Levi–Civita connection and geodesics on $P(n)$ with the trace metric

Question

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!

Answer 1

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.