# Position vector and tangent vector in Riemannian spaces

• mma
In summary, the equation of a geodesic connecting a point P with the origin in a Riemannian space is equal to the scalar product of the position vector and the tangent vector at P.
mma
In Euclidean vector spaces the derivative of the position vector of a running point of a
curve is the tangent vector of the curve.

In thehttp://www.matematik.lu.se/matematiklu/personal/sigma/Riemann.pdf" , on page 78 appears a vector which can be regarded as position vector in a Riemann space. This vector is the element of the tangent spce at the given point, it has the direction parallel to the direction of the geodesic connecting this point with the origin, and magnitude equal with the length of this geodesic. (Of course 'origin' means an arbitrarily fixed point of our Riemann space).

This vector is denoted by \hat \sigma on p. 78.

On this page when the author calculates the derivative of the norm of this position vector along a curve, I see that in the right side of the equation appears the tangent vector of the curve instead of the covariant derivative of the position vector.

Is it really true that in Riemann spaces also holds that the covariant derivative of the (above defined) position vector of a running point of a curve is the tangent vector of the curve? How can one see this?

Last edited by a moderator:
I am afraid that this was a bad thought and really there is nothing to do here (in the told equation of the lecture notes) with the covariant derivaive. I thought of it only because of the formal Leibnitz-rule with respect to the scalar product.

So, my question is simply: Why can we calculate the derivative of the norm-square of the "position vector" along a curve as the twice of the scalar product of the "position vector" and the tangent vector of the curve?

The basic idea is that the exponential map induces a local diffeomorphism between the tangent space and the Riemannian manifold. In Euclidean space, there is a natural identification between position vectors and tangent vectors, and normal coordinates are a technical nicety used to extend this (locally) to Riemannian manifolds. In general, a Riemannian connection allows you to compute the derivative of the norm squared by representing it as the inner product:

d/dt |c'(t)|^2 = d/dt <c'(t),c'(t)> = 2<D_t (c'(t)), c'(t)>

In normal coordinates, we have two nice things to work with. The first is that we have Euclidean coordinates, and the second is that the equation of a geodesic is particularly simple. Thus, certain calculation simplify.

Thanks! My doubts are because the metric pulled back by the exponential map isn't an euclidean metric on the tangent space (on the domain if the exponential map), i.e. the metric tensor of this metric can change point by point. So it's surprising for me that the derivative of the metric tensor doesn't appear in this derivation. But I see from your answer that I should make a closer acquaintance with these normal coordinates. I will do this.

I should certainly clarify that only the geodesics going through the origin in normal coordinates are straight lines. The fact that no derivative of the metric appears is precisely the condition that the connection is Riemannian.

But I still don't understand. I try to explain what is that isn't clear for me.

Let $$o \in M$$. Take a curve $$c(t)$$ in $$T_oM$$. Then for each $$t$$, $$c(t)$$ is a vector in $$T_oM$$, so $$\Vert c(t) \Vert = g_o(c(t),c(t))$$, hence

$$\frac{d\Vert c(t) \Vert^2}{dt} = 2g_o(\dot c(t),c(t))$$

where

$$\dot c(t) = \lim_{h \to 0} \frac{c(t+h) - c(t)}{h} \in T_oM$$

It's OK, but my question is just coming.

Let $$\Phi$$ be the exponential map between $$T_oM$$ and $$M$$. Then my question was that what ensures that

$$\frac{d\Vert d\Phi (\partial_{c(t),c(t)}) \Vert^2}{dt} = 2g_{\Phi(c(t))}(d\Phi(\partial_{c(t),\dot c(t)}),d\Phi(\partial_{c(t),c(t)}))$$

where $$\partial_{p,v} \in T_pT_oM$$ denotes the partial derivation at $$p \in T_oM$$ in the direction of $$v \in T_oM$$.

We can restrict the problem into $$T_oM$$ if we introduce the pullback metric $$\tilde g$$ on $$T_oM$$ as $$\tilde g_p(X_p, Y_p) := g_{\Phi(p)}(d\Phi(X_p), d\Phi(Y_p))$$
for any $$p \in T_oM$$ and $$X_p, Y_p \in T_pT_oM$$.

