# Tangent space to the sphere

kent davidge
While studying Relativity I decided to take over a concrete case. So I thought of (what I think is) the simplest case which is the Sphere ##S^2##. So I want to construct the tangent space to the sphere. I think for this I need to embbed it in ##R^3##.

I worked out similar problems in the early Calculus 2 classes (Now I'm doing Calculus 3). But I'm struggling to get it complete.

So what I've being doing:

##p## = point on the sphere of radius ##r##
##x## = general point on ##R^3##
##(x - p)## = first basis vector
##(x - p) \cdot p \stackrel{!}{=} 0##
##\Rightarrow x^1p^1 + x^2p^2 +x^3p^3 = r^2##
This should give the components of the first basis vector

For the second basis vector, I think, it's necessary that it be orthogonal to both the first and to ##p##. So

##(y - p) \cdot p \stackrel{!}{=} 0## and ##(y - p) \cdot (x - p) \stackrel{!}{=} 0##

I solved this a couple of times even using mathematica to make sure nothing's wrong, but it's just a matter of substitute into some values to see that this doesn't give right results.

What am I doing wrong?

kent davidge
I know GR/SR deals with four dimensions, but as I said, I just wanted to get insights by looking at lower dimensions.

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While studying Relativity I decided to take over a concrete case. So I thought of (what I think is) the simplest case which is the Sphere S2S2S^2. So I want to construct the tangent space to the sphere. I think for this I need to embbed it in R3R3R^3.
You never need an embedding space to construct the tangent space. This is the entire point of using manifolds.

PeterDonis and Pencilvester
Mentor
So I want to construct the tangent space to the sphere. I think for this I need to embbed it in R3R3R^3.
You don’t need to embed it in R3. On coordinate notation, if your coordinates in S2 are ##(\theta,\phi)## then the vectors ##\partial_{\theta}## and ##\partial_{\phi}## are a pair of basis vectors for the space

Pencilvester
kent davidge
You don’t need to embed it in R3. On coordinate notation, if your coordinates in S2 are
In this way, how do you know what are the boundaries of your coordinates? That is, up to where you can strech the same tangent space.

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hat is, up to where you can strech the same tangent space.
Every tangent space exists only at one point. The tangent spaces of different points are different vector spaces.

PeterDonis and kent davidge
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In this way, how do you know what are the boundaries of your coordinates? That is, up to where you can strech the same tangent space.
The tangent space is unique to each point of the manifold, it cannot be stretched at all.

kent davidge
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While studying Relativity I decided to take over a concrete case. So I thought of (what I think is) the simplest case which is the Sphere ##S^2##. So I want to construct the tangent space to the sphere. I think for this I need to embbed it in ##R^3##.

You might want to do this as a visual aid, but you don't have to.

I worked out similar problems in the early Calculus 2 classes (Now I'm doing Calculus 3). But I'm struggling to get it complete.

So what I've being doing:

##p## = point on the sphere of radius ##r##
##x## = general point on ##R^3##
##(x - p)## = first basis vector

x-p isn't a basis vector :( So you are losing me at this point. And it most likely means that your understanding of what a basis vector is isn't quite right.

If you're trying to do what I think you're doing, which is constructing an embedding of the tangent plane to compare to the embedding of the sphere, we can proceed as folllows.

The tangent plane at point p is a plane that includes p, and is tangent to the sphere at p. Being tangent to the sphere means that the radial 3-vector at point p is orthogonal to the tangent plane.

I don't see any symbol in your post for a point on the tangent plane. x is a general point. Let y be a point on the tangent plane that is tangent to some point p on the sphere. . Let O be the origin of the sphere. Then we can write the equations that a point (in our embedding space, which is 3 dimensoinal) on the tangent plane must satisfy

$$(y-p) \cdot (p-O) = 0$$

We then have an embedding of the 2-sphere in a 3-d embedding space space, and we have an embedding of the tangent plane(s) (there is a different tangent plane for every point p on the sphere) in the 3-d embedding space. The tangent plane is just the set of points y that satisfy the equation we worked out above.

Basis vectors are something else. On our embedding diagram, they are vectors that lie within the tangent plane - any vector that lies in the plane can be made into a basis vector, one needs two basis vectors to span the tangent plane.

kent davidge
kent davidge
Thanks. That's sort of what I was thinking of, but giving in the values, and also trying to explicitaly find and give values for the basis vectors. And the map would go like this ##\mathbb{S}^2 \longrightarrow (0, 2\pi] \times (0, 2\pi]##, and then make the embedding from ##(0, 2\pi] \times (0, 2\pi]## to ##\mathbb{R}^3##.
In such case the coordinates ##(\theta, \varphi)## are bounded by ##(0, 2\pi]##. Is this correct?

Mentor
There is no embedding. The tangent space is just the space spanned by ##\partial_{\theta}## and ##\partial_{\phi}##. There is no map to R3 because neither of the basis vectors are vectors in R3.

kent davidge
There is no embedding. The tangent space is just the space spanned by ##\partial_{\theta}## and ##\partial_{\phi}##. There is no map to R3 because neither of the basis vectors are vectors in R3.
Yes, but I was trying to visualize it as an embedding in ##\mathbb{R}^3##.

