If $R\mathrel{:=}\mathbb{R}[x_1,\dotsc,x_{n+1}]/(x_1^2+\dotsb+x_{n+1}^2-1)$ and $S^n\mathrel{:=}\operatorname{Spec}(R)$ is the real $n$-spere, a classical result of Borel and Serre says that the only real spheres with an almost complex structure is $S^2$ and $S^6$. An almost complex structure is an endomorphism of the tangent bundle $$J: T_{S^2} \rightarrow T_{S^2}$$ with $J^2=-\operatorname{Id}$. In the case of the real 2-sphere it follows the tangent bundle $T_{S^2}$ is a real algebraic vector bundle of rank 2.

Question 1: I'm looking for an example of a real algebraic (even dimensional) manifold $M$ with an almost complex structure $J:T_M \rightarrow T_M$, where $J$ is not algebraic. The problem of constructing a holomorphic structure on $S^6$ — is this still an open problem?

Note: If you let $k\mathrel{:=}\mathbb{R}$ and $K\mathrel{:=}\mathbb{C}$, it follows there is an isomorphism $$\operatorname{Spec}(K\otimes R)\cong \operatorname{Spec}(B)\mathrel{:=}S^2_K$$ with $$B\mathrel{:=}K[u,v,w]/(uv-(w^2+1)).$$

If $J_K$ is the pull-back of $J$ to $S^2_K$ it follows that $$\phi\mathrel{:=}\frac{1}{2}(I+iJ) \in \operatorname{End}(T_{S^2_K})$$ is an idempotent: $\phi^2=\phi$ and you get a direct sum $$T_{S^2_K} \cong L_1\oplus L_2$$ with $L_i \in \operatorname{Pic}(S^2_K)$. If $J$ is algebraic it follows $L_i$ are algebraic, and I'm interested in this decomposition.

In the case of the real 6-sphere $S^6$ it follows the endomorphism bundle $\operatorname{End}(T_{S^6})$ is a real algebraic vector bundle of rank $36$. We may consider the subvariety $$I(S^6)\mathrel{:=}\{J \in \operatorname{End}(T_{S^6}): J^2=-\operatorname{Id}\}$$ and the group scheme $G\mathrel{:=}\operatorname{GL}(T_{S^6})$. There is a canonical action $$\sigma: G \times I(S^6) \rightarrow I(S^6)$$ and a "parameter space" $I(S^6)/G$ parametrizing algebraic almost complex structures on $S^6$. Is this construction used in the study of the Hopf problem — the problem of constructing a holomorphic structure on $S^6$? If there is a holomorphic structure on $S^6$ — is this neccessarily algebraic? I ask for references.

Note: We may also consider the ring $R\mathrel{:=}C^{\infty}(S^6)$ and the $R$-module $T^{\infty}_{S^6}$ of smooth sections of $T_{S^6}$, and the projective $R$-module $\operatorname{End}_R(T^{\infty}_{S^6})$. We may consider a smooth almost complex structure $J\in \operatorname{End}_R(T^{\infty}_{S^6})\cong \Omega^{1,\infty}_R\otimes_R T^{\infty}_{S^6}$. We get a topological subspace $I^{\infty}(S^6)$ of smooth almost complex structures on $S^6$ equipped with an action of the group of smooth automorphisms of $T^{\infty}_{S^6}$. This is a topological subspace of a smooth vector bundle of rank $36$, and the orbit space parametrize all smooth almost complex structures on $S^6$. Hence if there is a holomorphic structure on $S^6$ it "lives" in this orbit space.

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    $\begingroup$ The existence of complex structure on $S^6$ is somewhat sensitive topic, but to my best knowledge the community consensus is that it's still very much an open problem. $\endgroup$
    – M.G.
    Apr 14, 2021 at 15:20
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    $\begingroup$ You can take any almost complex structure which is not integrable and is algebraic, and then patch in local coordinates with a locally defined complex structure, using bump functions, so you get integrability on an open set, but not everywhere. It is easiest to see using complex linear coframings. $\endgroup$
    – Ben McKay
    Apr 14, 2021 at 15:55
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    $\begingroup$ Take any symplectic manifold with $b_1$ odd, for example Thurston's example math.stonybrook.edu/~milivojevic/… This cannot be algebraic because $b_1$ would have to be even. Any symplectic manifold admits almost complex structure compatible in a rather nice way with the symplectic form. $\endgroup$ Apr 14, 2021 at 19:44
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    $\begingroup$ @LiviuNicolaescu: Every compact manifold is diffeomorphic to a real algebraic manifold. The question is not about complex algebraic manifolds, only real ones, so $b_1$ is not relevant. $\endgroup$
    – Ben McKay
    Apr 15, 2021 at 9:49
  • $\begingroup$ @M.G. - can you specify "sensitive topic"? $\endgroup$
    – user122276
    Apr 16, 2021 at 10:17

