Curvature of the sphere, how can I get it?

Question

I want to compute the curvature of the sphere. I have the following definition : The curvature is given by $$K_p(T_p\mathbb S^2)=\frac{R(X,Y,Y,X)}{\|X\wedge Y\|^2}$$ where $X,Y$ is a basis of $T_p(\mathbb S^2)$ and $R(X,Y,Z,W)=g(R_{XY}Z,W)$ where $g$ is the Riemanian metric, and $Z\longmapsto R_{XY}Z$ Rieman tenseur, i.e. $$R_{XY}Z=[\nabla _X,\nabla _Y]Z-\nabla _{[X,Y]}Z.$$

I think that a basis of $T_p(\mathbb S^2)$ is given by $$X=\frac{\partial }{\partial \theta}=(-\sin \varphi\sin \theta,\sin\varphi\cos\theta,0)$$ $$Y=\frac{\partial }{\partial \varphi}=(\cos\varphi\cos\theta,\cos\varphi\sin\theta,-\sin\varphi),$$

My try

On the sphere, we have $g=\mathrm d r^2+r^2\mathrm d \theta^2+r^2\sin^2\theta\mathrm d \varphi^2$ and since $r=1$ is constant, I would say that $\mathrm d r^2=0$, and thus $$g=\mathrm d \theta^2+\sin^2\varphi\mathrm d \varphi^2.$$ Now, $$\|X\wedge Y\|^2=g(X,X)g(Y,Y)-g(X,Y)^2=\sin^2\varphi.$$

Now, $$R(X,Y,Y,X)=g([\nabla _{\partial \theta},\nabla _{\partial \varphi}]\partial _\varphi-\nabla _{[\partial _\theta,\partial _\varphi]}\partial _\varphi,\partial _\theta).$$

For example $$\nabla _{\partial _\varphi}\partial _\varphi=\Gamma_{\varphi\varphi}^{\theta}\partial _\theta+\Gamma_{\varphi\varphi}^\varphi \partial _\varphi.$$

Question 1) Are the $\Gamma_{ij}^k$ numbers or function, i.e. what will give for example $\nabla _{\partial _\theta}\Gamma_{\varphi\varphi}^\varphi \partial _\varphi$ ? Will it be $$\partial _\theta (\Gamma_{\varphi\varphi}^\theta)\partial _\varphi+\Gamma_{\varphi\varphi}^\theta \nabla _{\partial _\theta}\partial _\varphi$$ or just $$\Gamma_{\varphi\varphi}^\theta \nabla _{\partial _\theta}\partial _\varphi ?$$

Question 2) Am I on the right way ? Is there an easier way ?

Answer 1

Ok here's my partial answer (partial in the sense that you seem incredibly capable of completing the connection coefficients yourself). It may constitute an easier way but then I'm a sucker for connection coefficients.

Setting the Scene

The three dimensional Euclidean metric in spherical coordinates is given by \begin{equation} ds^{2}=dr^{2}+rd \theta^{2}+r^{2} \sin^{2} \theta d \phi^{2} \end{equation} You note that this is the unit sphere, and hence you correctly observe that \begin{equation} ds^{2}=d \theta^{2}+\sin^{2} \theta d \phi^{2} \end{equation} As a matrix $$ g_{ij} = \begin{pmatrix} 1 & 0 \\ 0 & \sin^{2} \theta \end{pmatrix} $$ With inverse $$ g^{ij}= \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{\sin^{2} \theta} \end{pmatrix} $$ And now for the computation of the connection. by definition \begin{align} \Gamma^{i}_{jk} &= g^{im}\Gamma_{mij}\\ \Gamma_{mij} &= \frac{1}{2}\left(g_{mj, k}+g_{mk,j}-g_{jk, m} \right) \end{align} We can be smart here since the only non constant connection component is $g_{\phi \phi}$ which only depends on $\theta$ hence the only non-vanishing connection component is $g_{\phi \phi, \theta}$. now, from symmetry of the Christoffel symbols, we find \begin{align} \Gamma_{\phi \phi \theta} &= \Gamma_{\phi \theta, \phi} \\ &= \frac{1}{2}\left(g_{\phi \phi, \theta}+g_{\phi \theta, \phi}-g_{\phi \theta, \phi} \right) \\ &= \frac{1}{2}g_{\phi \phi, \theta} \\ &= \sin \theta \cos \theta \\ \implies \Gamma_{\theta \phi \phi} &= -\sin \theta \cos \theta \end{align} Since I showed that the metric is clearly diagonal, raising an index is a trivial task, viz, \begin{align} \Gamma^{\phi}_{\theta \phi} &= \Gamma^{\phi}_{\phi \theta} \\ &= g_{\phi \phi} \Gamma_{\phi \phi \theta} \\ &= \frac{\sin \theta \cos \theta}{\sin^{2} \theta} \\ &= \cot \theta \\ \Gamma^{\theta}_{\phi \phi} &= -\sin \theta \cos \theta \end{align} Now, the curvature tensor is defined by \begin{equation} R^{i}_{jkm} = \Gamma^{i}_{jm, k} - \Gamma^{i}_{jk, m}+\Gamma^{i}_{nk}\Gamma^{n}_{jm}- \Gamma^{i}_{nm}\Gamma^{n}_{jk} \end{equation} Notice that \begin{align} R_{ijkm} &= - R_{jikm} \\ &= -R_{ijmk} \end{align} Which means there is only one independent component. Thus \begin{align} R^{\theta}_{\phi \theta \phi} &= \Gamma^{\theta}_{\phi \phi, \theta} - \Gamma^{\theta}_{\phi \theta, \phi}+\Gamma^{\theta}_{n \theta}\Gamma^{n}_{\phi \phi}- \Gamma^{\theta}_{n \phi}\Gamma^{n}_{\theta \phi} \\ &= (-\sin \theta \cos \theta)_{, \theta}-0+0.\Gamma^{n}_{\phi \phi}-\Gamma^{\theta}_{n \phi}\Gamma^{n}_{\theta \phi}\\ &= (- \sin^{2}\theta + \cos^{2}\theta)-(-\sin \theta \cos \theta)\cot \theta \\ &= \sin^{2} \theta \end{align} You may wish to present the final answer as \begin{equation} R^{i}_{jkm} = \delta^{i}_{k}g_{jm}-\delta^{i}_{m}g_{jk} \end{equation} again, which yields \begin{align} R^{\theta}_{\phi \theta \phi} &= \delta^{\theta}_{\theta}g_{\phi \phi} - \delta^{\theta}_{\phi}g_{\theta \phi} \\ &= g_{\phi \phi} \\ &= \sin^{2} \theta \end{align}

I hope this helps you in your quest.