# Is this covariant derivative compatible with the Euclidean metric on $\mathbb{R}^2$?

I'm currently learning about metrics and covariant derivatives in differential geometry, and I can't figure out if the following form of covariant derivatives on $\mathbb{R}^2$ satisfy the conditions for "metric compatibility" or not (with respect to the usual Euclidean metric):

\begin{align} \nabla_X X &= \alpha Y \\ \nabla_X Y &= -\alpha X \\ \nabla_Y X &= \beta Y \\ \nabla_Y Y &= -\beta X \end{align}

Which, if I understand correctly, is equivalent to:

\begin{align} \nabla_X \tilde{Y} &= \alpha \tilde{X} \\ \nabla_X \tilde{X} &= -\alpha \tilde{Y} \\ \nabla_Y \tilde{Y} &= \beta \tilde{X} \\ \nabla_Y \tilde{X} &= -\beta \tilde{Y} \end{align}

Where $X$ and $Y$ are the standard basis vector fields, $\tilde{X}$ and $\tilde{Y}$ are the dual basis one-forms, and $\alpha$ and $\beta$ are arbitrary manifold functions $\mathbb{R}^2 \to \mathbb{R}$.

To the best of my understanding, this covariant derivative satisfies $\nabla g = 0$ (where $g$ is the Euclidean metric tensor) and is therefore a metric connection (but not the unique torsion-free Levi-Civita connection), because applying the tensor product rule gives:

\begin{align} \nabla_X g &= \nabla_X \left( \tilde{X} \otimes \tilde{X} + \tilde{Y} \otimes \tilde{Y} \right) \\ &= \left( \nabla_X \tilde{X} \right) \otimes \tilde{X} + \tilde{X} \otimes \left( \nabla_X \tilde{X} \right) + \left( \nabla_X \tilde{Y} \right) \otimes \tilde{Y} + \tilde{Y} \otimes \left( \nabla_X \tilde{Y} \right) \\ &= -\alpha \tilde{Y} \otimes {X} - \alpha \tilde{X} \otimes \tilde{Y} + \alpha \tilde{X} \otimes \tilde{Y} + \alpha \tilde{Y} \otimes \tilde{X} \\ &= 0 \end{align}

And similarly for $\nabla_Y$:

\begin{align} \nabla_Y g &= \nabla_Y \left( \tilde{X} \otimes \tilde{X} + \tilde{Y} \otimes \tilde{Y} \right) \\ &= \left( \nabla_Y \tilde{X} \right) \otimes \tilde{X} + \tilde{X} \otimes \left( \nabla_Y \tilde{X} \right) + \left( \nabla_Y \tilde{Y} \right) \otimes \tilde{Y} + \tilde{Y} \otimes \left( \nabla_Y \tilde{Y} \right) \\ &= -\beta \tilde{Y} \otimes {X} - \beta \tilde{X} \otimes \tilde{Y} + \beta \tilde{X} \otimes \tilde{Y} + \beta \tilde{Y} \otimes \tilde{X} \\ &= 0 \end{align}

However, this connection seems (although I am not sure) as though it would give rise to geodesics which are not straight lines when $\alpha$ and $\beta$ are nonzero, which contradicts other definitions of metric compatibility I have heard. So, my questions are:

1. Is the covariant derivative defined above compatible with the Euclidean metric on $\mathbb{R}^2$, or am I misunderstanding something about metric compatibility?
2. If it is compatible, what kind of behavior does parallel transport exhibit under this connection?
3. In particular, do nonzero $\alpha$ and $\beta$ produce geodesics which are not straight lines?
• Are you sure you want $\nabla_Y Y = -\beta Y$ and not $\nabla_Y Y = -\beta X$ (or something) in your fourth equation? Because I think it might make hella difference! – Robert Lewis Jul 17 '17 at 2:54
• I'm also curious as to whether $\nabla_Y Y = - \beta Y$ and $\nabla_Y {\tilde Y} =\beta {\tilde X}$ etc. are consistent – Robert Lewis Jul 17 '17 at 3:07
• $-\beta Y$ was a typo -- thanks for pointing it out! Fixed now. – exists-forall Jul 17 '17 at 3:13
• Now it makes hella more sense! Thanks! – Robert Lewis Jul 17 '17 at 3:14

1) By definition, the connection $\nabla$ is metric or compatible with the metric if it satisfies $\nabla g=0$.
2) Equivalently, the connection $\nabla$ is metric if and only if the parallel transport it induces preserves the Riemannian metric.
3) Geodesics with respect to a metric connection $\nabla$ are not, in general, the same as geodesics with respect to the Levi-Civita connection.