# Motivation for product structure

I guess I understand some reasons why we should care for complex structure on manifolds, but what is the reason why product structure is studied? Does it arise naturally somewhere?

The product structure on a smooth manifold $M$ is given by a (1,1)-tensor $E$ such that $E \neq \pm1$ and $E^2 = 1$ and the following integrability condition holds: $$E[X,Y] = [EX,Y] + [X,EY] - E[EX,EY]$$ Compare it to a complex structure $J$ such that $J^2 = -1$ and $$J[X,Y] = [JX,Y] + [X,JY] + J[JX,JY]$$

I'm still studying basic literature, one thing I do know is that on a Lie group it $E$ induces a double algebra structure on a correspoinding Lie algebra, and thus two foliations on the group, if I remember correctly; I guess it is generalized for an arbitrary manifold, so this 'splitting' of the tangent bundle and how it's compatible with other structures (complex, Hermitian, Kähler etc., e.g. a complex product structure is defined by the two respective structures such that $JE = -EJ$) is probably one of the reasons to study it, but I'm not sure this reason is good enough on its own.

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What's "product structure"? Do you mean a Poisson manifold? – Alon Amit Mar 13 '11 at 7:00
Or do you mean Lie group structure? – Eric O. Korman Mar 13 '11 at 7:07
@Alon Amit, @Eric: please see the update. – Alexei Averchenko Mar 13 '11 at 7:34

I don't know a lot about this but in the context of a complex manifold, $E$ is called a real structure. Searching the literature for that term should turn up a lot of material.
The conditions $JE = -EJ$ and $E^2 = 1$ make $E$ analogous to a conjugation operation. This allows you to split up your tangent space into real and complex parts corresponding to the +1 and -1 eigenvalues of $E$, respectively (note that $JE = -EJ$ means that $J$ is an isomorphism between the +1 and -1 eigenspaces).
Is the integrability necessary for $E$ to be a conjugation? Does it induce a $G$-structure? – Alexei Averchenko Mar 13 '11 at 8:11
I don't see why integrability should be necessary. I would guess that $E$ is integrable if (only if?) $J$ is since if $J$ is integrable we get our $z_i$ and $\bar z_i$ coordinates from which we can define $E$. I don't know why it would induce a $G$-structure. – Eric O. Korman Mar 13 '11 at 8:22
I know nothing about this, but just from reading this question I'm not even convinced that every $E$ comes from a $J$. It seems like there could be easy topological obstructions (probably $\pi_1$?) to taking a global square-root of this endomorphism? – Aaron Mazel-Gee Mar 13 '11 at 14:45