Let $(M,g)$ be a $2$-dimensional Riemannian manifold. Let $p\in M$ and let $r>0$ be such that the ball $B(p,r)\subset M$ is geodesic. Further let $(r,\theta)$ be polar coordinates in $T_p M$.
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Show that the pullback metric $(\exp_p|B(0,r))^* g$ can be represented in the form $dr^2+f(r,\theta)d\theta^2$ for a smooth function $f>0$.
My solution: Let $\overline g=(\exp_p|B(0,r))^* g$. Under the identification $T_0(T_p M)\cong T_p M$, the differential $d\exp_p:T_0(T_p M)\rightarrow T_p M$ is the identity map and $g_{xy}=\delta_{xy}$ in the coordinate chart $(x,y)=v\mapsto\exp_p(v)$. We compute \begin{align*} \overline g_{xx}+\overline g_{yy} &=\left( \left( \frac{\partial x}{\partial r} \right)^2 +\left( \frac{\partial y}{\partial r} \right)^2 \right)g_{rr} +2\left( \frac{\partial x}{\partial r}\frac{\partial x}{\partial\theta} +\frac{\partial y}{\partial r}\frac{\partial y}{\partial\theta} \right)g_{r\theta} +\left( \left( \frac{\partial x}{\partial\theta} \right)^2 +\left( \frac{\partial y}{\partial\theta} \right)^2 \right)g_{\theta\theta} \\ &=(\cos^2\theta+\sin^2\theta)g_{rr} +2r(\sin\theta\cos\theta-\cos\theta\sin\theta)g_{r\theta} +r^2(\cos^2\theta+\sin^2\theta)g_{\theta\theta}. \end{align*} Therefore $$\overline g=dr^2+r^2 d\theta^2.$$
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Show that the Gauss curvature of $(M,g)$ at a point $x=\exp_p(r,\theta)$ fulfills $$K(x)\geq-\frac{\partial^2}{\partial r^2}f(r,\theta)/f(r,\theta).$$
It seems to me that I've done the first part incorrectly, because I would think that $f$ somehow depends on $\theta$, or does it not do that because that dependence is absorbed by the exponential map?
I'm not sure how I would go about the second part. I know the formula $$K=g(\nabla_y\nabla_x\partial_x-\nabla_x\nabla_y\partial_x,\partial_y)/\det g,$$ but I'm still struggling with that formula.
It looks straight forward if I can compute the covariant derivatives, but I'm struggling with those...