Bernhard Riemann
1. The Metric
Two posts ago we found the gradient problem. Take a function on a manifold, differentiate it, and you get \(df\), which is a covector, a stack of contour lines. In ordinary vector calculus the gradient is a vector. There's no way to turn one into the other, because their components transform with inverse Jacobians and no chart-independent rule converts between them. Last time we built tensors of every type and still couldn't measure anything. Nothing so far assigns a length to a vector. So we take on the metric!
1.1 What it needs to give us
We want lengths and angles. In \(\mathbb{R}^3\) both come from the dot product. What does a dot product do? It eats two vectors and returns a number, linearly in each. We saw that! It’s a type \((0,2)\) tensor. So define a metric to be a \((0,2)\) tensor \(g\) that is\ Symmetric, \(g(X,Y) = g(Y,X)\), so \(g_{\mu\nu} = g_{\nu\mu}\).\
Non-degenerate, meaning the only vector orthogonal to everything is zero: if \(g(X,Y) = 0\) for all \(Y\) then \(X = 0\). Equivalently \(\det(g_{\mu\nu}) \neq 0\), so the matrix of components is invertible.\
Then define the squared length of a vector and the interval as
\[ |X|^2 = g_{\mu\nu} X^\mu X^\nu, \qquad ds^2 = g_{\mu\nu} dx^\mu dx^\nu \tag{1} \]
Notice we are choosing this. A manifold has no metric until we give it one, and different choices give different geometries on the same space. In that sense, the manifold is like the stage, but the metric is the physics on that stage. This is not that strange - we can also choose different inner products for a given space. That’s the same thing - the metric just acts as the generalized inner product. Also, nothing says \(|X|^2 > 0\). That gives us Riemannian geometry, which is the geometry of ordinary curved surfaces.
1.2 Raising and lowering
Now the gradient problem is not a problem. Given a vector \(X\), define a covector by feeding \(X\) into one slot and leaving the other open:
\[ X_\mu = g_{\mu\nu} X^\nu \tag{2} \]
That's a covector, since it eats a vector and returns a number, namely \(g(X, \cdot)\). And because \(g\) is non-degenerate, the map \(X \mapsto X_\mu\) is invertible. Write the inverse matrix as \(g^{\mu\nu}\), defined by
\[ g^{\mu\lambda} g_{\lambda\nu} = \delta^\mu_\nu \tag{3} \]
and it raises indices back:
\[ X^\mu = g^{\mu\nu} X_\nu \tag{4} \]
Just to check it is consistent. Lower then raise:
\[ g^{\mu\lambda} X_\lambda = g^{\mu\lambda} g_{\lambda\nu} X^\nu = \delta^\mu_\nu X^\nu = X^\mu \tag{5} \]
Back where we started. So the two operations are inverses, and vectors and covectors are now interchangeable. The gradient is now available. Take \(df\), which has components \(\partial_\mu f\), and raise:
\[ (\nabla f)^\mu = g^{\mu\nu} \partial_\nu f \tag{6} \]
That is the vector everyone calls the gradient. It was never a natural object! It's the differential with a metric applied. In \(\mathbb{R}^3\) with Cartesian coordinates \(g^{\mu\nu} = \delta^{\mu\nu}\), so raising does nothing really, and they’re basically the same. Also, raising one index of the metric with the inverse metric gives \(g^\mu{}_\nu = g^{\mu\lambda}g_{\lambda\nu} = \delta^\mu_\nu\). The mixed metric is just the Kronecker delta, which is why you never see it written.
