# The necessity of using tangent space as the vectors in general relativity

I’ve recently learnt what manifolds are to prepare myself for a course in GR. My relevant mathematical background is linear algebra (abstract, proof-ish course) and multivariable/vector calculus (course mostly focussed on computations). I use Knuth and Renteln as my main references.

The idea of an n-dimensional manifold is introduced as a combination of open sets whose union forms the manifold. Each such open set must have a continuous 1-to-1 map to an open set in n-dimensional Euclidean space; that is: each point within these open sets can be described as an n-tuple, just like vectors and points in ”regular” space can. Please correct me if I’m wrong here, I really want to get my definitions right.

Subsequently the tangent space is introduced as a vector space for each point on the manifold, whose elements are differential operators. I recognise that that this vector space is very much usable to describe vector fields on the manifold.

But as I read the definition of manifolds, I intuitively expected the displacement between points to be a good definition for vectors on my manifold. If our manifold consists of multiple charts, this would not be possible, since the space would not be closed: if we added the displacement between points to itself long enough we would eventually “exit” the open set in which the tuples are defined, which is by definition not possible. I’m still interested in knowing why in general relativity we use tangent spaces instead of conventional classic coordinate tuples.

So my first question is: if a manifold is describable by a single chart, can we define a vector space simply by taking displacements between points on that chart as n-tuple vectors? For instance, take the manifold $$\mathbf{r} = (x,y,z)^T = (f_1(u,v), f_2(u,v), f_3(u,v))^T$$: a 2D-manifold embedded in 3D space. If the functions $$f_i$$ are well behaved so that the manifold is smooth and has no sharp edges/crossings/etc, wouldn’t $$(u,v)$$ tuples form a nice vector space? You could add them, scale them, and fulfil all the properties of a vector space.

My second question is more to manage my own expectations for later: is this single-coordinate chart situation one that occurs at all in G.R.? If each/most manifold(s) in G.R. naturally requires two or more charts, then it would be senseless to take these displacement tuples as vectors. The reason I still ask is because I intuitively expected that a single chart should encompass all of space; I would be very surprised if I was working out a physics problem and I couldn’t describe my worldline in the same coordinate system everywhere.

So my first question is: if a manifold is describable by a single chart, can we define a vector space simply by taking displacements between points on that chart as n-tuple vectors?

Not in general. Consider the real line as a manifold $$M$$, equipped with the chart $$(\mathbb R, x)$$ where the chart map $$x$$ is defined by

$$x:M \rightarrow \mathbb R$$ $$p \mapsto x(p) = \tan^{-1}(p)$$

I cannot define a vector space from the coordinates of points on the manifold, because the range of the chart map is only $$x(\mathbb R)=(-\pi/2,\pi/2)$$.

You might say that this is simply a bad choice of chart (should that matter?), but note that if the Riemann curvature tensor of the manifold does not vanish everywhere, then it is not possible to construct a globally Euclidean chart.

is this single-coordinate chart situation one that occurs at all in G.R.? [...]I would be very surprised if I was working out a physics problem and I couldn’t describe my worldline in the same coordinate system everywhere.

Polar coordinates have singular behavior at the coordinate origin; spherical coordinates have singular behavior at the origin as well as the poles. Even manifolds as mundane as $$\mathbb R^n$$ generically require more than one chart if you use non-cartesian coordinates.

It can be shown that if a manifold possesses intrinsic curvature then it cannot be described by a single, globally Euclidean chart. In other words, while it is possible to find very special cases in which you could find a single chart of the kind you want, it is only possible when the space possesses no curvature.

Essentially, you could (on rare occasions) artificially shoehorn a vector space structure into your spacetime if you want, but it would be generally pointless. Don't bother.

As far as I’m aware vectors do not need to follow any transformation requirements (though it’s nice if they do) but only a set of eight definitions. (comment on other answer)

This is the mathematical definition of a vector space. The vectors that puppetsock is referring to are tangent vectors to a manifold; those objects (or rather, their components in a given chart), which are the ones you'll be dealing with in GR, do possess certain transformation properties which follow from the requirement that tangent vectors be chart-independent geometrical objects.

• “It can...no curvature.” This is exactly the confirmation I needed! It’s not that I don’t understand tangent space, it’s just that I found it so silly to work with if the displacement between points sufficed as well. Now I know that it doesn’t (is a proof of this a standard part of diff geometry courses?), so thanks a lot! One last tiny question: in my OP I wrote that I figured in a multiple-chart manifold we could never use coordinate displacements as vectors since eventually you’d “walk out” of the open space on which those coordinates are defined. Was that correct (if perhaps a tad loose)? – Heatherfield Jul 29 '19 at 19:06
• @Heatherfield That's the general idea - or at least a component of it. Loosely speaking, a manifold admits a global chart which you could equip with a vector space structure if and only if the manifold is a smooth deformation of $\mathbb R^n$ for some $n$. There are many manifolds for which this is not the case - any compact manifold, like a circle or sphere, would serve as an example. More importantly, your vector space structure would only make sense for that chart - if you wanted to switch to different coordinates (polar, cylindrical, etc) the structure would not carry over. – J. Murray Jul 29 '19 at 19:44
• "if the Riemann curvature tensor of the manifold does not vanish everywhere, then it is not possible to construct a globally Euclidean chart." What exactly do you mean by a "globally Euclidean chart"? It is certainly possible to have spaces diffeomorphic to $\Bbb R^n$ (i.e. manifolds that can be completely covered by a single chart) which nevertheless have non-trivial curvature. – Danu Jul 30 '19 at 15:30
• @Danu By globally Euclidean chart, I mean a chart whose range is $\mathbb R^n$ and in which the connection coefficients all vanish. It's a stronger statement than was asked for, but I interpreted subtext of the question to be "why can't we just equip spacetime with an affine structure like we do in nonrelativistic physics?" in which case I felt that things like the connection would be more relevant. – J. Murray Jul 30 '19 at 16:27
• OK. When in your terminology a manifolds admits a "globally Euclidean chart", the usual terminology is to say that it is isometric to Euclidean space. As you note, this really isn't the notion that the OP asked about, so it's potentially confusing to bring it up here. For instance, many Riemannian manifolds do not admit any "Euclidean chart" (i.e. local isometry to $\Bbb R^n$) even locally, though every smooth manifold has many charts (local diffeomorphisms to $\Bbb R^n$). – Danu Jul 30 '19 at 20:36

You can define the tangent space in sort of the way you have in mind, but it's not enough to make the displacements small enough to keep from crossing into a different chart. You need to make them infinitesimal in size, and we also want a definition that's coordinate-independent. For a mathematically rigorous treatment in this style, see Nowik and Katz, "Differential geometry via infinitesimal displacements," https://arxiv.org/abs/1405.0984 .

• Thanks, I’ll save it for later when I feel more prepared to tackle a subject using nonstandard analysis (since I barely understand usual nonstandard analysis as is). – Heatherfield Jul 29 '19 at 19:42

Vectors must transform in a specific manner under a coordinate transform. It's not clear that your $${u,v}$$ necessarily forms a vector.

• As far as I’m aware vectors do not need to follow any transformation requirements (though it’s nice if they do) but only a set of eight definitions. A copy of the list is available on the relevant wiki page: en.m.wikipedia.org/wiki/Vector_space. Does this mean that I’m not looking to just define vectors on my manifold, but I’m looking to define a special kind? That’d seem strange to me, while I know some textbooks define tangent vectors in terms of their transformation properties, the resources I use derive these transformation properties as a handy consequence, not as an intended one. – Heatherfield Jul 29 '19 at 18:32