Definition of a manifold

[RedX]

I always thought one could define a manifold as a collection of points with a distance function or metric tensor.

But in a layperson's book by Penrose, he defined a manifold as a collection of points with a rule for telling you if a function defined on the manifold is smooth. He says this is more general than a collection of points with a local structure such as a metric tensor.

Can someone give me an example of a manifold defined in this more general way? I always thought smoothness required a distance function.

Also I'm also a little confused about the definition of smoothness. Penrose says a sphere is smooth, but a cube is not. I thought smoothness referred to functions defined on the manifold, not the manifold itself. Can't you have an unsmooth function defined on a smooth manifold (e.g., a function that goes to infinity on the sphere), or a smooth function defined on an unsmoothed manifold (e.g., the constant function on the cube).

Also, isn't the definition of a collection of points with a metric tensor also the same as the definition of a metric space? So what is the difference between a manifold and a metric space?

 

[Studiot]

The point of manifolds such as Penrose is dealing with is the we want to be able to do calculus on them. This is where the smoothness comes in.

A simple example is the modulus function which can be regarded as a 1 dimensional manifold, but is not smooth since it is not differentiable at the origin. This is a simpler equivalent to your sphere/cube example.

 

[quasar987]

I always thought one could define a manifold as a collection of points with a distance function or metric tensor. [...] Also, isn't the definition of a collection of points with a metric tensor also the same as the definition of a metric space? So what is the difference between a manifold and a metric space?

Another name for a "distance function" d:X x X-->R on a set X is a "metric", but besides the similarity in the name, this has very little do to with the notion of a "metric tensor" on a manifold. The latter consists, in laymen's term, of choice of a scalar product function $<\cdot,\cdot>_p=g_p:T_pM \times T_pM\to \mathbb{R}$ on each tangent space $T_pM$ of the manifold M. In precise mathematical terms, it is a smooth section g of the tensor bundle $T^*M\otime T^*M$ such that for each point p in the manifold M, the bilinear form $g_p:T_pM x \times T_pM\to \mathbb{R}$ is symmetric and positive definite (i.e., it is a scalar product!)

But in a layperson's book by Penrose, he defined a manifold as a collection of points with a rule for telling you if a function defined on the manifold is smooth. He says this is more general than a collection of points with a local structure such as a metric tensor.

Can someone give me an example of a manifold defined in this more general way? I always thought smoothness required a distance function.

A manifold is a collection of points that looks locally like R^n (i.e., that is locally homeomorpohic to R^n). And in R^n, we know how to make sense of differentiability. So the "rule" for telling you if a function defined on the manifold is smooth that Penrose is alluding to is simply this: For a function f:M-->R, to determine if f is differentiable at a point p in M, let U be a little neighborhood of p and h:U-->R^n be a homeomorphism. Then we say f is differentiable at p if, by definition, the composite map f o h:R^n-->R is differentiable (in the usual sense).

However, there is an ambiguity with this definition because it could be that even though f o h is differentiable at p, for h':U-->R^n another homeomorphism, f o h' is not differentiable at p! So what now? Is f differentiable at p or not? According to h it is but according to h', it is not! So actually, we define a smooth manifold as a pair (M,A), where M is a set of points that look locally like R^n, and A is a collection of homeomorphisms A= {h:U-->R^n} where all the functions h agree with each other as to whether any given function f:M-->R is differentiable or not. Precisely, we ask that the functions h:U-->R in A be such that for any two h,h' of them, h^{-1} o h' and h'^{-1} o h be smooth.

If such a collection A can be found for the set of points M, then M is called a smooth manifold (because the notion of a "smooth map" f:M-->R can be defined unambiguously).

Also I'm also a little confused about the definition of smoothness. Penrose says a sphere is smooth, but a cube is not. I thought smoothness referred to functions defined on the manifold, not the manifold itself.

