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Originally posted on Maths Stack Exchange.


Let $V$ be a real vector space. An almost complex structure on $V$ is a map $J : V \to V$ such that $J^2 = -\mathrm{id}_V$. An almost complex structure gives $V$ the structure of a complex vector space by defining $(a+bi)v = av + bJ(v)$. The idea behind this definition is that $J$ represents multiplication by $i$. What if you instead consider multiplication by other complex numbers?

Let $n > 2$ be a fixed positive integer and let $J : V \to V$ be such that $J^n = -\mathrm{id}_V$ which I will refer to as an alternative almost complex structure. One can regard $J$ as trying to capture multiplication by $\zeta = \exp\left(\frac{\pi i}{n}\right)$. In particular, as $\{1, \zeta\}$ is linearly independent over $\mathbb{R}$, it is a basis for $\mathbb{C}$ as a real vector space. One can then endow $V$ with a complex structure by defining $(a + b\zeta)v = av + bJ(v)$.

Do alternative almost complex structures give rise to the same results as the standard almost complex structures? In particular, in the case where we extend them to bundle endomorphisms of the tangent bundle of a manifold.

If not, what fails? If so, is dealing with almost complex structures easier than dealing with alternative almost complex structures?


As Henry T. Horton asks about in a comment to the original post, I am interested in the integrability of alternative almost complex structures but also whether there are any difficulties when combining with other structures such as a symplectic form or a Riemannian metric.

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    $\begingroup$ For me (and everybody around me) a linear map $J\colon V \to V$ would be called a complex structure. Even if you have a vector bundle $E$, people call an endomorphism $J\in \Gamma(\mathrm{End}(E))$ with $J^2 = -\mathrm{id}_E$ a complex structure on $E$. The "almost" in "almost complex structure" is a complex structure on the tangent bundle that is not integrable. But it makes little sense to ask if a linear map between vector spaces or arbitrary vector bundles is integrable (or you must have a different definition of integrable than the one I know). $\endgroup$ – Klaus Niederkrüger Aug 12 '19 at 9:00
  • $\begingroup$ I have the impression that there is a bit of confusion here. If you have a complex structure $J$ on the tangent bundle $TM$ you can always find bundle charts for $TM$ such that $J$ looks like the standard complex multiplication in the chosen local bundle trivialization, but the difference between $TM$ and an arbitrary vector bundle is that there are natural (!) trivializations of $TM$ coming from the coordinate charts of $M$. Being integrable means that you can find a coordinate chart of $M$ such that the induced trivialization of $TM$ identifies the fibers with $\mathbb{C}^r$. $\endgroup$ – Klaus Niederkrüger Aug 12 '19 at 9:12
  • $\begingroup$ @KlausNiederkrüger: I was hesitant to use the phrase 'complex structure' because $V$ is also a manifold, and I didn't want to allude to a complex structure on that manifold (i.e. a collection of charts with biholomorphic transition maps). I have since come across the phrase 'linear complex structure' to describe $J : V \to V$ with $J^2 = -\operatorname{id}_V$ which I think is more appropriate than 'almost complex structure' (again, this could be confused with an almost complex structure on the manifold $V$). $\endgroup$ – Michael Albanese Sep 19 '19 at 23:43
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Let us first deal with linear algebra. Assume a matrix $J$ satisfies $J^k= -Id$. Then, there exists a poylnomial $P$ whose coefficients depend on the eigenvalues of your $J$ such that $P(J)$ is a complex structure. Moreover, if your matrix $J$ is a smooth (1,1)-tensor on a manifold then the polynomial is the same at all points of the manifold.

Now, concerning differential geometry, your alternative complex structure $J$ have the same eigenvalues (with multiplicities) at all points. Then, the polynomial $P$ is the same at all points. By [Lemma 6, arXiv:0904.0535], the complex structure $P(J)$ is integrable.

Moreover, if
the Nijenhujs tensor of $J$ vanishes,there exists a coordinate system such that the entries of $J$ are constants by Thompson, G.: The integrability of a field of endomorphisms. Math. Bohem. 127 (2002), no. 4, 605–611.

If the coefficients of your tensor are constant in a certain coordinate system, the Nijenhujs tensor of it vanishes.

Thus, as in the case or ordinary almost complex structure, the vanishing of the Nijenhujs tensor decides about integrability.

Now, about combining your alternative complex structure with another structures such as metric or symplectic structure, the only reasonable formula I can imaging is in [Section 5, arXiv:1103.3877] and it does not give anything very interesting. If you have other ways of combining, please tell us.

Sorry for bad news.

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  • $\begingroup$ I don't follow the part of your argument where you apply Lemma 6 of arXiv:0904.0535. What if $k=2$? Then the polynomial $P$ is the identity, and the eigenvalues of $J$ are the same at all points, but it certainly does not follow that $J$ is integrable. $\endgroup$ – YangMills May 25 '16 at 18:02
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Wait: if we ask for $J^3=−I$, then $J=−I$ is one solution, and that doesn't determine an almost complex structure, so Vladimir's answer doesn't quite cover all possible cases. If $J^k=−I$ and $k$ is even, then clearly Vladimir's answer is ok: $J$ commutes with a unique almost complex structure, and the whole story reduces to the story of that almost complex structure.

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    $\begingroup$ More generally, it is easy to prove that any real linear map $T$ with no real eigenvalues on a finite dimensional real vector space $V$ commutes with a unique almost complex structure $J$ so that all eigenvalues $\lambda$ of $T$ on the $J=\sqrt{-1}$-eigenspace of $J$ inside $V \otimes \mathbb{C}$ have positive imaginary part. $\endgroup$ – Ben McKay May 25 '16 at 15:50

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