Equip $H=\{(x,y):y>0, x,y \in \mathbb{R}\}$ with the metric $$ds^2=\frac{dx^2+dy^2}{y^2}.$$
(https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model). I want to show that the sectional curvature
$$ K(\partial_i,\partial_j) = \frac{\langle R(\partial_i,\partial_j)\partial_j,\partial_i\rangle}{\det(g)}=\frac{R_{ijji}}{\det(g)}$$
is $-1$. I know this is a simple calculation, but for some reason I'm off by a sign and it's driving me nuts. Here $g$ is the metric in matrix form:
$$g=(g_{ij})=
\begin{pmatrix}
\frac{1}{y^2} & 0 \\
0 & \frac{1}{y^2}
\end{pmatrix}.$$
Let $\{ \partial_1=\partial/ \partial x,\partial_2=\partial/ \partial y\}$ be the coordinate basis, and just consider the computation of $K(\partial_1,\partial_2).$ So I just need to compute $R_{1221}=g_{1m}R^m_{\ 221}=g_{11}R^1_{\ 221}$, where
$$R^i_{jkl}=\partial_k \Gamma^i_{jl}-\partial_l \Gamma^i_{jk}+\Gamma^p_{jl}\Gamma^i_{pk}-\Gamma^p_{jk}\Gamma^i_{pl}.$$
I already have the Christoffel symbols, namely
\begin{equation*}
\Gamma^1_{12}=\Gamma^1_{21}=-\frac{1}{y},\\
\Gamma^2_{12}=\Gamma^2_{21}=0,\\
\Gamma^1_{11}=\Gamma^1_{22}=0,\\
\Gamma^2_{11}=\frac{1}{y},\\
\Gamma^2_{22}=-\frac{1}{y}.
\end{equation*}
So
\begin{align*}
R^1_{\ 221} &= \partial_2 \Gamma^1_{21}-\partial_1 \Gamma^1_{22}+\Gamma^p_{21}\Gamma^1_{p2}-\Gamma^p_{22}\Gamma^1_{p1}\\
&= \partial_2 \left( -\frac{1}{y}\right)+\Gamma^1_{21}\Gamma^1_{12}-\Gamma^2_{22}\Gamma^1_{21}\\
&= \frac{1}{y^2}+\left(-\frac{1}{y}\right)^2-\left(-\frac{1}{y}\right)^2\\
&= \frac{1}{y^2}.
\end{align*}
But this gives $R_{1221}=1/y^4$, and hence $K(\partial_1,\partial_2)=1$.
Please tell me what I'm doing wrong here or what formula is wrong.
Best Answer
The sectional curvature for a Riemannian manifold, $(M,g)$ with respect to an orthonormal basis for $P=\text{span}_{\mathbb{R}}(e_1,e_2)\subset T_pM$ at a point $p\in M$ is given by: $$K(P)=\langle R(e_1,e_2)e_2,e_1\rangle.$$ Recall that the components of the Riemann curvature tensor take the form: $$R_{\kappa\lambda\mu\nu}=\langle R(e_\mu,e_\nu),e_\lambda,e_\kappa\rangle.$$ So, $\langle R(e_1,e_2)e_2,e_1\rangle=R_{1212}.$ You calculated $R_{1221}$, so if you use the symmetries of the Riemann curvature tensor in its last two indices you find the missing negative sign.