I hope the text above makes it clear that there is a notion of smoothness for manifolds and that some manifolds are smooth (a set A as above exists: smoothness of maps can be defined consistently) while others are not (no such A as above exists: smoothness of maps cannot be defined consistently). An example of a manifold that is not smooth is the cube.

 

[Bacle]

Quasar Wrote:
"
Originally Posted by RedX
Also I'm also a little confused about the definition of smoothness. Penrose says a sphere is smooth, but a cube is not. I thought smoothness referred to functions defined on the manifold, not the manifold itself.

I hope the text above makes it clear that there is a notion of smoothness for manifolds and that some manifolds are smooth (a set A as above exists: smoothness of maps can be defined consistently) while others are not (no such A as above exists: smoothness of maps cannot be defined consistently). An example of a manifold that is not smooth is the cube.

"

But there seems to be something strange here; specifically, this result:


Given topological spaces X,Y , if there exists a homeomorphism h between X and Y

and Y (equiv. X ) admits a smooth structure, then so does X. We basically use the

homeomorphism h to pull-back the smooth structure on Y into the space X.

In this case (use X= Cube , Y=Closed Ball , and smoothly deform the cube into the

closed ball) that would imply that a cube can be given a smooth structure.


I think the best we can say is that the cube is not a smooth submanifold of

R^3.

 

[quasar987]

Yes you're absolutely right... I didn't think when I wrote that, I just assumed "Penrose is right."

So, just to summarize:
1. The cube does not admit a smooth manifold structure; not because "not such A as in post #2 exists", but rather, because there is a topological obstruction: the cube is not locally homeomorphic to R^3 (there are boundary points).

2. There is a more general notion of manifold, which is that of a manifold with boundary, in which the space is only required to be locally homeomorphic of an open subset of the "half-space" $\mathbb{H}^n:=\{(x_1,...,x_n) \in \mathbb{R}^n | x_n\geq 0\}$. And there is a notion of smoothness for manifolds with boundary which is the natural generalization of smoothness for ordinary manifolds. The closed ball is an example of a smooth manifold with boundary, and the cube being homeomorphic to the closed bass, it admits the structure of a smooth manifold with boundary. However with the smooth structure, the cube C, as a subset of R^3, is not an embeded submanifold. That is, it is not the image by a smooth embedding of any smooth manifold with boundary. The problem is the "corner points" of C at which differentiability fails.

3. It is worth mentioning also that there is an even more general notion of smooth manifold which is that of smooth manifold with corners an example of which is the cube. And as such, the cube now is a smooth embedded submanifold of R^3. (This follows immediately from the definition of smooth manifold with corners once you've put the obvious smooth manifold with corners structure on the cube).

 

[Studiot]

The problem is the "corner points" of C at which differentiability fails.

That's what I said in more babyish terms.

 

[Tinyboss]

Yeah, as Bacle said, we can "transfer" a smooth structure on a manifold M to any topological space N which is homeomorphic to M. Just compose the homeomorphisms. Is this from "The Road to Reality", page 181? I haven't looked at this book since I really started studying math...now that I read this part again, it seems like a terrible example.

According to the textbook (John Lee, Intro to Smooth Manifolds) I'm working from now:

In all dimensions greater than 3 there exist compact manifolds that admit no smooth structure, but every topological manifold (compact or not) of dimension 3 or less has a smooth structure (unique up to diffeomorphism in fact). So I guess Penrose is out of luck if he wants to give an intuitive example of a manifold that isn't smooth. ;-)

 

[RedX]

Thanks everyone. Yeah this is from page 181 of "The Road to Reality".

I'll think about all these answers, and see if the library has that textbook you're using.

 

[Bacle]

As a general comment, I think we can also say that the cube, with the standard embedding in R^n, does not have the submanifold property, meaning that we
cannot find local charts (specifically in the corner points ) (Phi,U) , sending the
neighborhood U to a subspace neighborhood in R^n. I remember seeing this in
Boothby's book. I will look it up , to make it more precise.

 

[zhentil]

Yes you're absolutely right... I didn't think when I wrote that, I just assumed "Penrose is right."

