Is the Vector Space geometric?

[FallenApple]

Or is it something separate that acts on a geometric space? So we know that the Euclidean space is a vector space. But is it geometric? I ask this because in group theory, the group elements are the operators acting on another set, but clearly we see that this doesn't mean that the group elements are the set that it is acting on. For example, the permutation group elements are permutes a set of n objects. But those objects could anything.

So for the Euclidean space, do we need a preexisting geometric grid, and then say that we have separate vector space that operates on the points within the geometric space filled out between and on those geometric gridlines?

I mean, if we take away the object that it is acting on, the geometric grid, we just have vectors left, which is just abstract scaling of abelian group elements by field elements. That is, ((V,+),(F,+', *'), *) where * is the group action of F on V. Even if inner product is defined, I don't see how we get geometry from this. V is just a set of group objects themselves that can be viewed as operators which act on something else.

 

[fresh_42]

You are confusing various concepts and put them in a mixer in my opinion. For a geometry, we need a measure, usually an inner product which allows to measure angles and distances. Whether the underlying object is a vector space is at prior not of interest. E.g. we can do geometry on a curved manifold like earth.

Next a group does not have to operate on any set. Of course it always operates on itself via conjugation, but this is not how groups should be seen in the first place.

If a group operates on a certain set, then we have a representation of the group on named set, i.e. we try to find out properties of group elements by their action on this set. There is neither a vector space nor a geometry anywhere near. Of course there are cases of group actions on those spaces.

A vector space, if equipped with an inner product can serve as a space, where we can do geometry, i.e. we have angles and distances. As Euclidean spaces have a Euclidean metric, they are an example. There is no such thing as a geometric grid. It is the possibility to measure which counts.

 

[WWGD]

Or is it something separate that acts on a geometric space? So we know that the Euclidean space is a vector space. But is it geometric? I ask this because in group theory, the group elements are the operators acting on another set, but clearly we see that this doesn't mean that the group elements are the set that it is acting on. For example, the permutation group elements are permutes a set of n objects. But those objects could anything.

So for the Euclidean space, do we need a preexisting geometric grid, and then say that we have separate vector space that operates on the points within the geometric space filled out between and on those geometric gridlines?

I mean, if we take away the object that it is acting on, the geometric grid, we just have vectors left, which is just abstract scaling of abelian group elements by field elements. That is, ((V,+),(F,+', *'), *) where * is the group action of F on V. Even if inner product is defined, I don't see how we get geometry from this. V is just a set of group objects themselves that can be viewed as operators which act on something else.

In my experience you have a geometry defined once you have an inner-product in your space. If your vector space is isomorphic as a vector space to $\mathbb R^n$ then you can use the isomorphism to pullback/pushforward the inner-product from there. Once you have this, you can talk about angle, orthogonality, length, etc. witrhout it, I don't see how.

 

[WWGD]

You are confusing various concepts and put them in a mixer in my opinion. For a geometry, we need a measure, usually an inner product which allows to measure angles and distances. Whether the underlying object is a vector space is at prior not of interest. E.g. we can do geometry on a curved manifold like earth.

Next a group does not have to operate on any set. Of course it always operates on itself via left multiplication, but this is not how groups should be seen in the first place.

If a group operates on a certain set, then we have a representation of the group on named set, i.e. we try to find out properties of group elements by their action on this set. There is neither a vector space nor a geometry anywhere near. Of course there are cases of group actions on those spaces.

A vector space, if equipped with an inner product can serve as a space, where we can do geometry, i.e. we have angles and distances. As Euclidean spaces have a Euclidean metric, they are an example. There is no such thing as a geometric grid. It is the possibility to measure which counts.

Sorry, Herr Frisch, I just parroted what you wrote. Should I delete?

 

[fresh_42]

Sorry, Herr Frisch, I just parroted what you wrote. Should I delete?

Why? Because you invented to wit-rhout something?
Those topologists ...

 

[FallenApple]

You are confusing various concepts and put them in a mixer in my opinion. For a geometry, we need a measure, usually an inner product which allows to measure angles and distances. Whether the underlying object is a vector space is at prior not of interest. E.g. we can do geometry on a curved manifold like earth.

Next a group does not have to operate on any set. Of course it always operates on itself via conjugation, but this is not how groups should be seen in the first place.

If a group operates on a certain set, then we have a representation of the group on named set, i.e. we try to find out properties of group elements by their action on this set. There is neither a vector space nor a geometry anywhere near. Of course there are cases of group actions on those spaces.

A vector space, if equipped with an inner product can serve as a space, where we can do geometry, i.e. we have angles and distances. As Euclidean spaces have a Euclidean metric, they are an example. There is no such thing as a geometric grid. It is the possibility to measure which counts.

That makes sense. So a geometry rely on concepts of length/angle. That is why the mapping VxV->F is needed, for example, the dot product. And you can have a metric space without regard for an identity element, closure, associativity, inverses etc. So no regard is needed for symmetry between points on a raw metric space. It just so happens that the usual space we think of has symmetries because the set of points that is manifold has a symmetry which is a representation of the group on that manifold. If one can traverse continuously back and forth on a manifold, then the act of traversal has to form a group because it satisfies the group axioms.

