simple calculation with Christoffel symbols on Poincare half-plane

Question

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.

Answer 1

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.

Answer 2

When using the definition of $R(X,Y)Z$ ending with $-\nabla_{[X,Y]}Z$, the sectional curvature $K(X,Y)$ is defined as

$$ K(X,Y) = \frac{\langle R(X,Y)Y,X\rangle}{\langle X,X \rangle \langle Y,Y \rangle-\langle X,Y \rangle^2}$$

for $X,Y$ linearly independent. We derive the coordinate expression for $R_{ijk}^{\ \ \ \ \ l}$ in terms of the Christoffel symbols. Following Lee's Riemannian Manifolds, we take the components $R_{ijk}^{\ \ \ \ \ l}$ to be defined by $R(\partial_i,\partial_j)\partial_k=R_{ijk}^{\ \ \ \ \ l} \partial_l$, and the Christoffel symbols $\Gamma^k_{\ ij}$ to be defined by $\nabla_{\partial_i}\partial_j=\Gamma^k_{\ ij}\partial_k$. Applying the coordinate dual basis element $dx^l$ to both sides of the defining equation for $R_{ijk}^{\ \ \ \ \ l}$ gives \begin{align*} R_{ijk}^{\ \ \ \ \ l} &= dx^l(R(\partial_i,\partial_j)\partial_k) \\ &= dx^l(\nabla_{\partial_i}\nabla_{\partial_j}\partial_k-\nabla_{\partial_j}\nabla_{\partial_i}\partial_k-\nabla_{[\partial_i,\partial_j]}\partial_k)\\ &= dx^l(\nabla_{\partial_i}\nabla_{\partial_j}\partial_k-\nabla_{\partial_j}\nabla_{\partial_i}\partial_k)\\ &= dx^l(\nabla_{\partial_i}(\Gamma^m_{\ jk}\partial_m)-\nabla_{\partial_j}(\Gamma^m_{\ ik}\partial_m))\\ &= dx^l((\partial_i\Gamma^m_{\ jk})\partial_m+\Gamma^m_{\ jk}\nabla_{\partial_i}\partial_m-(\partial_j \Gamma^m_{\ ik})\partial_m-\Gamma^m_{\ ik}\nabla_{\partial_j}\partial_m)\\ &= dx^l((\partial_i\Gamma^m_{\ jk})\partial_m+\Gamma^m_{\ jk}\Gamma^p_{\ im}\partial_p-(\partial_j \Gamma^m_{\ ik})\partial_m-\Gamma^m_{\ ik}\Gamma^p_{\ jm}\partial_p)\\ &= (\partial_i\Gamma^m_{\ jk})dx^l(\partial_m)+\Gamma^m_{\ jk}\Gamma^p_{\ im}dx^l(\partial_p)-(\partial_j \Gamma^m_{\ ik})dx^l(\partial_m)-\Gamma^m_{\ ik}\Gamma^p_{\ jm}dx^l(\partial_p)\\ &= \partial_i\Gamma^l_{\ jk}+\Gamma^m_{\ jk}\Gamma^l_{\ im}-\partial_j \Gamma^l_{\ ik}-\Gamma^m_{\ ik}\Gamma^l_{\ jm}. \end{align*}

Now for the coordinate basis $\{ \partial_1=\partial/ \partial x,\partial_2=\partial/ \partial y\}$, we can write $$ K(\partial_i,\partial_j) = \frac{\langle R(\partial_i,\partial_j)\partial_j,\partial_i\rangle}{\det(g)}=\frac{R_{ijji}}{\det(g)},$$ where $\det(g)=1/y^4$. Clearly, $K(\partial_i,\partial_j)=0$ for $i=j$. For $i \neq j$, we have $$ K(\partial_1,\partial_2)=y^4 R_{1221} \quad \text{and} \quad K(\partial_2,\partial_1)=y^4 R_{2112}.$$ By symmetry properties of the components $R_{ijkl}$, $R_{1221}=R_{2112}$, so in fact $K(\partial_1,\partial_2)=K(\partial_2,\partial_1)$. Since $$R_{1221}=g_{1m}R^{\ \ \ \ \ m}_{122}=g_{11}R^{\ \ \ \ \ 1}_{122},$$ we use the coordinate expression for the components $R_{ijk}^{\ \ \ \ \ l}$ derived at the outset to compute $R^{\ \ \ \ \ 1}_{122}$ as follows: \begin{align*} R^{\ \ \ \ \ 1}_{122}&=\partial_1\Gamma^1_{\ 22}+\Gamma^m_{\ 22}\Gamma^1_{\ 1m}-\partial_2 \Gamma^1_{\ 12}-\Gamma^m_{\ 12}\Gamma^1_{\ 2m}\\ &=\Gamma^2_{\ 22}\Gamma^1_{\ 12}-\partial_2\left(-\frac{1}{y}\right)-\Gamma^1_{\ 12}\Gamma^1_{\ 21}\\ &=\left(-\frac{1}{y}\right)^2-\frac{1}{y^2}-\left(-\frac{1}{y}\right)^2\\ &=-\frac{1}{y^2}. \end{align*} Therefore, $R_{1221}=-1/y^4$, and so $K(\partial_1,\partial_2)=K(\partial_2,\partial_1)=-1.$