# chart of 3-ball

• Oct 26th 2007, 03:26 AM
mrandersdk
chart of 3-ball
if I have the unit ball in R^3, that is

{(x,y,z) | x^2+y^2+z^2 <= 1 } = B

and i have to find a chart, can then just use the chart

(B,f)

where f(x,y,z) = (x,y,z) for (x,y,z) in B. That is my first question. My next question is if I can just do this, what about if I identify (x,y,z) with (-x,-y,-z), then i gues my chart have to illustrate this by sending (x,y,z) and (-x,-y,-z) to the same point, ie.

f(-x,-y,-z) = f(x,y,z)

is that correct that the way to interpreted that to point are the same, that the chart have to do that. Ps. I don't need anyone to give me the charts, (i like to try and find them myself) but I just eanted to make sure I'm going the right way.

• Oct 26th 2007, 10:56 AM
ticbol
f(-x,-y,-z) is the point on the other octant opposite to where f(x,y,z) is.
Or, f(-x,-y,-z) is the mirror image of f(x,y,z) relative to the center of B.
• Oct 26th 2007, 11:01 AM
mrandersdk
I don't understand what you are saying, I have the point (x,y,z) and then I say that this is equivalent to (-x,-y,-z). I know where the points are, but do my charts on the manifold have to fullfil:

f(x,y,z) = f(-x,-y,-z)

so I just can't use f(x,y,z)=(x,y,z) because that doesn't fulfill this requirment.
• Oct 26th 2007, 11:12 AM
ticbol
I also don't understand what you are saying that f(-x,-y,-z) is equal to f(x,y,z). Or why point (x,y,z) is equivalent to point (-x,-y,-z). Why are two different points equivalent to each other.
• Oct 27th 2007, 01:19 AM
mrandersdk
ok, maybe my question is a bit unclear. I have two manifolds one (M1) consisting of the unit ball i R^3, and one (M2 )consisting of the unit ball in R^3 where opposite points on the surface are equivalent. That is

(x,y,z) ~ (-x,-y,-z)

for points on the surface, so thats why they are equivalent, because the manifold is constructed in that way.

My question is, for the first manifold is a chart just

f: M1 -> R^3, where f(x,y,z) = (x,y,z) for (x,y,z) in M1

And is the reason why I can't use this chart for M2 that it doesn't fullfil

f(x,y,z) = f(-x,-y,-z)

and my chart have to do this, because these points are equivalent.
• Oct 28th 2007, 01:31 AM
Opalg
Quote:

Originally Posted by mrandersdk
if I have the unit ball in R^3, that is

{(x,y,z) | x^2+y^2+z^2 <= 1 } = B

and i have to find a chart, can then just use the chart

(B,f)

where f(x,y,z) = (x,y,z) for (x,y,z) in B.

The mapping f is fine, but you need to be careful about its domain. You cannot use the whole of B as the domain, because a chart has to be a homeomorphism, and f is not 1-1 on B. So you need to subdivide B into overlapping regions on which f is 1-1, and use these for the domains of the charts.

For the manifold M2 you can use the same charts as for M1, but the way that they fit together (given by the transition functions) will be different.
• Oct 28th 2007, 01:55 AM
mrandersdk
Not quite sure why my f is not 1-1 on the first manifold. But I can however see that this is the problem on M2 because f sends (x,y,z) and (-x,-y,-z) to different points, so f^-1(x,y,z) do not equal f^-1(-x,-y,-z), but that have to be the same point.
• Oct 28th 2007, 04:35 AM
Opalg
Quote:

Originally Posted by mrandersdk
Not quite sure why my f is not 1-1 on the first manifold. But I can however see that this is the problem on M2 because f sends (x,y,z) and (-x,-y,-z) to different points, so f^-1(x,y,z) do not equal f^-1(-x,-y,-z), but that have to be the same point.

Oops, let's try again. (I shouldn't have attempted that first comment, early on a Sunday morning!)

What I failed to notice is that the manifolds M1 and M2 are surfaces, so they are two-dimensional, not three-dimensional. So the charts should be maps from B to R^2, not to R^3. For example, you could use the map f(x,y,z)=(x,y). Since (x,y,z) is in B, there are two values of z that go to the same point in R^2, namely $\displaystyle z=\pm\sqrt{1-x^2-y^2}$. That is why the map cannot be defined on the whole of B. In fact, you need to have each chart map defined on just one hemisphere of B.
• Oct 28th 2007, 04:38 AM
mrandersdk
you where right the first time, they are not surfaces, it is the unit ball in R^3, hence

{(x,y,z) | x^2+y^2+z^2 <= 1}

but I still can't se why my f for M1 isn't 1-1?
• Oct 28th 2007, 05:15 AM
Opalg
Quote:

Originally Posted by mrandersdk
you were right the first time, they are not surfaces, it is the unit ball in R^3, hence

{(x,y,z) | x^2+y^2+z^2 <= 1}

but I still can't se why my f for M1 isn't 1-1?

[No doubt about it, I can't think straight on a Sunday morning.]

Okay, in that case the identity map will do fine for M1. It is 1-1, and B is just embedded as a submanifold of R^3.

I'm inclined to back off at this point, before saying something else stupid. But if you want to get a geometric insight into what M2 looks like, then you should think about the two-dimensional analogue. You can visualise this as follows. Take a disc of paper, cut it along a radius, and fold it over on itself so as to make a paper hat, with two thicknesses, in which each point is glued to the one that was originally opposite to it on the disc.
• Oct 30th 2007, 11:47 AM
mrandersdk
thanks for your help. I have played a little with some charts and i think i have found some that works, but while i played with the charts I came to think about something:

If I have two spaces (not manifolds yet) M1 and M2, if I then make M1 to an manifold, with an atlas {(U_i,f_i)}, and I can find a bijection F: M2 -> M1, can I then make M2 to a manifold by making the atlas

{(F(U_i),g_i = f_i*F)} * means composition

I need to know this, because I have made the unit ball with the identification (x,y,z) ~ (-x,-y,-z) to a manifold, and found a bijection from that to SO(3).

My asignment is to show that the unit ball with the identification (x,y,z) ~ (-x,-y,-z) and SO(3) can be given a natural manifold structure, and they are diffeomorfic. So I thought i was done if I can just doo this, but even if this works I don't know if this is a natural way to do it, find this statement 'natural' a bit weak.