I'm getting ready for a Differential Geometry exam and after trying to carry out the exercises from last year's final exam, I have come up with several questions.
In the first question of the exam, we are given two pairs of vector fields, $\{X_1,Y_1\}$ and $\{X_2,Y_2\}$, defined as $$X_1=(1+y^2)\frac{\partial}{\partial x}, Y_1=\frac{\partial}{\partial y}$$ $$X_2=(1+x^2)\frac{\partial}{\partial x}, Y_2=\frac{\partial}{\partial y}.$$ For the first section of the question we have to show that there exist connections $\nabla_1$ and $\nabla_2$ in $\mathbb{R}^2$ such that the pairs of vector fields are respectively parallel. My question is, do I have to compute all of the possible combinations, $\nabla_{1{X_1}}Y_1, \nabla_{1{X_1}}X_1, \nabla_{1{Y_1}}Y_1$, etc. impose that they are equal to zero, find the Christoffel symbols and therefore say that both connections will be uniquely defined? Or maybe there exists an easier, less calculistic way to show it?
For the second section, we must show that there does not exist any Riemannian metric in $\mathbb{R}^2$ such that it has $\nabla_1$ as its Levi-Civita connexion. If the previous section is true, then all I have to do is calculate the torsion $T^{\nabla_1}(X_1,Y_1)$ using the previously calculated Christoffel connection and, if it is not zero, then there will not exist such Riemannian metric. Is there any other way to approach this problem?
Also, for the third section I have to give an expression of all the Riemannian metrics in $\mathbb{R}^2$ such that $\nabla_2$ is its Levi-Civita connexion. As far as I can see, all I have to do is impose $\nabla g=0$, but I really don't know how to apply this for those two vector fields, $\{X_2,Y_2\}$.
Then I have to calculate Riemann's curvature tensor but I haven't had any problems with that so far.
Let $M$ be some $n$-dimensional manifold and let $(X_1,\dots,X_n)$ be a global frame of $TM$ (assuming it exists). That is, at each point $p \in M$, the tangent vectors $(X_1(p), \dots, X_n(p))$ for a basis of $T_pM$. Then there is a unique connection $\nabla$ on $M$ for which the vector fields $X_i$ are parallel ($\nabla X_i = 0$). To see why this is true, assume first that such a connection exists. Any vector field $X$ on $M$ can be written uniquely as $X = \zeta^i X_i$ where $\zeta^i \colon M \rightarrow \mathbb{R}$ are smooth functions. Writing $Y = \eta^j X_j$, we see that we must have
$$ \nabla_{X}(Y) = \nabla_{\eta^i X_i}(\zeta^j X_j) = \eta^i \nabla_{X_i}(\zeta^j X_j) = \eta^i (d(\zeta^j)(X_i) X_j + \zeta^j \nabla_{X_i}(X_j)) \\ = \eta^i d(\zeta^j)(X_i) X_j.$$
Now, you can readily define a connection by the above formula and verify that it is indeed a connection. The point is that the frame $X_i$ doesn't have to be a coordinate frame, it can be any arbitrary frame of $TM$. Now, how can we determine if there exists a metric $g$ on $M$ such that $\nabla$ (defined as above) is the Levi-Civita connection of $g$? We have two conditions:
$$ g(X,Y)|_{p} = g(\zeta^i X_i, \eta^j X_j) = g_{ij} \zeta^i(p) \eta^j(p) $$
then this is readily seen to be a Riemannian metric with respect to which, $\nabla$ is $g$-metric.
In your case, both $(X_1,Y_1)$ and $(X_2,Y_2)$ are global frames of $T\mathbb{R}^2$ so there exists a unique connection $\nabla_i$ making the frame $(X_i,Y_i)$ $\nabla_i$-parallel. Since $[X_1,Y_1] \neq 0$, the connection $\nabla_1$ cannot be a metric connection with respect to any Riemannian metric $g$ on $\mathbb{R}^2$. Since $[X_2,Y_2] = 0$, the connection $\nabla_2$ will be a metric connection with respect to any Riemannian metric $g$ that is "constant" on $\mathbb{R}^2$ in the sense described above.