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kent davidge
tangent space is unique to each point
I don't understand. Let's take ##\mathbb{S}^1## for simplicity. It, minus one of its points, is bijective to ##\mathbb{R}##, which is a vector space. So Why should we insist in using a different vector space for each of its points, when we can use a same vector space for all but one of its points?

weirdoguy
Why should we insist in using a different vector space for each of its points, when we can use a same vector space for all but one of its points?

We can because tangent bundle of a line is trivial. In general it is not and all tangent spaces are distinct because they are "attached" to different points of manifold. I think that your problem here is that you do not distinguish tangent space (vector space which is quite abstract in general) from tangent plane (affine subspace of 3D Euclidean space) to an embedded manifold. Visualising is good, but you must think more abstract! Especially in this case, because in 3-dimensional Euclidean space we can identify with each other way to many things which are different in general. Or even don't exist, like tangent plane to a general unembedded manifold.

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kent davidge
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I don't understand. Let's take ##\mathbb{S}^1## for simplicity. It, minus one of its points, is bijective to ##\mathbb{R}##, which is a vector space. So Why should we insist in using a different vector space for each of its points, when we can use a same vector space for all but one of its points?
If you just consider ##\mathbb{S}^1## minus one of its points to be equivalent to ##\mathbb{R}##, wouldn't you lose the curvature information? To retain curvature information, wouldn't you have to keep track of the point-by-point tangent bundle and its connection to other nearby tangent bundels?

PS. If you do want to embed ##\mathbb{S}^1##, you should embed it in ##\mathbb{R}^2##. Then you can at least talk about it being a circle.

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kent davidge
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Let's take S1S1\mathbb{S}^1 for simplicity. It, minus one of its points, is bijective to RR\mathbb{R}, which is a vector space.
That is pointless since there is no curvature in a 1D manifold. But S2 minus one point is similarly bijective to R2 which is a vector space.

However, we are not interested in establishing random mathematical bijections, we are interested in doing physics. Current physics is based on finding relationships between fields and their first and second derivatives.

Consider a 2D spherical manifold like the surface of the Earth minus the North Pole and fluid represented by a scalar density field and a vector velocity field. At Paris there is a vector whose bijection corresponds to Rome, but there is no sense in which the resulting velocity has anything to do with Rome. There is also no meaningful sense in which you could add Paris and Rome although such an operation is defined on the bijection vector space.

kent davidge
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I don't understand. Let's take ##\mathbb{S}^1## for simplicity. It, minus one of its points, is bijective to ##\mathbb{R}##, which is a vector space. So Why should we insist in using a different vector space for each of its points, when we can use a same vector space for all but one of its points?

I think the point is that there is no particular meaning to "pointing in the same direction" for vectors at different locations. An extreme case is a non-orientable manifold such as a Mobius strip. If you take a Mobius strip and start drawing arrows on it and you go all the way around trying to draw arrows pointing in the "same" direction, then when you get back to where you started, the arrow will be pointing in the opposite direction of the first arrow. What this shows is that on some manifolds, there is no meaningful notion of arrows pointing in the "same" direction, except locally.

Now, a Mobius strip isn't even curved, from the point of view of differential geometry. The fact that it's not a strip of 2D Euclidean space can only be discovered by going all the way "around". Within any bounded region of the Mobius strip that doesn't go all the way around, you can give a meaning to "same direction" that works in that region. A manifold with curvature is in an even worse situation---you can't even give a consistent meaningful notion of arrows pointing in the same direction in a small region (unless it's infinitesimally small). On a curved surface (such as the surface of the Earth), if you march around a closed loop, and try to maintain an arrow pointing in the same direction, when you get back to where you started, it will have shifted directions a little. A clear example is to start at the North Pole, with an arrow pointing south along the line of 0 degrees longitude. March down to the equator, and your arrow is still pointing south along the line of 0 degrees longitude. Now, go East along the equator to the line of 180 degrees longitude. Go back north to the North pole. At this point, your arrow will be pointing along the line of 180 degrees longitude, exactly the opposite direction from when you started.

Using the same vector space at every point in your manifold just doesn't make sense. In the Mobius strip example, having the same vector space for each point means at every point, having a pair of directions that you can call the x-axis (pointing in the direction "around" the strip) and another direction that you can call the y-axis (pointing from one edge toward the other). But if you try to do that, then at some point, your notion of the y-axis must shift discontinuously.

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kent davidge and FactChecker
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Now, a Mobius strip isn't even curved, from the point of view of differential geometry.
This would depend on the connection you put on it. It is certainly true that it is possible to pick a connection that has zero curvature.

That is pointless since there is no curvature in a 1D manifold. But S2 minus one point is similarly bijective to R2 which is a vector space.
To expand a bit on this: The lowest dimension in which the curvature tensor can be non-zero is two. This is due to the (anti)symmetries of the tensor.

kent davidge
kent davidge
Thank you all for the answers. I will give myself some time to think about them...