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Simply, we discuss things on $C^{2n}$ ($or R^{2n}$) but not on manifolds. Given an anayltic (or real analytic) vector field $V$ on $C^{2n}$ ($or R^{2n}$), with a zero at the origin . And $V_{0}$ is the linear part of $V$. Suppose one has some other conditions, i.e., usually one of the conditions is that $V_0$ associate with a matrix that is semi-simple. Is there any analytic (real analytic) symplectomorphism $f:(C^{2n},0)\rightarrow (C^{2n},0)$ (or $f:(R^{2n},0)\rightarrow (R^{2n},0)$), such that $f_{*}V=V_0$?

I know that Henri Poincare had the result of linearization of a real analytic vector field on $R^n$, and Sternberg has some theory for linearizing smooth vector fields.

BTW, naturally one can ask the similar question for odd dimensional spaces.

For smooth case of our problem, the answer is yes and I have found some literature. And for (real) analytic case, I guess it also has a well-known result, am I right? Please help me to make sure of it.

Latest update. 13. May. 2013

I found a monograph, "Lectures on Analytic Differential Equations" wrote by Yulij IIyashenko and Sergei Yakowenko, I think it helps me to understand the linearization theory. However, thank you for all your kind helps.

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  • $\begingroup$ Have a look at Arnold's book Mathematical Methods of Classical Mechanics. In Appendix 7 he discusses normal forms of Hamiltonian systems in a neighborhood of a stationary point. $\endgroup$
    – Liviu Nicolaescu
    Commented Apr 28, 2013 at 13:41
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    $\begingroup$ That's Henri Poincaré, not "Henry". $\endgroup$
    – Loïc Teyssier
    Commented Apr 28, 2013 at 17:59
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    $\begingroup$ It's not clear why you are considering coordinate changes only in the group of symplectomorphism if you don't also assume that $V$ is Hamiltonian with respect to the given symplectic structure. It's clear from first principles that no assumption on the linear part $V_0$ (even Poincaré's conditions as explained by Kofi below) will suffice for linearization if you are only allowing changes that preserve a given symplectic structure. The reason is that the local symplectomorphisms depend only on $1$ function of $2n$ variables while the vector fields depend on $2n$ functions of $2n$ variables. $\endgroup$
    – Robert Bryant
    Commented Apr 29, 2013 at 0:38

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Since you don't say what your 'some other conditions are', it is hard to know what result you are claiming. However, it is not difficult to show that, for $\mathbb{R}^4$ endowed with the symplectic form $\omega = dp\wedge dx + dq\wedge dy$, the vector field $$ Z = x\ \partial_x + (2y{+}x^2)\ \partial_y - (p{+}2xq)\ \partial_p - 2q\ \partial_q $$ is Hamiltonian and yet it cannot be smoothly (let alone real-analytically) linearized near the origin. The reason is the presence of $C^1$ integral curves of $Z$ passing through the origin that are not $C^2$, let alone smooth. For example, the curve $(x,y,p,q)=(x,x^2\ln|x|,0,0)$ is an integral curve of $Z$. (For the linearized vector field (which is also Hamiltonian) $$ Z_0 = x\ \partial_x + 2y\ \partial_y - p\ \partial_p - 2q\ \partial_q\ , $$ every integral curve passing through the origin is smooth.)

