How do charts on differentiable manifolds have derivatives without a metric?

[jackferry]

TL;DR Summary

How do you define a derivative on a manifold with no metric?

I was reading about differentiable manifolds on wikipedia, and in the definition it never specifies that the differentiable manifold has a metric on it. I understand that you can set up limits of functions in topological spaces without a metric being defined, but my understanding of derivatives suggests that you need a metric in both the domain and the codomain, in order to come up with a rate of change which you are finding the limit of. Is there a more general definition of the derivative that is being used here?

 

[fresh_42]

You use the metric in the Euclidean space of the chart. Differentiation is a local property, so it takes place within a single chart: so switch to the chart, differentiate, and turn back to the manifold.

 

[jackferry]

I realized that I thought the chart had to be differentiable, but it is the transition map between charts that must be differentiable, which makes more sense

 

[fresh_42]

I like the example $\det \, : \,GL(2,\mathbb{R}) \longrightarrow \mathbb{R}^*$. The regular matrices build a manifold, the determinant is a smooth mapping into the multiplicative group of real numbers, which again is a manifold. Hence we have a smooth map between manifolds, both with easy charts.

Now try to calculate $D_I\det$ the derivative of the determinant at the point $\vec{x}=I$ the identity matrix.

 

[member 587159]

I realized that I thought the chart had to be differentiable, but it is the transition map between charts that must be differentiable, which makes more sense

Yes, I posted this but other people decided that this post was not relevant to your answer so they hided it in another thread.

The point is you always go back to Euclidean space and take your derivatives there, then you go back to your manifold.

 

[fresh_42]

Yes, I posted this but other people decided that this post was not relevant to your answer so they hided it in another thread.

Sorry, my fault. I wasn't careful enough as I split the threads when the discussion derailed into obscure topologies:
https://www.physicsforums.com/threa...er-fields-other-than-r-c.978845/#post-6246358

 

[Geometry]

Hello, to answer to the first question, if you've got a topological space $M$ such that any $x \in M$ admit a neighborhood homeomorph to an open space of $\mathbb{R}^{n}$, then the following assertion are equivalent :

a)$M$ is an Hausdorff space with a countable basis of open.
b)$M$ has a metric that induce its topology and is a separable space.

But you don't need of that to speak about derivation : you can look here page 212 : http://etananyag.ttk.elte.hu/FiLeS/downloads/_01_Csikos_Differential_geometry.pdf .

I got better links but those are in french.

Have a good day.

 

[member 587159]

Hello, to answer to the first question, if you've got a topological space $M$ such that any $x \in M$ admit a neighborhood homeomorph to an open space of $\mathbb{R}^{n}$, then the following assertion are equivalent :

a)$M$ is an Hausdorff space with a countable basis of open.
b)$M$ has a metric that induce its topology and is a separable space.

But you don't need of that to speak about derivation : you can look here page 212 : http://etananyag.ttk.elte.hu/FiLeS/downloads/_01_Csikos_Differential_geometry.pdf .

I got better links but those are in french.

Have a good day.

This is a highly untrivial result for which you need Riemannian metric, but interesting nonetheless.

 

[lavinia]

The idea of differentiability does not require a metric. Derivatives only need the idea of differentiating a function of one variable and this only depends upon the idea of absolute value on the real line or in the complex plane.

Calculus is defined on differentiable manifolds. These do not have metric relations. A metric is an additional structure that gives the manifold a geometry.

 

[Geometry]

Math_QED : you can show it without Riemannian structure (in fact you can show the assertions above are equivalent to "$M$ is homeomorph to a closed set of $L^{2}(\mathbb{N})$").End of digression.

 

[WWGD]

Cant we, don't we deal with differentiability in algebraic terms in Differential Galois theory?

 

[fresh_42]

You can define differentiability as the application of a derivation, which is a linear operator which obeys the Leibniz rule. This gives immediately the derivative of polynomials which Galois theory deals with.

However, the discussion within the limits of this thread would urge me to perform another spin-off.

End of digression.

 

[WWGD]

You can define differentiability as the application of a derivation, which is a linear operator which obeys the Leibniz rule. This gives immediately the derivative of polynomials which Galois theory deals with.

However, the discussion within the limits of this thread would urge me to perform another spin-off.

Yes, but I have not seen this done in the setup of manifolds and how it meshes with topology although there is some mention of germs using derivations when ambient space is not available.