1.3 The sphere
Last post I mentioned the round metric in stereographic coordinates without proof. Let's actually get it, since we have everything needed. The sphere sits in \(\mathbb{R}^3\) where the length of a small displacement is \(ds^2 = dx^2 + dy^2 + dz^2\). So we need \(x, y, z\) in terms of the chart coordinates \(u, v\). We got \(u = x/(1-z)\) and \(v = y/(1-z)\), and then \(z = (s-1)/(s+1)\) where \(s = u^2+v^2\). Then \(1 - z = 2/(s+1)\), so
\[ x = \frac{2u}{s+1}, \qquad y = \frac{2v}{s+1}, \qquad z = \frac{s-1}{s+1} \tag{7} \]
Now differentiate. Writing \(D = s+1\) and using \(\partial s/\partial u = 2u\):
\[ dx = \frac{2\,du}{D} - \frac{2u \cdot 2u\,du + 2u \cdot 2v\,dv}{D^2} = \frac{2}{D^2}\left[ (D - 2u^2)du - 2uv\,dv \right] \tag{8} \]
and by the same computation with \(u\) and \(v\) exchanged,
\[ dy = \frac{2}{D^2}\left[ -2uv\,du + (D - 2v^2)dv \right] \tag{9} \]
For \(z\), differentiating \((s-1)/(s+1)\) gives \(dz = 2\,ds/D^2\), so
\[ dz = \frac{4}{D^2}\left( u\,du + v\,dv \right) \tag{10} \]
Now square and add. The \(du^2\) coefficient is
\[ \frac{4}{D^4}\left[ (D-2u^2)^2 + 4u^2v^2 + 4u^2 \right] \tag{11} \]
Expand the bracket, using \(D = 1 + u^2 + v^2\) so that \(D - 2u^2 = 1 - u^2 + v^2\):
\[ (1 - u^2 + v^2)^2 + 4u^2v^2 + 4u^2 \tag{12} \]
The first term expands to \(1 + u^4 + v^4 - 2u^2 + 2v^2 - 2u^2v^2\). Adding \(4u^2v^2 + 4u^2\) gives
\[ 1 + u^4 + v^4 + 2u^2 + 2v^2 + 2u^2v^2 = (1 + u^2 + v^2)^2 = D^2 \tag{13} \]
So the \(du^2\) coefficient is \(4/D^2\). By symmetry the \(dv^2\) coefficient is the same, and the cross terms cancel, since the \(du\,dv\) contributions from \(dx^2\) and \(dy^2\) are \(-2uv(D-2u^2)\) and \(-2uv(D-2v^2)\) scaled, and adding the \(dz^2\) contribution \(8uv\) kills them. We get:
\[ ds^2 = \frac{4}{(1+u^2+v^2)^2}\left( du^2 + dv^2 \right) \tag{14} \]
Two things we get. First, the metric is a positive function times \(du^2 + dv^2\), so angles measured in the chart are the true angles on the sphere, and the stereographic projection is conformal. And the factor shrinks as \(u^2+v^2\) grows, so regions that look enormous far out on the chart are actually tiny patches near the north pole. The chart distorts sizes wildly and angles not at all.
1.4 Signature
The components \(g_{\mu\nu}\) form a symmetric matrix, so at any point it can be diagonalised, and by rescaling the basis the diagonal entries can be made \(\pm 1\). How many of each is the signature, and Sylvester's law of inertia says it can't change from point to point on a connected manifold. All plus signs is Riemannian. Lengths are positive, and this is the geometry of surfaces. One minus and the rest plus is Lorentzian:
\[ \eta_{\mu\nu} = \operatorname{diag}(-1, +1, +1, +1) \tag{15} \]
Now \(|X|^2\) can have either sign or vanish for nonzero \(X\). Vectors split into three classes that no coordinate change can mix: timelike with \(|X|^2 < 0\), spacelike with \(|X|^2 > 0\), and null with \(|X|^2 = 0\). The null vectors form a cone at each point, and it's the light cone. In fact we get a nice connection to group theory here. The Lorentz group was defined there as the matrices preserving this \(\eta\), exactly as \(O(n)\) was the matrices preserving \(\delta\). So the metric was the invariant object there.
1.5 What the metric gives us
Lengths of curves. Integrate \(ds\) along a path.\
Angles. From \(\cos\theta = g(X,Y)/(|X||Y|)\), in the Riemannian case.\
Volumes. The factor \(\sqrt{|\det g|}\) makes integration chart-independent, and this is where the Jacobian from the earlier post reappears.\
Geodesics. The straightest paths are those extremizing length, and the metric determines them.\
Curvature. Everything in the Riemann tensor is built from the metric and its derivatives.\
Which is the claim underneath general relativity. One symmetric \((0,2)\) tensor field on a four-dimensional manifold, and that is all of gravity! Einstein really was… Einstein! Clever guy. Now we have a manifold, tensors, and now a metric. What we still can't do is differentiate. \(\partial_\mu T^\nu\) is not a tensor, because differentiating the transformation law produces second-derivative terms from the product rule. So the most basic operation in physics (asking how a field changes from point to point) has no meaning yet. Fixing it needs a way to compare vectors in different tangent spaces.