# Why do we use the tangent and cotangent space?

## Main Question or Discussion Point

So i was considering minkowski space which is a 4-d manifold, why is that we use the tangent and cotangent space, to construct tensors on the space?

The definition of a manifold says that the space is locally homeomorphic to Euclidean space. So is the tangent space and cotangent space defined at each point used as a substitute for this homeomorphic euclidean space?

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Matterwave
Gold Member
Vector spaces must be linear spaces. You have to be able to add and subtract vectors, etc. Curved spaces, which, in general, spacetimes are, are not linear (no good method to define how to add 2 curved arrows), which is why you have to use the tangent space and cotangent space.

Since Minkowski spacetime is flat, you don't have to use the tangent and cotangent construction on it. Probably whichever text you are reading uses them in Minkowski spacetime so the transition to curved spacetime is not so jarring.

pervect
Staff Emeritus
So i was considering minkowski space which is a 4-d manifold, why is that we use the tangent and cotangent space, to construct tensors on the space?

The definition of a manifold says that the space is locally homeomorphic to Euclidean space. So is the tangent space and cotangent space defined at each point used as a substitute for this homeomorphic euclidean space?
The tangent and cotangent spaces transform differently. If you take a simple example of one variable where x' = f(x), one type of vector transforms by being multiplied by $\partial x' / \partial x$, the other type transforms by being multipled by the inverse, $\partial x / \partial x'$. So, if you say that x' = 2x, in one sort of vector the components double, in the other sort the components are cut in half.

Tangent and cotangent vectors are duals. If you start off by defining a basic vector, the dual vector space is the same dimension as the original vector space and a dual vector is a linear map from vectors to scalars.

The dual of a dual vector is the same (transforms in the same manner) as the original vector space.

The dot product of two vectors is a scalar, and is linear. You can regard the dot product as defining a map from a vector to a dual vector. I.e. if a and b are vectors, then the existence of the dot product of (a.b) defines a map from a to (a map from b to a scalar), i.e. a map from a to a dual vector, as a map from a vector to a scalar is by definition a dual vector.

The tangent and cotangent spaces transform differently. If you take a simple example of one variable where x' = f(x), one type of vector transforms by being multiplied by $\partial x' / \partial x$, the other type transforms by being multipled by the inverse, $\partial x / \partial x'$. So, if you say that x' = 2x, in one sort of vector the components double, in the other sort the components are cut in half.

Tangent and cotangent vectors are duals. If you start off by defining a basic vector, the dual vector space is the same dimension as the original vector space and a dual vector is a linear map from vectors to scalars.

The dual of a dual vector is the same (transforms in the same manner) as the original vector space.

The dot product of two vectors is a scalar, and is linear. You can regard the dot product as defining a map from a vector to a dual vector. I.e. if a and b are vectors, then the existence of the dot product of (a.b) defines a map from a to (a map from b to a scalar), i.e. a map from a to a dual vector, as a map from a vector to a scalar is by definition a dual vector.
Just to follow up a little on what pervect has so ably pointed out here, I've sketched a space-time diagram showing an example of the utility of the dual spaces. Here we show how you can regard the spaces for any two observers moving relative to each other as dual. You can see how a metric comes out of this relationship along with the time coordinates Lorentz transformation (time dilation).

DrGreg
Gold Member
Sorry, bobc2, but pervect's use of the term "dual" has a specific technical meaning for tensors which isn't the informal way you are using it.

pervect
Staff Emeritus
One of my textbooks represents vectors in the usual way (litle arrows with a head and tail), and dual vectors as stacked parallel planes. (Stacked planes works for 3d spaces - in general if you have an n-dimensonal space, the "planes" are n-1 dimensonal surfaces).

If you have a vector and a dual vector, because the dual vector maps a vector to a scalar, you also have an associated scalar. This scalar is the number of planes that the vector pierces (including fractional piercings).

Most any linear algebra textbook should have some discussion of vectors and their duals. In engineering notation, vectors and duals are represented with row and column vectors. In more advanced tensor notation, they are representedwith subscripts and superscripts instead.

I understand what the tangent and contangent spaces are and how you can find the basis for them. The cotangent space being the dual space corresponding to the tangent space. Tangent vectors transform contravariantly and cotangent vector transform covariantly.

I'm not sure why they are used, what is the relation between the local Euclidean space at a point on the manifold and the tangent/cotangent spaces at a point?

DrGreg
Gold Member
In your opening question you specifically referred to Minkowski space, which we use in special relativity in the absence of gravity. Minkowski space is isomorphic to its own tangent spaces, so you don't really need to use tangent spaces in this case.

When you move on to curved spacetime in general relativity, the manifold and its tangent spaces are different: the manifold is curved but each tangent space is flat. Each tangent space is a vector space (you can add 4-vectors and multiply them by scalars) but the manifold is not.

The tangent space arises when you differentiate; for example if $x^\alpha(\lambda)$ is a worldline parameterised by $\lambda$, then $dx^\alpha/d\lambda$ lies in the tangent space.

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Minkowski space is isomorphic to its own tangent spaces
Is there any way of showing this?

I've just read that the tangent/cotangent space is isomorphic to the local Euclidean space so would this imply that Minkowski space is itself locally isomorphic to Euclidean space?

Is there any way of showing this?

I've just read that the tangent/cotangent space is isomorphic to the local Euclidean space so would this imply that Minkowski space is itself locally isomorphic to Euclidean space?
Yes. That is true.

Sorry, bobc2, but pervect's use of the term "dual" has a specific technical meaning for tensors which isn't the informal way you are using it.
The "informal" presentation does not change the concept. The technical meaning is the same as pervect's.

The1337gamer said:
so would this imply that Minkowski space is itself locally isomorphic to Euclidean space?
Yes. That is true.
Isomorphic as vector spaces, as affine spaces, and as smooth manifolds (diffeomorphic), but not as pseudo-Riemannian manifolds (i.e. not isometric)?

Matterwave
Gold Member
Is there any way of showing this?

I've just read that the tangent/cotangent space is isomorphic to the local Euclidean space so would this imply that Minkowski space is itself locally isomorphic to Euclidean space?
I basically covered this in my first post, perhaps you didn't see it?

lavinia
Gold Member
So i was considering minkowski space which is a 4-d manifold, why is that we use the tangent and cotangent space, to construct tensors on the space?

The definition of a manifold says that the space is locally homeomorphic to Euclidean space. So is the tangent space and cotangent space defined at each point used as a substitute for this homeomorphic euclidean space?
The tangent space of a manifold at a point is a vector space of linear operators on functions that are differentiable at that point. The points of the manifold are not.

The cotangent space at a point is the space of virtual displacements at that point. The points of the manifold are not.

For instance the Minkowski metric is a tensor defined in terms of products of virtual displacements.