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kiuhnm
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Is it possible to introduce the concept of a gradient vector on a manifold without a metric?
kiuhnm said:Is it possible to introduce the concept of a gradient vector on a manifold without a metric?
This is correct only if you assume that the connection is metric compatible and torsion free. Having a metric compatible connection is not sufficient to uniquely identify it as the Levi-Civita connection.Alex Petrosyan said:A big problem is that when you do have a metric, the definition of the connection is unique, and you can't uniquely extend it to the cases where the metric tensor can't be defined.
Alex Petrosyan said:There fixed it.
lavinia said:BTW: A continuous manifold may not admit an idea of differentiation. By continuous is meant that the transition functions are continuous but not necessarily differentiable. There are continuous manifolds that do not admit an idea of differentiation.
stevendaryl said:Is there a simple example of a manifold that is not differentiable?
lavinia said:I don't know. The first example discovered was 10 dimensional and was reported in this paper, It uses some heavy algebraic topology.
https://web.math.rochester.edu/people/faculty/doug/otherpapers/kervaire.pdf
I would be an interesting project to study this paper together if you like.
stevendaryl said:That would be interesting, but I was only asking if there was some simple example that could give an intuitive reason that some manifolds can't have a differentiable structure. It sounds like the example is not simple.
lavinia said:I don't know of one.
This is different. The 7 sphere as a topological manifold admits an idea of differentiation. It is just that globally across the entire sphere two differentiable structures may not be diffeomorphic.stevendaryl said:John Baez gave a tutorial showing how the 7-sphere (I think that was it) allowed multiple notions of "differentiable". I sort of understood it at the time.
Well, it's not unusual to switch between a vector and its dual. And in the end, a one form is a vector, too! Notations are always a compromise, e.g. what is ##D_pf(v)\,##? The answer depends on what is considered to be the variable here, and notation does not say this. In this sense, the error is already to write a function as its image: ##f(x)## instead of ##x \mapsto f(x)\,.## And I do not see a fundamental difference between ##\nabla## and ##\operatorname{grad}## - these are only symbols and meaning has still to be attached. That one of them is a vector and the other a covector, is already your individual choice. It isn't part of the syntax.martinbn said:I personally think that the notation ##\nabla f## for the gradiant of ##f## is not a good one. A better one is ##\text{grad} f##. The first one is the same as the notation for the covariant differential of ##f##, which is a one form, not a vector.
A gradient vector without a metric is a mathematical concept that describes the direction and magnitude of the steepest ascent of a scalar function on a manifold without using a metric tensor. It is a way to generalize the concept of a gradient to curved spaces.
A traditional gradient vector is defined using a metric tensor, which is a mathematical object that measures distances and angles on a manifold. A gradient vector without a metric, on the other hand, is defined without using a metric tensor and is applicable to curved spaces.
Gradient vectors without a metric have applications in fields such as differential geometry, general relativity, and machine learning. They can be used to describe the curvature and geometry of space-time, as well as to optimize functions in non-Euclidean spaces.
A gradient vector without a metric is calculated using the concept of a covariant derivative, which is a way to differentiate vector fields on a manifold without using a metric tensor. This allows for the calculation of gradients in curved spaces.
One of the main challenges with using gradient vectors without a metric is that they can be difficult to visualize and interpret, especially in higher dimensions. Additionally, the lack of a metric tensor can make it challenging to define and measure distances and angles on the manifold.