Differentiation on Smooth Manifolds without Metric

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dx

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Hi,

I'm confused about what differentiation on smooth manifolds means. I know that a vector field [tex] v [/tex] on a manifold [tex] M[/tex] is a function from [tex] C^{\infty}(M) [/tex] to [tex] C^{\infty}(M) [/tex] which is linear over [tex] R [/tex] and satisfies the Leibniz law. This should be thought of, I'm told, as a 'derivation' on smooth functions on the manifold, i.e. differentiation. Intuitively, a vector field is a field of little arrows on the manifold, and its action on a function is just the directional derivative of the function in the direction of the tangent vector at that point.

What I don't quite understand is what it means to differentiate on a space without any notion of distance. To take a simpler example, consider a vector space [tex] R^2 [/tex] (not the euclidean vector space with metric, [tex] E^2 [/tex]). What is a smooth function on a vector space? What is differentiation of smooth functions on a vector space without any other structure?
 

morphism

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But there is a notion of distance, at least locally. After all, an n-manifold is locally like R^n. This is the point of defining a smooth structure on a manifold: it gives us a notion of differentiability. Think about this as you would a topology: without a topology, it doesn't make sense to talk about continuity.
 

dx

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But there is a notion of distance, at least locally. After all, an n-manifold is locally like R^n. This is the point of defining a smooth structure on a manifold: it gives us a notion of differentiability. Think about this as you would a topology: without a topology, it doesn't make sense to talk about continuity.
What is distance on [tex] R^n [/tex]? There are many possible ways to define a metric on [tex] R^n [/tex]; it doesn't in the mathematical sense come equipped with one.
 

quasar987

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When no metric on R^n is specified, it it implicit that we mean the euclidean one...

In particular, when we say a manifold is a locally euclidean topological space, we mean is it locally homeomorphic to R^n, given the topology of the euclidean norm.
 
Assume you have a two-dimensional smooth manifold [itex]M[/itex], a smooth function [itex]f[/itex] and a vector (field) [itex]\mathbf{v}[/itex]. And you would like to calculate the directional derivative of [itex]f[/itex] at some point [itex]p[/itex].

Since it is a manifold you have a coordinate system [itex](x,y)[/itex] around the point and since the function is smooth you also have the partial derivates [itex]\partial f/\partial x,\, \partial f/\partial y[/itex] in this coordinate system. A directional derivative [itex]L_vf[/itex] is then a linear combination of these partial derivatives. If the vector has coordinates [itex](dx,dy)[/itex] in the coordinate system (strictly: in the induced tangent space coordinate system), then the directional derivative is
[tex]L_vf=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy[/itex].
 

dx

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When no metric on R^n is specified, it it implicit that we mean the euclidean one...
As far as I can tell, the definition of a manifold doesn't involve any metric, implicit or explicit. It is only required that it is locally like the vector space [itex] R^n [/itex]. Anyway, I figured out what my problem was. I was thinking of the directional derivative as [tex] \nabla f(a) \cdot v [/tex] instead of as

[tex] \nabla_v f(a) = \lim_{t \rightarrow 0} \frac{1}{t}[ f(a + tv) - f(a) ][/tex].

The latter definition is covaraint with respect to linear transformations (homeomorphisms are locally linear transformations) and requires only vector space structure, while the former needs a notion of dot product.
 
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quasar987

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A manifold is required to be locally like R^n as a topological space, not as a vector space! (What would that mean anyway?)

And the topology implicit in this definition is the standard one, in which distances are measured according to the law of Pythagoras.
 

Hurkyl

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A manifold is required to be locally like R^n as a topological space, not as a vector space! (What would that mean anyway?)
And a differentiable manifold is required to be locally like R^n as a differentiable manifold. And a smooth manifold is required to be locally like R^n as a smooth manifold.

And the topology implicit in this definition is the standard one, in which distances are measured according to the law of Pythagoras.
The standard topology doesn't really say much about distances.
 

dx

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And the topology implicit in this definition is the standard one, in which distances are measured according to the law of Pythagoras.
The topology of a topological space X is defined by the set of open sets T, no notion of distance is needed.
 

quasar987

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True, but I find it important to keep in mind that a manifold is a topological space locally like R^n with the topology produced by the familiar euclidean metric. So a manifold is a possibly alien being, but locally familiar.
 

morphism

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I'm not sure I buy that the standard topology on R^n doesn't have much to do with distances. It does come from a metric after all (from a norm even, and any norm at that).
 

mathwonk

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why not just read the definition in any book on manifolds?
 

dx

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I'm not sure I buy that the standard topology on R^n doesn't have much to do with distances. It does come from a metric after all (from a norm even, and any norm at that).
You can use the Euclidean metric to define the open sets, but once you have the open sets you forget about the metric. The set of open sets T is what defines the topology of the space. There are many metrics which produce the same topology, so the standard topology on R^n has nothing to do with the Euclidean metric in particular. The point is that we only need an idea of whether two points are "nearby", not how far apart they are.
 

mathwonk

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even on the real line derivatives have nothing to do with a metric, since you divide deltay by delta x. i.e. changing the metric by a scalar multiple does not affect it. it does relate to the affine structure.

