Let $S^{2n-1}\subset\mathbb{C}^{n}$, and denote by $\langle\,\cdot\,,\,\cdot\,\rangle$ the Hermitian product. Then $$ \mathcal{C}_p:=\{\xi\in T_pS^{2n-1}\mid\langle p,\xi\rangle=0\},\quad p\in S^{2n-1}, $$ defines a contact structure on $S^{2n-1}$, i.e., a 1-codimensional completely non-integrable distribution (in terms of contact forms, $\mathcal{C}$ may be thought of as the conformal class $[\theta]$ uniquely defined by $\mathcal{C}=\ker\theta$). I will call it standard.

QUESTION: How many contact structures $\mathcal{C}$ one can equip the sphere $S^{2n-1}$ with, in such a way that the corresponding Lie group $\mathrm{Cont}\,(S^{2n-1})$ of contactomorphisms contains a finite-dimensional Lie subgroup $G$ acting transitively on $S^{2n-1}$?

REMARK: By "how many" I mean, of course, up to equivalence via diffeomorphisms. Moreover, if this will facilitate the answer, some extra topological (e.g., compactness, simply-connectedness) and/or algebraic (e.g.,semi-simplicity) property can be added to $G$.

Just to motivate the question, observe that, if $\mathcal{C}$ is the above standard structure, then $G$ can be taken as the unitary group $\mathrm{U}(n)$, so that there is - at least - the equivalence class of the standard contact structure. I'd like to know whether there are others.

SIDE QUESTIONS. Even without an answer to the main question, perhaps some clues/references concerning the topics below will help me:

  • how many contact structures there are on odd-dimensional spheres?
  • in how many ways one can construct spheres as homogeneous spaces?


According to R. Bryant's remark, there is a unique "up to equivalence" homogeneous contact structure on the odd-dimensional sphere. I'm trying now to understand why.

First, I lack the notion of "up to equivalence" in the context of homogeneous spaces. Here it goes my own intuition.

Let $M$ be a smooth manifold which is homogeneous w.r.t. two (in principle) different Lie groups $G$ and $\widetilde{G}$, i.e., $$M=\frac{G}{H}=\frac{\widetilde{G}}{\widetilde{H}}.$$

DEFINITION. The two structures of homogeneous manifolds are equivalent iff there exists $\phi\in\textrm{Hom}\,(G,\widetilde{G})$ such that: 1) $\phi(\widetilde{H})\subseteq \widetilde{H}$ and 2) $[\phi]\in\textrm{Diff}\, M$, where $\frac{G}{H}\stackrel{[\phi]}{\longrightarrow}\frac{\widetilde{G}}{\widetilde{H}}$ is the induced map.

SIDE QUESTION: is this definition correct? does it have some relevant application? does it fit in some category-theoretic approach to homogeneous spaces?

Second, assuming that the above definition is the correct one, how to prove that $S^{2n-1}=\frac{SU(n)}{SU(n-1)}$ is the unique homogenous manifold structure on $S^{2n-1}$?

So, given another structure $S^{2n-1}=\frac{G}{H}$, all boils down to prove that there is a group homomorphism $\phi:G\to SU(n)$, such that $\phi(H)\subseteq SU(n-1)$ and $[\phi]$ is a diffeomorphism.

SIDE QUESTION: is this the right way to tackle with the problem? can infinitesimal arguments be used instead? is there any book/paper where this sort of problems are dealt with?

Finally, going back to the key topic of this post, suppose that $(M,\mathcal{C})$ and $(M,\widetilde{\mathcal{C}})$ are two different homogeneous contact structure on the same manifold $$ M=\frac{G}{H}=\frac{\widetilde{G}}{\widetilde{H}}, $$ i.e., $\mathcal{C}$ is $G$-invariant and $\widetilde{\mathcal{C}}$ is $\widetilde{G}$-invariant.

Infinitesimally, this means that $$ \mathfrak{g}=\mathcal{C}_o\oplus\mathbb{R}Z \oplus\mathfrak{h};\quad \widetilde{\mathfrak{g}}=\widetilde{\mathcal{C}}_o\oplus\mathbb{R}\widetilde{Z} \oplus\widetilde{\mathfrak{h}} $$ where $Z$ and $\widetilde{Z}$ are the Reeb vector fields, and $$ \mathcal{C}_o\oplus\mathbb{R}Z = \widetilde{\mathcal{C}}_o\oplus\mathbb{R}\widetilde{Z} = T_oM $$ is the tangent space at the origin. Since there a linear transformation $h\in\mathrm{Aut}\, T_oM$ such that $h(\mathcal{C}_o)=\widetilde{\mathcal{C}}_o$, the local diffeomorphism $$ \widehat{h}:=\widetilde{\exp}\,\circ h\circ \exp^{-1} $$ sends $\mathcal{C}$ to $\widetilde{\mathcal{C}}$.

