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Can one define a vector space structure on a Riemannian manifold ##(M,g)##?! By this I mean, does it make a sense to write ##x+y## where ##x,y## are arbitrary points on ##M##?
No. But do check out the exponential map for something close to what you're attempting.Can one define a vector space structure on a Riemannian manifold ##(M,g)##?! By this I mean, does it make a sense to write ##x+y## where ##x,y## are arbitrary points on ##M##?
Why can't you pullback the vector space structure from ## \mathbb R^n?## : for ##x,y \in M## , andNo. But do check out the exponential map for something close to what you're attempting.
Yes, that is a harder question, I dont know if it can be done or not.Locally sure, but I guess he means a global vector space structure.
It can't. There are topological obstructions.Yes, that is a harder question, I dont know if it can be done or not.
But isnt vector space structure independent of topology ( if you ignore the inner-product generating a norm and "inducing" a normed space/vector space structure)?It can't. There are topological obstructions.
Sorry, I dont get your point. Do you mean there is no compatibility as _normed spaces_ , alone, tho there is as pure vector spaces? Could you cite the topological obstruction?In that case, every manifold has the same cardinality of ##\mathbb{R}##, so you can find a bijection.
I see, so it can be done for contractible manifolds then? Or is/are there some other obstruction(s)?The topological construction is contractibility.
If you don't care for topology, then you can find a bijection and then just transport the structure.
Here is an exercise. Prove that a manifold that is a vector space over the real numbers is diffeomorphic to Rn.Can one define a vector space structure on a Riemannian manifold ##(M,g)##?! By this I mean, does it make a sense to write ##x+y## where ##x,y## are arbitrary points on ##M##?
I couldn't get your point on this quote..how can one define the distance on M?!I don't think (at least I don't see how) the Riemannian metric here plays a role.
Yeah.. I think things are starting to be a bit clearer .. however, I couldn't understand what does really a Riemannian vector space look like?!Here is an exercise. Prove that a manifold that is a vector space over the real numbers is diffeomorphic to Rn.
- If you only require addition and not scalar multiplication, then the manifold is called a Lie group. If the addition is commutative, then the manifold can be an n dimensional torus.
Your question will not work if you replace vector space by Lie group. The condition of being a Lie group is very restrictive. For instance the only closed 2 dimensional surface that is a Lie group is the torus.
Yeah.. I think things are starting to be a bit clearer .. however, I couldn't understand what does really a Riemannian vector space look like?!
Not sure what you mean by Riemannian vector space - but Rn together with an inner product is a Riemannian manifold that is also a vector space. Algebraically, one sees the connection between geometry and vector algebra through the bilinearity of the inner product.Yeah.. I think things are starting to be a bit clearer .. however, I couldn't understand what does really a Riemannian vector space look like?!
I'm interested in a Riemannian manifold in which ##x+x'## and ##\alpha.x## where ##\alpha## is a scalar from some field have a meaningful definition..so I thought that this kind of riemannian manifold should has an additional vector space structure..if I understand it correctly, one can locally consider it..Not sure what you mean by Riemannian vector space - but Rn together with an inner product is a Riemannian manifold that is also a vector space. Algebraically, one sees the connection between geometry and vector algebra through the bilinearity of the inner product.
Geometrically, the inner product assigns lengths to vectors and angles to pairs of vectors. Without the inner product there is no geometry.
- Technically, a Riemannian metric is defined on the tangent bundle of the manifold. But in Rn the tangent space and the vector space are canonically identified.
In any local coordinate system that does not map surjectively onto all of Rn e.g. normal coordinates for almost all manifolds, the scalars that keep you inside the coordinate system are limited(really big scalars throw you outside) and some vectors when added will be outside of the coordinate system as well. The exponential mapping is only a local diffeomorphism. That means that it fails to be 1-1 after a while (for sufficiently large radius in the tangent space) and it may have singularities( called conjugate points). In these cases the addition on the manifold is not well-defined. Try defining addition of the south pole of a sphere to any other point using the exponential map at the north pole.I'm interested in a Riemannian manifold in which ##x+x'## and ##\alpha.x## where ##\alpha## is a scalar from some field have a meaningful definition..so I thought that this kind of riemannian manifold should has an additional vector space structure..if I understand it correctly, one can locally consider it.. normal coordinates doesn't mean the same thing?!
The distance is dealt with by treating ##(M,g)## as a length space: the distance between two points equals the length of the shortest path between the points.I couldn't get your point on this quote..how can one define the distance on M?!
What it comes down to is this: You want your n-manifold to be a finite-dimensional (n -dimensional) inner-product space, with the inner-product given by the Riemannian metric. But there is a result that every finite-dimensional normed space (over a fixed field) of dimension n is isomorphic to ##\mathbb R^n ## . This means your manifold, if given an inner-product normed space condition, will have to be homeomorphic to ## \mathbb R^n ##, since the (metric) topology generated by the norm will be homeomorphic to that of ##\mathbb R^n ##.Yeah.. I think things are starting to be a bit clearer .. however, I couldn't understand what does really a Riemannian vector space look like?!
Depending on what you really mean by this question, there are various answers. (Though it is not clear how you might want the Riemannian structure to relate, if at all, to the vector space structure.)Can one define a vector space structure on a Riemannian manifold ##(M,g)##?! By this I mean, does it make a sense to write ##x+y## where ##x,y## are arbitrary points on ##M##?