Tangent spaces and local approximations of manifolds

Question

I'm approaching differential geometry from a physicist's perspective in the hope of understanding GR more thoroughly.

I've been told that, intuitively, the tangent space $T_{p}M$ to a point $p$ on a manifold $M$ is "the best linear approximation to the manifold $M$ at that point". What is meant by this?

Is it meant in the sense that the tangent vectors at that point provide the best linear approximation of functions on the manifold at that point? Does this extend for a sufficiently small neighbourhood around a given point?

In the context of GR, is this a mathematical implementation of the equivalence principle, in the sense that the $T_{p}M$ is flat and so the laws of physics are those of special relativity (SR) on $T_{p}M$. The laws of physics on $M$ are therefore SR for a sufficiently small neighbourhood of $M$ around a given point?

Answer 1

I've been told that, intuitively, the tangent space $T_{p}M$ to a point $p$ on a manifold $M$ is "the best linear approximation to the manifold $M$ at that point". What is meant by this?

This statement is meaningful if your manifold is embedded in an higher dimensional Euclidean space $\mathbb{R}^{n}$. It is very much like a linear tangent to a $1D$ curve. For example, see this illustration from Wikipedia of a tangent plane to a sphere

Lets look at manifolds embedded in $\mathbb{R}^{3}$. If, for example, you can write your manifold as $z=f\left(x,y\right)$ around $p$, and you expand this function into its Taylor series

$$z=f\left(x,y\right)\approx f\left(p\right)+\nabla f\left(p\right)\cdot\left(\left(x,y\right)-p\right)+\dots$$

then

$$z=f\left(p\right)+\nabla f\left(p\right)\cdot\left(\left(x,y\right)-p\right)$$

is the tangent place. $T_{p}M$ is this plane as a vector space.

In the context of GR, is this a mathematical implementation of the equivalence principle, in the sense that the $T_{p}M$ is flat and so the laws of physics are those of special relativity (SR) on $T_{p}M$. The laws of physics on $M$ are therefore SR for a sufficiently small neighbourhood of $M$ around a given point?

Yes. It means that by changing coordinate system at a point, your metric can be transformed into the Minkowski metric $\eta_{\mu\nu}$.

Answer 2

This interpretation is not a good one when we talk about manifolds from a intrinsic point of view (i.e., not embedded in Euclidean space), and understanding the intrinsic point is part of the requirements and power of differential topology and also GR.

One good point of view is the following:

Suppose you are at a point $p$ on a manifold: imagine the surface of the Earth, for instance. How would you be able to distinguish directions intrinsically, using only information from Earth itself? You can try looking at the curves passing through $p$. There are some which are obviously distinct from the others in terms of directions, but there are some which may be getting very close to themselves. They are not equal as curves, but it seems like they are "tangentially" equal. Well, now you are in trouble, because if you wanted to measure if they are tangentially equal or not directly, you would need to know what are the "geometrical" tangent vectors: an information which you may not have. You could, if you knew you were inside an Euclidean space, but the fact is that you don't. What now?

Let's be more primitive about what we mean by directions. Our view of tangents is fairly intricate if you think about it. Let's discard it for a while. Imagine you and me are at the point $p$, and then we agree to start moving. I feel that it is getting hotter, and you feel that is getting colder. Well, certainly, whatever definition of direction we use, this should mean that we've gone to different directions. Now let's come back to the point $p$, and start moving again. This time, we did not sense such a difference on heat, but instead on pressure. We must have gone different directions again. Heat, pressure etc are all things intrinsic to Earth. This idea may go somewhere.

Now, let's take a small leap of faith: we propose, as inhabitants of the Earth, that there are sufficiently many measurements (heat, pressure etc) that we can do in order to fully distinguish directions.

Let's mathematize the discussion above. We are at a point $p \in M$. A measurement is a smooth function $f:M \to \mathbb{R}$. We are following curves $\gamma: I \to M$ such that $\gamma(0)=p$. Now, my curve then determines an object $D_{\gamma}$ which takes the measurements and give me how they are varying when I go through $p$, i.e., it is an object of the form $$D_{\gamma}: C^{\infty}(M) \to \mathbb{R}$$ $$f \mapsto (f \circ \gamma)'(0). $$ This is, by definition, a tangent vector: the operator $D_{\gamma}$.

And you can check that there are sufficiently many smooth functions on the manifold $M$ to guarantee that we can distinguish the tangent space as a $n$-dimensional vector space as expected.

Answer 3

The following is not really a mathematical answer, but here we go anyway, to clear up some misconceptions.

There is no such thing as "laws of physics on $T_pM$ are those of special relativity", for the following reason. The vectors in the space $T_pM$ is only supposed to give you some directions to point towards. The actual spacetime is just the manifold $M$. More precisely, assuming you already know what an event (a point in $\Bbb R^4$ spacetime) is in the context of special relativity, an event in the context of general relativity is a point on the manifold $M$. The vectors/points on $T_pM$ are not events by definition.

The following statement is one I am not even sure about (I haven't really studied general relativity/pseudo-Riemannian geometry yet). You should ask an expert to double-check.

Since there is the notion of curvature in GR but not in SR (or more precisely, the curvature in SR is simply zero everywhere), it cannot be the case that laws of physics on a small open set in $M$ is the same as those of SR unless the curvature on the small open set is zero.