SIDE QUESTION: can this $\widehat{h}$ be used to prove Bryant's claim on the uniqueness of the homogeneous contact structure on $S^{2n-1}$? more generally, is there a criterion to patch together these diffeomorphisms and obtain a global diffeomorphism, thus proving the uniqueness of homogeneous contact structure on any homogeneous manifold? if not, are there examples of non-equivalent homogeneous contact structures on the same manifold?

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    $\begingroup$ These papers and the references they contain may be useful arxiv.org/pdf/math/0403358v2.pdf math.washington.edu/~duchamp/preprints/warsaw.pdf $\endgroup$ Jun 24, 2014 at 15:20
  • $\begingroup$ @FabriceBaudoin I wasn't aware of Bland and Duchamp's works: thank you for pointing them out. They'll certainly help in facing my first side question. $\endgroup$ Jun 25, 2014 at 6:58
  • $\begingroup$ @G_infinity Since you mentioned Alekseevsky below, I recalled that there is a short paper by him: Contact homogeneous spaces, Funktsional. Anal. i Prilozhen. 24 (1990), no. 4, 74--75. English translation in Funct. Anal. Appl. 24 (1990), no. 4, 324–325 (1991). See the MathSciNet entry. $\endgroup$ Jun 26, 2014 at 15:19

3 Answers 3


Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least.

Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$.

By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well.

By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n{+}1),\quad \mathrm{SU}(n{+}1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ (thought of as column vectors of height $\tfrac12(n{+}1)$ with quaternion entries) and uses the same formula as before, but now, one allows $J$ to be scalar multiplication (on the right) by any unit imaginary quaternion. Upon conjugating by an element of the subgroup $\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$ consisting of multiplication on the right by a unit quaternion, any two of these contact structures can be identified, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure identified in the first three cases.

Thus, there are really only four cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the four groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$.

This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this.

For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold. (There might be others; I haven't checked.)

  • $\begingroup$ Should $G = \mathrm{SU}(n + 1, 1)$ be $G = \mathrm{SU}(n, 1)$? Otherwise the rank-counting seems strange. $\endgroup$
    – LSpice
    Jan 2, 2019 at 3:16

Let me address the first side question, i.e. Question: How many contact structures are there on odd-dimensional spheres?

An answer: Usually there are at least infinitely many inequivalent contact structures on $S^{2n+1}$.

1.The case of $S^{3}$:

The classification is complete only in the case of the three-sphere $S^3$ due to [Eliashberg][1]. There are essentially "countably many plus one" of them.

Here I talk about classification with respect to isotopies of contact structures: two contact structures are equivalent if there exists an isotopy of the underlying manifold sending one contact structure to the other. By Gray stability this is the same as existence of a homotopy of contact distributions from one to the other (the homotopy has to be through contact distibutions not just whatever distributions).

Now in dimension 3 there is a dichotomy -- either the contact manifold $(M, \xi)$ contains an overtwisted disc, in which case $\xi$ is called overtwisted, or not, in which case $\xi$ is called tight. By Eliashberg there exists up to isotopy a unique tight contact structure on $S^3$, the standard one. Moreover, every homotopy class of plane distributions on $S^3$ contains a unique overtwisted contact structure. There are $(\pi_{3}(S^2) \cong \mathbb{Z})$-many plane distributions on $S^3$ and so there are $\mathbb{Z}$-many inequivalent overtwised contact structures on $S^{3}$.

All in all, there $(\mathbb{Z}+1)$-many inequivalent contact structures on $S^3$.

2.Higher-dimensional spheres:

Only very recently a higher-dimensional analogue of an overtwisted disc has been discovered by [Borman-Eliasberg-Murphy][2] (this work has not yet been refereed, as far as I know). It states that every homotopy class of almost contact structures contains a unique (again up to isotopy) overtwisted contact struture.

An almost contact structure on a manifold is a co-dimension one distribution with a (linear) complex struture on it. In particular, the underlying distribution of a contact structure is an almost contact structure.

Homotopy classes of almost contact structures on $S^{2n+1}$ are classified by the homotopy group $\pi_{2n+1}(\mathrm{SO}(2n+2)/\mathrm{U}(n+1))$. In either the Ding-Geiges paper suggested by Fabrice Baudoin (in the first comment) or in the B-E-M paper [2] you can read that $$\pi_{2n+1}(\mathrm{SO}(2n+2)/\mathrm{U}(n+1)) = \begin{cases}\mathbb{Z}/n!\mathbb{Z}, & n=4k, \\ \mathbb{Z}, & n=4k+1, \\ \mathbb{Z}/\frac{n!}{2}\mathbb{Z}, & n=4k+2, \\ \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}, & n=4k+3.\end{cases} $$

So there are at least that many inequivalent (overtwsited) contact structures on $S^{2n+1}$.

