Let $X$ be an analytic vector field on a smooth manifold. Is it true that the flow $\Phi_t:M\to M$ associated to that vector field is also analytic?
2026-03-29 12:05:41.1774785941
Is the flow of an analytic vector field also analytic?
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As requested: First of all (as Mike correctly noted), to have a well-defined notion of a real-analytic vector field (or any real-analytic tensor) on the manifold $M$, you need not only a smooth manifold but a manifold equipped with a real analytic structure. Secondly, the equation $\frac{d}{dt}\Phi_t(m)= X(m)$, $m\in M$, $X$ is a vector field on $M$, may not have a short term solution without further hypothesis. One way to guarantee existence of a short term solution is to assume compactness of $M$. Given this, Cauchy - Kovalevskaya theorem implies that the 1st order ODE $\frac{d}{dt}\Phi_t(m)= X(m)$ has a real analytic solution on $M$ for $t\in [0,T]$ and some $T>0$ which will the required real analytic flow on $M$. (To apply the theorem, use it one chart at a time and then use uniqueness of solution to guarantee that the local solutions yield a global solution.) Note that compactness of $M$ also implies existence of a long term solution, i.e. a solution (a flow) which exists (and is real-analytic) for all values of $t\ge 0$.