- #1
shooride
- 36
- 0
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##?
shooride said: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##?
micromass said:No. But do check out the exponential map for something close to what you're attempting.
Yes, that is a harder question, I don't know if it can be done or not.micromass said:Locally sure, but I guess he means a global vector space structure.
WWGD said:Yes, that is a harder question, I don't know if it can be done or not.
But isn't vector space structure independent of topology ( if you ignore the inner-product generating a norm and "inducing" a normed space/vector space structure)?micromass said:It can't. There are topological obstructions.
Sorry, I don't 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?micromass said:In that case, every manifold has the same cardinality of ##\mathbb{R}##, so you can find a bijection.
micromass said:The topological construction is contractibility.
If you don't care for topology, then you can find a bijection and then just transport the structure.
shooride said: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?!WWGD said: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?!lavinia said: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.
shooride said: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?!
shooride said: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..lavinia said: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.
shooride said: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.shooride said: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 ##.shooride said: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.)shooride said: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##?
A Riemannian manifold is a mathematical space that has a smooth, curved structure and allows for the measurement of distances and angles between points. It is named after the mathematician Bernhard Riemann and is a fundamental concept in differential geometry.
Riemannian manifolds are important in mathematics because they allow for the study of curved spaces and enable the development of theories such as general relativity. They also have applications in fields such as physics, computer science, and statistics.
"##x+y## on a Riemannian manifold" refers to a mathematical operation that takes place on a curved space, where the addition of two vectors results in a new vector that is tangent to the manifold at the point of addition. This operation is known as the Riemannian sum and is used to define the metric structure of the manifold.
The metric structure of a Riemannian manifold is defined by a metric tensor, which is a mathematical object that assigns a distance between any two points on the manifold. This distance is calculated using the tangent vectors at each point and measures the curvature of the manifold at that point.
Riemannian manifolds have various applications in fields such as physics, computer vision, and machine learning. For example, they are used in general relativity to model the curvature of spacetime and in computer vision to analyze shapes and objects in images. They are also used in machine learning algorithms, such as manifold learning, to find low-dimensional representations of high-dimensional data.