Time derivative of squared distance function under evolving metric

Question

Suppose we have an evolving familly of Riemannian manifolds $\{M, g(t)\}_t$ indexed by time $t$. Typically a flow. We fix a point $x$ in $M$ and consider the following function : $$ f_x(t;y) = d_{g(t)}^2(x, y),$$ where $d_{g(t)}$ denotes the distance function on $M$ w.r.t. the metric $g(t)$. I would like to compute the time derivate of $f_x$ but I am not sure I am doing the right things. I am in the situation where $y$ depends on $t$ as $y(t) = \exp_{x}^{(t)}(v)$. A priori I should have something of that type $$ \frac{d}{dt} f_x(t; y(t)) = \partial_t f_x(t; y(t)) + \partial_i f_x(t; y(t))\partial_t y^i(t),$$ where $\partial_i = \frac{\partial}{\partial y^i(t)}$ in a chosen basis of $T_xM$. One can show that the differential of the squared distance function is given by $$df_x(t; y)(\cdot) = g_x(t)\left(2\left(\exp_y^{(t)}\right)^{-1}(x), \cdot\right).$$ I think that, since in a normal basis around $x$ at time $t$, we have $y^i(t) = v^i(t)$, we have $\partial_t y^i(t) = \dot{v}^i(t)$. The second term is then of the form $$ \partial_i f_x(t; y(t))\partial_t y^i(t) = g_{ij}(t)2v^j(t)\dot{v}^i(t) = \frac{d}{dt} \left(g(t)\left[v, v\right]\right) - \left(\partial_t{g}_{ij}\right)\left[v^i(t), v^j(t)\right]. $$ Is it true that $$ \left(\partial_t{g}_{ij}\right)\left[v^i(t), v^j(t)\right] = \left(\partial_t g\right)\left[v, v\right] ? $$

Answer 1

EDIT: I misread the question. Answer below has been edited to allow the endpoint $y$ to depend on $t$.

Here's another approach to computing the variation of distance: Given $t$, let $\gamma_t: [0,1] \rightarrow M$ be the constant speed geodesic with respect to $g_t$ from $x$ to $y$. Therefore, $$g_t(\gamma'_t(s),\gamma'_t(s)) = d_t^2(x,y)\text{, for each }s\in[0,1].$$ $\gamma_t$ also satisfies the geodesic equation $$\nabla^t_{\gamma'_t}\gamma'_t = 0.$$

Let $T(s) = \left.\partial_t\right|_{t=0}\gamma_t(s)$. Observe that since $\gamma_t(0)=x$ and $\gamma_t(1)=y$ for any $t$, $T(0) = T(1) = 0$. Then \begin{align*} \left.\frac{d}{dt}\right|_{t=0}d_t^2(x,y)& = \left.\frac{d}{dt}\right|_{t=0} \int_{s=0}^{s=1}g_t(\gamma'_t(s),\gamma'_t(s))\,ds\\ &= \int_{s=0}^{s=1}\dot{g}_0(\gamma_0',\gamma_0') + 2g_0(\nabla^0_T\gamma_0',\gamma_0')\,ds, \end{align*} where \begin{align*} \dot{g}_0 &= \left.\partial_t\right|_{t=0}g_t\\ T &= \left.\partial_t\right|_{t=0}\gamma_t. \end{align*} The vector field $T$ is a Jacobi field along $\gamma_0$ such that $T(0) = 0$ and $T(1) = y'(0)$. It also satisfies $$ \nabla_T\gamma_0' = \nabla_{\gamma'_0}T, $$ and therefore \begin{align*} \int_{s=0}^{s=1}g_0(\nabla^0_T\gamma_0',\gamma_0')\,ds &= \int_{s=0}^{s=1}g_0(\nabla^0_{\gamma_0'}T,\gamma_0')\,ds\\ &= \int_{s=0}^{s=1} \frac{d}{dt}(g_0(T,\gamma_0')) - g_0(T,\nabla_{\gamma_0'}\gamma_0')\,ds\\ &= g_0(T(1),\gamma_0'(1)). \end{align*} Therefore, the change in distance is the sum of the change due to the change in the metric and the change due to how the endpoint $y$ moves: \begin{align*} \left.\frac{d}{dt}\right|_{t=0}d_t^2(x,y)& = \int_{s=0}^{s=1}\dot{g}_0(\gamma_0',\gamma_0')\,dx + 2g_0(y'(0),\gamma_0'(1)). \end{align*} This can be written in terms of the exponential map as follows: \begin{align*} \int_{s=0}^{s=1}\dot{g}_0(\exp^{(0)}_x(sv))((d\exp^{(0)}_x)_*(sv), (d\exp^{(0)}_x)_*(sv))\,dx + 2g_0(y'(0), (d\exp^{(0)}_x)_*(v)). \end{align*}