Why the unit vector is represented as a partial derivative in GR?

Question

Can someone give a good intuitive explanation why we represent the unit vector as a partial derivative in GR and what does it mean?

Answer 1

In general a manifold M (which is the topological space that describes spacetime mathematically) is not a vector space; it does not admit the linear structure and that's because a manifold is defined as a union of patches that contain mappings from the manifold to a point in the space $R^n $ (also the coordinates are given by projection mappings to each one of the $R^n $ "axes", but let's not consider this here).

Now, on defines a think called tangent vector by two equivalent(as can be shown) ways:

1. Either as a collection of equivalent curves passing from an element of the manifold M. For this we consider the real line $R $ and we define a mapping from the line to the manifold- we say that a tangent vector is a class of equivalent curves with equivalent relation the fact that are tangent because in general there are many curves that pass from the point in the manifold. Note also that the curve is defined as the map itself and not the value(image) it takes on the manifold- it is the process.
2. Or either as a derivation, that is as a map from the functions on M to the real numbers, with the properties $$u(f+g)=uf+ug, f,g \in F(M), $$ u the derivation $$u(rf)=ru(f), r\in R $$ $$u(fg)=u(f)g+fu(g) $$, it satisfies the Leibniz property.

Finally we also define the tangent space as the set of all tangent vectors and it can be shown that this is a vector space( the sum of two tangent vectors at a point gives a third in the the same point). In the alternative second perspective, the space of derivations can be defined as a vector space.

Now comes the important part: One can think of a directional derivative on M as the usual derivative of the function that defines the curve that defines the tangent vector. So, a "directional derivative" is $$u(f):=\frac{df(\sigma(t)}{dt}|_{t=0} $$ and where $\sigma $ is the curve with the set $[\sigma] $ of all curves that pass from the same point being the tangent vector by definition.

But also, one can define a "partial derivative" by thinking of the derivation definition and the projection maps $\phi^{\mu} $ of the maps that define the manifold from charts $(U,\phi) $ : $$\frac{\partial}{\partial x^{\mu}}_{p} f := \frac{\partial}{\partial \phi^{\mu}}f \cdot \phi^{-1}|_{\phi (p)} .$$ This is a partial derivative in the usual sense because it is defined locally on the local representatives, that is by the functions that connect the charted elements of M with the real numbers. This is also a way of understanding that the definition of a manifold gives it the euclidean, linear structure locally- a sentence much used for conceptualizing a manifold.

Finally, a vector field is defined as an assignment of tangent vector to all point in M and it also carries a vector space structure. The fact that it can be shown that the space of all vector fields Vfld has the structure if a Lie algebra can be related to the notion that a vector field may be thought of as a generator of infinitesimal transformations on the manifold, of the infinite dimensional group of diffeomorphisms, but here there are many mathematical subtleties irrelevant for a first reading.

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• A nice introductory text from where I've read most of the above is C.J. Isham's Modern Differential Geometry for Physicists.
• Also, another reference is of Bernard Schutz : Geometrical methods of differential physics
• and Differential Manifolds and Theoretical Physics, by W.D.Curtis and F.R.Miller

Answer 2

We'd like to say that a (unit) tangent vector is a direction on a manifold. But we can only define and distinguish directions because there must be something different about different points on the manifold, that is, we have a non-constant 'testing' function. So, the vector is the direction in which we differentiate functions defined on manifold. Hence, the notation of partial derivative.