# Why are vectors defined in terms of curves on manifolds

Tags:
1. Mar 5, 2015

### "Don't panic!"

What is the motivation for defining vectors in terms of equivalence classes of curves? Is it just that the definition is coordinate independent and that the differential operators arising from such a definition satisfy the axioms of a vector space and thus are suitable candidates for forming tangent spaces on a manifold? Or is there a deeper (or different) reason?

2. Mar 5, 2015

### Fredrik

Staff Emeritus
There's another coordinate dependent definition: Let $\mathcal F$ be the vector space of all smooth $f:M\to\mathbb R$, and define $T_pM$ as the set of all linear $v:\mathcal F\to\mathbb R$ such that $v(fg)=v(f)g(p)+f(p)v(g)$ for all $f,g\in\mathcal F$.

This definition is much easier to work with than the other one. The only problem with it is that it makes you think "Why are these things called tangent vectors at p?" The best way to answer this question is probably to show that this $T_pM$ is isomorphic to the $T_pM$ defined the other way.

3. Mar 5, 2015

### "Don't panic!"

Is there much motivation for the definition in terms of curves though (is it similar to what I put or is there another reason)?

4. Mar 5, 2015

### Fredrik

Staff Emeritus
The motivation, as I see it, is that while both of these definitions associate a vector space with each point in the manifold, it's only the definition based on equivalence classes of curves that makes it easy to understand why it makes sense to think of the vectors as tangent vectors of the manifold at the specified point. The stuff you have described in other threads are those technical details that make it easy to understand why we think of these things as tangent vectors.

5. Mar 5, 2015

### "Don't panic!"

I confused myself on the matter as someone asked me whether the curve is introduced to give the notion of "direction" to a vector, but I thought that in general one can define a vector without needing the notion of magnitude and direction, these are additional structures that only have meaning if one defines an inner product and a norm (although the former somewhat implies the latter) on the vector space?!

6. Mar 5, 2015

### Fredrik

Staff Emeritus
The definition of "vector space" doesn't mention magnitude or direction. It just lists the properties of the addition and scalar multiplication operations. You need a norm to assign a magnitude to each vector, and you need an inner product to associate an angle with a pair of vectors, but you don't need a norm or inner product to say that every vector determines a direction. A vector v spans the 1-dimensional subspace $\{rv|r\in\mathbb R\}$, and the subset that consists of those rv with r>0 can certainly be thought of as a direction. Also, the number |r| is a measure of how "big" a vector in that subspace is.

7. Mar 5, 2015

### Matterwave

Another way to think about why an inner product is not necessary to define a "direction" is that two vectors $v,u$ can be said to point "in the same direction" if and only if $\{v=cu\left.\right| c\in\mathbb{R},c>0\}$. An inner product just adds a way to form a scalar from 2 vectors.

8. Mar 6, 2015

### dx

At a given point on the curve, it locally has a certain direction, that's obvious. The 'magnitude' of the vector comes from the parameter of the curve. Remember that we are talking about parametrized curves. You can have different curves with the same direction at that point, but their parameters may be varying at different rates.

9. Mar 6, 2015

### "Don't panic!"

This is what I thought was the case for vector spaces in general, but is it possible then to have a notion of direction for each vector in any vector space?

That's what I thought, but is a vector introduced in this way such that the equivalence class of curves with the same derivative at a given point define a vector 'pointing in a particular direction' on the manifold at that point? Is it that the differential operator that such an equivalence class of curves defines encodes all information about the possible directions in which a function can pass through a given point and how 'quickly' it can do so, hence capturing the definitive properties of a tangent vector at that point?!

Last edited: Mar 6, 2015
10. Mar 6, 2015

### dx

That's right. The tangent vector encodes the direction, and the rate, which are common to all elements of the equivalence class. An equivalence class is just a set of objects whose elements have something in common. It is not a particularly good way to understand what 'tangent vector' means, although it may be convenient for mathematical definition.