So, just to summarize:
1. The cube does not admit a smooth manifold structure; not because "not such A as in post #2 exists", but rather, because there is a topological obstruction: the cube is not locally homeomorphic to R^3 (there are boundary points).
.

I'm a bit confused by this, since the cube is certainly homeomorphic to a sphere. Where the manifold is embedded has nothing to do with whether it possesses a smooth structure; one would be hard-pressed to define a "cube" without referencing an ambient space.

The (topological) embedding of a 2-sphere into R^3 (as the cube) is not a smoothly embedded submanifold with the natural smooth structure on R^3, since one has to make derivatives vanish at the corners. I think this is what people are trying to say.

 

[quasar987]

In this thread, cube has always been meant to mean "solid cube". As in, the thing homeomorphic to the closed ball.

 

[zhentil]

In this thread, cube has always been meant to mean "solid cube". As in, the thing homeomorphic to the closed ball.

My apologies. I was getting a bit confused by all the talk of "corners." In any event, the obstruction for the cube is the same as the solid ball - topologically, we can't tell them apart. The corners have nothing to do with it, since topologically, we can't "see" corners.

 

[George Jones]

Penrose says a sphere is smooth, but a cube is not.

In this thread, cube has always been meant to mean "solid cube". As in, the thing homeomorphic to the closed ball.

Unfortunately, this is not what Penrose means.

For an intuitive notion of what a 'smooth' manifold is, think of a sphere as opposed to a cube (where, of course, in each case I am referring to the surace and not the interior.)

The (topological) embedding of a 2-sphere into R^3 (as the cube) is not a smoothly embedded submanifold with the natural smooth structure on R^3, since one has to make derivatives vanish at the corners. I think this is what people are trying to say.

Yes, I think Penrose means something like this. On page 181, Penrose talks about 2-dimensional surfaces, and he doesn't introduce general manifolds until a little later.

 

[quasar987]

Ok, well empty cube or solid cube, the situation is pretty much the same:

The empty cube (as it is homeomorphic to the 2-sphere) can be endowed with a smooth structure, but the empty cube sitting in R^3 (usual topological embedding) is not a smooth submanifold because differentiability fails at the "corners".

The solid cube (as it is homeomorphic to the 3-ball) can be endowed with the structure of a smooth manifold with boundary, but the solid cube sitting in R^3 (usual topological embedding) is not a smooth submanifold with boundary because differentiability fails at the "corners".

 

[Studiot]

I return to what I said in post#2

A topological space is not the same as a metric space.
In particular the defining axioms do not require or even mention a metric.

Penrose wants a calculus on his space, so he needs a metric, so he has to distinguish.
You need to be careful applying results from topological spaces to less general ones.

 

[Couchyam]

Three equivalent definitions of a manifold:

A set of points M $$\subset$$ $$\Re^{n}$$ is a k dimensional manifold if;

1) $$\forall$$x$$\in$$M, $$\exists$$ an open neighborhood W such that

p$$\in$$W and

$$\forall$$x $$\in$$ M$$\cap$$W,

x is in the graph of a smooth function g: V$$\Rightarrow$$$$\Re^{n-k}$$,
where V $$\subset$$ $$\Re^{k}$$ is an open set. (We are allowed to choose any k integers 1$$\leq$$i$$_{1}$$$$\leq$$n etc..

you can think of this definition as just saying that the "manifold" can be expressed as the image of a smooth function from $$R^{k}$$ to $$R^{n}$$ with a few more restrictions (the difference between this definition and the one mentioned above is that in the above definition, the remaining coordinates remain in the graph of the function, while in the parametric description this isn't necessarily the case).