1 Answer 1


Let $M=SU_3$, the compact semisimple Lie group.

By request of the OP, for those unfamiliar with Maurer-Cartan form, let me define it. Write each point of $SU_3$ as a matrix $g$. Left translation by $g^{-1}$ takes $g$ to $I$, so takes $T_g SU_3$ to $T_I SU_3$. Consider any tangent vector $A\in T_g SU_3$ as a $3\times 3$ complex matrix, since $SU_3$ sits in the vector space of $3\times 3$ complex matrices. We have a linear map $A\in T_g SU_3\mapsto g^{-1}A\in T_I SU_3$, just the derivative of left translation (as the derivative of linear map on a vector space is that same linear map, and then restrict from all $3\times 3$ matrices to the linear subspace $T_g SU_3$). We define the left invariant Maurer Cartan form $\omega$ to be the linear map which takes each tangent space $T_g SU_3$ to $T_I SU_3$ by $\omega_g(A)=g^{-1}A$. So $\omega$ is a $1$-form valued in $\mathfrak{su}_3=T_I SU_3$. (This is all very standard, in every Lie group textbook.) Naturally, one writes this form not as $\omega$ but as $g^{-1}dg$.

As requested: for more information about the Maurer-Cartan form (for any Lie group, real or complex or $p$-adic):

Chevalley, Claude, Theory of Lie Groups. I., Princeton Mathematical Series, vol. 8. Princeton University Press, Princeton, N. J., 1946. ix+217 pp. 152 section V.IV.

Sharpe, R. W., Differential geometry. Cartan's generalization of Klein's Erlangen program. With a foreword by S. S. Chern. Graduate Texts in Mathematics, 166. Springer-Verlag, New York, 1997. xx+421 pp. ISBN: 0-387-94732-9, p. 96.

Write the left invariant Maurer--Cartan form $g^{-1}dg$ as $\omega=(\omega_{\mu\bar\nu})$ for $\mu,\nu=1,2,3$. The coframing $\omega_{1\bar{1}}+i\omega_{2\bar{2}},\omega_{1\bar{2}},\omega_{1\bar{3}},\omega_{2\bar{3}}$ is complex linear for a unique left invariant complex structure. Take a function $f(g)=e^{-|g_{1\bar{2}}|^2/\varepsilon}$, or your favourite nonzero nonalgebraic function. The coframing $\omega_{1\bar{1}}+if\omega_{2\bar{2}},\omega_{1\bar{2}},\omega_{1\bar{3}},\omega_{2\bar{3}}$ continues to have linearly independent real and imaginary parts, so is still complex linear for a unique almost complex structure. But since $f$ is not holomorphic and is not constant, there is a nonzero Nijenhuis tensor. You could also take $f$ to be instead some smooth function with support in some small compact set, and then you get complex structure on some open set turning smoothly almost complex, not complex, on some other open set, so clearly not algebraic or even analytic.

To clarify, as requested: as $SU_3$ is a Lie group, its tangent bundle is trivial as a smooth real vector bundle. Consider the linear map $\Omega \colon TSU_3\to \mathbb{C}^4$ given by $\Omega(v)=(\omega_{1\bar{1}}+i\omega_{2\bar{2}},\omega_{1\bar{2}},\omega_{1\bar{3}},\omega_{2\bar{3}})$ and $\Omega_{\varepsilon} \colon TSU_3\to \mathbb{C}^4$ given by $\Omega(v)=(\omega_{1\bar{1}}+if\omega_{2\bar{2}},\omega_{1\bar{2}},\omega_{1\bar{3}},\omega_{2\bar{3}})$. Let $J_0$ be the standard almost complex structure on $\mathbb{C}^4$. Let $J(v)=\Omega^{-1}(J_0\Omega(v))$. Let $J_{\varepsilon}(v)=\Omega_{\varepsilon}^{-1}(J_0\Omega_{\varepsilon}(v))$. These are my almost complex structures: elements of $\operatorname{End}T_{SU_3}$. You can compute from the Maurer-Cartan equations that $J_0$ is a complex structure, and that $J_{\varepsilon}$ has zero Nijenhuis tensor (so arises from a complex structure) on any open set where $f=0$, but not near any point where $df\ne 0$.

