# How should I think about 1-motives?

By definition, a 1-motive over an algebraically closed field $$k$$ is the data

$$M = [X\stackrel{u}{\to}G]$$

where $$X$$ is a free abelian group of finite type, $$G$$ is a semi-abelian variety over $$k$$, and $$u:X \to G(k)$$ is a group morphism.

As far as I understand, this was one of the earlier attempts of coming up with a workable notion of a mixed motive. As evidence, Deligne shows that 1-motives over $$\mathbb{C}$$ are equivalent to mixed Hodge structures (without torsion) of type $$\{(0,0),(0,-1),(-1,0),(-1,-1)\}$$.

The problem is that I don't really understand the intuition behind this definition. I assume that Deligne had concrete examples in mind when he came up with this formulation. Pehaps some extensions of $$H^1(A)$$, where $$A$$ is an abelian variety?

How should I interpret the data in the definition of a 1-motive? My naïve view on motives is that they should be compatible systems of realizations with "geometric origin", and extensions thereof. How to reconcile these two points of view? Perhaps what I lack is simply a bag of toy examples to help me think about it.

• It might help to think about cohomology and the Picard group of a singular open curve over $k$. For example, if $X$ is the curve obtained by glueing two points of $\mathbb{G}_m$, then $\text{Pic}^0X$ can be written as the cokernel of some morphism $\mathbb{Z} \rightarrow \mathbb{G}_m$. More generally, one can associate to any open semi stable curve a $1$-motive.
– Jef
Jan 12 at 14:21
• If you look at section 10 of Deligne's Théorie de Hodge III, he discusses this in some detail. He includes the example that Jef mentioned, and also discusses the $\ell$-adic realization. Jan 12 at 16:32
• "compatible systems of realizations with "geometric origin"" is not a very good idea, actually.:) It just gives one more realization. Jan 12 at 19:51
• As Mikhail suggested, thinking of motives as "compatible systems of realizations with "geometric origin"" is not a very good idea. Better to think of them as objects of some mysterious $\mathbb{Z}$ (or $\mathbb{Q}$)-linear category that admit lots of (concrete) realizations. Now, that's exactly what 1-motives are. OK, since I don't wish to play the points game, I'll refrain from commenting, since it seems to offend everyone.
– user475150
Jan 14 at 19:24

I don't fully understand the hostility to thinking of motives in terms of compatible systems. It's true that a motive is not just a compatible system of realizations (or else we would call it a "compatible system" and not a motive) but it's also true that studying compatible systems has always been a good way to get intuition for many aspects of the theory of motives. Maybe a better thing to say is that a motive should be a geometric recipe which produces a realization in any existing or future cohomology theory.

You asked how to reconcile this perspective on 1-motives with the definition of 1-motives. I guess this means how to derive a compatible system of realizations from a map $$X \to G$$.

The realization of the 1-motive $$X \to G$$ in any cohomology theory should be the extension of $$H^1(G)$$ by $$H^1(X^\vee)$$. (More properly I guess this is the dual of the realization).

You ask if this is an extension of $$H^1(A)$$. The group $$G$$ is always an extension of an abelian variety $$A$$ by a torus $$T$$, and $$H^1(G)$$ is an extension of $$H^1(T)$$ by $$H^1(A)$$. So we see that we are potentially extending $$H^1(A)$$ in both directions.

Which extension? The idea is we should morally get the $$H^1$$ of the quotient space $$G/X$$. Since this space is not very well-behaved, we can simulate this, in an arbitrary cohomology theory, as follows:

We observe that any class in $$H^1(G)$$ (ideally, completely represented by a torsor) is multiplicative in the sense that its pullback to $$G \times G$$ under the multiplication map is equal to the sum of its pullbacks under the two projections. For torsors, we can fix an isomorphism of torsors realizing this.

An element of $$H^1( X \to G)$$ is a class in $$H^1(G)$$ together with a trivialization of its pullback to $$H^1(X)$$ which is multiplicative in the sense that the pullback of this trivialization to $$X \times X$$ along the multiplication is equal to the sum of its pullbacks under the two projections (using the above isomorphism to define this sum).

You can follow this recipe in any theory and get a concrete description of the extension.

Furthermore, in every reasonable cohomology theory you will get a clue to the idea that this construction produces all the extensions with "geometric origin". For example, in $$\ell$$-adic cohomology theories, you will see that extensions of $$H^1(G)$$ by $$X^\vee$$ correspond to a Selmer group, and extensions arising from this construction are the image of the map from the group of rational points to the Selmer group. So as long as the Tate-Shafarevich group is finite, there will not be room to fit any more extensions.