If we don't need a Riemannian metric to compare length of vectors, why do we use metrics to measure curvature?

Question

I read that, in the absence of a Riemannian metric tensor field, we can still measure how much a vector changes when parallel transported around a curve by comparing the initial and final vectors.

Then why do we need Rimeannian metrics to define lengths and measure curvature?

Answer 1

I am assuming your knowledge of tensor analysis is on the level of Weinberg, which I know you have read or are reading. When I say "usual" or "familiar," I am probably referring to material present in that text.

The curvature tensor is not actually defined in terms of the metric, but rather the affine connection. The general definition of the affine connection is

Definition 1. An affine connection on a manifold $M$ is a mapping $\nabla$ which assigns to every pair $X$, $Y$ of $C^\infty$ vector fields on $M$ another $C^\infty$ vector field $\nabla_X Y$ with the following properties:

1. $\nabla_XY$ is $\mathbb{R}$-bilinear in $X$ and $Y$

2. If $f\in\mathcal{F}(M)$ [set of $C^\infty$ functions], then $$\nabla_{fX}Y=f\nabla_XY,\quad \nabla_X(fY)=f\nabla_XY+X(f)Y$$

Such a mapping $\nabla$ is also called a covariant derivative.

Note, no mention of the metric. Next, we define the so-called Christoffel symbols.

Definition 2. Set, relative to a chart $(U,\{x^i\}_{1\le i\le n})$, $$\nabla_{\partial_i}\partial_j=\Gamma^k_{\;ij}\partial_k$$ The $n^3$ functions $\Gamma^k_{\;ij}$ are called the Christoffel symbols of the connection $\nabla$ in the given chart.

From this we can derive the usual coordinate transformation of the Christoffel symbols. Alternatively, if for every chart there are $n^3$ functions that transform like the Christoffel symbols, we may define a connection in terms of them.

You may verify that the Levi-Civita connection coefficients $$\Gamma^k_{\;ij}=\frac{1}{2}g^{kl}(\partial_ig_{jl}+\partial_jg_{il}-\partial_lg_{ij})$$ satisfy this requirement and thus define a connection, the so-called Levi-Civita connection. Ignore this for now.

We further verify that our connection agrees with the usual covariant derivative. Given a covector $\omega$ we define the tensor $\nabla X\in \mathcal{T}_1^1(M)$ by $$\nabla X(Y,\omega)=\langle \omega,\nabla_YX\rangle$$ $\nabla X$ is the covariant derivative of $X$. Set up a chart as before. Let $$X=\xi^i\partial_i,\quad \nabla X=\xi^i_{;j}dx^j\otimes\partial_i$$ Then one has $$\xi^i_{;j}=\nabla X(\partial_j,dx^i)=\langle dx^i,\nabla_{\partial_j}(\xi^k\partial_k)\rangle=\langle dx^i,\xi^k_{,j}\partial_k+\xi^k\Gamma^s_{\;jk}\partial_s\rangle=\xi^i_{,j}+\Gamma^i_{\;jk}\xi^k$$

This is of course just (4.6.4) in Weinberg.

Lastly, we define curvature. Let $\mathcal{X}(M)$ denote the set of smooth vector fields on $M$.

Definition 3. The Riemann curvature tensor is the mapping $R:\mathcal{X}(M)\times \mathcal{X}(M)\times \mathcal{X}(M)\longrightarrow \mathcal{X}(M)$ $$R(X,Y)Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$$

It may be shown that the components of this tensor are $$R^i_{\;jkl}=\Gamma^i_{\;lj,k}-\Gamma^i_{\;kj,l}+\Gamma^s_{\;lj}\Gamma^i_{\;ks}-\Gamma^s_{\;kj}\Gamma^i_{\;ls}$$

Now we tackle parallel transport.

Definition 4. Let $\gamma$ be a curve in $M$ and let $X$ be a vector field along $\gamma$. Let $t$ denote the curve parameter and define the tangent vector as $$\dot\gamma(t)f=\frac{d}{dt}(f\circ\gamma)(t)$$ for all $f\in\mathcal{F}(M)$. The parallel transport operator $\tau_{t,s}$ is defined by $$\nabla_{\dot\gamma}X(\gamma(t))=\left.\frac{d}{ds}\right|_{s=t}\tau_{t,s}X(\gamma(s))$$

A tensor is said to be parallel transported if $$\nabla_{\dot\gamma}T=0$$

Parallel transport says nothing about arc length. In fact, the definition of the metric is that it measures arc length via the formula $$L[\gamma]=\int_\gamma ds=\int_a^b\sqrt{g(\dot\gamma,\dot\gamma)}\,dt$$ In fact, we cannot measure the length of vectors without $g$. This is because we need to contract the vector with its corresponding covector to get the length squared. The isomorphism $T_p(M)\longrightarrow T^*_p(M)$ is furnished by the metric.

As you know, we can apparently measure curvature using parallel transport. We parallel transport a vector along an "infinitesimal" loop and the difference is supposedly proportional to the curvature. This is detailed in section 6.3 of Weinberg. Here I reproduce some key points. Let $\gamma:[0,1]\longrightarrow M$ be a closed path, with $\gamma(0)=p$. We displace an arbitrary vector $v_0\in T_p(M)$ parallel along $\gamma$ and obtain the field $v(t)=\tau_tv_0\in T_{\gamma(t)}(M)$. Using the parallel transport equation, we have the exact relation $$\Delta v^i=-\oint_\gamma \omega^i_{\;j}v^j$$ where $\omega^i_{\;j}$ denotes the spin connection. Let $\sigma$ be the surface enclosed by $\gamma$. Then, assuming $\sigma$ is small and using Stokes' theorem, we have the approximation $$\Delta v^i\simeq -\frac{1}{2}R^i_{\;jkl}(p)v_0^j\int_\sigma dx^k\wedge dx^l$$ The fundamental problem with this is that as $\sigma\rightarrow\emptyset$, this is supposed to become exact, but then the integral also vanishes! So, as we have partially shown, it is heuristically possible to measure curvature without using the metric. However, this will always be an approximation. Taking the Christoffel symbols calculated with the metric and putting them into Definition 3 gives an exact description of curvature.

Note that fundamentally the curvature and even the covariant derivative are defined without mention of the metric. In GR, however, $M$ is a pseudo-Riemannian manifold. Thus, there exists a unique Levi-Civita connection $\nabla$ such that

1. $\nabla g=0$ (Weinberg Eq. 4.6.16)

2. $\nabla_XY-\nabla_YX=[X,Y]$ (Weinberg Eq. 4.5.6)

In GR texts the original definitions of these concepts is fudged because we never need anything other than the Levi-Civita connection.

Answer 2

ocelo7 answers correctly but I want to boil down to just what you asked. To have parallel transport in this sense you need a connection. Then, as you as you read, you can measure how much a vector changes in parallel transport around a loop without using a metric. But the measure will be a vector, not a length, and that's why it does not need a metric. And that measure suffices to define curvature.

As to why we need Riemannian metrics to define lengths, it is because the metric is the measure of length.