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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.

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You've changed notation there a bit. Is $G=X$? And I'm not sure all your pullbacks end up over the same space - $m_i$ and $m_{ij}$ $i,j=1,2,3$ don't have the same codomain. –  David Roberts Dec 6 '10 at 0:04
    
The $m_\underline{i}$ really do have the same codomain. For example, in coordinates, $m_3(x,y,z)=z=p_3(x,y,z)$ and $m_{2,3}(x,y,z)=y \cdot z$, just as described by 'summing the coordinates whose indices are in $\underline{i}$'. –  Eric Peterson Dec 6 '10 at 0:51
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Dear jlk: It is more natural to tack on an additional tensor factor of $0^{\ast}(L)^{-1}$ where $0:G^3 \rightarrow G$ is the constant map to the origin. When working over a field this has no serious effect, but in the relative setup (which makes various Yoneda-type arguments more effective) it is clearly necessary. –  BCnrd Dec 6 '10 at 1:14
    
@BCnrd: Thank you for the comment. @David Roberts: Thanks. Fixed. –  jlk Dec 6 '10 at 1:18
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Not every component of $X$ necessarily has a $k$-rational point (the component group could be non-constant). Can you work out something reasonable when $X$ is a finite etale commutative group? If not, there seems little point in considering the more general case. –  BCnrd Dec 6 '10 at 3:41

2 Answers 2

up vote 4 down vote accepted

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.

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@t3suji: thanks! –  jlk Dec 6 '10 at 21:12

Uniqueness may fail if $X$ is a finite group: any two cube structures "differ" by a "quadratic" map $X\to \mathbb{G}_m$. If, say, $X=\mathbb{Z}/n\mathbb{Z}$ and $\zeta$ is an $n$-th root of unity, then $m\mapsto \zeta^{m^2}$ is such a map.

This of course implies non-uniqueness whenever the component group of $X$ admits nontrivial quadratic maps to $\mathbb{G}_m$.

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@Laurent Moret-Bailly: Thanks! –  jlk Dec 6 '10 at 21:15

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