Manifolds can be expressed graphically, parametrically, and as the level curve to some function from $$\Re^{n}$$ to $$\Re^{n-k}$$ (or equivalently as the preimage of 0 for some other function)

There are however some important technical issues to be aware of when classifying a shape as a manifold, and the restrictions of the definitions are important. Note: this is obviously for smooth manifolds

 

[DrRocket]

Another name for a "distance function" d:X x X-->R on a set X is a "metric", but besides the similarity in the name, this has very little do to with the notion of a "metric tensor" on a manifold. The latter consists, in laymen's term, of choice of a scalar product function $<\cdot,\cdot>_p=g_p:T_pM \times T_pM\to \mathbb{R}$ on each tangent space $T_pM$ of the manifold M. In precise mathematical terms, it is a smooth section g of the tensor bundle $T^*M\otime T^*M$ such that for each point p in the manifold M, the bilinear form $g_p:T_pM x \times T_pM\to \mathbb{R}$ is symmetric and positive definite (i.e., it is a scalar product!)

In this case, where the inner product is positive-definite, you have a classical Riemanian manifold, and can use that bilinear form to define a distance function between points. It is a fundamental theorem of Riemannian geometry that the distance function so defined also defines the underlying topology of the manifold, which is therefore a topological metric space. That is the connection between the metric tensor and a topological metric.

In the case of a pseudo-Riemannian manifold (as in the general theory of relativity), in which the metric tensor is only assumed to be non-degenerate, the connection no longer holds.

 

[RedX]

In this case, where the inner product is positive-definite, you have a classical Riemanian manifold, and can use that bilinear form to define a distance function between points. It is a fundamental theorem of Riemannian geometry that the distance function so defined also defines the underlying topology of the manifold, which is therefore a topological metric space. That is the connection between the metric tensor and a topological metric.

In the case of a pseudo-Riemannian manifold (as in the general theory of relativity), in which the metric tensor is only assumed to be non-degenerate, the connection no longer holds.

So are you saying that for Riemannian space (i.e., positive definite distance), the metric contains the complete information about the manifold (i.e., all the coordinate charts and the set of points)? So if you are given a metric, then that uniquely determines the manifold?

Also, I was wondering if the metric tensor can be defined as a way to convert vectors in the tangent space to vectors/forms in the cotangent space instead of being defined as a bilinear operation that takes two tangent vectors or two cotagent vectors into a distance? Because as I understand it there is a natural dot product if you have a vector (a,b,c) and a covector (x,y,z): ax+by+cz. This holds even if your manifold is not equipped with a metric tensor (assuming you define your vector basis as the partial derivatives in your coordinate chart and your covector basis as the gradient of the coordinates). So is a metric more about converting vectors to covectors than it is about defining a distance?

Lastly, I know that mathematicians like to have positive-definite distance functions, but in physics as you mentioned that requirement has to be relaxed. Does that mean most mathematicians don't work with distance functions that can be negative?

 

[Bacle]

Dr. Rocket wrote:

"In this case, where the inner product is positive-definite, you have a classical Riemanian manifold, and can use that bilinear form to define a distance function between points. It is a fundamental theorem of Riemannian geometry that the distance function so defined also defines the underlying topology of the manifold, which is therefore a topological metric space. That is the connection between the metric tensor and a topological metric.

In the case of a pseudo-Riemannian manifold (as in the general theory of relativity), in which the metric tensor is only assumed to be non-degenerate, the connection no longer holds. "

I am a little confused here: the inner-products are defined in each of the tangent
spaces, and, AFAIK, they induce a metric on each tangent space in the usual way,
i.e., for any vector X_p in T_pM , ||X||=<X,X>_p.


But I thought the global metric in M was given by the shortest path ( seeing

the manifold as a Length Space.) , by the formula:

d(x,y)=inf. L {{ all paths joining x,y }}

where L is a length function satisfying some properties. But I cannot see how

L can be defined by using the Metric Tensor. Would you Comment.?