Because this metric turns $$\Phi$$ into an isometry between $$T_oM$$ and $$M$$, the question is now that what ensures that

$$\frac{d\tilde g_{c(t)}(\partial_{c(t),c(t)}, \partial_{c(t),c(t)})}{dt} = 2 \tilde g_{c(t)}(\partial_{c(t),\dot c(t)},\partial_{c(t),c(t)})$$

Because of te properties of the exponential map, this equation is equivalent with

$$\frac{d\tilde g_{0}(\partial_{0,c(t)}, \partial_{0,c(t)})}{dt} = 2 \tilde g_{c(t)}(\partial_{c(t),\dot c(t)},\partial_{c(t),c(t)})$$

My problem is that the LHS of this equation is independent of $$\tilde g_{c(t)}$$ (in fact, it is $$\frac{dg_o(c(t),c(t))}{dt}$$ ) and so are $$\partial_{c(t),\dot c(t)}$$ and $$\partial_{c(t),c(t)}$$ on the RHS. If the equality holds for a specific $$\tilde g_{c(t)}$$ ,
then when I change this $$\tilde g_{c(t)}$$ then the equality must go wrong, isn't it?

Last edited:
zhentil said:
d/dt |c'(t)|^2 = d/dt <c'(t),c'(t)> = 2<D_t (c'(t)), c'(t)>

I think that c'(t) is $$\partial_{c(t),c(t)}$$, and D_t (c'(t)) is $$\nabla_{\dot c(t)}c'(t)$$ in my notation.

In this case the question is that why would be

$$\nabla_{\dot c(t)}c'(t) = \partial_{c(t),\dot c(t)}$$ ?

The notation is making it difficult, but I'll attempt to explain. You said that for each t, c(t) is a vector. This is certainly not correct - only c'(t) is a vector. Once this is clarified, the rest is not terribly bad. If you want to take the derivative of c'(t) along c(t), you need a connection. There's no canonical way to identify the vector spaces in question. My guess is that
$$\nabla_{\dot c(t)} \dot c(t) = \partial_{c(t),\dot c(t)}$$
is simply the translation of my notation into your notation, if I'm not mistaken.

zhentil said:
The notation is making it difficult, but I'll attempt to explain. You said that for each t, c(t) is a vector.

Sorry for the unsatisfactory notation and explanation. Originally I've really spoken about an arbitrary Riemann space $$M$$, and, in general, of course, the points of a Riemann space aren't vectors. But in post #6 I've restricted our investigations to one special Riemann space, namely to the tangent space of M at its point $$o$$, i.e. to $$T_oM$$, which I endowed by the metric pulled back from $$M$$ to $$T_oM$$ by the exponential map. I took the $$c(t)$$ curve in $$T_oM$$, i.e. in a vector space, hence for each $$t$$, $$c(t)$$ is an element of a vector space, i.e. it is a vector.

zhentil said:
There's no canonical way to identify the vector spaces in question. My guess is that
LaTeX Code: <BR> \\nabla_{\\dot c(t)} \\dot c(t) = \\partial_{c(t),\\dot c(t)} <BR>
is simply the translation of my notation into your notation

Because I made $$T_oM$$ a Riemann space, I must make a distiction between the vectors that are the elements of $$T_oM$$ from the tangent vectors of $$T_oM$$. Of course, we can identify the elements of $$T_oM$$ with the tangent vectors at any point $$p \in T_oM$$ (i.e. with the elements of $$T_pT_oM$$). Each vector $$v \in T_oM$$ defines a derivation (i.e. a tangent vector) at $$p \in T_oM$$ by

$$\left. {f \mapsto \frac{df(p + sv)}{ds}}\right|_{s = 0}$$.

where, for the sake of the next usage, I rewrote the usual parameter $$t$$ to $$s$$. (This is the directional derivation of $$f$$ at $$p$$ in the direction of $$v$$ )

This tangent vector is denoted by $$\partial_{p,v}$$. Specially, in the case of $$p=c(t)$$ and $$v=\dot c(t)$$

$$\partial_{c(t),\dot c(t)}$$ is the derivation $$f \mapsto \frac{df(c(t) + s \dot c(t))}{ds}$$

This derivation (= tangent vector) has nothing to do with the metric, while $$\nabla$$ is defined by the connection defined by the metric. So it is hard to believe that they are the same.