The complex structure on $SU_3$ is homogeneous under left action of $SU_3$. It is not Kähler, as $b_2(SU_3)=0$, so all closed 2-forms are exact, and hence no symplectic form on $SU_3$. In particular, $SU_3$ does not have a complex structure under which it could become a Kähler manifold, and a fortiori is not a complex algebraic variety. There are infinitely many nonbiholomorphic complex structures on $SU_3$ invariant under left $SU_3$ action.

Note that $SU_3$, in any complex structure, is not a complex Lie group, i.e. its multiplication is not holomorphic. Proof: Indeed $SU_3$ is compact. Take a compact complex Lie group $G$. Take a complex linear finite dimensional representation of $G$. The representation is a map to complex matrices, which form an affine space. Complex affine space has no compact complex subvarieties except points. So $G$ has discrete image. So $G$ has discrete adjoint representation. Hence the identity component of $G$ is an abelian group, compact, so a complex torus. So compact complex Lie groups are precisely finite group extensions of complex tori. In particular, since $SU_3$ is connected and nonabelian and compact, it is not a complex Lie group for any complex structure. If its multiplication were holomorphic for some complex structure, then the holomorphic implicit function theorem would ensure that its inverse operation was also holomorphic, so it would be a complex Lie group.

For more information about this and other $SU_3$-invariant complex structures on $SU_3$ (and all other compact and simply connected homogeneous complex manifolds) see

Hsien-Chun Wang, Closed manifolds with homogeneous complex structure, American Journal of Mathematics, vol. 76, no. 1 (January 1954), pp. 1-32.

Phillip Griffiths, On certain homogeneous complex manifolds, Proceedings of the National Academy of Sciences of the United States of America, vol. 48, no. 5 (May 15, 1962) pp. 780-783.

  • $\begingroup$ I ask you to write this answer in terms of the tangent bundle $T_M$, its vector bundle of endomorphisms $End(T_M)$, possibly using the Serre-Swan theorem and the module of smooth sections of $T_M$. $\endgroup$
    – user122276
    Apr 15, 2021 at 13:10
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    $\begingroup$ From your comments, my impression is that my description of an example is not to your taste. I have never seen an almost complex structure described using the Serre-Swan theorem or in terms of vectors fields as forming a module, so I will need some idea of how you prefer to write almost complex structures if I am to provide a clearer description. Perhaps you can give an example of almost complex structures from your preferred perspective. $\endgroup$
    – Ben McKay
    Apr 15, 2021 at 15:40
  • $\begingroup$ as a differential geometer you are used to using local coordinates, with background in algebra/geometry I use sheaves of functions and sheaves of sections of vector bundles. Hence an almost complex structure in my language is an endomorphism $J$ of the tangent bundle (which is a projective $R$-module of rank $2$) with square $=-Id$. If in a local trivialization $J$ is defined using "rational functions", $J$ is algebraic. If $J$ is locally defined using smooth functions it is smooth. $\endgroup$
    – user122276
    Apr 15, 2021 at 17:27
  • $\begingroup$ The real Lie group $SU(3)$ is 8-dimensional and its real tangent bundle is trivial of rank 8: $T_{SU(3)}\cong SU(3) \times \mathbb{R}^8$. It seems to me that $\Omega_{\epsilon}$ can not be inverted as you claim - it is a map from a real Lie group to a 6-dimensional real vector space. Is there a mistake? $\endgroup$
    – user122276
    Apr 16, 2021 at 9:47
  • $\begingroup$ Right: it is 8-dimensional. Fixed $\endgroup$
    – Ben McKay
    Apr 16, 2021 at 9:50

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