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 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.