1

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 $$\Gamma_{k,i}^j = g(\nabla_{e_k}e_i,e_j) = -g(e_i,\nabla_{e_k} e_j)= - \Gamma_{k,j}^i$$

Arctic Char
  • 16,007
Nuke_Gunray
  • 2,816
  • Christoffel symbols are defined relatively to a coordinate system $(x^1,\ldots,x^n)$. If $(e_1,\ldots,e_n)$ denotes the tangent frame to that system, one usually does not have $g(e_i,e_j)= \delta_{i,j}$: this can only happen in a flat manifold. So what you are stating is usually false. – Didier Jun 19 '22 at 11:17
  • @Didier Yeah sorry I was a bit sloppy with notation. I meant for the basis to be a orthonormal basis field, i.e. I defined them to be $g(e_i,e_j) = \delta_{i,j}$ – Nuke_Gunray Jun 19 '22 at 11:27
  • Then it is true, but "Christoffel symbols" is then not the right terminology (nor "$\Gamma$"). What you are deriving are the connection form in the method of moving frame – Didier Jun 19 '22 at 11:44
  • @Didier Interesting. But where exactly is the difference? From what I understand, the Christoffel symbol $\Gamma^k_{i,j}$ for a chosen basis $(e_1,...e_n)$ of the tangent space is just the $k$-th component of $\nabla_{e_i} e_j$, i.e. after taking $g$ on both sides we get $g(e_k,\nabla_{e_i} e_j) = \Gamma^k_{i,j}\cdot g_{kk}$. In my case, $g_{ij} = \delta_{i,j}$, i.e. $g_{kk} = 1$, since the basis was chosen to be orthonormal. Why shouldn't I use the notation with the $\Gamma$? – Nuke_Gunray Jun 19 '22 at 13:10
  • @Nuke_Gunray usually the symbol $\Gamma$ and the name "Christoffel symbols" is reserved for the case when the basis comes from a coordinate chart. It's just a terminology convention. – Jackozee Hakkiuz Jun 19 '22 at 21:24

1 Answers1

3

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).

Matheus Andrade
  • 6,773
  • This is true for an orthonormal frame, which usually does not come from a coordinate patch, so there is no associated Christoffel symbols – Didier Jun 19 '22 at 11:18
  • @MatheusAndrade thanks! Just goes to show one shouldn't write questions here at 5am while falling into bed ^^. I omitted a lot of details, thanks for reminding me. – Nuke_Gunray Jun 19 '22 at 11:29
  • @Didier you're right, I'll edit my answer. – Matheus Andrade Jun 20 '22 at 02:25