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Symmetry of Connection 📂Geometry

Symmetry of Connection

Definition1

An affine connection $\nabla$ on a differentiable manifold $M$ is said to be symmetric if it satisfies the following.

$$ \nabla_{X}Y - \nabla_{Y} X = \left[ X, Y \right] \quad \forall X,Y \in \mathfrak{X}(M) $$

Here $\mathfrak{X}(M)$ is the set of vector fields on $M$, and $[ \cdot, \cdot]$ is the Lie bracket.

Explanation

Let’s take Euclidean space as an example. Consider a coordinate system $(U, \mathbf{x})$ of $\mathbb{R}^{n}$. If we denote this as,

$$ \nabla_{X_{i}}X_{j} - \nabla_{X_{j}}X_{i} = [X_{i}, X_{j}] = X_{i}X_{j} - X_{j}X_{i} = \dfrac{\partial ^{2}}{\partial x_{i}x_{j}} - \dfrac{\partial ^{2}}{\partial x_{j}x_{i}} = 0 \\ \implies \nabla_{X_{i}}X_{j} = \nabla_{X_{j}}X_{k} $$

Furthermore, since $\nabla_{X_{i}}X_{j} = \Gamma_{ij}^{k}X_{k}$,

$$ \nabla_{X_{i}}X_{j} - \nabla_{X_{j}}X_{i} = \Gamma_{ij}^{k}X_{k} - \Gamma_{ji}^{k}X_{k} = (\Gamma_{ij}^{k} - \Gamma_{ji}^{k})X_{k} = 0 $$

Therefore, $\Gamma_{ij}^{k} = \Gamma_{ji}^{k}$ holds.


  1. Manfredo P. Do Carmo, Riemannian Geometry (Eng Edition, 1992), p54 ↩︎