but thats because there you want the derivative to be a number. on a manifold a derivative is a linear map between to abstract vector spaces of equivalence classes of curves, and no metric is needed.

or of you prefer, one does refer to a metric on R^n but the specific metric does not matter, only the result of the process, which can be done with a different metric, equivalent up to diffeomorhism.

really you should read a standard book on this.
 

mathwonk

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play with this: a manifold is a topological ,space equipped with a family of real valued functions on each open set that behave like diffble functions (closed under lin ear combinations..). Moreover assume that each point p has a nbhd U and a homeomorphism of that nbhd with a ball B around the origin in R^n making the family of diffble functions on U correspond to the family of all smooth functions on B.

then define a map f from one manifold to another to be diffble if for each point p in the domain, the pullback of every function diffble near f(p) is diffble near p.

for such a function we will define the derivative as follows: first we define a tangent vector at p as an equivalence class of parametrized curves through p, where two curves are equivalent iff composing both of the m with every smooth function at p gives two functions defined near t=0 and having the same derivative at 0.

then check that diffble functions between manifolds preserve equivalence of curves, and define the derivative at p of f, as the map induced by f on equivalence classes of curves at p.

hows that? is the metric sufficiently obscured in that definition for your taste?
 
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[tex] \nabla_v f(a) = \lim_{t \rightarrow 0} \frac{1}{t}[ f(a + tv) - f(a) ][/tex].
Well, yes, but the notation [itex]a+tv[/itex] is a bit problematic. You are adding a vector to a point, so strictly speaking you are going off on a tangent. :smile: Instead you can restate the definition by first choosing any curve [itex]\gamma(t)[/itex] such that [itex]\gamma(0)=a,\, \gamma'(0)=v[/itex] and then defining the derivative to be

[tex] \nabla_v f(a) = \lim_{t \rightarrow 0} \frac{f(\gamma(t)) - f(\gamma(0))}{t}. [/tex]

That said, you are in good company using this kind of notation; Arnold speaks of "bent vectors" in some place and even Cartan are doing similar things. So maybe there are some hidden ideas to be understood here.

But note that although the concept of directional derivative is perfectly valid without a metric, you need a metric to make it really useful. This is since then you can normalize the direction vector which makes it possible to compare the derivative value in different directions. You might even calculate the direction of, say, the steepest descent.

Without a metric the only interesting direction is the one for which [itex]\nabla_v f(a)=0[/itex], since this will always evaluate to zero even if v is rescaled.
 
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dx

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Well, yes, but the notation [itex]a+tv[/itex] is a bit problematic. You are adding a vector to a point, ...
Actually, I was working on a chart, so i was adding a vector to a vector. In fact, this definition doesn't depend on the which chart you use, since the transition functions are smooth, and therefore locally linear transformations.
 

dx

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even on the real line derivatives have nothing to do with a metric, since you divide deltay by delta x.
That's not correct if you're using the standard definition of a derivative as the limit of a difference quotient. [tex] \Delta y [/tex] doesn't depend on the metric on the domain of the function, but [tex] \Delta x[/tex] does, so changing the metric changes the derivative. If you use the covariant definition of derivative, then the metric doesn't matter. See post #6.

play with this: a manifold is a topological ,space equipped with a family of real valued functions on each open set that behave like diffble functions (closed under lin ear combinations..). Moreover assume that each point p has a nbhd U and a homeomorphism of that nbhd with a ball B around the origin in R^n making the family of diffble functions on U correspond to the family of all smooth functions on B. ...
The meaning of the part in bold is what I was confused about, and it was resolved in post #6.
 
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Hurkyl

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As a notational aside....

Suppose you want to specify in an expression a tangent vector v along with the point P upon which it's based. I think the most common way to do it is something like (P, v), but I've found that coopting + for this purpose is extremely useful.

Definition: If P is a point on a manifold, and v is a tangent vector at P, then the expression P+v is the corresponding point of the tangent bundle.

This notation has various useful features. My favorite is, given a differentiable map f of smooth manifolds, you have this relationship between the pushforward on the tangent bundle and the derivative:
[tex]f_*(P + v) = f(P) + f'(P) v[/tex]
and, generally speaking, notationifies the intuition that tangent vectors capture the idea of an infinitessimal.

I vague recall getting some use out of this for lie groups/algebras, but I don't remember exactly how it worked out.
 

mathwonk

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i said on the real line, which meant both x and y were points of the real line, hence both delta x and delta y are intervals on the same affine line, so their ratio does not depend on the length chosen just the ability to compare lengths, so it is correct as stated.
 
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Differentiability does not require the idea of a metric just the idea of a directional derivative.

In a coordinate chart that's easy just differentiate the function along a curve. The only thing you need to know is that there is a smooth curve that fits any tangent direction at a point. This is not hard.

In calculus it is often said that the derivative of a function in a direction is its gradient dot product the direction. This is true but uses the Euclidean metric on Euclidean space to define the gradient. On a manfold with a metric there is also an idea of a gradient of a function and this definition still works. However, the metric is not necessary. One just takes direction derivatives and calls it a day.

The directional derivative of the function,f, is just the differential of f applied to the direction, df(v). There is no metric here.
 

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