Are you familiar with the fact that in normal coordinates, the Christoffel symbols vanish at the origin? This may be the cause of your confusion. In your derivation, you're evaluating everything at the origin, correct? In normal coordinates at the origin, both the Christoffel symbols and the first partials of the metric vanish.

zhentil said:
In your derivation, you're evaluating everything at the origin, correct?

Of course not. The scalar product is taken always at c(t), and your covariant derivative is taken also there, isn't it?

In general, p is taken to be the origin in normal coordinates, with what you've written. If p is not the origin in normal coordinates, you can still talk about the vector field, but in general it will be more complicated, since the Christoffel symbols don't vanish (equivalently, the first partials of the metric don't vanish). Once you get away from the origin, there is no guarantee that the geodesics between two points will be the straight lines in normal coordinates.

Based on the link you gave, the author is assuming that p is the origin in normal coordinates.

I am afraid that it still isn't clear, what is my problem. I try to give a specific example in 2 dimensions. Let use polar coordinates. The usual Euclidean metric tensor in polar coordinates is:
$$g = \begin{pmatrix}1 & 0 \\ 0 & r^2 \\ \end{pmatrix}$$

Suppose that our pullback metric is: $$\tilde {g} = \begin{pmatrix}1 & \phi \\ \phi & r^2 \\ \end{pmatrix}$$

The geodesics through the origin with this metric are the same straight lines as in the case of $$g$$. But, if we take $$c(t)$$ or, with the notation of the lecture notes, $$\sigma(t)$$ , for example

$$\sigma(t) = (d,t)$$ where d is a constant, then

$$\frac{d}{dt}\left|\hat\sigma(t)\right| = 0$$

but

$$\tilde {g}(\dot \sigma(t), \hat \sigma(t)) = t$$.

So it seems that the equality

$$\frac{d}{dt}\left|\hat\sigma(t)\right| = \frac{ \tilde {g}(\dot \sigma(t), \hat \sigma(t))}{\left| \hat \sigma(t) \right |}$$

in the lecture notes doesn't hold in this case. Is this really an error in the lecture notes, or I miss something?

Last edited:
mma said:
or I miss something?

Of course I missed.
The metric I provided isn't positive definite, i.e. it isn't a Riemannian metric.

## 1. What is a position vector in Riemannian spaces?

A position vector is a mathematical object that represents the location or position of a point in a Riemannian space. It is typically denoted by a lowercase letter with an arrow on top, such as r. In Riemannian geometry, position vectors are used to describe the curvature and distance of a space.

## 2. How is a tangent vector defined in Riemannian spaces?

A tangent vector is defined as a vector that is tangent to a curve or surface at a specific point on that curve or surface. In Riemannian spaces, tangent vectors are used to describe the direction of motion or change at a particular point. They are denoted by a lowercase letter with a prime symbol, such as r'.

## 3. Can you explain the relationship between position and tangent vectors in Riemannian spaces?

In Riemannian spaces, the position vector r and the tangent vector r' are closely related. The magnitude of the tangent vector r' at a specific point is equal to the rate of change of the position vector r at that point. In other words, the tangent vector represents the direction and speed of movement along the position vector.

## 4. How are position and tangent vectors used in Riemannian spaces?

Position and tangent vectors are essential tools in Riemannian geometry, as they allow us to describe the curvature and distance of a space. They are used to define important concepts such as geodesics (the shortest paths between points), curvature tensors (which measure the curvature of a space), and the Riemannian metric (which gives the distance between points).

## 5. Can position and tangent vectors be used in other mathematical contexts?

Yes, the concept of position and tangent vectors can be extended to other mathematical contexts beyond Riemannian spaces. For example, they are also used in vector calculus to describe the direction and rate of change of a function at a specific point. Additionally, they play a crucial role in differential geometry, where they are used to study the properties of smooth curves and surfaces in various types of spaces.

• Differential Geometry
Replies
21
Views
623
• Differential Geometry
Replies
6
Views
2K
• Differential Geometry
Replies
5
Views
2K
• Differential Geometry
Replies
7
Views
3K
• Differential Geometry
Replies
12
Views
3K
• Differential Geometry
Replies
5
Views
1K
• Differential Geometry
Replies
4
Views
2K
• Differential Geometry
Replies
9
Views
5K
• Differential Geometry
Replies
11
Views
3K
• Differential Geometry
Replies
63
Views
8K