement

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)

Homework Equations

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?

 

[micromass]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

So I have the Jacobian

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

I don't get the same Jacobian

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

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 $(0,0,0,0)$.

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?

 

[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

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

T_pF : T_pM → T_F(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.

 

[micromass]

Shoot!

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

I still don't agree.

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

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.

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

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

T_pF : T_pM → T_F(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.

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

 

[BrainHurts]

sorry my equation was wrong

 

[micromass]

sorry my equation was wrong

OK, I agree with the Jacobian now. But please don't edit any posts after you already received a reply. Just reply with the correct equation.

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

 

[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?

 

[micromass]

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?

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

 

[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.