# Differentiation on Smooth Manifolds without Metric

1. Nov 27, 2008

### dx

Hi,

I'm confused about what differentiation on smooth manifolds means. I know that a vector field $$v$$ on a manifold $$M$$ is a function from $$C^{\infty}(M)$$ to $$C^{\infty}(M)$$ which is linear over $$R$$ 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 $$R^2$$ (not the euclidean vector space with metric, $$E^2$$). What is a smooth function on a vector space? What is differentiation of smooth functions on a vector space without any other structure?

2. Nov 27, 2008

### morphism

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.

3. Nov 27, 2008

### dx

What is distance on $$R^n$$? There are many possible ways to define a metric on $$R^n$$; it doesn't in the mathematical sense come equipped with one.

4. Nov 27, 2008

### quasar987

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.

5. Nov 28, 2008

### OrderOfThings

Assume you have a two-dimensional smooth manifold $M$, a smooth function $f$ and a vector (field) $\mathbf{v}$. And you would like to calculate the directional derivative of $f$ at some point $p$.

Since it is a manifold you have a coordinate system $(x,y)$ around the point and since the function is smooth you also have the partial derivates $\partial f/\partial x,\, \partial f/\partial y$ in this coordinate system. A directional derivative $L_vf$ is then a linear combination of these partial derivatives. If the vector has coordinates $(dx,dy)$ in the coordinate system (strictly: in the induced tangent space coordinate system), then the directional derivative is
$$L_vf=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy[/itex]. 6. Dec 3, 2008 ### dx As far as I can tell, the definition of a manifold doesn't involve any metric, implicit or explicit. It is only required that it is locally like the vector space $R^n$. Anyway, I figured out what my problem was. I was thinking of the directional derivative as [tex] \nabla f(a) \cdot v$$ instead of as

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

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.

Last edited: Dec 3, 2008
7. Dec 3, 2008

### quasar987

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.

8. Dec 3, 2008

### Hurkyl

Staff Emeritus
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.

The standard topology doesn't really say much about distances.

9. Dec 3, 2008

### dx

The topology of a topological space X is defined by the set of open sets T, no notion of distance is needed.

10. Dec 3, 2008

### quasar987

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.

11. Dec 3, 2008

### morphism

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).

12. Dec 3, 2008

### mathwonk

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

13. Dec 4, 2008

### dx

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.

14. Dec 4, 2008

### mathwonk

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.

15. Dec 4, 2008

### Doodle Bob

...what he said. In fact, you can get a free copy of one of those standard books here: http://www.math.harvard.edu/~shlomo/ (Advanced Calculus) Check out p. 373.

16. Dec 5, 2008

### mathwonk

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?

Last edited: Dec 5, 2008
17. Dec 6, 2008

### OrderOfThings

Well, yes, but the notation $a+tv$ is a bit problematic. You are adding a vector to a point, so strictly speaking you are going off on a tangent. Instead you can restate the definition by first choosing any curve $\gamma(t)$ such that $\gamma(0)=a,\, \gamma'(0)=v$ and then defining the derivative to be

$$\nabla_v f(a) = \lim_{t \rightarrow 0} \frac{f(\gamma(t)) - f(\gamma(0))}{t}.$$

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 $\nabla_v f(a)=0$, since this will always evaluate to zero even if v is rescaled.

Last edited: Dec 6, 2008
18. Dec 6, 2008

### dx

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.

19. Dec 6, 2008

### dx

That's not correct if you're using the standard definition of a derivative as the limit of a difference quotient. $$\Delta y$$ doesn't depend on the metric on the domain of the function, but $$\Delta x$$ 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.

The meaning of the part in bold is what I was confused about, and it was resolved in post #6.

Last edited: Dec 6, 2008
20. Dec 6, 2008

### Hurkyl

Staff Emeritus
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:
$$f_*(P + v) = f(P) + f'(P) v$$
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.