The operation of traversing on a manifold is a local operation. Just like how addition is a local operation because it operates between two elements. But the geometric structure is the entire set taken as a whole, satisfying angle and local consecutiveness using metrics for the distance concept. So a geometry is more of a global concept that is comprised of the net of metric values, which are undirected edges of a graph and hence local, that make up the structure? And the symmetry on the manifold is global, but the operators that comprise that symmetry is local and the graph that makes up the net of the edges that represent operators could be directed.

But are there metric spaces that do not contain a representation of any group? I think we can always treat d(x,y) as an operator. For example, the isometry of a metric space to itself is a group. So that means that while a general metric space can represent the isometry group, there are metric spaces such as certain manifolds in which there could be multiple symmetries.

 

[WWGD]

That makes sense. So a geometry rely on concepts of length/angle. That is why the mapping VxV->F is needed, for example, the dot product. And you can have a metric space without regard for an identity element, closure, associativity, inverses etc. So no regard is needed for symmetry between points on a raw metric space. It just so happens that the usual space we think of has symmetries because the set of points that is manifold has a symmetry which is a representation of the group on that manifold. If one can traverse continuously back and forth on a manifold, then the act of traversal has to form a group because it satisfies the group axioms.

The operation of traversing on a manifold is a local operation. Just like how addition is a local operation because it operates between two elements. But the geometric structure is the entire set taken as a whole, satisfying angle and local consecutiveness using metrics for the distance concept. So a geometry is more of a global concept that is comprised of the net of metric values, which are undirected edges of a graph and hence local, that make up the structure? And the symmetry on the manifold is global, but the operators that comprise that symmetry is local and the graph that makes up the net of the edges that represent operators could be directed.

But are there metric spaces that do not contain a representation of any group? I think we can always treat d(x,y) as an operator. For example, the isometry of a metric space to itself is a group. So that means that while a general metric space can represent the isometry group, there are metric spaces such as certain manifolds in which there could be multiple symmetries.

Somewhat oversimplifying, geometry is local : curvature, angles, etc. while topology is global.

 

[fresh_42]

Somewhat oversimplifying, geometry is local : curvature, angles, etc. while topology is global.

Or simply look at the words: geo-metry = measurements of earth, topo-logy = science of locations (topoi = location).
I think both have local as well as global properties, but geometry is about measurements (angles) and topology about the intrinsic nature of a space (open sets). I would even say that topology is local at its heart, as everything is about neighborhoods of certain points.

 

[WWGD]

Maybe we can look at spaces that are homeomorphic but not diffeomorphic to tease out the differences between the two concepts.

 

[FallenApple]

Somewhat oversimplifying, geometry is local : curvature, angles, etc. while topology is global.

I'm thinking of a situation where there is an annulus embedded in 2 space. If I pick a circular path inside the annulus that is concentric to the center, then I don't think there is a way for me to tell if its an annulus unless I consider the embedding space. If I pick two points, x,y, on that circular path and opposite to each other by 180 degrees, which I would know from using the inner product, then the Euclidean metric would have a length that cuts into and across the void ( a section of the embedding space that isn't part of the annulus) to the other side. But if a 2d ant that exists on the annulus can't traverse the void, he/she could only reach the other point via chaining of local neighborhoods, which is a topological. I think this chaining of neighborboods can also be connected to modular arithmetic, which seems to be global. But to know the shape of object in the embedding space, wouldn't one need the Euclidean metric to observe the length cutting through the upper bound in the radial direction represented by the inner ring? Perhaps in the ant's universe, traversing that upper bound via chained union of open sets is impossible simply because the union of open sets are open sets, which would seem to require a maximal union of open sets to contain the boundary.

 

[WWGD]

Or simply look at the words: geo-metry = measurements of earth, topo-logy = science of locations (topoi = location).
I think both have local as well as global properties, but geometry is about measurements (angles) and topology about the intrinsic nature of a space (open sets). I would even say that topology is local at its heart, as everything is about neighborhoods of certain points.

But notice how any transformation that deform things in any small neighborhood distorts the geometry ( angles, lengths) but not necessarily the topology ( connectedness, compactness). EDIT: I wonder if the two groups ; of isometries, homeomorphisms, overlap at something other than the identity.

 

[fresh_42]

But notice how any transformation that deform things in any small neighborhood distorts the geometry ( angles, lengths) but not necessarily the topology ( connectedness, compactness).

Yes, that's true. I had the separation axioms in mind.
"We call it a draw." (the Black Knight)

 

[lavinia]

Maybe we can look at spaces that are homeomorphic but not diffeomorphic to tease out the differences between the two concepts.

A smooth manifold also needs to be given a metric in order to have a geometry.

 

[WWGD]

A smooth manifold also needs to be given a metric in order to have a geometry.

