Is the function F: R^3 -> R^4 defined by F(x,y,z) = (sqrt(x), x,y,z) if x >= 0 and (sqrt(-x), x,y,z) if x < 0 smooth on all of R^3?
I'm thinking it is not, because taking the square root messes up smoothness at x=0, is this correct? Thanks.
Is the function F: R^3 -> R^4 defined by F(x,y,z) = (sqrt(x), x,y,z) if x >= 0 and (sqrt(-x), x,y,z) if x < 0 smooth on all of R^3?
I'm thinking it is not, because taking the square root messes up smoothness at x=0, is this correct? Thanks.
Right.
Fernando Revilla
OK. The problem is I am trying to find a smooth bijection F: S^2 (unit sphere in R^3) --> U where U is the subset of R^4 defined as:
U = {(x,y,z,w) in R^4 such that x^2 + y = 0 and y^2 + z^2 + w^2 = 1}
All bijections I have tried involve taking a square root in the first component so I have not managed to find a smooth one.
You may find it easier to visualise what to do if you consider the same problem in one dimension lower: Find a smooth bijection F: S^1 (unit sphere in R^2) --> U where U is the subset of R^3 defined as:
U = {(x,y,z) in R^3 such that x^2 + y = 0 and y^2 + z^2 = 1}.
Here, $\displaystyle x^2 + y = 0$ is a cylindrical parabola and $\displaystyle y^2 + z^2 = 1$ is a circular cylinder. Their intersection looks like a distorted circle. In two dimensions, the circle $\displaystyle x^2+y^2=1$ is smoothly mappable to the curve $\displaystyle x^4+y^2=1$ (which also looks like a distorted circle). If you can produce a smooth map $\displaystyle (x,y)\mapsto (z,w)$ implementing this equivalence, then the map $\displaystyle (x,y)\mapsto(z,-z^2,w)$ will be a smooth bijection from the circle to the set U.
If you succeed in doing that, you should have no trouble jacking up the dimension by 1 to solve the given problem.
Thanks a lot. For the smooth map from the circle to the curve, I drew them both on the plane and for (x,y) in S1 defined r(x,y) to be the point on the curve that meets the straight line starting from the origin and going through (x,y). Is this the smooth map you had in mind?