Spacetime as a Lorentzian Manifold

Posted on November 1, 2025 by Aronno Mirdha

One of the central tasks in differential geometry is to make precise the notion of length and angle on a smooth manifold. Unlike $\mathbb R^n$, a general manifold comes with no preferred inner product. The metric tensor is not something that “appears” automatically; rather, it is a geometric structure we deliberately introduce. The goal of this post is to derive the metric tensor from first principles in a clean, logically economical way.

We begin with a smooth $n$-dimensional manifold $M$. At each point $p\in M$, we want to measure lengths of infinitesimal displacements through $p$. Infinitesimal displacements are modeled by tangent vectors, so the correct object to define is an inner product on each tangent space $T_pM$.

A precise and coordinate-free definition of tangent vectors is the following. A tangent vector at $p$ is a derivation at $p$: a linear map $X:C^\infty(M)\to\mathbb R$ satisfying the Leibniz rule $X(fg)=f(p)X(g)+g(p)X(f)$. The collection of all such derivations forms a real vector space $T_pM$ of dimension $n$.

Now introduce a coordinate chart $(U,\varphi)$ with $\varphi=(x^1,\dots,x^n)$ and $p\in U$. For each coordinate function, define a derivation by

$$ \left.\frac{\partial}{\partial x^i}\right|_p (f) := \frac{\partial (f\circ\varphi^{-1})}{\partial u^i}\Big|_{\varphi(p)}. $$

These derivations form a basis of $T_pM$. Every tangent vector $X\in T_pM$ can therefore be written uniquely as

$$ X = X^i \left.\frac{\partial}{\partial x^i}\right|_p $$

for real coefficients $X^i$. These coefficients depend on the chosen coordinates, but the vector $X$ itself does not. At this point, nothing like length exists yet. To measure length, we must specify an inner product on each tangent space.

A Riemannian metric on $M$ is a rule that assigns to each point $p\in M$ a bilinear map

$$ g_p : T_pM \times T_pM \to \mathbb R $$

such that:

  1. $g_p$ is symmetric: $g_p(X,Y)=g_p(Y,X)$
  2. $g_p$ is positive definite: $g_p(X,X)>0$ for all $X\neq 0$
  3. The assignment $p\mapsto g_p(X_p,Y_p)$ is smooth whenever $X,Y$ are smooth vector fields

This definition is entirely coordinate-free. The metric is simply a smoothly varying inner product on tangent spaces.

Now fix a coordinate chart $(x^1,\dots,x^n)$. Since $g_p$ is bilinear, it is completely determined by its values on basis vectors. Define

$$ g_{ij}(p) := g_p\!\left(\left.\frac{\partial}{\partial x^i}\right|_p,\left.\frac{\partial}{\partial x^j}\right|_p\right). $$

These functions $g_{ij}$ are smooth, symmetric in $i,j$, and vary from point to point. They are the components of the metric tensor in coordinates.

Given two tangent vectors

$$ X = X^i \frac{\partial}{\partial x^i}, \qquad Y = Y^j \frac{\partial}{\partial x^j}, $$

bilinearity immediately gives

$$ g_p(X,Y) = g_{ij}(p)\, X^i Y^j. $$

This is the familiar coordinate expression of the metric. Importantly, this is not a definition but a representation of the underlying geometric object $g$.

From this formula, the squared length of a tangent vector $X$ is

$$ |X|^2 = g_{ij}(p)\,X^i X^j. $$

Thus, the metric tensor generalizes the Euclidean dot product by allowing the coefficients $g_{ij}$ to vary with position and coordinates.

It is instructive to check how these components transform. If we change coordinates from $x^i$ to $\tilde x^a$, then the basis vectors transform via the chain rule:

$$ \frac{\partial}{\partial \tilde x^a} = \frac{\partial x^i}{\partial \tilde x^a}\frac{\partial}{\partial x^i}. $$

Substituting into the definition of the metric components yields

$$ \tilde g_{ab} = g_{ij}\frac{\partial x^i}{\partial \tilde x^a}\frac{\partial x^j}{\partial \tilde x^b}. $$

This transformation law is exactly what characterizes $g_{ij}$ as the components of a $(0,2)$-tensor field. The metric tensor is therefore not merely a matrix-valued function but a genuine tensorial object on the manifold.

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