User justin shih - MathOverflow

Question about specifying complex 1-motives

2012-02-01T01:02:21Z

A 1-motive over a field $k$ is an algebraic torus $T$, an abelian variety $A$, a group scheme $G$ that's an extension of $A$ by $T$, a finitely generated free abelian group $L$, and a group homomorphism $L \longrightarrow G(k)$.

I'm currently reading a paper of Carlson, and I want to use his construction to identify something that came up in a problem that I'm working on. However, on the first page of that paper he defines a complex group scheme, but appears to leave out the requirement that $T$ and $G$ be group schemes. Later on (in section 4), he constructs the trace motive, and consistent with his definition, appears to only define the $\mathbb{C}$-points. I'm missing something -- but I don't really know what.

Does the fact that the groups in the trace motive come from group schemes somehow follow from some general nonsense about $\mathbb{C}$? Is it long and unenlightening to write down? Or am I just completely misunderstanding the paper?

Answer by Justin Shih for If $k[S]$ is noetherian, is S finitely generated?

2010-11-04T19:14:38Z

No, at least in the case that $k$ is infinite, and not prime - for an indeterminate $t$ take $S$ generated by ${t, t/a, t/a^2, ...}$ with $0 \ne a \in k$ of infinite (multiplicative) order not in the prime field. Then $k[S] = k[t]$ but $S$ is not finitely generated.

Explicit examples of resolution of (projective) 3-folds over k?

2010-10-19T18:21:43Z

I'm looking for examples of explicit resolutions of (projective) 3-folds over a field k (char 0), with isolated singularities, or at least with smooth singular locus. I've looked in various books and online, but the examples they present have only been for curves and surfaces, for which resolution of singularities is much less complicated. It would also be nice if the exceptional divisor were sufficiently nice, say, with smooth components that intersect transversally (or if the exceptional divisor were itself smooth).

Are there any well-known/easy examples in dimension 3? Because resolution quickly becomes complicated as dimension increases, I'd imagine that examples become harder to come by, although I'm sure many exist!

In general, I seem to not be very good at finding references/papers relevant to specific things I'm looking for. Search engines typically turn up a lot of irrelevant papers, and thumbing through a bunch of books seems time-inefficient. Search engines have turned up a few papers, but I suspect that I can do better than what I've found thus far.

I realize that adding this second part about looking for references may detract from the specific issue I have right now, but I'd rather get better at finding things than ask here when I can't find something.

Calculations of Pic^0, Pic, NS of surfaces

2010-09-29T23:37:46Z

I'm looking for an example in the literature where $\mbox{Pic}^0(X)$, $\mbox{Pic}(X)$, and $NS(X)$ of a projective surface $X$ over a field are calculated. I want them for an example I'm trying to work out, so ideally $X$ would be relatively simple, perhaps a cubic hypersurface in $\mathbb{P}^3$, or something along those lines. I know it's out there, but googling and browsing arXiv and MathSciNet haven't quite panned out.

Answer by Justin Shih for Finding the codomain of a monoid homomorphism

2010-01-15T10:58:25Z

As has been stated, $M \rightarrow G$ factors via the group completion $\hat{M}$. Furthermore, since $G$ is abelian, the map $\hat{M} \rightarrow G$ factors via the abelianization $\hat{M}/[\hat{M},\hat{M}]$ of $\hat{M}$. Since the composite of the maps $M \rightarrow \hat{M} \rightarrow \hat{M}/[\hat{M},\hat{M}]$ is already known, $f$ is uniquely determined by the map $\hat{M}/[\hat{M},\hat{M}] \rightarrow G$, so we may assume $M$ to be an abelian group.

Since $f = 0$ gives us no information, we may assume that $f$ is onto. Suppose we have a $G$ so that $f$ factors via it, and let us write $f = pg$, where $g \colon M \rightarrow G$ and $p \colon G \rightarrow \{0,1\}$.

By $f = pg$ we must necessarily have $\ker(g) \subset \ker(f)$, so that $g$ is isomorphic to the canonical projection $M \rightarrow M/N$, where $N$ is a subgroup of $\ker(f)$. In addition, $|G|$ must be even. (If $f$ is onto, $p$ must take on the values $0$ and $1$ equally, since $|G| = 2|ker(p)|$.)

I claim this is all $f$ tells you. Given $f$, we've just shown that we must have that $g$ is isomorphic to modding out by a subgroup of $\ker(f)$, and that $|G|$ is even.

Conversely, given any map $g \colon M \rightarrow G$, whose kernel is a subgroup of $\ker(f)$, and such that $|G|$ is even, then the image of $\ker(f)$ will be of index 2 in $G$, so we can just compose this map with the map that mods out the image of $\ker(f)$ in $G$.

Edit: Oops, $G \rightarrow \{0,1\}$ is only a set map! Well then, let me at least try to contribute to the discussion! In this case, we can still assume $M$ to be an abelian group. If you can recover $G$ (assuming $G = M$ is not allowed), then certainly the kernel of $M \rightarrow G$ cannot have any subgroups, hence must be a cyclic group of order $p$. In the case where $M = \mathbb{Z}_+$ , then $\hat{M} = \mathbb{Q}_+$ which is torsion free, so no maps $f$ exist which allow you to recover $G$ entirely, unless we allow $G = \hat{M}$.

Comment by Justin Shih

2012-02-04T02:56:10Z

Hmm... I'm an algebraic geometer, so I don't know that much complex geometry. Why is that extension analytic again? A lot of the 1-motive he defines directly from the divisor groups.

Comment by Justin Shih

2010-11-04T19:32:36Z

Right. So then can we make an increasing chain of ideals by picking $s_1 \in S$, setting $I_1 = (s_1)$, and then by induction picking $s_{k+1} \in $S\backslash\I_k$ and setting $I_{k+1} = I_k + (s_{k+1})$? This chain eventually terminates, which I think means that $S$ is finitely generated?

Comment by Justin Shih

2010-11-04T19:27:09Z

Oh, I see now. I guess I'm taking $k[t]$ and then finding a sub-semigroup inside there, instead of taking a semigroup $S$ and then making $k[S]$.

Comment by Justin Shih

2010-11-04T19:25:14Z

Sorry, I guess I mean the semigroup generated by the $t/a^n$. I've edited my post to reflect this change. And I'm missing something, but I don't quite understand what you mean by the elements of $S$ are by definition linearly independent?

Comment by Justin Shih

2010-10-19T19:27:29Z

If $I$ is the value of the integral, why can't you just take $s(x) = I^2/(Au(x))$? This is probably not the answer you were looking for, so can you be more specific?

Comment by Justin Shih

2010-01-15T11:50:23Z

Whoops! I thought that "monoid homomorphism" in the title referred to the function $f$ instead of the map $M \rightarrow G$.