This seems to bring up a difference. Once a metric is given, the topology is determined fully, as the metric topology with basic open sets.

 

[lavinia]

But notice how any transformation that deform things in any small neighborhood distorts the geometry ( angles, lengths) but not necessarily the topology ( connectedness, compactness). EDIT: I wonder if the two groups ; of isometries, homeomorphisms, overlap at something other than the identity.

Isometries are homeomorphisms. So the isometry group of a Riemannian manifold is contained in the group of homeomorphisms.

 

[lavinia]

This seems to bring up a difference. Once a metric is given, the topology is determined fully, as the metric topology with basic open sets.

One difference is that a manifold with no smooth structure can not have a Riemannian metric because its tangent bundle is not defined. One would have to give it a metric in a different way.

 

[WWGD]

One difference is that a manifold with no smooth structure can not have a Riemannian metric because its tangent bundle is not defined. One would have to give it a metric in a different way.

Aren't non-smnoothable manifolds a sort of exception? I know once given a smooth structure you can just pull back the metric from the ambient Euclidean space.

 

[lavinia]

Aren't non-smnoothable manifolds a sort of exception? I know once given a smooth structure you can just pull back the metric from the ambient Euclidean space.

A non-smoothable manifold can not be given a Riemannian metric. It would have to be given a geometry by some other means, some distance measure that respects the topology.

 

[WWGD]

Isometries are homeomorphisms. So the isometry group of a Riemannian manifold is contained in the group of homeomorphisms.

Oh, yes, Duh myself. The condition $d(x,y)=d'(f(x),f(y))$ gives away the continuity of the inverse , which clearly exists as we can show it is a bijection.

 

[WWGD]

A non-smoothable manifold can not be given a Riemannian metric. It would have to be given a geometry by some other means, some distance measure that respects the topology.

Yes, this is what I was referring to this; that while this is true, they are, informally , it seems, an exception.

 

[WWGD]

Oh, yes, Duh myself. The condition $d(x,y)=d'(f(x),f(y))$ gives away the continuity of the inverse , which clearly exists as we can show it is a bijection.

Still, an example in the opposite direction, a homeo that is not an isometry would be just any deformation of the manifold.

 

[lavinia]

But notice how any transformation that deform things in any small neighborhood distorts the geometry ( angles, lengths) but not necessarily the topology ( connectedness, compactness). EDIT: I wonder if the two groups ; of isometries, homeomorphisms, overlap at something other than the identity.

There are examples of Riemannian manifolds whose group of connection preserving diffeomorphisms is trivial. Sort of unbelievable.

 

[lavinia]

Still, an example in the opposite direction, a homeo that is not an isometry would be just any deformation of the manifold.

Right. The group of homeomorphisms is way larger than the group of isometries.

 

[WWGD]

There are examples of Riemannian manifolds whose group of connection preserving diffeomorphisms is trivial. Sort of unbelievable.

Things become weird quickly once you go beyond surfaces, beyond n=2.

 

[lavinia]

Things become weird quickly once you go beyond surfaces, beyond n=2.

Yes. In dimensions 3 I know of a manifold whose group of affinities is finite.

 

[WWGD]

Yes. In dimensions 3 I know of a manifold whose group of affinities is finite.

Is it a " theoretical" one or "naturally-occurring"?

 

[WWGD]

There are other curiosities that tell how the two worlds are apart , like nowhere-differentiable homeomorphisms.

 

[lavinia]

Is it a " theoretical" one or "naturally-occurring"?

It is called the Hansche-Wendt manifold. I think it has some use in Physics. It is a compact 3 manifold and can be given a flat Riemannian metric much like the Klein bottle or the torus. It is easy to describe. If you like we can go through it.

 

[WWGD]

It is called the Hansche-Wendt manifold. I think it has some use in Physics. It is compact and can be given a flat Riemannian metric much like the Klein bottle or a torus. It is easy to describe. If you like we can go through it.

Sure, just some other time, please, I had a 12-hr shift today .

 

[martinbn]

A smooth manifold also needs to be given a metric in order to have a geometry.

What if it s given a connection, wouldn't that count as geometry?

 

[lavinia]

What if it s given a connection, wouldn't that count as geometry?

What if it s given a connection, wouldn't that count as geometry?

Yes - in a more general sense.

- One could have a connection that is compatible with a metric but is not a Levi_Civita connection

- One could have a connection on the tangent bundle that is not compatible with any metric. In this case I do not see how distance relations can be derived - but not sure. Still one has curvature and parallel translation,

This thread though seems to assume metric relations of some kind.

 

[lavinia]

@WWGD Here is Kervaire's original paper where he describes a manifold that does not admit any differentiable structure.

http://web.math.rochester.edu/people/faculty/doug/otherpapers/kervaire.pdf

 

[WWGD]

@WWGD Here is Kervaire's original paper where he describes a manifold that does not admit any differentiable structure.

http://web.math.rochester.edu/people/faculty/doug/otherpapers/kervaire.pdf

Excellent, Lavinia, thanks, will check it out asap.