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  • $\begingroup$ Yeah, you are right, it is not always possible to do the linearization. I just want to suppose one had sufficient "good" conditions, whether there is a general result to the problem. I don't think it is an open problem, though it is hard for me to find the right literature. $\endgroup$
    – AlpS
    Commented Apr 28, 2013 at 15:40
  • $\begingroup$ In the (compplex) analytic setting, Kofi gave you the usual conditions under whihch the problem is not open. Otherwise, it is, even for vector fields in $\mathbb C^2$. In particular when the quotient of the eigenvalues is negative real (or zero), the quotient space of vector fields with given linear part modulo germs of analytic changes of coordinates is in general infinite dimensional. $\endgroup$
    – Loïc Teyssier
    Commented Apr 28, 2013 at 18:01
  • $\begingroup$ @Loïc: After reading over the OP's question again, I realized that the OP did not require that $V$ be Hamiltonian with respect to a symplectic structure (maybe this was an oversight, though). Hamiltonian vector fields never satisfy even the positivity conditions on the eigenvalues of $V_0$, since the eigenvalues of such $V_0$ are symmetric about $0$. One still might naïvely hope for symplectic linearization of Hamiltonian vector fields when $V_0$ is sufficiently 'generic', but simple examples in dimension $2$ show that this is hopeless; the moduli space is infinite dimensional in this case. $\endgroup$
    – Robert Bryant
    Commented Apr 29, 2013 at 10:50
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The result of Poincaré is in fact not about the real case but about the complex case. Suppose a vector field of the form $$ X = \sum_i\lambda_i x^i \partial_i + \text{higher order terms}$$ Then there is an analytic linearization diffeomorphism in a neighborhood of the origin, provided $$ \text{all} ~\lambda_i~ \text{lie in the same half plane about the origin}$$ and $$ \lambda_i \neq \sum_j m_i \lambda_j ~ \text{for any non-negativ integral} ~m_i~ \text{such that} \sum_i m_i> 1.$$

This result is cited in "Local Contractions and a Theorem of Poincaré" by Shlomo Sternberg.

Note that the semisimplicity of your matrix is not sufficient; I think you cannot get away without some diophantic conditions on the eigenvalues as stated above. This maybe seems somewhat strange at first, but it is not hard to construct counterexamples in the other case.

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  • $\begingroup$ Are you sure that you cannot guarantee that the linearization germ is real analytic in the real analytic setting when you don't necessarily require $V_0$ to be diagonal (all other hypothesis beeing met)? $\endgroup$
    – Loïc Teyssier
    Commented Apr 28, 2013 at 10:45
  • $\begingroup$ @Loïc: The vector field $X = x\ \partial_x + (2y{+}x^2)\ \partial_y$ in the $xy$-plane cannot be smoothly linearized because its integral curves near the origin are not smooth. (They are $x=0$ and $y=(c{+}\ln|x|)x^2$ for a constant $c$, and these latter curves are only $C^1$ at the origin.) Perhaps the OP should check whether the Hamiltonian prolongation of $X$ to the corresponding vector field on the cotangent bundle of the $xy$-plane can be smoothly linearized. At first glance, this seems doubtful. $\endgroup$
    – Robert Bryant
    Commented Apr 28, 2013 at 12:39
  • $\begingroup$ @Robert: the conditions stated by Kofi exclude your example, that was what I referred to in "(all other hypothesis beeing met)" from my comment. I concur that your example is not linearizable, though. Yet the vector fields beginning by the same 2-jets as you describe are real-analytically equivalent to the Poincaré-Dulac model $x\partial_x+(2y+x^2)\partial_y$ (as I'm sure you well know and only mention it for other readers). $\endgroup$
    – Loïc Teyssier
    Commented Apr 28, 2013 at 13:24
  • $\begingroup$ Yes, it needs some resonance condition as you said. But I just want to know if one had some sufficient "good" conditions, is it possible to have a general answer to our problem "linearization of a real analytic vector field in even or odd dimensional spaces, with the linearizing map being real analytic symplecto-morphism or contact transformation"? $\endgroup$
    – AlpS
    Commented Apr 28, 2013 at 15:34
  • $\begingroup$ I'm no specialist on symplectomorphisms. Yet regarding real-analyticity I can assure you that it works. Indeed, assuming the above "good" conditions you only have to show that the formal process bringing the vector field to its linear form is convergent. This is usually done with the use of majorizing series. Since the formal linearization is obtained by recursively solving a linear system, if all your data is real then your formal transform also is. $\endgroup$
    – Loïc Teyssier
    Commented Apr 28, 2013 at 16:46
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It might be worth mentionning works on linearization/normal-forms of Poisson structures, in particular around a degenerate singular point.

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  • $\begingroup$ Thank you very much. I will have a look immediately. Though I don't know french, I will try. $\endgroup$
    – AlpS
    Commented Apr 28, 2013 at 15:36

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