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Let $M$ be an even-dimensional Lie group. Is it true that $M$ admits a complex structure?

I can see that there exists an almost complex structure, since the tangent bundle is trivial. But this is not sufficient Trivality of tangent bundle and complex structure.

So I'm wondering if adding the Lie group structure is enough to guarantee integrable. Maybe the Nijenhuis tensor simplifies in some way?

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  • $\begingroup$ Well, I can see that it admits an almost complex structure since the tangent bundle is trivial. But I don't see how to compute the Nijenhuis tensor. $\endgroup$ – user308973 Sep 26 '16 at 22:27
  • $\begingroup$ I see that your previous questions are related to this. It may be better to explain the motivation for this question as it is currently on the way to being closed. I admit the problem is harder than I initially imagined (I was thinking almost complex, not complex). $\endgroup$ – Michael Albanese Sep 26 '16 at 22:29
  • $\begingroup$ Why on way to being closed? Feel like this is a good question. Specific and interesting. Is the answer obvious? $\endgroup$ – user308973 Sep 26 '16 at 22:33
  • $\begingroup$ A complex structure $J$ on the real lie algebra $\mathfrak{g}$ extends to a (so-called left-invariant) complex structure of the Lie group $G$ if and only if $[JX,JY] = [X,Y]+J[JX,Y]+J[X,JY]$ for all $X,Y\in\mathfrak{g}$. Is the question whether every Lie Group has such a left-invariant complex structure, or any? I think the latter is incredibly hard... $\endgroup$ – Ben Sep 26 '16 at 22:34
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    $\begingroup$ Aha, that's a different question. You should add context... Hint: It's possible (and not too hard) to show that $G$ has a left-invariant complex structure if and only if the complexified Lie-algebra $\mathfrak{g}_{\mathbb{C}}$ has a complex sub-algebra $\mathfrak{g}_0$ such that $\mathfrak{g}_{\mathbb{C}} = \mathfrak{g}_0\oplus \overline{\mathfrak{g}_0}$. $\endgroup$ – Ben Sep 26 '16 at 22:41
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This answers the question only tangentially; namely, I'll address whether every Lie group has an invariant complex structure. The answer to this question is no. Let's recall the standard construction. Say we have a real Lie group $G$ of dimension $2n$ and a complex structure on its Lie algebra $\mathfrak{g} = \mathrm{Lie}(G) \cong T_1G$, that is, an $\mathbb{R}$-linear endomorphism $J_1\colon T_1G\to T_1G$ such that $J_1^2 = -\mathrm{id}_{T_1G}$. We want to give $G$ the structure of an almost complex manifold that restricts to $J_1$ on $T_1G$ and such that the left-multiplication, $\ell_g\colon G\to G$, $h\mapsto gh$, has $\mathbb{C}$-linear differential (is almost-holomorphic?) for each $g\in G$. It's easy to see that there is only one such almost complex structure $J\colon TG\to TG$; more concretely, over $g\in G$ it is given by the map $J_g := (\ell_{g})_*\circ J_1\circ (\ell_g^{-1})_*\colon T_gG\to T_gG$. In particular, if $X\colon G\to TG$ is the left-invariant vector field associated with $v\in T_1G$, i.e., $X_g = (\ell_g)_*(v)$, then $(JX)_g = (l_g)_*\circ J_1(v)$ and so $JX$ is just the left-invariant vector field associated with $J_1(v)$.

The question now is: when is this almost complex structure integrable?

The Newlander–Nirenberg theorem states that an almost complex structure $I\colon TG\to TG$ is integrable if and only if its Nijenhuis tensor $$[X,Y]+I[IX,Y]+I[X,IY]-[IX,IY]$$ vanishes for all vector fields $X,Y$. Since the Nijenhuis tensor only depends on the point-wise values of the vector fields, it is seen to be trivial as soon as it vanishes on all left-invariant vector fields. Thus, by construction:

Proposition— Let $J$ be a complex structure on $\mathfrak{g}$. Then the left-invariant almost complex structure on $G$ induced from $J$ is integrable if and only if $[JX,JY] = [X,Y]+J[JX,Y]+J[X,JY]$ for all $X,Y\in\mathfrak{g}$.

We see that there is an obstruction to the integrability of an invariant almost complex structure, but this obstruction is seen entirely on the level of Lie algebras. Since every finite-dimensional real Lie algebra is the Lie algebra of some simply connected Lie group, it suffices to construct a Lie algebra which fails to admit a $J$ such that $J^2 = -\mathrm{id}_{\mathfrak{g}}$ and $[JX,JY] = [X,Y]+J[JX,Y]+J[X,JY]$ for all $X,Y\in\mathfrak{g}$.

Examples can be found, e.g., in work of Goze and Remm, Non existence of complex structures on filiform Lie algebras, arXiv:math/0103035v2. In fact, by Proposition 4, the following Lie algebras provide examples:

Given $n\geq 2$ let $\mathfrak{l}_{2n}$ be the $2n$-dimensional Lie algebra generated by $x_1,x_2,\dots ,x_{2n}$ and Lie bracket such that $[x_1,x_i] = x_{i+1}$ for all $i=2,3,\dots,2n-1$ and $[x_i,x_j] = 0$ for all other $i,j$. The strategy here is to show that if there were such an $J$ for $\mathfrak{l}_{2n}$, then $J(x_{2n})$ had to be central; but the center is generated by $x_{2n}$, hence $J(x_{2n}) = ax_{2n}$. But then $a^2 = -1$, a contradiction.

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In the case of compact Lie groups, it is true that an even-dimensional Lie group always admits a complex structure. See Example 4.32 in Félix, Oprea, Tanré, Algebraic Models in Geometry (available here https://www.maths.ed.ac.uk/~v1ranick/papers/tanre.pdf).

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