 

[Bacle]

Dr. Rocket wrote, in Part:


"Also, I was wondering if the metric tensor can be defined as a way to convert vectors in the tangent space to vectors/forms in the cotangent space instead of being defined as a bilinear operation that takes two tangent vectors or two cotagent vectors into a distance? Because as I understand it there is a natural dot product if you have a vector (a,b,c) and a covector (x,y,z): ax+by+cz. This holds even if your manifold is not equipped with a metric tensor (assuming you define your vector basis as the partial derivatives in your coordinate chart and your covector basis as the gradient of the coordinates). So is a metric more about converting vectors to covectors than it is about defining a distance? "

I think there are such maps, even for Pseudo-Riemannian Manifolds; they are
called 'Flat' , and 'Sharp'.

 

[Bacle]

Sorry For Previous:

I was referring to RedX's post, not Dr. Rocket's.

 

[quasar987]

So are you saying that for Riemannian space (i.e., positive definite distance)

Positive-definite bilinear map. The metric tensor is not a distance function on the manifold; it is a mean of assigning lengths and angles to tangent vectors.

So are you saying that for Riemannian space (i.e., positive definite distance), the metric contains the complete information about the manifold (i.e., all the coordinate charts and the set of points)? So if you are given a metric, then that uniquely determines the manifold?

Given a metric tensor, you can assign a length (norm) to tangent vectors. You can then calculate the length of a path between points by integrating the speed of a parametrization, just as you are used to do in R^n. And then you can define a distance function d:M x M-->R on the manifold (which by the way, is always non negative by definition) by setting d(p,q)="length of the shortest path between p and q". Now as we know any distance function induces a topology and it turns out that the topology induced by the distance function above coincides with the topology of the manifold.

Also, I was wondering if the metric tensor can be defined as a way to convert vectors in the tangent space to vectors/forms in the cotangent space instead of being defined as a bilinear operation that takes two tangent vectors or two cotagent vectors into a distance?

Again, the metric tensor does not assign a distance to vectors, only a norm and an angle (between two vectors). Is is a dot/scalar product on each tangent space before being a norm or a distance function.

Your question is quite vague. What do you mean by "a way to convert vectors to covectors"? The way the metric tensor does the tangent-cotangent pairing is like so. Given p in M, the fact that the bilinear form $g_p:T_pM\times T_pM\rightarrow \mathbb{R}$ is nondegenerate (meaning for all nonzero $v\in T_pM$, there exists another vector $w\in T_pM$ such that $g_p(v,w)\neq 0$, namely w=v in this case!) is equivalent to the map $T_pM\rightarrow T^*_pM:v\mapsto g_p(v,\cdot)$ being an isomorphism (indeed, nondegeneracy means precisely that the kernel of this map is trivial). So actually any field of nondegenerate bilinear maps on a manifold induces an isomorphism between tangent and cotangent spaces. In particular, this property does not characterize metric tensors.

 

[quasar987]

Dr. Rocket wrote:

"In this case, where the inner product is positive-definite, you have a classical Riemanian manifold, and can use that bilinear form to define a distance function between points. It is a fundamental theorem of Riemannian geometry that the distance function so defined also defines the underlying topology of the manifold, which is therefore a topological metric space. That is the connection between the metric tensor and a topological metric.

In the case of a pseudo-Riemannian manifold (as in the general theory of relativity), in which the metric tensor is only assumed to be non-degenerate, the connection no longer holds. "

I am a little confused here: the inner-products are defined in each of the tangent
spaces, and, AFAIK, they induce a metric on each tangent space in the usual way,
i.e., for any vector X_p in T_pM , ||X||=<X,X>_p.


But I thought the global metric in M was given by the shortest path ( seeing

the manifold as a Length Space.) , by the formula:

d(x,y)=inf. L {{ all paths joining x,y }}

where L is a length function satisfying some properties. But I cannot see how

L can be defined by using the Metric Tensor. Would you Comment.?

$$L(\gamma)=\int_{t_0}^{t_1}\|\dot{\gamma}(t)\|dt=\int_{t_0}^{t_1}\sqrt{g_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t))}dt$$

(the exact analogue of the case of R^n with g the euclidean dot product)

 

[DrRocket]

Dr. Rocket wrote:

"In this case, where the inner product is positive-definite, you have a classical Riemanian manifold, and can use that bilinear form to define a distance function between points. It is a fundamental theorem of Riemannian geometry that the distance function so defined also defines the underlying topology of the manifold, which is therefore a topological metric space. That is the connection between the metric tensor and a topological metric.