A tangent vector to a parametrized curve ζ: λ → M is simply a name for the local behavior of this function near some value of its parameter λ, say λ = 0. This is denoted (d/dλ)λ = 0.

Last edited: Mar 6, 2015
11. Mar 6, 2015

### "Don't panic!"

So as each equivalence class of curves defines a vector at a given point on a manifold, as they capture the notion of passing through a particular point, in a particular direction (at that point) at a particular 'speed' (i.e. they have the same derivative at that point), also as such equivalence classes satisfy the axioms of a vector space we can identify them with tangent vectors at that point. Would this be a correct summary?

12. Mar 6, 2015

### Hawkeye18

Going back to the original question, the motivation for the definition of tangent vectors in terms of equivalence classes of curves is best understood if one considers embedded manifolds. It should be intuitively clear to a physicist that when you "drive" on an embedded manifold, your velocity is always tangential to the manifold. This lead to a definition to a tangent vector as a possible "velocity", and when you formalize this definition, you arrive to the definition in terms of equivalence classes of curves.
That is a "physical" definition of a tangent vector. There is also a natural "geometric" definition, namely a vector $\mathbf v$ is tangent to an embedded manifold $M$ at a point $\mathbf p$ if $\operatorname{dist}(\mathbf p + t \mathbf v, M)=o(t)$. And it can be shown that for embedded manifolds "physical" and "geometric" definitions of the tangent space give the same object.

The above "physical" definition of the tangent space does not require knowing how the manifold is embedded (as soon as you know how to check that the curve on the manifold is smooth, and 2 curves have the same "velocity"), so it can be transferred to an abstract manifold. In contrast, the "geometric" definition requires knowing the embedding, and it cannot be easily transferred the the case of abstract manifolds.

So, going back to the motivation: the tangent space defined in terms of "velocities" not only satisfies the axioms of a vector space, but for an embedded manifolds it gives you the "real" tangent space (tangent space for an embedded manifold defined geometrically is a real, natural and a well defined object).

13. Mar 6, 2015

### lavinia

I would point out that the idea of tangent is valid for simple good old Euclidean space and just as Hawkeye18 described, the idea is both physical and geometric. In classical vector calculus, no manifolds required, the directional derivative of a function with respect to a tangent vector at a point is its derivative along a curve whose velocity equals that vector at that point. One proves that away from the point the curve can do whatever it wants, the directional derivative is the same. This suggests the idea of an equivalence class of curves.

But Hawkeye18 has pointed out another thing that is important when thinking about tangent spaces to manifolds. That is that the collection of all possible tangent vectors on a manifold, itself has a structure - a topology - and is not simply an collection of local operators on functions. For instance the tangent spaces to an embedded 2-sphere form a non-trivial manifold in and of themselves. In fact, the tangent circle bundle to the 2-sphere is homeomorphic to the 3 dimensional real projective space, or if you like, to the rotation group, SO(3).

Last edited: Mar 6, 2015
14. Mar 6, 2015

### "Don't panic!"

Is this done by showing that the directional derivative at a point is independent of the curve one chooses to parametrise it by? In this sense can one say that the definition in this way is coordinate independent as each curve defines a 1-dimensional coordinate system (that parametrises the curve i.e. $\gamma : (-\varepsilon, \varepsilon)\rightarrow M\;,\;\;t\mapsto\gamma (t)$, where $t\in (-\varepsilon, \varepsilon)$) for a submanifold or the given manifold, but as the definition is as an equivalence class it does not depend on any particular curve and thus does not depend on any of the particular 1-dimensional coordinate systems introduced?!

15. Mar 6, 2015

### lavinia

This is what the Chain Rule says. In Euclidean space, if c(t) is a curve and f(x) is a function, then f(c(t) is a composed function, F(t), and the Chain Rule says that

F'(t) = df(c'(t))

So only c'(t) matters and not the rest of c(t).

In physics this is often written as the dot product, grad(f).c'(t), where f is a scalar field such as a potential.

Last edited: Mar 6, 2015