This week I tried to learn a proof of Beltrami's theorem, but hit an obstacle which may have to do with some misunderstanding in basic concepts. So I'll try to give some of those first before presenting my question.
Terms and Background
"Geodesically equivalent up to reparameterization" is said "projectively equivalent" by Matveev, but the longer form emphasizes the discrepancy I'm grappling with.
A curve $\gamma$ of $(M, \nabla)$, $\nabla$ some affine connection on $M$, is called a geodesic if the covariant derivative along $\gamma$ satisfies $\nabla_{\dot{\gamma}}\dot{\gamma} = 0$. $\gamma$ is called a geodesic-up-to-reparameterization if there exists a differentiable function $h : J \to I$ from an open interval to the domain of definition of $\gamma$, such that $\gamma \circ h$ is a geodesic. Such regular curves are characterized as those such that there exists some differentiable $\alpha : I \to \mathbb{R}$ so that $\nabla_{\dot{\gamma}}\dot{\gamma} = \alpha\dot{\gamma}$.
Connections $\nabla, \bar{\nabla}$ on $M$ are called geodesically equivalent if they determine the same geodesics and are called projectively equivalent if they determine the same geodesics-up-to-reparameterization. So the Levi-Civita connection induced by the Beltrami-Klein model should be projectively equivalent to the standard connection on $\mathbb{R}^2$ but I believe not geodesically equivalent.
It is well known that the difference tensor $A(X, Y) = \bar{\nabla}_{X}Y – \nabla_{X}Y$ of two connections is antisymmetric if and only if they are geodesically equivalent. $\delta_j^i$ denotes the Kronecker delta function. Matveev writes that two connections $\nabla = (\Gamma_{jk}^i)$ and $(\bar{\nabla}) = (\bar{\Gamma}_{jk}^i)$ are projectively equivalent if and only they there exists a $1$-form $\phi$ so that $\bar{\Gamma}_{jk}^i – {\Gamma}_{jk}^i = \phi_k\delta_{j}^i + \phi_j\delta_k^i$, or equivalently such that $A(X, Y) = \phi(X)Y + \phi(Y)X$. Theorem 94.1 of Kreyszig is essentially the similar statement that a diffeormorphism $F$ from a neighborhood of surface $(S, g, \nabla)$ to another $(\bar{S}, \bar{g}, \bar{\nabla})$ preserves images of geodesics if and only if $$\bar{\Gamma}_{jk}^i -\Gamma_{jk}^i = \delta_{j}^i\frac{\partial{}h}{\partial{}\bar{u}^k} + \delta_{k}^i\frac{\partial{}h}{\partial{}\bar{u}^j}$$
where $\bar{u}$ are the coordinates on the image of the diffeormorphism and $h = \frac{1}{6}\log\frac{|(F_\ast g)_{ij}|}{|\bar{g}_{ij}|}$.
Question
By these definitions, if connections $\nabla, \bar{\nabla}$ are geodesically equivalent then they should be projectively equivalent. So geodesically equivalent connections have antisymmetric difference tensor, but also symmetric difference tensor by Matveev or Kreyszig's characterizations. But this would imply geodesically equivalent connections must be equal, contradicting the facts that connections with antisymmetric difference tensor are geodesically equivalent and you can generate connections with nontrivial and antisymmetric difference tensor by adding a nontrivial antisymmetric tensor to a connection, obtaining a new connection.
Matveev shows that the existence of such a $1$-form implies projective equivalence but he doesn't prove the converse. I believe I can parse the analogous converse proved in Kreyszig however, and it seems to go through fine. Their statements are not exactly the same but I imagine they're effectively so. So will somebody point out the flaw in my understanding?
Additionally perhaps some more nontrivial examples of pairs of connections satisfying these properties could elucidate things. Can anybody provide or point out resources with (elementary hopefully) examples?
Best Answer
I may have found my mistake, but confirmation with any additional insight would be helpful. If we restrict the statement to "symmetric connections are projectively equivalent if in only if there exists some such $1$-form" then the apparent contradiction I raised does not apply.
Kreyszig is working with the Levi-Civita connection on each surface. This is sensible because he uses the characterization of projective equivalence to prove Beltrami's theorem. This is a statement involving the Gaussian curvature, and the Gaussian curvature is defined using the metric.
Matveev seems to suggest early in his slides that he's talking about symmetric connections, but then in other places drops the word symmetric (like in his statement of Theorem 1, which involves projective equivalence, which is defined in terms of apparently arbitrary connections). I think he is always talking about symmetric connections and these are either typos, or a common practice I'm not keen on yet.