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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?
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.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.
There fixed it.
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.
Is there a simple example of a manifold that is not differentiable?
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.
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.
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.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.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.