I am trying to understand the theory of cubical structures and am interested in knowing if a *disconnected* commutative group variety whose identity component is a semi-abelian variety satisfies the theorem of the cube.


Recall that if $X$ is an abelian variety and $L$ is a line bundle on $X$ that is rigidified along the identity sectionn, then the 
[Theorem of the Cube](http://en.wikipedia.org/wiki/Theorem_of_the_cube) implies that the line bundle


$$
\Theta(L) := m\_{123}^{\*}(L) \otimes m\_{12}^{\*}(L^{-1}) \otimes m\_{13}^{\*}(L^{-1}) m\_{23}^{\*}(L^{-1}) \otimes m\_{1}^{\*}(L) \otimes m\_{2}^{\*}(L) \otimes m\_{3}^{\*}(L)
$$


admits a unique trivialization that makes


$$
\Lambda(L) := m^{\*}(L) \otimes p\_1^{\*}(L^{-1}) \otimes p\_{2}^{\*}(L^{-1})
$$


into a symmetric biextension of $X \times X$ by $\mathbb{G}\_{m}$.


Here $m$ denotes the multiplication map, $p\_i$ the projection maps, and 
$m\_{\underline{i}}$ the morphism $X \times X \times X \to X$ given by summing the
coordinates whose indices are in $\underline{i}$.


More generally, Breen proved this fact remains true when $X$ is a semi-abelian variety.


Does this statement remain true if we allow $X$ to have non-trivial component group?


If not, what is a example of a rigidified line bundle that does not have canonical cubical structure.


Does the theorem remain valid you rigidify along $1$ fixed point on every component (rather along the identity element)?


**Added** As BCnrd notes, over a more general base the formulas should be modified by adding the term $0^{\*}(L^{\otimes \pm 1})$, which should be thought of as $m\_{\emptyset}^{\*}(L^{\otimes \pm 1})$. This (rigidified) line bundle is trivial when the base if a field, but not in general.



Failure of Theorem of the Cube?
Here is an explanation why connectedness is important. Let's work over ${\mathbb C}$. The Theorem of the Cube can be stated as follows: If $s:X\to X$ is a shift by a fixed element $g\in X$, then $s^\*L\otimes L^{-1}$ satisfies the Theorem of the Square. The reason it holds is because $s^\*L\otimes L^{-1}$ is topologically trivial, which in turn is true because $g$ can be continuously deformed into $0$. The last step depends on $X$ being connected. 


For an explicit counterexample, take $X=E\times ({\mathbb Z}/{2\mathbb Z})$ where $E$ is an elliptic curve, and let $L$ be a line bundle whose degrees on the two components differ. I am not sure how rigidifying at any number of points is going to change this.