 

[lavinia]

This is a highly untrivial result for which you need Riemannian metric, but interesting nonetheless.

I don't see why one needs a Riemannian metric. A Hausdorff space with a countable basis need not be a manifold. So there may not even be any Riemannian metric possible.

 

[WWGD]

The idea of differentiability does not require a metric. Derivatives only need the idea of differentiating a function of one variable and this only depends upon the idea of absolute value on the real line or in the complex plane.

Calculus is defined on differentiable manifolds. These do not have metric relations. A metric is an additional structure that gives the manifold a geometry.

I am not sure I follow you, maybe we're talking about different things.I think they are referring to the/a distance function. Most manifolds are , in that sense, metrizable by being 2nd countable and normal by Urysohn's lemma on metrization. And I think you need only mild conditions to pullback a Riemannian metric.

 

[lavinia]

I am not sure I follow you, maybe we're talking about different things.I think they are referring to the/a distance function. Most manifolds are , in that sense, metrizable by being 2nd countable and normal by Urysohn's lemma on metrization. And I think you need only mild conditions to pullback a Riemannian metric.

If you mean by Riemannian metric a smoothly varying positive definite quadratic form on the tangent spaces, then the manifold must be a differentiable manifold and calculus is already defined.

There is an entire subject called Differential Topology which uses calculus to study the topology of smooth manifolds. No metric is required.

These manifolds can be given many different Riemannian metrics. But one does not need a metric to define the notion of differentiability.

 

[WWGD]

If you mean by Riemannian metric a smoothly varying positive definite quadratic form on the tangent spaces, then the manifold must be a differentiable manifold and calculus is already defined.

There is an entire subject called Differential Topology which uses calculus to study the topology of smooth manifolds. No metric is required.

These manifolds can be given many different Riemannian metrics. That was not my point. My point was that one does not need a metric of any kind to define the notion of differentiability.

But can you do Calculus without a (distance)metric? For differentiability I believe the minimal conditions are having an uncountable ordered topological field F where 0 is a limit point of F-{0}; this allows the differential quotient limit to make sense. This narrows it down, I believe, to the Reals, Complexes and p-adics all of which are metrizable. And, yes, we work indirectly in $\mathbb R^n, \mathbb R^m$ , but ., at least in the Reals, when we talk about f(x+h)-f(x) , I believe we use a metric. EDIT: As in a previous post, I guess we can do a lot algebraically by differential Galois theory and derivations; maybe we can fully ignore the metric.

 

[lavinia]

But can you do Calculus without a (distance)metric? For differentiability I believe the minimal conditions are having an uncountable ordered topological field F where 0 is a limit point of F-{0}; this allows the differential quotient limit to make sense. This narrows it down, I believe, to the Reals, Complexes and p-adics all of which are metrizable. And, yes, we work indirectly in $\mathbb R^n, \mathbb R^m$ , but ., at least in the Reals, when we talk about f(x+h)-f(x) , I believe we use a metric. EDIT: As in a previous post, I guess we can do a lot algebraically by differential Galois theory and derivations; maybe we can fully ignore the metric.

As I pointed out, a metric on the base field is used. But this is not a metric on the manifold.

 

[WWGD]

As I pointed out, a metric on the base field is used. But this is not a metric on the manifold.

Yes, most of the hard work is done in the base field and pushed forward and pulled back, I agree on that. And piecewise at that.

 

[lavinia]

By The Way:

The proofs of the existence of a Riemannian metric on a smooth manifold that I have seen use a Partition of Unity to piece together arbitrarily chosen metrics on a covering of the manifold by coordinate patches. This covering is required to be what is called a Good Cover which means that each point on the manifold lies in only finitely many of the patches. The existence of a Good Cover is trivial for compact manifolds and not hard to prove in general.

Once one has a Riemannian metric, there is a distance function because one can calculate the lengths of piecewise smooth paths using integration. With some proving the conclusion is that there is an induced metric and it determines the topology of the manifold.

This technique is specific to smooth manifolds and does not generalize, not even to manifolds that have no smooth structure.

 

[WWGD]

By the length of a path you mean the inf of the length over all paths? But, yes, I think paracompact + 2nd countable guarantees existence of partition of unity that glues together pullback of the metric from R^n in each chart. I think this is not too demanding of conditions.