# RP2 into R4 embedding

1. Sep 30, 2009

### sin123

P.S. This was supposed to go into the homework forum. I had windows open for both forums and I ended up using the wrong one.

Let F be a map from S2 in R3 into R4, given by

$$F(x,y,z) = (x^2 - y^2, xy, xz, yz) [ = (a,b,c,d)]$$

Eventually I am supposed to show that this is an embedding of the real projective plane, but first I am asked to verify that the image of this map is a manifold at all. And that proved trickier than it looked like.

I know two ways of verifying that something is indeed a manifold.

1) Find local diffeomorphisms, taking a neighborhood of the manifold into R4 such that points on the manifold land in a copy of R2 inside R4

2) Show that the manifold is the level set of some function, where the derivative of that function has full rank at every point inside the level set.

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2) Didn't seem to work because I always ended up squaring too many terms to get to a constant. The Jacobian matrix ended up degenerate in too many places to do those by hand, and I also ended up including points in the level set that are not part of the manifold.

Now I am trying to go with approach (1). I can write down rules for the required functions, but I am not sure how to find neighborhoods that work.

For example, if I start with a point p=F(x,y,z) where z2 is not 0 or 1/2, I get

$$a = \frac{2c^2 - 2 d^2}{1 \pm \sqrt{1 - 4 (c^2 + d^2)}}, b = \frac{2cd}{1 \pm \sqrt{1 - 4 (c^2 + d^2)}}$$

where I use a plus in the denominator if z2 < .5 and a minus if z2 > .5. Once I have a and b in terms of c and d, I could pick a map as follows:

$$g(a,b,c,d) = (a - \frac{2c^2 - 2 d^2}{1 \pm \sqrt{1 - 4 (c^2 + d^2)}}, b - \frac{2cd}{1 \pm \sqrt{1 - 4 (c^2 + d^2)}}, c,d)$$

This map returns (0,0,c,d) iff (a,b,c,d) was on the manifold in a small enough neighborhood of p. My problem is that I need to pin down an open set in R4 that does not accidentally contain a point on the manifold that is "on a different part" of the manifold than p.

Any suggestions? How do I find and verify my neighborhoods?

P.S. I also tried working in polar coordinates but I had a very hard time tracking cases and solving for variables in a couple of instances.

2. Oct 6, 2009

### wofsy

Direct computation seems tedious and difficult. Why not show that there are local inverses.

For instance, if z, x,and y are all non-zero and x and y have the same sign,one can find two inverse open regions on the sphere. The second coordinate tells you that they both have the same sign and the last two allow you to solve for their squares. Similarly, you get two inverse regions if x and y have opposite sign. these 4 cover most of the image. Since F is smooth and has these local inverses (you need to check that the local inverses are smooth) the image is a manifold in these regions. Now go for the cases where z is zero and one of x and y are zero.