Tangent vectors in affine spaces

In summary, affine spaces can be seen as smooth manifolds using the natural topology and affine coordinate charts as an atlas. The tangent vector of a curve in a point p can be defined as a derivation, and in the case of affine spaces, it can also be defined as an element of the underlying vector space V. The relationship between these two definitions can also be described using coordinates. However, the use of a norm in this definition is unnecessary and can be avoided. The directional derivative on affine spaces only requires a topology and does not need a metric. The total derivative of an M \rightarrow \mathbb{R} function can be defined without a metric by using a functional that maps tangent vectors to the directional
  • #1
mma
245
1
Affine spaces can be regarded as smooth manifolds if we take the natural topology and affine coordinate charts as atlas. So, if M is an n-dimensional affine space, then the tangent vector of a curve [tex] C: [0,1] \rightarrow M [/tex]in a point [tex]p = C(t_0)[/tex] can be defined as a derivation (as in any smooth manifold):
[tex]\dot C(t_0): C^\infty(M) \rightarrow \mathbb{R}, f \mapsto \frac{d f(C(t))}{d t}\bigg|_{t=t_0} [/tex]

On the other hand, for M is an affine space, the tangent vector of the curve C in the point [tex]p = C(t_0)[/tex] can be defined as an element of the underlying vectorspace V:
[tex]C'(t_0) = \lim_{t \to t_0} \frac {C(t) - C(t_0)}{t - t_0}[/tex]

[tex]C'(t_0)[/tex] and [tex]\dot C(t_0)[/tex] is related simply as
[tex]\dot C(t_0)(f) = \frac{d f(C(t))}{d t}\bigg|_{t=t_0} =f'(C(t_0))C'(t_0)[/tex]
where [tex]f'(C(t_0))[/tex] is the derivative of f at [tex]C(t_0)[/tex], that is a linear functional on V which satisfy:

[tex]f(C(t))-f(C(t_0))= f'(C(t_0)) (C(t) - C(t_0)) + \mathcal{O}(\|{C(t) - C(t_0)\|^2})[/tex]

But this works only if a norm is also defined on V. Evidently, the relation between [tex]C'(t_0)[/tex] and [tex]\dot C(t_0)[/tex] can also described using coordinates and then showing that the relation is independent from the coordinates chosen.

My question is: How can be related [tex]C'(t_0)[/tex] to[tex]\dot C(t_0)[/tex] without using norm or coordinates?
 
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  • #2
Perhaps [tex]f'(a)[/tex] can be defined alternatively on affine spaces as follows.

Let [tex]\Phi: M \rightarrow\mathbb{R}^n [/tex] be an affine chart on [tex]M[/tex] with origin [tex]a \in M [/tex], i.e.,
[tex] p - a = \sum_{i=1}^{n} \mathrm{proj}_i(\Phi(p))e_i [/tex]
[tex](p \in M)[/tex]
with some basis [tex]\{e_i\} [/tex] of [tex] V [/tex].
Clearly, this is equivalent with a linear map [tex]\tilde \Phi [/tex] between [tex]V [/tex] and [tex]\mathbb{R}^n[/tex]: [tex]\Phi(p) = \tilde{\Phi}(p-a)[/tex].
Let [tex]g := f \circ \Phi^{-1} : \mathbb{R}^n \rightarrow \mathbb{R}[/tex],
and let [tex]g'(0) = A [/tex], the derivative of [tex]g[/tex] at [tex]\Phi^{-1}(a) = 0 [/tex] in the usual sense.

We define now [tex]f'(a):= A \circ \tilde \Phi [/tex]. This definition doesn't require a norm on [tex]V[/tex] and can easily shown that the definition is unambiguous, i.e. independent of the choice of the [tex]\Phi[/tex] chart.
 
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  • #3
Of course the original problem persists here. We have chosen a coordinate chart for defining the derivative and then see that the definition is independent of our choice.

Of course we could have used the original definition of the derivative using an arbitrarily chosen norm on V and then we could have shown that the any norm yields the same result.

But why do we use something in a definition what is really irrelevant? Couldn't we avoid such objects? What would Mr. Occam say about this?
 
  • #4
Tangent vectors define directional derivatives. Directional derivatives are derivations.
 
