I'm trying to verify the formula for the Levi–Civita connection on $P(n)$ (the manifold of positive-definite symmetric matrices) with the trace metric defined by $g_X(A, B) = \operatorname{tr}(X^{-1}AX^{-1}B)$ for $A, B\in T_XP(n)\cong S(n)$ (the vector space of symmetric matrices) against the known formula for the geodesics. The formula for the connection is given here as $$\nabla_AB = -\frac{1}{2}(AX^{-1}B+BX^{-1}A)$$ and the geodesics $\gamma : \mathbb{R}^+\to P(n)$ are given for $\gamma(0) = X$ and $\gamma'(0) = A$ as $$\gamma(t) = X^{1/2}\exp(tX^{-1/2}AX^{-1/2})X^{1/2}$$ By the definition of the geodesics on a Riemannian manifold, the curve $\gamma$ must have zero acceleration at all points, or $\nabla_{\dot{\gamma}}\dot{\gamma}\equiv 0$. We have \begin{align*} \dot{\gamma}(t) &= X^{1/2}(X^{-1/2}AX^{-1/2})\exp(tX^{-1/2}AX^{-1/2})X^{1/2} \\ &= AX^{-1/2}\exp(tX^{-1/2}AX^{-1/2})X^{1/2} \\ &= X^{1/2}\exp(tX^{-1/2}AX^{-1/2})X^{-1/2}A \end{align*} Then, \begin{align*} \gamma(t)^{-1}\dot{\gamma}(t) &= X^{-1/2}\exp(-tX^{-1/2}AX^{-1/2})X^{-1/2}X^{1/2}\exp(tX^{-1/2}AX^{-1/2})X^{-1/2}A \\ &= X^{-1}A \end{align*} Note that this also allows us to conclude that the geodesic is correctly parametrized by length, as $$g_{\gamma(t)}(\dot{\gamma}(t), \dot{\gamma}(t)) = \operatorname{tr}(X^{-1}AX^{-1}A) = g_X(A, A)$$ I don't quite see how this will give us the required zero acceleration $\nabla_{\dot{\gamma}}\dot{\gamma}\equiv 0$. Am I misinterpreting something here? Thanks!
Levi–Civita connection and geodesics on $P(n)$ with the trace metric
differential-geometrygeodesicriemannian-geometry
Related Solutions
The $\frac{\partial}{\partial x^{i}}$ represents the chosen basis of the tangent space at a point. In your case, Y could act on some function g as (omitting the $\Sigma$, using Einstein summation notation) : Y(g) = $f^{i}\frac{\partial g}{\partial x^{i}}$ and that gives you a number at that point which the tangent space is around. Because the partial is linear and acts on scalars in the usual way (i.e. it doesn't), this gives basis for a vector space. So if we use the ideas from Hawking and Ellis (1973) 'The large scale structure of space-time' we have the definition $\nabla_{X}Y$ as the covariant derivative of Y in the direction of X at p. So let's work with this and let's choose an arbitrary basis, {$E_{i}$}. Then let $Y = Y^{i}E_{i}$ and let $X = X^{i}E_{i}$ (capital Y is a little more natural for me for components of Y). The covariant derivative of Y is $\nabla Y = \nabla_{i}Y^{j}E^{i}\otimes E_{j}$ where I use the fact that $\nabla_{i}$ with the lowered index is the natural use of the covariant derivative, because we can calculate partials, e.g. $\nabla_{i}G_{j} = \frac{\partial}{\partial x^{i}}G_{j} - \Gamma^{k}_{ij}G_{k}$. We let $E^{i}$ be the basis for the cotangent space so that <$E^{i},E_{j}$> = $\delta^{i}_{j}$ then projecting $\nabla Y$ in to X we have $<X^kE_{k},\nabla_{i}Y^{j}E^{i}\otimes E_{j}>$ = $X^{k}\nabla_{i}Y^{j}<E_{k},E^{i}> E_{j}$ = $X^{k}\nabla_{i}Y^{j}\delta^{i}_{k} E_{j}$ = $X^{i}\nabla_{i}Y^{j} E_{j}$ = $\nabla_{X}Y$, and as you would in a normal vector equation, solve the components of the equation. As we're on a manifold, our natural basis is the one of partials, subbing $\frac{\partial}{\partial x^{i}}$ for $E_{i}$. This should give you the ODE.
I hope this helps, I'm sorry if I've missed something. This is my interpretation of what I've seen in various books on mathematical relativity (the area which I'm trying to become good at). I'd highly recommend checking out 'The large scale structure of space-time' by Hawking and Ellis (1973) and the introduction to 'General Relativity' by Wald (1984). These give a somewhat more practical approach, but that can be useful for building a level of intuition.
Edit
Upon re-reading, I think I went on a tangent too much. But I'll leave the answer anyway.
As stated the answer is negative: the velocity of geodesics has constant length, but an integral curve for a general vector field does not have to.
The question becomes interesting if you formulate it by asking that the support of the curve is the support of a geodesic. But even in this case the answer is negative. Indeed, consider a vector field over a compact manifold, vanishing at some points $p_-$ and $p_+$, $p_-\neq p_+$, and suppose that there is a flowline $\gamma$ limiting to $p_-$ and $p_+$ at $-\infty$ and $+\infty$ respectively. The support of $\gamma$ cannot be the support of a geodesic, as a geodesic on a compact manifold is a curve that has either infinite length or is diffeomorphic to $\mathbb S^1$.
Best Answer
In expression you have written for $\nabla_AB$ the right side is incorrect; As XT Chen points out, it does not depend at all on the the derivative of $B$ and doesn't satisfy the Leibniz rule. The right side is in fact equal to the Christoffel symbol $\Gamma$ of the connection, $$ \Gamma(A,B)|_X=-\frac{1}{2}(AX^{-1}B+BX^{-1}A) $$ The Christoffel symbol is related to the covariant derivative by $$ \nabla_AB=\partial_AB+\Gamma(A,B) $$ Where $\partial$ is the derivative induced by the coordinates. In this case, it can be computed as $$ (\partial_AB)|_X=\frac{d}{dt}(B|_{X+tA})|_{t=0} $$ Notice that in the spacial case that that the coordinate representation of $B$ is constant, we have $\nabla_AB=\Gamma(A,B)$. In general, however, the derivative term needs to be included. The line from the linked paper is seems to discussing just such a vector field with constant coefficients, defined by $Y=Y^\alpha E_\alpha$, with $Y^\alpha$ constant.
As an aside, I humbly take issue with some of the linked paper's treatment of covariant derivatives. The notion of "restricting" $\nabla$ to a bilinear map $T_pM\times T_pM\to T_pM$ doesn't make sense, generally speaking. If these topics are unfamiliar to you, I would recommend looking at a more standard reference first.