Algebraic intuition of the Levi-Civita connection

Question

I was wondering about the difference between the algebraic approach to the Levi-Civita connection and the purely geometric one. Let me explain myself a little better.

So, I'm just entering the world of algebraic geometry, and I came across this tweet which shows a method of building the Levi-Civita connection by way of rings, modules and derivations. Then I read further and saw this question whose answer is with manifolds, tangent bundles and connections. I'm wondering whether the intuition I've built up until this point is valid or not; that the algebraic approach is basically using the "abstract structure" of the space to deduce the connection, and the manifold route studies the geometric object itself. Sorry if I can't explain myself in clear terms.

I tried to figure this out by writing the two methods in parallel and trying to relate the two, but I got lost in a sea of complicated math which I don't understand yet. Also, if anyone can point me in the direction of relevant literature that would be much appreciated!

Answer 1

While there may seem to be a disconnect when using coordinates and Christoffel symbols, many modern texts will develop the notion of vector fields and connections in a way which maps directly onto the algebraic definitions. The important correspondences are the following:

1. The set of smooth functions $M\to\mathbb{R}$, denoted $C^\infty M$, forms a commutative ring under pointwise multiplication.

2. Let $\mathfrak{X}M$ be the $C^\infty M$-module of smooth vector fields on $M$. Every element $X\in\mathfrak{X}M$ corresponds to a "directional derivative" map $X:C^\infty M\to C^\infty M$. In fact, this map is a derivation, and $\mathfrak{X}M$ is isomorphic to the module of derivations of $C^\infty M$. Additionally, the Lie bracket of vector fields corresponds to the canonical Lie bracket of derivations.

3. Every $(0,2)$ tensor field $T$ can act on a pair of vector fields, giving a map $T:\mathfrak{X}M\times\mathfrak{X}M\to C^\infty M$. Such a map corresponds to a tensor field iff it is $C^\infty M$-bilinear. In fact, the space of $(0,2)$ tensor fields is isomorphic to the space of $C^\infty M$-bilinear forms $\mathfrak{X}M\times\mathfrak{X}M\to C^\infty M$. Furthermore, such a tensor field is symmetric/nondegenerate in the geometric sense iff the corresponding $C^\infty M$-bilinear form is symemetric/nondegenerate in the algebraic sense.

These give an algebraic definition of vector fields and covariant $(0,2)$ tensor fields in terms of the ring $C^\infty M$. Of course, these geometric objects can be defined in many other ways, some of which make their algebraic properties less obvious. For a much more detailed discussion, see for instance Lee's Introduction to Smooth Manifolds (in particular the chapters on vector fields and tensors).

With the above correspondences established, one can see that the algebraic and geometric treatments of connections are quite similar:

A connection on the module of derivations $\mathfrak{D}$ of a commutative ring $R$ is a map $\nabla:\mathfrak{D}\times\mathfrak{D}\to\mathfrak{D}$, with the following properties (throughout, let $X,Y,Z\in\mathfrak{D}$, and $f,g\in R$):

• $R$-linearity in the first argument: $\nabla_{fX+gY}Z=f\nabla_XZ+g\nabla_YZ$
• additivity in the second argument: $\nabla_X(Y+Z)=\nabla_XY+\nabla_XZ$
• Leibniz rule: $\nabla_X(fY)=X(f)Y+f\nabla_XY$

They can also have some additional properties:

• A a connection $\nabla$ is said to be torsion free if it satisfies $\nabla_XY-\nabla_YX=[X,Y]$
• Given an $R$-bilinear form $h:\mathfrak{D}\times\mathfrak{D}\to R$, we say that $\nabla$ is compatible with $h$ if $X(h(Y,Z))=h(\nabla_XY,Z)+h(X,\nabla_XZ)$.

An affine connection on a smooth manifold $M$, meanwhile, is defined exactly as above, but with $R=C^\infty M$ and $\mathfrak{D}=\mathfrak{X}M$. Further, given a Riemannian metric $g$ on $M$ we say that $\nabla$ is compatible with $g$ if it is compatible with the corresponding $C^\infty M$-bilinear form $g:\mathfrak{X}M\times\mathfrak{X}M\to C^\infty M$.