Using Koszul's formula to compute $\nabla_X Y$, where $\nabla$ is the Levi-Civita connection

Question

Main question: I am following the answer in this question: computing Riemannian connection and Killing fields (very basic). He calculates $$ g(\nabla_X(Y),X) = 0 \\ g(\nabla_X(Y),Y) = 0 \\ g(\nabla_X(Y),Z) = 1 $$ where $\nabla$ is the Levi-Civita connection. Then he deduces that putting the above together yields $\nabla_X(Y)=Z$. Can anyone elaborate on how to get to that conclusion?

Any help is much appreciated.

---

I have two additional questions, but the answers might be clear, as soon as I grasp the above. Here goes anyway: In the specific problem I am working on, I look at $\mathcal{H}^3=\{(x,y,z)\in\mathbb{R}^3:z>0\}$ equipped with the Riemannian metric $$g=\frac{dx^2+dy^2+dz^2}{z^2}$$ and vector fields $$ A= z\partial_x ,\quad B=z\partial_y,\quad C=z\partial_z. $$

1. I have calculated $$ g(\nabla_AB,A)=g(\nabla_AB,B)=g(\nabla_AB,C)=0\;, $$ $\nabla$ still being the Levi-Civita connection. Would this mean that $\nabla_AB=0$?

2. I also have $$ g(\nabla_AC,A)=-1\\ g(\nabla_AC,B)=g(\nabla_AC,C)=0\;. $$ What does it mean that the first inner product is negative? Would I get $\nabla_AC=-A$?

Thank you in advance.

Answer 1

For the second part of your question: your conclusions are correct and this is because $A,B,C$ form a orthonormal basis. See here. Most of the things you know from linear algebra about orthonormal bases can be used here, including that for a ONB $e_1,\ldots, e_n$ and any vector $v$ the following is true:

$$ v=\sum_{i=1}^n \langle v,e_i \rangle e_i. $$

I assume that your first question has an answer in a similar direction, but I was to lazy to check that $X,Y,Z$ form an ONB. :)

Added: Since the question came up, let me elaborate a little bit. Yes its true that $A,B,C$ form a basis of the smooth vectorfields, since the $\partial_x,\partial_y,\partial_z$ form a basis. What are the $\partial_*$? Well you take a chart $\varphi:M\to \mathbb{R}^n$ around a point $x$ and a open set $U$ on which $\varphi$is defined. Then fix the canonical basis $e_1,\ldots,e_n$ of $\mathbb{R}^n$ and define for every $y\in U$:

$$ \partial_{x_i}(y):=d\varphi^{-1}_{\varphi(y)}(e_i), $$

this will give you a vector on $T_yM$ and indeed a vectorfield on $U$. Indeed, since $\varphi$ is a diffeomorphism this will give you locally a basis of your smooth vectorfields.

Now in your concrete situation you have that $\mathcal{H}^3$ is a open subset of $\mathbb{R}^3$, so you can take the identity as a chart and you indeed have a global chart and therefore the $\partial_*$ form a global basis. The calculations in the link above give a hint in how to compute that the $A,B,C$ are indeed a ONB for your given metric $g$.

You should try to compute the $\partial_*$ and then use that in $\mathbb{R}^n$ the Levi-Civita connection is given by $\nabla_X(Y)=X(Y)$ to double check your calculations and get a better feeling for what is going on.