In this question , I don't know why the Lie bracket is needed to compute Ricci tensor. In my view , knowing the metric is enough to compute Ricci tensor.But if just using metric ,I can't get the same result of THW. I am not familiar with Lie algrebra.


1 Answer 1


One explanation comes from using the Koszul Formula to the calculate the connection.

Let $(M, g)$ be a Riemannian manifold. Recall that the Levi-Civita connection $\nabla$ is defined uniquely by requiring that \begin{align*} 2 g\left(\nabla_{X}Y, Z\right) = &X\cdot g(Y, Z) + Y\cdot g(Z, X) - Z\cdot g(X, Y)\\ &g(\left[X, Y\right], Z) - g([X, Z], Y) - g([Y, Z], X) \end{align*} for all vector fields $X$, $Y$, and $Z$ on $M$. Note that the notation of $\cdot$ is used to indicate the directional derivative of the scalar function determined by $g(-, -)$ restricted to two vector fields in the indicated direction.

One can then compute the connection coefficients (or Christoffel symbols) relative to a given frame $E_{1}, E_{2}, \ldots, E_{n}$ by writing $\nabla_{E_{i}}E_{j} = \Gamma_{ij}^{k}E_{k}$ and then substituting into the Koszul formula and solving for $\Gamma_{ij}^{k}$.

Computing the connection coefficients in this way is simplified greatly in two particular instances.

  1. The chosen frame $E_{1}, E_{2}, \ldots, E_{n}$ is a local-coordinate frame, in which case the Lie brackets all vanish.
  2. $(M)$ is a Lie group and $g$ is a left invariant metric. In this instance, one can select the frame $E_{1}, E_{2}, \ldots , E_{n}$ to be a left invariant frame so that both $g$ and the frame are determined at the identity of $G$, i.e., at the Lie algebra of $G$. In this instance, the expressions $g(E_i, E_j)$ are constant, so the directional derivatives in the first line on the right hand side of the Koszul formula are all zero. When restricted to basis/frame vectors, the second line of the Koszul formula then reduces to an expression involving the $g_{ij}$ of the metric and the structure constants of the Lie algebra. The connection coefficients $\Gamma_{ij}^{k}$ will then depend solely on those two sets of quantities. Note that this can be further simplified by requiring that the left invariant frame is an orthonormal frame for $g$.

It should also be noted that there are other ways to compute the Ricci curvature in the setting of a Lie group.


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