Unproven statement in a book

[facenian]

I have a problem with an statement in the book Differential Geometry by Lipschultz(Schaum's outlines)
This book difines a coordinate patch as: A coordinate patch of class $C^m$ in $S\subset R^3$ is a mapping $\vec{x}=\vec{x}(u,v)$ of an open set $U\subset R^2$ into S such that:
(i) $\vec{x}:U\subset R^2\rightarrow S\subset R^3$ is of class $C^m$ on U
(ii) $\vec{x}_u\wedge\vec{x}_v\neq 0$ for all (u,v) in U.
(iii) $\vec{x}$ is 1-1 and bicontinuous on U, that is, its inverse is also continous.
Then he defines a simple surface in $R^3$ as a set S of points and a collection $\mathcal{B}$ of coordinates patches satisfying:
(1) $\mathcal{B}$ covers S
(2) Every coordinate pacht is the intersecction of an open set O in $E^3$ with S, $\vec{x}(U)=O\cap S$

Notice that condition (2) is not demanded for a coordinate patch. Later in the book one can find the following satatement:
(3) It can be shown that if $\vec{x}=\vec{x}(u,v)$ is a coordinate patch on a simple surface S and P is a point on $\vec{x}=\vec{x}(u,v)$, then there exists a spherical neighborhood S(P) in $E^3$ such that the intersecction of S(P) with the surface S is contained in the pathc $\vec{x}=\vec{x}(u,v)$.

Then he uses (3) to show that every patch on a simple surface is the intersection of an open set in $R^3$ with S which is easy to prove. However he does not provide a proof for (3).

Does someone know how (3) can be proven in the context of these definitions?

 

[Stephen Tashi]

With the caveat that I know practically nothing about differential geometry, I have some comments on material as presented in the post.

This book difines a coordinate patch as: A coordinate patch of class $C^m$ in $S\subset R^3$ is a mapping ...

So a coordinate patch is kind of mapping.

Then he defines a simple surface in $R^3$ as a set S of points and a collection $\mathcal{B}$ of coordinates patches satisfying:
(1) $\mathcal{B}$ covers S
(2) Every coordinate pacht is the intersecction of an open set O in $E^3$ with S, $\vec{x}(U)=O\cap S$

So a coordinate patch is now an open set? Perhaps what is meant is that the codomain of each coordinate patch is an open set O.

Notice that condition (2) is not demanded for a coordinate patch.

However it is demanded for each coordinate patch in $\mathcal{B}$.

However he does not provide a proof for (3).

If he is proving something about a "simple surface", the simple surface is defined by a collection of coordinate patches $\mathcal{B}$, each of which does satisfy (2).

An awkward consequence of such a definition of "simple surface" is that two different ways of defining a collection of patches for the unit sphere would, technically, result in two different simple surfaces.

 

[facenian]

So a coordinate patch is kind of mapping.
So a coordinate patch is now an open set? Perhaps what is meant is that the codomain of each coordinate patch is an open set O.

Yes, you are right. It is an unfortunate consequence of the nomenclature he uses. I think that it is implicit that both terms refer to two different concepts depending on the context used.

However it is demanded for each coordinate patch in $\mathcal{B}$.

Yes, that's right. Had he demanded by definition that ech coordinate patch should have this property the problem would had been resolved.

If he is proving something about a "simple surface", the simple surface is defined by a collection of coordinate patches $\mathcal{B}$, each of which does satisfy (2).
An awkward consequence of such a definition of "simple surface" is that two different ways of defining a collection of patches for the unit sphere would, technically, result in two different simple surfaces.

That is my mistake. He does not define a simple surface in that way. The definition states: "Then S together with the totality of coordinate patches of Class S is a simple surface of class $C^m$ in $E^3$
And again the problem I'm having is this: according to his definition the coordinates patches that are not in $\mathcal{B}$ do not have property (2) by definiton but then he proves they do using (3). This is easy to do however he do not provide a proof for (3).

I would be nice to hear a some opinion from a mathematician how already knows this book.

 

[lavinia]

What proof have you tried?

