Let $X$ be a real analytic vector field defined on $\mathbb{R}^n$. Assume the origin $0 \in \mathbb{R}^n$ is a zero of $X$. Assume, furthermore, that we know that the center-stable manifold (in the sense of [Al Kelley - The stable, center-stable, center, center-unstable, unstable manifolds](https://doi.org/10.1016/0022-0396(67)90016-2)) of $X$ at $0$ is actually *stable*. In this context, this means that every trajectory starting in the center-stable manifold close enough to $0$ converges to $0$. The center-stable manifold being stable implies that it is unique (also in Kelley's article above).



> 
> Do we know in this situation whether the center-stable manifold is analytic?
> 
> 
> 


If not,



> 
> Are there any known sufficient conditions that allow one to conclude that the center-stable manifold is analytic?
> 
> 
> 



Analyticity of central stable manifolds
Quick answer to the first question: no, there is no reason why it should be analytic. Take *e.g.* the parametric vector field (written as a Lie derivative)$$X(x,y):=-x^3\partial\_x+(y+\alpha x)\partial\_y~~~,~\alpha\in\mathbb{R}.$$ Here the center manifold is a $C^\infty$ regular curve through the origin, tangent there to the vector $\left[1,-\alpha\right]$. If I understood correctly what the center-stable manifold being stable means, then $X$ satisfies that property.


If $\alpha=0$, then clearly the center manifold is the line $\left\{y=0\right\}$, hence analytic. Yet for every nonzero value of $\alpha$ it cannot be analytic. The easiest way to see this is to express the center manifold as the graph $\left\{y=-\alpha x+f(x)\right\}$, and this can be done explicitly by solving the underlying (affine) differential equation $x^3y'=y+\alpha x$. On the other hand, one computes formally the Taylor series of $f$ at $0$ by looking for a $\hat{f}(x)=\sum\_{n>1} a\_n x^n\in\mathbb{R}[[x]]$ solving the same differential equation. Both objects are unique and, if $f$ were to be analytic, then the Taylor series $\hat{f}$ would be convergent and its sum would coincide with $f$. However it is not difficult to show that $a\_n$ diverges like $\mathrm{cst}\sqrt{n!}$, hence the Taylor series at $0$ has a null convergence radius.