However, these are not necessarily all the contact structures. [Ustilovsky][3] constructed infinitely many inequivalent and non-overwisted contact structures on the spheres $S^{4m+1}$. So for example on $S^{5}$ there is a unique overtwisted contact structure but infinitely many non-overtwisted ones.

To conclude, you can watch Eliashberg's Banff talk on these recent discoveries (but it is not specifically about spheres).

[1]: Y.Eliasberg, Contact 3-manifolds twenty years since J.Martinet's work, link to the paper on Numdam.

[2]: M.S.Borman, Y.Eliashberg, E.Murphy, Existence and classification of overtwisted contact structures in all dimensions, arXiv.

[3]: I.Ustilovsky, Infinitely Many Contact Structures on $S^{4m+1}$, Internat. Math. Res. Notices 1999, no. 14, 781–791.

  • $\begingroup$ I'll try to polish this answer later... $\endgroup$ Jun 25, 2014 at 17:54
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    $\begingroup$ Concerning overtwisted higher-dimensional contact structure, there is also a much earlier proposition by Klaus Niederkrüger (the plastikstufe replacing the overtwisted disk, no superseeded by the notion of a bLob). $\endgroup$ Jun 25, 2014 at 18:44
  • $\begingroup$ @BenoîtKloeckner Quoting from the B-E-M paper, page 3: "We note that there were many proposals for defining the overtwisting phenomenon in dimension greater than three. We claim that our notion is stronger than any other possible notions, in the sense that any exotic phenomenon, e.g. a plastikstufe, can be found in any overtwisted contact manifold." $\endgroup$ Jun 25, 2014 at 23:31
  • $\begingroup$ @OldřichSpáčil : I did not know this overtwisted/tight dichotomy - thank you for pointing it out. I'm going to give a look at the references you've suggested! $\endgroup$ Jun 26, 2014 at 7:18
  • $\begingroup$ @OldřichSpáčil: i read the introduction of this paper too, but that the given notion is stronger than preceding ones does not mean that it is the first time such an analogue has been defined (or discovered), right? $\endgroup$ Jun 26, 2014 at 10:02

To answer your second side question, constructing a sphere as a homogeneous space is the same as saying that some Lie group acts transitively on it. The list of such is known as the Berger classification, and it coincides with the list of possible holonomy groups of Riemannian structures plus two extra cases. See http://en.wikipedia.org/wiki/Holonomy#The_Berger_classification

The list is quite short, so you can probably go through all cases and look for contact structures.

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    $\begingroup$ Actually, the Berger classification gives the list of compact groups acting transitively on spheres, but there are many non-compact groups that act transitively on spheres as well, and to really reduce this question to the Berger classification, you need to know that every Lie group acting transitively on the $n$-sphere contains a compact subgroup that acts transitively. (This is false for $n=1$, by the way, since the simply-connected cover of $\mathrm{SL}(2,\mathbb{R})$ acts transitively on $S^1=\mathbb{RP}^1$ and yet has no nontrivial compact subgroups.) $\endgroup$ Jun 24, 2014 at 13:59
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    $\begingroup$ @Robert : it is true that a compact simply connected homogeneous space of a Lie group (which one can assume connected) is a homogeneous space of a maximal compact subgroup. This is due to Deane Montgomery "Simply connected homogeneous spaces" PAMS 1950 $\endgroup$
    – BS.
    Jun 24, 2014 at 16:53
  • $\begingroup$ @BS: Thanks for this reference. I have run across this statement before, but I never knew a reference or to whom to attribute it. $\endgroup$ Jun 24, 2014 at 18:30
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    $\begingroup$ For $S^5$ the only compact $G$ are those you mentioned and $G=U(3)$. In general only $U(n),SU(n)$ and $SO(2n)$ will act on $S^{2n-1}$, while variations of $Sp(n)$ act on $S^{4n-1}$. In addition $Spin(9)$ acts on $S^{15}$ and $Spin(7)$ on $S^7$. Your remark about symmetric spaces refers to another use of the Bergman list, we are not interested in holonomy, only the fact that they act transitively on spheres. $\endgroup$ Jun 25, 2014 at 9:48
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    $\begingroup$ To follow up on Henrik's observations: $\mathrm{Spin}(7)$ does not preserve any contact structure on $S^7$ and $\mathrm{Spin}(9)$ does not preserve any contact structure on $S^{15}$. $\mathrm{Sp}(n)$ preserves a $2$-sphere of contact structures on $S^{4n-1}$, which are all equivalent under the action of $\mathrm{Sp}(n)\mathrm{Sp}(1)$ and, indeed, are equivalent to the unique contact structure on $S^{4n-1}$ that is preserved by $\mathrm{SU}(2n)$. Thus, there is, up to equivalence, a unique homogeneous contact structure on $S^{2n-1}=\mathrm{U}(n)/\mathrm{U}(n{-}1)$ for each $n>1$. $\endgroup$ Jun 27, 2014 at 14:22

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