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  • #5
Hurkyl said:
Tangent vectors define directional derivatives. Directional derivatives are derivations.

Exactly. My problem about it is the following. The definition of the directional derivative on affine spaces needs a metric. But it doesn't matter what metric we use. The metric can be arbitrary: any choice of it leads to the same directional derivative. So the definition of the directional derivative really doesn't need a metric. Still we use this unnecessary metric in the definition. But if it is unnecessary, why do we use it?

How can we define directional derivatives without any reference of any metric?
 
  • #6
mma said:
Exactly. My problem about it is the following. The definition of the directional derivative on affine spaces needs a metric.
It doesn't need a metric; it only needs a topology.
 
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  • #7
OK, C' needs only topology. And directional derivative too.
But how do you define the total derivative of an [tex]M \rightarrow \mathbb{R} [/tex] function without metric (norm)? That is, the linear functional what maps the directional derivative of on [tex]M \rightarrow \mathbb{R} [/tex] function to the tangent vectors.
 
  • #8
I just realized the directional derivative doesn't even use the topology on M! (Because it's defined as the derivative of a particular real-valued function of the reals) But that's getting into irrelevant detail.

mma said:
OK, C' needs only topology. And directional derivative too.
But how do you define the total derivative of an [tex]M \rightarrow \mathbb{R} [/tex] function without metric (norm)?
Defining [itex]d_p f[/itex] is easy; a functional is just a fancy sort of function, and you can define it 'pointwise'. e.g.

[tex](d_p f)(v) := (\nabla_v f)(p)[/tex]

Which is exactly what the English definition says: it's the functional that maps tangent vectors to the directional derivative of f at p in that direction.

(which I think is what you were trying to say here:)
the linear functional what maps the directional derivative of on [tex]M \rightarrow \mathbb{R} [/tex] function to the tangent vectors.
 
  • #9
Hurkyl said:
I just realized the directional derivative doesn't even use the topology on M! (Because it's defined as the derivative of a particular real-valued function of the reals) But that's getting into irrelevant detail.

Really. Only C' requires the topology.

Hurkyl said:
Defining [itex]d_p f[/itex] is easy; a functional is just a fancy sort of function, and you can define it 'pointwise'. e.g.

[tex](d_p f)(v) := (\nabla_v f)(p)[/tex]

Which is exactly what the English definition says: it's the functional that maps tangent vectors to the directional derivative of f at p in that direction.

(which I think is what you were trying to say here:)

Really. We have to take only

[tex](\nabla_v f)(p) : = \frac{df(p+tv)}{dt}\bigg|_{t=0} [/tex]

Thank you, Hurkyl!
 
  • #10
The differential of a function requires no norm. Its value on a tangent vector is the derivative of the function composed with any curve whose velocitiy equals that vector.

It is important to realize that metrics are not part of the definition of calculus. What they allow you to do is to interpret differentials (1 forms) as vectors (gradients)
 

What are tangent vectors in affine spaces?

Tangent vectors in affine spaces are vectors that are parallel to the tangent plane of a point on a curve or surface in space. They represent the direction of change at a specific point and are used to calculate the rate of change or gradient of a function.

How are tangent vectors different from normal vectors?

Tangent vectors are different from normal vectors in that they are defined at specific points on a curve or surface, while normal vectors are perpendicular to the surface at that point. Tangent vectors represent the direction of change, while normal vectors represent the direction of the surface's normal.

What is the importance of tangent vectors in affine spaces?

Tangent vectors are essential in understanding the behavior of curves and surfaces in space. They allow for the calculation of rate of change and gradients, which are important concepts in fields such as physics, engineering, and mathematics. Tangent vectors also play a crucial role in differential geometry and calculus.

How are tangent vectors calculated?

Tangent vectors can be calculated by taking the derivative of a curve or surface at a specific point. In calculus, the tangent vector is represented by the first derivative of a function. In differential geometry, the tangent vector is calculated using the directional derivative of a function along a specific direction.

Can tangent vectors be negative?

Yes, tangent vectors can be negative. The direction of a tangent vector is determined by the direction of change, not its magnitude. Therefore, a tangent vector can be negative if it is pointing in the opposite direction of the positive direction on a curve or surface.

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