Is $\Gamma_{k,i}^j = - \Gamma_{k,j}^i$ correct? (For an isometric $\nabla$) I have a small question about Riemannian Geometry - I am pretty sure it's something very easy, but I'd still like to ask for confirmation, since I haven't found a smiliar result when googling for it.
Assume we have a $n$-dimensional Riemannian manifold $(M,g)$ and a orthonormal basis field $\{e_1,...e_n\}$. If we denote by $\nabla$ the Levi-Civita-connection (or any isometric connection for that matter) and for $x\in M$ let $X\in T_xM$ be any vector, then we have due to isometry,
$$g(\nabla_X e_i, e_j) + g(e_i,\nabla_X e_j) = X(\underbrace{g(e_i,e_j)}_{\equiv 0}) = 0,$$
i.e. when choosing $X = e_k$, we get $$\Gamma_{k,i}^j = g(\nabla_{e_k}e_i,e_j) = -g(e_i,\nabla_{e_k} e_j)= - \Gamma_{k,j}^i$$
 A: Minus a few tiny mistakes/wrong use of terminology, you are right (the symbol $\Gamma$ and name of "Christoffel symbols" is usually reserved for frames induced from coordinate charts, and those being orthonormal is very rare - but switch your $\Gamma$ for, say, $\Lambda$, and everything works out the same). Isometries have nothing to do with the property you're using (I mean, where are you seeing any isometries there?), the property $\nabla g = 0$ (which is equivalent to the one you used) is usually called compatibility with the metric (or preserving the metric). Also, $$g(e_i, e_j) = 0$$ is wrong. Remember that we are fixing $i$ and $j$, and $g(e_i, e_j) = 1$ if $i = j$ and it's $0$ otherwise. It just can't always be zero (unless you were working with a metric on a $0$-dimensional manifold which has zero dimensional tangent space, but that makes no sense). What is true is that $g(e_i, e_j)$ is a constant function: as I just mentioned, it is either $0$ or $1$, depending on how you fixed $i$ and $j$. So it is constant and hence it is indeed always true that $X(g(e_i, e_j)) = 0$ (for all vector fields $X$, of course).
