Is a (smooth) manifold allowed to have different dimensions in different points.

[alemsalem]

obviously in one coordinate neighborhood it can't..
I'm thinking of a line which smoothly develops into a surface : -----<<

what particular properties would this object have..

Thanks :)

 

[quasar987]

No, this can't happen: the property of a manifold to be locally n-euclidean around a given point is both open and closed, so if a manifold is locally n-euclidean in a nbhd of a point, then it is locally n-euclidean everywhere in the connected component containing that point.

So a manifold can't morph from 2-d to 1-d. Of course, it can have a connected component of dimension 2 and another of dimension 1.

 

[lavinia]

obviously in one coordinate neighborhood it can't..
I'm thinking of a line which smoothly develops into a surface : -----<<

what particular properties would this object have..

Thanks :)

manifolds can have boundaries and corners.

 

[klackity]

You couldn't have a "line which smoothly develops into a surface".

Suppose for the sake of contradiction such a manifold, M, existed. Then about some point p, there would be a 1 dimensional coordinate chart, and about some point q, a 2 dimensional coordinate chart.

Now, by "smoothly develops into a surface", I assume you mean that the manifold is path-connected. So let g(t) be a curve connecting p and q. i.e., g(0) = p and g(1) = q.

Now, as [0,1] is compact, its image g([0,1]) is compact in M. However, about every point x in the image g([0,1]) there is an open neighborhood U(x) which is homeomorphic to either R^1 or R^2. As the image g([0,1]) is compact, only a finite number of these open sets U(x) suffice to cover g([0,1]).

But then we have reached a contradiction, because in that finite subcollection of open sets, there must be two partially overlapping neighborhoods, one homeomorphic to R^1, the other to R^2. This is a contradiction, because their intersection (or any open set for that matter) cannot possibly be homeomorphic to both R^1 and R^2

 

[alemsalem]

Thanks...

 

[Bacle]

I think this would lead you to an invariance of domain problem, i.e., you would end up with a copy of R^n homeomorphic to a copy of R^m for m=/n , on chart overlaps.

 

[lavinia]

obviously in one coordinate neighborhood it can't..
I'm thinking of a line which smoothly develops into a surface : -----<<

what particular properties would this object have..

Thanks :)

One can certainly create surfaces from a one parameter family of lines. They would have a straight line through every point. Depending on how this is done, the surface could have points of self intersection, cusps, places where it it not a 2 dimensional manifold.

For instance the straight lines in the direction of a curve in Euclidean 3 space's unit normal might span a surface that fails to be 2 dimensional at some points..