Note that axiom (iii) says that for any patch, the inverse image an open set in the plane is an open set in the image of the patch. That is: if $U→X(U)$ is a patch and $V⊂U$ is an open subset of $U$, then $X(V)$ is open in $X(U)$. (This is because $X$ is 1-1 and $X^{-1}$ is continuous as a map from $X(U)$ into the plane.) This means that there is an open set in $R^3$ whose intersection with $X(U)$ is equal to $X(V)$. But a priori, this open set might intersect the surface $S$ in points outside of $X(U)$. You must show that the open set can be chosen to intersect only $X(U)$.

 

[facenian]

You must show that the open set can be chosen to intersect only $X(U)$.

That is correct, and to show this he uses (3). After proving (3) it is easy to show that $X(U)$ is the intersection of S with an open set of $E^3$. So what I can not pove is (3) and curiously enough the book does not prove it either.

 

[lavinia]

That is correct, and to show this he uses (3). After proving (3) it is easy to show that $X(U)$ is the intersection of S with an open set of $E^3$. So what I can not pove is (3) and curiously enough the book does not prove it either.

Yes I know. U

Unless I am missing something:

Suppose the point $p$ is in a chart $V→Y(V)$. It is also in a chart $X→X(U)$ in $\mathcal B$ since these charts cover the entire surface . One has an open set whose intersection with $Y(V)$ is $Y(V)$ and another open set whose intersection with the surface is $X(U)$. Their intersection is an open neighborhood of $p$ whose intersection with the surface is $X(U)∩Y(V)$.

- Suppose one did not assume that the inverse of a patch is continuous. Is it still possible to prove the theorem?

 

[facenian]

Suppose the point $p$ is in a chart $V→Y(V)$. It is also in a chart $U→X(U)$ in $\mathcal B$ since these charts cover the entire surface . One has an open set whose intersection with $Y(V)$ is $Y(V)$ and another open set whose intersection with the surface is $X(U)$. Their intersection is an open neighborhood of $p$ whose intersection with the surface is $X(U)∩Y(V)$.

I don't think it will do because from these arguments we have: $O_1\cap Y(V)=Y(V)$ and also $O_2\cap S=X(U)$ so $(O_1\cap O_2)\cap Y(V)\cap S=Y(V)\cap X(U)$
However what we need is $$(O_1\cap O_2)\cap S\subset Y(V) $$

It is very supicious that the book only states "it can be shown that" because it normaly includes this kind of detail.

 

[lavinia]

I don't think it will do because from these arguments we have: $O_1\cap Y(V)=Y(V)$ and also $O_2\cap S=X(U)$ so $(O_1\cap O_2)\cap Y(V)\cap S=Y(V)\cap X(U)$
However what we need is $$(O_1\cap O_2)\cap S\subset Y(V) $$

It is very supicious that the book only states "it can be shown that" because it normaly includes this kind of detail.

$O_2∩S=X(U)$ and $O_1∩O_2$ is a subset of $O_2$ so Its intersection with $S$ is contained in $X(U)$. It is also a subset of $O_1$ so its intersection with $Y(V)$ is contained in $Y(V)$. So the intersection of $O_1∩O_2$ with $S$ is a subset of the intersection of $X(U)$ with $Y(V)$ which is a subset of $Y(V)$.

 

[facenian]

$O_2∩S=X(U)$ and $O_1∩O_2$ is a subset of $O_2$ so Its intersection with $S$ is contained in $X(U)$.

Yes, we have $O_2\cap S=X(U)\rightarrow O_1\cap O_2\cap S=O_1\cap X(U)\subset X(U)\ldots (1)$

It is also a subset of $O_1$ so its intersection with $Y(V)$ is contained in $Y(V)$.

Also correct, because $O_1\cap Y(V)=Y(V)\rightarrow (O_2\cap O_1)\cap Y(V)=O_2\cap Y(V)\subset Y(V)\ldots (2)$

So the intersection of $O_1∩O_2$ with $S$ is a subset of the intersection of $X(U)$ with $Y(V)$ which is a subset of $Y(V)$.

This last statement does not follow from (1) and (2), i.e., we cannot assert $(O_1\cap O_2)\cap S\subset X(U)\cap Y(U)$

 

[lavinia]

I see. Yes you are right.