# On which the function is a diffeomorphism to its image?

1. Jan 7, 2007

### cristo

Staff Emeritus
Are the following functions $\mathbb{R}^2\rightarrow\mathbb{R}^2$ diffeomorphisms. If not is there an open set containing the origin on which the function is a diffeomorphism to its image?

1. $$(x,y)\mapsto(x+y^3, y)$$
2. $$(x,y)\mapsto(x+x^3,x)$$

I have the definition of a diffeomorphism: $f:X\rightarrow Y$ is a diffeomorphism if f-1 exists, and f and f-1 are $C^{\infty}$ maps.

Where f is a $C^{\infty}$ map iff coordinates of a point $y \in Y$ are infinitely differentiable functions of $f^{-1}(y)\in X$.

So, for 1. I have tried the following:

$$f:\mathbb{R}^2\rightarrow \mathbb{R}^2: (x,y)\mapsto(x+y^3, y) =(u,v) \\ \Rightarrow x+y^3=u , y=v \Rightarrow x=u-v^3, y=v$$

So $f^{-1}: (u,v)\mapsto (u-v^3,v)=(x,y)$, and thus f-1 exists and both f and f-1 are $C^{\infty}$, therefore f is a diffeomorphism.

Is this right? I've tried a similar method for Q2, but to no avail!

Any help would be much appreciated!

2. Jan 7, 2007

### andytoh

Solution 1 looks good, but you should mention the bijectivity of f (isn't that part of the definition of diffeomorphism that you were given?) Because f inverse is defined on all of R^2, then f is surjective.

Are you sure you typed out Question 2 correctly? The image of R^2 under the map in question 2 is a curve, which is a 1-dimensional manifold. Diffeomorphisms preserve dimension of manifolds, so if you typed the question correctly, the answer would have to be no.

Note: You don't have to work out explicitly the inverse. Just calculate the Jacobian matrix and determine it has rank 2. For the second question, the Jacobian is

(1+3x^2 0)
(1 0)

which has rank 1 for all (x,y) in R^2, so the mapping cannot be a diffeomorphism.

For the first question, the Jacobian is

(1 3y^2)
(0 1)

which has rank 2 for all (x,y) in R^2. This means that the map is everywhere a local diffeomorphism. You went further and found a global inverse.

Last edited: Jan 7, 2007
3. Jan 7, 2007

### cristo

Staff Emeritus
Ok, thanks.

The question is definitely typed out correctly, so the answer must be no.

Thanks for this; it's definitely a better method than struggling to find the inverse!

However, how do I deduce the rank of the Jacobian matrix? I've not studied linear algebra for a while, and am quite rusty! Would I be right in thinking that the rank of a matrix is the number of linearly independent columns in the matrix- or am I completely off the mark here?

(Also, should the first entry in your first matrix above not be 1+3x^2)?

4. Jan 7, 2007

### andytoh

I've editted my first response. It is always good to find the explicit inverse if you can. Resort to the Jacobian if you cannot find an explicit inverse or want to prove that it is not a diffeomorphism. Remember to consider all possible values for x and y when determining the rank of the Jacobian. Sometimes two rows may look linearly independent but then they are linearly dependent for certain values of x and y. Fortunately those traps don't occur in your questions.

5. Jan 7, 2007

### cristo

Staff Emeritus
Ok, thanks for your help. I have one more question along the same lines; here's my attempt: $$(x,y)\mapsto(x^2+yx,y)=(u,v) \\ \Rightarrow x^2+yx=u, y=v \Rightarrow y=v, x(x+v)=u$$

So, the inverse function cannot be calculated. The Jacobi matrix is $$\left( \begin{array}{cc}2x+y&x\\0&1\end{array}\right)$$

This is not a diffeomorphism, since for the origiin and the points (1/2,-1) and (-1/2,1) the rows are linearly dependent (and hence the rank of the matrix is 1 for these points)

Is this correct? Thanks again!

Last edited: Jan 7, 2007
6. Jan 7, 2007

### andytoh

The Jacobian matrix has rank 1 only when 2x+y=0. So the map is a local diffeomorphism on all of R^2 except for the line {(x,y):2x+y=0}.

Now back to the original question, since the line 2x+y=0 goes through the origin, then there is NO local diffeomorphism at the origin. However it is a local diffeomorphism everywhere except on that line.

Last edited: Jan 7, 2007
7. Jan 7, 2007

### cristo

Staff Emeritus

8. Jan 7, 2007

### andytoh

HOWEVER, the map is not one-to-one (a diffeomorphism has to be one-to-one and onto). For example, (-1,0) and (1,0) both map to (1,0), so the map is NOT a diffeomorphism on R^2 - {(x,y):2x+y=0}. It is only LOCALLY a diffeomorphism on R^2 - {(x,y):2x+y=0}. This is the trouble that results when you cannot find an explicit inverse.

Last edited: Jan 7, 2007
9. Jan 7, 2007

### cristo

Staff Emeritus
Ok, but since the question asks for an "open set including the origin for on which the function is a diffeomorphism to its image". The set R^2-{(x,y):2x+y=0} includes the origin, and so does not satisfy this criterion.

10. Jan 7, 2007

### andytoh

This whole issue of local vs global confuses me too. For example, let's look at
f(x,y,z)=(x^2, y^2, z^2, yz, xz, xy), with domain {R^3: x,y,z>0}. Now f is one-to-one and clearly differentiable. The question is whether it is a diffeomorphism (not just a local diffeomorphism) from
{R^3: x,y,z>0} to f({R^3: x,y,z>0}). The Jacobian is

2x 0 0
0 2y 0
0 0 2z
0 x y
z 0 x
0 x y

which is of rank 3 since x,y,z>0. So f is locally a diffeomorphism at every point of {R^3: x,y,z>0}? So is f a diffeomorphism on all of {R^3: x,y,z>0} to its image? Can anyone answer?

Last edited: Jan 7, 2007
11. Jan 7, 2007

### andytoh

I said from {R^3: x,y,z>0} to f({R^3: x,y,z>0}), so the image is a 3-dimensional submanifold of R^6, so dimension is preserved and f is bijective, and so the question is valid.

Last edited: Jan 7, 2007