It is well known that every smooth and every complex manifold equipped with a group action by a (compact) Lie group $G$ admits a stratification by orbit types.
I would like to know if there is a similar orbit type stratification in every one of the following cases:

1.) smooth symplectic manifolds;

2.) holomorphic symplectic manifolds;

3.) Kaehler manifolds;

Most importantly, I would like to know if the strata are smoothly/holomorphically symplectic resp. Kaehler. I would appreciate if you can give some references/literature.

  • 3
    $\begingroup$ @Hassan Jolany: Thank you for the references and your answer. Does this stratification go through in the case of a complex manifold equipped with a non-degenerate closed (2,0)-form? Do you know if it goes through for Kaehler manifolds? $\endgroup$ Oct 12, 2017 at 11:14
  • 5
    $\begingroup$ @Flavius Aetius Your question depends on the meaning of non-degeneracy of symplectic (2,0)-form: You may find your answer here: Fehér, L. M.; Némethi, A.; Rimányi, R. Degeneracy of 2-forms and 3-forms. Canad. Math. Bull. 48 (2005), no. 4, 547–560. $\endgroup$
    – user21574
    Oct 12, 2017 at 11:27
  • 6
    $\begingroup$ You may check for simplest example(I learnt long times ago from someone): $G^\mathbb C≃T^∗G$ ,which is Kähler manifold and the symplectic quotient $G^{\mathbb C}//G$ is a stratified singular Kähler space. The quotient may be identified with the $ \mathbb T^{\mathbb C}/W$, where $\mathbb T^{\mathbb C}$ is the complexification of a maximal torus $\mathbb T$ in $G$ and $W$ is the Weyl group. The dimension of orbits of this stratification varies $\endgroup$
    – user21574
    Oct 12, 2017 at 11:55
  • 8
    $\begingroup$ @Flavius Aetius There is a nice survey paper of See Theorem1.3 numdam.org/article/AMBP_2006__13_2_209_0.pdf . I think you can ask your question of the author of this paper. Ask of Dr.Baohua FU $\endgroup$
    – user21574
    Oct 13, 2017 at 3:15
  • 6
    $\begingroup$ I think your question is very important.There is a very important conjecture of Fu-Yau . Conjecture 6.5 arxiv.org/pdf/math/0111089.pdf. Every birational contraction from a smooth projective symplectic variety is necessarily strictly semismall. Equivalently, it can always be symplectically stratified. $\endgroup$
    – user21574
    Oct 13, 2017 at 3:28

1 Answer 1


I have a partial answer to my own question that I posed 5 days ago. It turns out that at least in the case of a smooth symplectic manifold you can perform an orbit type stratification. Here is why:

Asume $X$ is a symplectic manifold equipped with a symplectic form $\omega$. Assume furthermore that a compact Lie group acts by symplectomorphisms on $X$. Then if we ignore the symplectic form for a while, $X$ is a smooth $G$-manifold. It therefore naturally decomposes in so-called orbit types $X_{(H)}:=\{x\in X: (Stab(x))=(H)\}$, where $H$ is a closed Lie subgroup of $G$ and the brackets $(~~)$denote the conjugacy class. These are in general not connected, but their connected components, $X_{(H)}^{i}$, have the structure of locally closed submanifolds. These connected components are referred to as orbit type strata of $X$ and satisfy a bunch of useful conditions, the most important of which is the frontier condition. These conditions are irrelevant here, so I will not mention them further. The smooth manifold $X$ is decomposed in a disjoint union of orbit type strata. Interestingly, each orbit type stratum $X_{(H)}^{i}$, coincides with some connected component $X_{H}^{j}$ of the isotropy type $X_{H}:=\{x\in X| Stab(x)=H\}$. Here is where the symplectic picture comes into play. By Lemma 27.1 in the book "Symplectic techniques in physics" by Guillemin and Sternberg the tangent space at the every point $x$ in $X_{H}$, $T_xX_H$ is a symplectic subspace of the symplectic tangent space at the point $x$ in all of $X$, $(T_xX, \omega_x)$. Hence $X_H$ is in fact a symplectic sub-manifold of $X$. Hence each connected component $X_{H}^{j}$ will be a symplectic submanifold. Thus, the orbit types $X_{(H)}^{i}$ will be symplectic submanifolds. The additional conditions, that strata need to satisfy, are topological in nature so I don't believe that the symplectic form $\omega$ will affect them. In light of what I tried to explain in the above, I believe that a smooth symplectic manifold $X$ endowed with a group action by a compact symplectic Lie group, can be stratified by symplectic orbit types. Moreover, I am quite certain that the above argumentation goes through in the case of a complex symplectic manifold, since in that case we have in advance a complex stratification by holomorphic orbit types. The only thing missing is that the strata admit a closed, non-degenerate $(2, 0)$-form. This however can be demonstrated in a similar fashion as in the book of Guillemin and Sternberg.

  • $\begingroup$ Flavius : Interesting $\endgroup$
    – user21574
    Oct 18, 2017 at 13:46
  • $\begingroup$ Hassan Jolany: I am assuming the orbit type stratification of a smooth/complex G-manifold. This, as mentioned in my original question, always exists. Then in case we have a smooth/holomorphic symplectic form what remains to be done is to show that the orbit type strata are in fact smoothly/holomorphically symplectic. I hope I did not miss something. I would recommend Symplectic techniques in physics by Guillemin and Sternberg. $\endgroup$ Oct 18, 2017 at 14:25
  • $\begingroup$ I know this book. I had read half of this book 5 years ago. It is nice book and the book written friendly $\endgroup$
    – user21574
    Oct 18, 2017 at 14:27
  • $\begingroup$ Ok, the relevant part is section II, chapter 27 Group actions and foliations. The only drawback is that everything is done in the smooth category. $\endgroup$ Oct 18, 2017 at 14:29
  • $\begingroup$ @FlaviusAetius I think, there is a small error in your argument. The statement "each orbit type stratum $X^i_{(H)}$ coincides with some connected component $X^j_H$ of the isotropy type $H$" does not hold. The orbit type strata are $G$-invariant, while only the normalizer of $H$ acts on the isotropy type submanifold. Moreover, the tangent space to the isotropy type submanifold is not the space of $H$-invariant vector (you still have an additional direction in the orbit direction), see for example Prop. 2.4.6 in Ortega & Ratiu. $\endgroup$ Oct 21, 2017 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy