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Let $\mathbf V \colon [0,T] \times \mathbb R^d \to \mathbb R^d$ (for $T>0$) be a given, bounded smooth vector field and let $\mathbf X=\mathbf X(t,x)$ be its flow, i.e. the unique solution to the initial-value problem \begin{equation} \begin{cases} \frac{\partial}{\partial t} \mathbf X(t,x) = \mathbf V(t,\mathbf X(t,x)) & \text{ in } (0,T) \times \mathbb R^d \\ \mathbf X(0,x) = x \quad \text{ for all } x \in \mathbb R^d.   \end{cases} \end{equation}

A well-known result in standard ODE's theory says that $$\tag{1} \nabla_x \mathbf X(t,x) = \exp\bigg( \int_0^t \nabla \mathbf V(s,\mathbf X(s,x))\,ds\bigg). $$

Is there an analogous formula to (1) for ODEs driven by (smooth) vector fields on Riemannian manifolds? In particular, does this formula involve somehow the geometry of the Riemannian manifold? A rather precise question could be: consider the $C^1$ norm of $\mathbf X$ (or even its Lipschitz constant) w.r.t. space variable $x$: does it depend on some known tensors on the manifold (e.g. curvature)?

I have gone through books in differential geometry/differential topology (e.g. Lee, Lang) and they prove that $\mathbf X$ is smooth but do not compute explicitly the derivative.

References are very much welcome. Thanks.

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    $\begingroup$ I suggest migrating this math.stackexchange.com $\endgroup$ – Deane Yang Oct 29 '17 at 16:42
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    $\begingroup$ Anyway, the answer is do everything with respect to local co-ordinates. $\endgroup$ – Deane Yang Oct 29 '17 at 17:27
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    $\begingroup$ When solving for integral curves (the flow) of a vector field on a manifold, no geometric structure (connection, metric, etc.) is needed at all. What you want to check is that any solution with respect to one set of local coordinates is also a solution with respect to any other set of local coordinates. $\endgroup$ – Deane Yang Oct 29 '17 at 19:32
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    $\begingroup$ This is probably covered in any differential geometry or differential manifolds textbook. I suggest looking at Lee's book, Introduction to Smooth Manifolds. $\endgroup$ – Deane Yang Oct 29 '17 at 19:34
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    $\begingroup$ A family of diffeos $\phi\colon M\to M$ lifts to a family $T\phi\colon TM \to TM$, the Jacobian of $\phi$, which is linear on the fibers of $TM$. The analog of the determinant here is the induced map $\Lambda^n T\phi\colon \Lambda^n TM\to \Lambda^n TM$ ($n=\dim M$). Differentiating both $T\phi$ and $\det T\phi$ with respect to the family parameter gives vector fields $TM\to T(TM)$ and $\Lambda^n TM\to T(\Lambda^n TM)$. The relation between the latter two should be a kind of "trace", though I don't know it's precise nature off hand. $\endgroup$ – Igor Khavkine Oct 29 '17 at 23:04
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I think you may be interested in this remarkable, recent paper by E. Brué and D. Semola (see in particular Theorem 3.11 which answers to your question in a much more general setting).

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As was pointed out by Deane Yang and Igor Khavkine in the comments, this feels like a fact that should be "looser" than Riemannian geometry. Indeed, as I will show below, your formula makes sense in the broader setting of smooth manifolds equipped with a volume form. I don't know a reference for this fact.

If $(M, \Omega)$ is a manifold equipped with volume form, ${\bf V}\in \operatorname{Vect}(M)$ is a vector field on $M$, and ${\bf X}_t:M\to M$ is the flow of ${\bf V}$, then:

1) One has $J_t\in C^\infty(M)$, (i.e., a smooth $J:\mathbb{R}\times M\to M$), with $J_0\equiv 1$, given by, $$J_t(x)=(\Omega_x)^{-1}\otimes \Omega_{{\bf X}_t(x)}\otimes \Lambda^n(d{\bf X}_t|_x),$$ well-defined since $\Lambda^n(d{\bf X}_t|_x):\Lambda^n( T_xM)\to \Lambda^n (T_{{\bf X}_t(x)}M)$ as Igor said.

2) One has a well-defined $\operatorname{div}({\bf V})$, namely (up to a sign, I forget whether $\pm$) $$\operatorname{div}({\bf V})=d(\iota_{\bf V}\Omega)/\Omega.$$

So the "hopefully-an-identity" $$\left.\frac{d}{dt}\right\rvert_{t=0}J_t=\operatorname{div}({\bf V})$$ (for simplicity I state just the $t=0$ formula) is well-formed. You can then prove it by choosing local co-ordinates in which $\Omega\equiv 1$. (There is surely an intrinsic proof, too, but I can't think of one now.)

Afterthought, 30 Oct: I imagine the intrinsic proof consists of identifying both sides with the Lie derivative $(\mathscr{L}_{\bf V}\Omega)/\Omega$.

Update, 30 Oct, in response to questions of OP in comments: If I calculate correctly, then yes, the formulas \begin{align*} \frac{dJ_t}{dt}(x)&=\operatorname{div}({\bf V})|_{{\bf X}_t(x)} J_t(x)\\ J_t(x)&=\exp\left(\int_0^t\operatorname{div}({\bf V})|_{{\bf X}_s(x)}ds\right) \end{align*} are also correct.

And yes, although the formula for "$\tfrac{d}{dt}\left[\det\left(d{\bf X}_t\right)\right]$" depends only on the volume form, I think one needs a Riemannian metric $g$ in order to make sense of "$\tfrac{d}{dt}\left[d{\bf X}_t\right]$" (really, the $g$-covariant derivative $D_t$ of the section $d{\bf X}_t$ of the bundle $T_x^*M\otimes TM$ along the curve ${\bf X}_t(x)$). Probably $$D_t\left[d{\bf X}_t\right]|_{t=0}=\nabla {\bf V},$$ where $\nabla$ is the Levi-Civita connection of $g$. (Please check this calculation before relying on it!)

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    $\begingroup$ There is a corollary of this fact which, I think, is better-known: the flow of ${\bf V}$ preserves a volume form if and only if ${\bf V}$ is divergence-free with respect to that volume form. $\endgroup$ – macbeth Oct 30 '17 at 7:30
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    $\begingroup$ @macbeth , Life is more complicated on the variational case : see my comments here mathoverflow.net/questions/277168/… we don't have uniqueness. $\endgroup$ – user21574 Oct 31 '17 at 1:10
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    $\begingroup$ @user111164 You're welcome. Unfortunately I don't know references for these formulas (and therefore, by the way, I disagree with the designation of the question as "off-topic"). If there exists a book on dynamical systems specifically on Riemannian manifolds, that might be a good bet. The formulas might also turn up in a book on Morse theory, at least in the case $V=\nabla f$. $\endgroup$ – macbeth Oct 31 '17 at 13:07
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    $\begingroup$ @DeaneYang There was a flag asking to consider reopening this question. In view of your apology in a comment above, do you think the question should be reopened? $\endgroup$ – Todd Trimble Feb 9 '18 at 12:41
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    $\begingroup$ I've voted to reopen. $\endgroup$ – Deane Yang Feb 9 '18 at 16:10

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