Find $R\left(\frac{\partial}{\partial\rho}, \frac{\partial}{\partial \theta}\right)\frac{\partial}{\partial \rho}$ in $(\mathbb{R}P^2, g)$

Question

Let $R$ be the curvature tensor of $(\mathbb{R}P^2, g)$ where in local coordinates the metric $g$ can be written as $$g^{\varphi}= \frac{1}{(\rho^2+1)^2} d\rho^2 + \frac{\rho^2}{\rho^2+1}d\theta^2$$ I am asked to find $$R\left(\frac{\partial}{\partial\rho}, \frac{\partial}{\partial \theta}\right)\frac{\partial}{\partial \rho}$$ and I am not sure that my calculations are correct. First we have that the Christoffer symbols are $$\Gamma_{11}^1 = \frac{-2\rho}{\rho^2+1},\; \Gamma_{22}^1 = -\rho,\; \Gamma_{12}^2 = \Gamma_{21}^2 = \frac{1}{\rho(\rho^2+1)}$$ The curvature tensor is given by $$R(X,Y)Z=-\nabla_X\nabla_YZ + \nabla_Y\nabla_XZ +\nabla_{[X,Y]}Z$$ so, first \begin{align*} \nabla_{\frac{\partial}{\partial\theta}}\frac{\partial }{\partial \rho} & = \Gamma_{ij}^k \frac{\partial}{\partial x^k} = \Gamma_{21}^1\frac{\partial}{\partial \rho} + \Gamma_{21}^2 \frac{\partial}{\partial \theta}\\ & = \frac{1}{\rho(\rho^2+1)}\frac{\partial}{\partial \theta} \end{align*} therefore \begin{align*} -\nabla_{\frac{\partial}{\partial \rho}}\nabla_{\frac{\partial}{\partial\theta}}\frac{\partial }{\partial \rho} & = -\nabla_{\frac{\partial}{\partial \rho}} \left(\frac{1}{\rho(\rho^2+1)}\frac{\partial}{\partial \theta}\right)\\ & = -\frac{\partial}{\partial \rho} \frac{1}{\rho(\rho^2+1)^2} - \nabla_{\frac{\partial}{\partial \theta}}\frac{\partial}{\partial \rho}\\ & = \frac{3 \rho^2+1}{\rho^2(\rho^2+1)^2} - \frac{1}{\rho(\rho^2+1)}\frac{\partial}{\partial \theta} \end{align*} finally \begin{align*} \nabla_{\frac{\partial}{\partial \theta}}\nabla_{\frac{\partial}{\partial\rho}}\frac{\partial }{\partial \rho} & = \nabla_{\frac{\partial}{\partial \theta}} \left( \Gamma_{11}^1\frac{\partial}{\partial \theta} + \Gamma_{11}^2 \frac{\partial}{\partial \theta}\right)\\ & = \nabla_{\frac{\partial}{\partial \rho}}\left(-\frac{2\rho}{\rho^2+1}\frac{\partial}{\partial \theta}\right)\\ & = \frac{\partial}{\partial \theta}\left(-\frac{2\rho}{\rho^2+1}\frac{\partial}{\partial \theta}\right) + \nabla_{\frac{\partial}{\partial \theta}}\frac{\partial}{\partial \theta}\\ & = \Gamma_{22}^1\frac{\partial}{ \partial \rho} + \Gamma_{22}^2\frac{\partial}{\partial \theta}\\ & = -\rho \frac{\partial}{\partial \rho} \end{align*} and since $\left[ \frac{\partial}{\partial \rho}, \frac{\partial}{\partial \theta}\right] = 0$ then $$R\left(\frac{\partial}{\partial\rho}, \frac{\partial}{\partial \theta}\right)\frac{\partial}{\partial \rho} = \frac{3 \rho^2+1}{\rho^2(\rho^2+1)^2} - \frac{1}{\rho(\rho^2+1)}\frac{\partial}{\partial \theta} -\rho \frac{\partial}{\partial \rho}$$

Is this correct? Thank you.

Answer 1

If $\rho=\tan\ t$, then $\rho'=\sec^2t\ dt$ so that $$ g = dt^2+\sin^2 t\ d\theta^2$$

Answer 2

Try this idea and tell me if it work $$R(X,Y,X,Y)=g(R(X,Y)Y,X)$$ then use the inverse of $[g_{ij}]$ to get first slot.