# General relativity -- How are components of $ds^2$ specified?

Suppose we take $M$ to denote the spacetime manifold. Suppose also that $ds^2 = g_{\mu \nu} dx^\mu dx^\nu$. I have some confusions with regards to the metric and the line elements.

My main confusion is at which points in the manifold are $ds^2$ defined? Is it correct that if $p \in M$, then it only makes sense to fix $ds^2$ for each tangent space $T_pM$, or is it fixed for each $v \in T_pM$? More specifically:

1. If there is some position dependent line element, such as the Schwartzschild line element, then $g_{\mu \nu}= g_{\mu \nu}(t,r, \theta, \phi)$. Are these $(t,r, \theta, \phi)$ the coordinates of some $(t,r, \theta, \phi) \in M$, or are they the coordinates of some $(t,r, \theta, \phi) \in T_pM$?

2. Are the differentials $dx^\mu(a)$ evaluated at each $a \in M$ or $a \in T_pM$?

3. Since all vectors are elements in some tangent space, it seems to me that for 4-dimensional spacetime we would need to specify 8 coordinates to make precise which vector we are talking about. The first 4 coordinates would specify the tangent space, and the last 4 coordinates would specify the vector inside of the tangent space. However, when I try and do exercises, only 4 coordinates are given. Are the remaining 4 coordinates given implicitly?

1) When we call something coordinates in this context, they always refer to coordinates in $M$. A tensor field is a function from $M$ to a tensor bundle, so the coefficients $g_{\mu\nu}$ are functions of points of $M$. If you coordinatize $M$, then you can view the coefficients as functions of the coordinates. So the coordinates $(t,r,\vartheta,\varphi)$ are labels for points of $M$, NOT $T_pM$.

2) The differentials are covectors, which, like vectors are associated with points of $M$. So $dx^\mu(p)$ is evaluated at a point of $M$.

3) You are right, and that is what the previous answer (the one you dismissed) referred to. To give an arbitrary vector of $M$, you need to specify 8 coordinates, coordinates of a point $p$, and the components of a vector at $p$.

You can say that $M$ is a 4-dimensional manifold, yo it can be coordinatized by 4 coordinates, and we have the tangent bundle $TM$, which can be seen as the set of all vectors of $M$. The tangent bundle is a 8 dimensional manifold, because a point of $TM$ (which is a vector of $M$) can be specified by giving the tuple $(p,v)$, where explicitly $p$ is described by the manifold coordinates and $v$ is described by the vector components taken with respect to the manifold coordinates.

When you do exercises, it depends on the exercise. If you are given a vector in the form $V^\mu=(\cos(\varphi),r^2e^{-r}\sin(\vartheta),0,0)$, for example, then you are actually given a vector field. Here the manifold points are implicit. If you want to ask what is the value of $V$ at the point $(t,r,\vartheta,\varphi)$, you insert the desired values of the coordinates into the functional expression, then the vector you get is the vector that is located at the point $p$ whose coordinates you have given.

If you are given a vector $v^\mu=(1,2,-5,4)$ for example, then either this is a vector field that has constant components in you coordinate system, or the exercise needs to specify a point $p$ of $M$ where this vector is located. Either case, you actually have 8 numbers. 4 for a point and 4 for components.

Lastly, here's a note that might clear some of your confusion:

What do we mean under coordinates?

In the context of vector spaces, we often say that if $e_1,...,e_n$ is a basis of $V$, then if we give $x=x^1e_1+...x^ne_n$, then the numbers $x^1,...,x^n$ are the coordinates of the vector $x$. It is because if we view $V$ as a space of positions, then technically speaking this "linear algebraic coordinate system" is also a coordinate system in the manifold sense.

But in physics we usually have two kinds of vectors. One that points between two points (position/separation vectors) and one that has a direction and magnitude, but which sits comfortably at a single point. The latter case, we almost never say that "coordinates of a vector", we say "components of a vector", because the latter kind of vector can never denote a spatial position.

In differential geometry, we only have one kind of vector, and its the latter case, eg. vectors at points.

So in differential geometric language, when we say coordinates, we mean labels that denote points of the manifold. When we say components, we mean the expansion coefficients of pointwise-defined linear objects (vectors, tensors, spinors, connections etc.) in some given basis.

These two concepts can be unified, precisely by the use of bundles. Imagine we have a point $p$ whose coordinates are $(x^0,x^1,x^2,x^3)$ and a vector $v$ at the point $p$, whose components are $(v^0,v^1,v^2,v^3)$.

Then from the point of view of "bundle-less differential geometry", the labels $(x^0,x^1,x^2,x^3)$ are coordinates, but the labels $(v^0,v^1,v^2,v^3)$ are components.

On the other hand, if we switch to the tangent bundle $TM$, then we can say that the 8-component tuple $(x^0,x^1,x^2,x^3,v^0,v^1,v^2,v^3)$ are coordinates of the tangent bundle, because fro mthe point of view of the tangent bundle, the vector $v$ at $p\in M$ is a point of $TM$, and not a vector of $TM$.

A point $p=(t,r, \theta, \phi)$ can be seen as an element of $M$. Note that $T_pM$ denotes the tangent space at $p$ whose elements are tangent vectors, so there is no sense $p$ is an element of $T_pM$.

One can say however the tuple $(p, v)$ consisting of a point and a vector is an element of the tangent bundle $TM$. However, in most mathematical literature, we often identify $T_pM$ with its image under the canonical injection $v \mapsto (p,v)$ and context may determine what notation is most appropriate to consider for an element of the tanegnt bundle.

• Was my question unclear? I am not asking if p is in TpM. Commented Apr 13, 2017 at 21:39