# Connections and Riemannian metrics Application

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.

For the first question: given $\mathcal{M}$ a smooth manifold, if you have an atlas $\{(U_{\alpha},\mathbf{x}_{\alpha})\}_{\alpha \in I}$ such that for all $\alpha , \beta \in I$ with $U_{\alpha} \cap U_{\beta} \neq \varnothing$ the jacobian matrix of $\mathbf{x}_{\beta} \circ \mathbf{x}_{\alpha}^{-1}$ is constant, you can define a unique connection $\nabla$ such that its components are null on every chart, i.e., for any smooth vector fields $X$ and $Y$ \begin{align*} \nabla_{X} Y = \sum_{i,k=1}^{n} X^{k} \, \partial_{k} Y^{i} \, \partial_{i} \qquad (\Gamma^{i}_{\; jk}=0) \end{align*} in $U_{\alpha}$, where the $\partial_{i}$ are the coordinate vectors fields of the chart $(U_{\alpha},\mathbf{x}_{\alpha})$, as the components of the vector fields.

Since your vector fields are globlaly defined and $\mathbb{R}^{2}$ admite global charts, you can check if there is a global chart such that they are its coordinate fields. You can already see that for $X_{1}$ and $Y_{1}$ this is not possible because they do not commute. If you can't find such a chart then you just do as you said and calculate all the derivatives.

The second part of your problem is pretty much just calculating the torsion as you said.

For the third part, if you could find a chart such that $X_{2}$ and $Y_{2}$ are its coordinate fields, you know that in this chart the Christoffel symbols are null, so write then in terms of the components of the metric and try finding the restrictions you need for the metric in this chart.

• Note that this doesn't explain why you can find a connection $\nabla_1$ with respect to which $(X_1,Y_1)$ are parallel (because they do not satisfy your condition). Commented Dec 29, 2016 at 21:57
• @levap your answer is definitely more complete. I wasn't sure if the same thing worked for non-coordinate frames, I just had a guess it would. I'm still studying this things, so I wasn't going to say something I'm not sure. Commented Dec 30, 2016 at 2:19

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:

1. First, $\nabla$ should be torsion free so $\nabla_X Y - \nabla_Y X = [X,Y]$ for all vector fields $X,Y$ on $M$. In particular, choosing $X = X_i, Y = X_j$ we see that we must have $[X_i,X_j] = 0$ for all $1 \leq i,j \leq n$. In other words, the frame $(X_1,\dots,X_n)$ must be a coordinate frame or else, the unique connection that renders $X_i$ parallel won't be torsion-free.
2. Now, let $g$ be any Riemannian metric on $M$ and write $g_{ij} := g(X_i,X_j)$ for the entries of $g$ with respect to the frame $(X_i)$. If $g$ is metric, we must have $$d(g(X_i,X_j))(X_k) = g(\nabla_{X_k} X_i, X_j) + g(X_i, \nabla_{X_k} X_j) = 0$$ for all $1 \leq i,j \leq n$. If $(X_i)$ is a coordinate frame and we choose coordinates $x^i$ such that $\frac{\partial}{\partial x^i} = X_i$, this translates into the condition $$\frac{\partial g_{ij}}{\partial x^k} = 0$$ for all $1 \leq i, j, k \leq n$. If $M$ is connected, this means that each $g_{ij}$ must be constant! Going the other way around, if we choose some fixed positive definite matrix $g = (g_{ij}) \in M_n(\mathbb{R})$ and define a Riemannian metric by the formula

$$g(X,Y)|_{p} = g(\zeta^i X_i, \eta^j X_j) = g_{ij} \zeta^i(p) \eta^j(p)$$

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.

• There is only one thing that I don't understand. Why is it that the condition of $\nabla g=0$ translates into $\frac{\partial g_{i,j}}{\partial x_k}=0$?
• In $(2)$, I wrote the condition that $g$ is metric with respect to $\nabla$ as $Xg(Y,Z) = d(g(Y,Z))(X) = g(\nabla_X Y, Z) + g(Y, \nabla_X Z)$ (this is equivalent to $\nabla g = 0$). If we take $X = \frac{\partial}{\partial x^k}, Y = \frac{\partial}{\partial x^i}, Z = \frac{\partial}{\partial x^j}$, the right hand side is $\frac{\partial g_{ij}}{\partial x^k}$. Commented Jan 5, 2017 at 21:26