In the case of a pseudo-Riemannian manifold (as in the general theory of relativity), in which the metric tensor is only assumed to be non-degenerate, the connection no longer holds. "

I am a little confused here: the inner-products are defined in each of the tangent
spaces, and, AFAIK, they induce a metric on each tangent space in the usual way,
i.e., for any vector X_p in T_pM , ||X||=<X,X>_p.


But I thought the global metric in M was given by the shortest path ( seeing

the manifold as a Length Space.) , by the formula:

d(x,y)=inf. L {{ all paths joining x,y }}

where L is a length function satisfying some properties. But I cannot see how

L can be defined by using the Metric Tensor. Would you Comment.?

The metric tensor defines an innerproduct on the tangent bundle, so an innerproduct on tangent vectors at each point, i.e. on each tangent space. So, given a smooth path you can calculate the tangent vector at each point, and using the metric determine its norm. The integral of the norm of the tangent vectors over the path is the length of the path. The distance between two points is the minimum of the lengths of the paths joining the two points. That defines a metric in the topological sense. That metric can be shown to yield the underlying topology of the manifold.

Edit: I see that quasar 987 has answered this question in the same way with a bit more detail.

 

[DrRocket]

So are you saying that for Riemannian space (i.e., positive definite distance), the metric contains the complete information about the manifold (i.e., all the coordinate charts and the set of points)? So if you are given a metric, then that uniquely determines the manifold?

In the case of Riemannian manifolds one does impose the condition that the metric preserve inner products under parallel translation. Given that. then the metric tensor determines a topological metric that defines the underlying topology.

If you now have another manifold that is isometric to the original manifold as metric spaces, then they are certainly homeomorphic. So, yes, in that sense the metric determines the manifold in the category of topological spaces. If you look in the category of smooth manifolds then you have another kettle of fish, since the map may not preserve differentiable structures and there are exotic structures on some manifolds.

But be careful. There are spaces that are locally isometric that are quite different topologically. A right circular cylinder is locally isometric to the plane, but not homeomorphic.

Also, I was wondering if the metric tensor can be defined as a way to convert vectors in the tangent space to vectors/forms in the cotangent space instead of being defined as a bilinear operation that takes two tangent vectors or two cotagent vectors into a distance? Because as I understand it there is a natural dot product if you have a vector (a,b,c) and a covector (x,y,z): ax+by+cz. This holds even if your manifold is not equipped with a metric tensor (assuming you define your vector basis as the partial derivatives in your coordinate chart and your covector basis as the gradient of the coordinates). So is a metric more about converting vectors to covectors than it is about defining a distance?

A metric is a bilinear form, a 2-tensor field, on the tangent bundle. You can define other tensor fields that would do what you suggest.

Lastly, I know that mathematicians like to have positive-definite distance functions, but in physics as you mentioned that requirement has to be relaxed. Does that mean most mathematicians don't work with distance functions that can be negative?

Mathematician study both cases.

The case in which the metric tensor is positive-definite is called Riemannian geometry.

The case in which the metric tensor is merely assumed to be non-degenerate is called semi-Riemannian geometry or pseudo-Riemannian geometry. But just to cloud things, the motivation for the study of Riemannina geometry has included a large dose of general relativity, so some authors include the semi-Riemannian case when they write of Riemannian geometry. The only sure-fire way to handle the terminology is to make certain that the definitions are clear up front.

In the semi-Riemanian case the construction of a topological metric from the metric tensor fails because it is possible to have distince points joined by a path of zero length. That is exactly the case in general relativity for the world lines of photons -- so-called null geodesics.

Caveat: I am an analyst, not a geometer so you will have no trouble asking questions that I cannot answer.