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Let $G$ be a commutative connected algebraic group over $\mathbb{C}$. A theorem of Serre says that there exists an exact sequence $$1\to \mathbb{G}_a^n\times \mathbb{G}_m^m\to G\to A\to 1,$$ where $A$ is an abelian variety. (See here, for example.)

I wonder what are some examples of such groups $G$ where this extension is non-trivial. That is, such that $G\ncong \mathbb{G}_a^n\times \mathbb{G}_m^m\times A$.

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    $\begingroup$ If you allow this to be split after passing to a finite covering, you can take $G_m$ times an elliptic curve and mod out by a nontrivial "diagonal" finite cyclic subgroup. Then it's not split (although a finite cover splits). This little game can't be performed with $G_a$. $\endgroup$
    – YCor
    Commented Nov 9, 2021 at 14:59
  • $\begingroup$ The total space of the universal torsor over the relative $\text{Pic}^0$ of any smooth projective variety with nontrivial Picard gives an example. $\endgroup$
    – Jason Starr
    Commented Nov 10, 2021 at 11:06
  • $\begingroup$ Dear @JasonStarr would you mind explaining more? $\endgroup$
    – Gabriel
    Commented Nov 10, 2021 at 13:07
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    $\begingroup$ Doesn't any line bundle on an abelian variety, which is algebraically equivalent to $0$ but not trivial give an example? (by taking the associated $\theta$ group). $\endgroup$
    – Damian Rössler
    Commented Nov 10, 2021 at 15:01
  • $\begingroup$ Yes, I was trying to say what @DamianRoessler said. There are nontrivial invertible sheaves over Abelian varieties, and these give examples. There is even a “universal torsor” defined over the product of the Abelian variety with its relative $\text{Pic}^0$. $\endgroup$
    – Jason Starr
    Commented Nov 10, 2021 at 15:28

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One example is provided by the generalized Jacobian. For a smooth projective curve $C$ and a divisor $D$ on $C$, the generalized Jacobian is defined to be the moduli space parameterizing pairs consisting of a line bundle of degree $0$ on $C$ together with a trivialization of that line bundle over $D$.

This admits a map to the usual Jacobian, whose kernel is a product of $\mathbb G_m$s and $\mathbb G_a$s depending on the multiplicity of points of $D$.

If this were trivial, than we could define a section of this map, which would give for an arbitrary line bundle a canonical trivialization at each point of $D$. That would mean that for each point $x\in D$, the line bundle on the usual Jacobian whose fiber at a point $L \in J$ is the fiber of $L$ at $x$ would admit a section and thus be trivial. But from the duality theory of abelian varieties, these line bundles are different for distinct points $x\in C$, so they cannot all be trivial as long as $D$ contains two or more points.

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  • $\begingroup$ Dear @WillSawin, that's incredible. Since I never studied generalized Jacobians, do you have a simple example of it in mind? $\endgroup$
    – Gabriel
    Commented Nov 9, 2021 at 14:38
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    $\begingroup$ @Gabriel I guess the simplest example would be the generalized Jacobian associated to two points on a curve of genus $1$, which is an extension of that elliptic curve by $\mathbb G_m$. As a scheme, it is some nontrivial degree-0 line bundle on the elliptic curve with the zero-section removed, but it carries a group structure. $\endgroup$
    – Will Sawin
    Commented Nov 9, 2021 at 14:53
  • $\begingroup$ Can you get both non-split extensions with kernel $G_a$ and $G_m$? $\endgroup$
    – YCor
    Commented Nov 9, 2021 at 14:55
  • $\begingroup$ @Ycor The argument I gave shows that extension by $\mathbb G_m$ produced by generalized Jacobians when the divisor $D$ contains two different points are nontrivial, but I believe taking the derivative of this argument shows that extensions by $\mathbb G_a$ produced by generalized Jacobians when $D$ contains a point of multiplicity $\geq 2$ are nontrivial. $\endgroup$
    – Will Sawin
    Commented Nov 9, 2021 at 14:59
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    $\begingroup$ @YCor If the multiplicities are one then difference between any two points of $D$ Is torsion in the Jacobian then this torsor will split after passing to a finite cover. If this is not true, it won't split. A non-split extension by $\mathbb G_a$ never becomes split after passing to a finite cover, I don't think. $\endgroup$
    – Will Sawin
    Commented Nov 9, 2021 at 15:05

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