The Levi-Civita Connection

Compatibility

Let $M$ be a smooth Riemannian manifold with metric $g$. A connection $\nabla$ on $M$ is said to be compatible with the metric $g$ if for every pair of vector fields $X$ and $Y$ on $M$, and every vector $Z\in \mathfrak{X}(M)$, $Z\left(g(X,Y)\right)=g\left({\nabla }_{Z}X,Y\right)+g\left(X,{\nabla }_{Z}Y\right).$ We can denote this condition as $\nabla g\equiv 0$

Remark

It is equivalent to say that $\nabla g=0$. However, this is not a good notation because it does not make sense to apply the affine connection on a $(0,2)$-tensor. It is meaningful once we define that $\nabla g:=({\nabla }^{\ast }\otimes {\nabla }^{\ast })g,$where ${\nabla }^{\ast }$ is the dual connection of $\nabla$.

Levi-Civita Connection

An affine connection defined on $(M,g)$ is called a Levi-Civita connection if it is compatible with the metric $g$ and it is torsion free. That is $T(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]=0$.

Koszul Formula

For a Levi-Civita connection $\nabla$ defined on $M,g$ satisfied the following formula $g({\nabla }_{X}Y,Z)=\frac{1}{2}(Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)+g([X,Y],Z)-g([Z,X],Y)-g([Y,Z],X))$

Proof

The condition $\nabla g\equiv 0$ implies that for any vector fields $X,Y,Z\in \mathfrak{X}(M)$, the directional derivative of the inner product satisfies: $Zg(X,Y)=g({\nabla }_{Z}X,Y)+g(X,{\nabla }_{Z}Y)(\text{C3})$ By cyclically permuting the vector fields $(X,Y,Z)$, we obtain two additional identities: $Xg(Y,Z)=g({\nabla }_{X}Y,Z)+g(Y,{\nabla }_{X}Z)(\text{C1})$ $Yg(Z,X)=g({\nabla }_{Y}Z,X)+g(Z,{\nabla }_{Y}X)(\text{C2})$ Since the connection is torsion-free, we can swap the lower indices of the covariant derivative at the expense of a Lie bracket: ${\nabla }_{Y}X={\nabla }_{X}Y-[X,Y]$ ${\nabla }_{X}Z={\nabla }_{Z}X+[X,Z]$ ${\nabla }_{Y}Z={\nabla }_{Z}Y+[Y,Z]$ To isolate the connection term ${\nabla }_{X}Y$, we compute the linear combination $(\text{C1})+(\text{C2})-(\text{C3})$: $\begin{array}{rl}Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)=& g({\nabla }_{X}Y,Z)+g(Y,{\nabla }_{X}Z)\\ & +g({\nabla }_{Y}Z,X)+g(Z,{\nabla }_{Y}X)\\ & -g({\nabla }_{Z}X,Y)-g(X,{\nabla }_{Z}Y)\end{array}$ Substituting the torsion-free relations into the right-hand side to group terms containing ${\nabla }_{X}Y$: $\begin{array}{rl}=& g({\nabla }_{X}Y,Z)+g(Y,{\nabla }_{Z}X+[X,Z])\\ & +g({\nabla }_{Z}Y+[Y,Z],X)+g(Z,{\nabla }_{X}Y-[X,Y])\\ & -g({\nabla }_{Z}X,Y)-g(X,{\nabla }_{Z}Y)\end{array}$ Expanding and using the symmetry of the metric tensor $g(\cdot ,\cdot )$, the un-bracketed terms involving ${\nabla }_Z$ cancel out cleanly: $g(Y,{\nabla }_{Z}X)-g({\nabla }_{Z}X,Y)=0$, $g({\nabla }_{Z}Y,X)-g(X,{\nabla }_{Z}Y)=0$ Gathering the remaining terms yields: $Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)=2g({\nabla }_{X}Y,Z)+g(Y,[X,Z])+g([Y,Z],X)-g(Z,[X,Y])$ Using the anti-symmetry of the Lie bracket (e.g., $[X,Z]=-[Z,X]$), we can rearrange this into the final expression. Isolating $g({\nabla }_{X}Y,Z)$ gives the : $g({\nabla }_{X}Y,Z)=\frac{1}{2}(Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)+g([X,Y],Z)-g([Z,X],Y)-g([Y,Z],X))$

Fundamental Theorem of Riemannian Geometry

Every Riemannian manifold admits a unique Levi-Civita connection. In particular, the unique Levi-Civita connection has the following coordinate expression: ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{kl}({\partial }_i{g}_{jl}+{\partial }_j{g}_{il}-{\partial }_l{g}_{ij}).$

Let $(M,g)$ be a Riemannian manifold equipped with the Levi-Civita connection $\nabla$. In a local coordinate chart with coordinate basis $\{{\partial }_i\}$ and dual basis $\{d{x}^i\}$, the \textit{connection 1-forms} ${\omega }_j^i$ are defined by: ${\nabla }_X{\partial }_j={\omega }_j^i(X){\partial }_i$ for any vector field $X$. \textbf{Relation to Christoffel Symbols}: Expanding ${\omega }_j^i$ in the coordinate dual basis $\{d{x}^k\}$, the components are precisely the Christoffel symbols: ${\omega }_j^i={\Gamma }_{kj}^{i}d{x}^k$ Since the Levi-Civita connection is torsion-free, the Christoffel symbols are symmetric in their lower indices (${\Gamma }_{kj}^i={\Gamma }_{jk}^i$), so this can also be written as: ${\omega }_j^i={\Gamma }_{jk}^{i}d{x}^k$Matrix Form and Cartan’s Structural Equations: We can organize these 1-forms into a matrix $\omega =({\omega }_j^i)$, where $i$ is the row index and $j$ is the column index. $d(d{x}^i)+{\omega }_j^i\wedge d{x}^j=0⟹{\omega }_j^i\wedge d{x}^j=0$. Let ${\omega }_{ij}=g_{ik}{\omega }_j^k$, then the metric compatibility condition reads: $d{g}_{ij}={\omega }_{ij}+{\omega }_{ji}$ (Note: If $\{{\theta }^i\}$ is an \textit{orthonormal} frame rather than a coordinate frame, then $g_{ij}={\delta }_{ij}$, making the matrix ${\omega }_{ij}$ strictly anti-symmetric: ${\omega }_{ij}=-{\omega }_{ji}$).