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Stream: theory: mathematics

Topic: Galois representations

John Baez (Apr 12 2024 at 20:03):

Calegari has a nice introductory paper... or at least it starts out pitched at my level:

• Motives and L-functions.

In Section 1.5 he starts by reviewing the Weil conjectures, and in Lemma 1.5.1 he points out an interesting relation between Galois representations and zeta functions. But before we even get to that, I'm confused about something.

John Baez (Apr 12 2024 at 20:06):

He starts with a smooth projective variety $X$ over $\mathbb{Q}$ and then later starts talking about $X(\mathbb{F}_q)$ where $q = \ell^m$ and $\ell$ is prime.

John Baez (Apr 12 2024 at 20:12):

This makes it sound like you can unambiguously talk about the points of $X$ over a finite field when $X$ starts out defined over $\mathbb{Q}$. I don't see how. I can do it if $X$ is defined over $\mathbb{Z}$ for that. But he doesn't say anything about that. He doesn't say anything about choosing a model for $X$, which I gather is a way of defining it over $\mathbb{Z}$.

John Baez (Apr 12 2024 at 20:13):

Is he just being relaxed and sloppy, or am I confused about something here?

Morgan Rogers (he/him) (Apr 13 2024 at 08:05):

He introduces $X$ in section 1.3 as an algebraic variety "cut out by polynomial equations with rational coefficients", where it makes more sense to look at what happens over different fields.

Morgan Rogers (he/him) (Apr 13 2024 at 08:11):

I imagine he is doing the analogous thing for projective varieties, not fixing the variety and finding points of it but fixing the presentation of the variety in terms of equations and allowing the field in which those equations are interpreted to vary. This will be a valid thing to do for "most" characteristics of field, although if he had chosen integer coefficients things would have been less ambiguous.
That's my reading of it, I'm pretty sure my AG is weaker than yours though, John.

John Baez (Apr 13 2024 at 09:31):

My point is that if you have a bunch of polynomial equations with rational coefficients, there's no systematic way to define their solutions in a finite field $\mathbb{F}_q$, basically because there's no homomorphism $\mathbb{Q} \to \mathbb{F}_q$. This problem applies both to affine and projective varieties.

John Baez (Apr 13 2024 at 09:33):

If you choose a way to "clear denominators" and write your equations using integer coefficients, then you can talk about their solutions in any commutative ring $R$, since there's always a unique homomorphism $\mathbb{Z} \to R$.

John Baez (Apr 13 2024 at 09:34):

But I believe different ways to clear denominators can give non-isomorphic results: choosing a way to do it is called choosing a 'model' of our original affine or projective variety over $\mathbb{Q}$.

John Baez (Apr 13 2024 at 09:38):

There is sometimes a 'best' model of a variety over $\mathbb{Q}$, called the Neron model, defined by a universal property. Elliptic curves and more generally abelian varieties have such a model. But I believe not all varieties have a Neron model. (This is just my impression; I have never really studied Neron models.)

John Baez (Apr 13 2024 at 09:41):

I have been avoiding the word "scheme" above, to avoid acting like I know more than I do, but perhaps you'd like me better if I said it more often! The link to Neron model talks in terms of schemes. A model of a scheme $X$ over $\mathbb{Q}$ is a scheme over $\mathbb{Z}$ which upon base change along $\mathbb{Z} \to \mathbb{Q}$ gives $X$. The Neron model, if it exists, is initial in a certain rather obvious sense. But the article says not every scheme has a Neron model. Since it says all abelian varieties do, in a somewhat excited tone of voice (in the quiet way that mathematicians are excited), I conclude that not all varieties do.

John Baez (Apr 13 2024 at 09:52):

All this makes it impossible for me to take Calegari seriously when he starts with a smooth projective variety $X$ over $\mathbb{Q}$ and comes up with formula for the number of its points over a finite field! He's an infinitely better algebraic geometer than me so I think maybe a gear slipped when he was trying to write things in a simple way for beginners like me.

Chris Grossack (they/them) (Apr 13 2024 at 13:42):

Does it matter that you might get different answers over $\mathbb{Z}$? We're immediately reducing mod $p$, and I think once you reduce you get the same answer unless you're going out of your way to choose a bad way to clear denominators

Chris Grossack (they/them) (Apr 13 2024 at 13:45):

Write $N$ for the least common multiple of all the junk appearing in the denominator of $f$, which is the obvious thing to try. Then for any $M$ (as long as $p$ doesn't divide $NM$) $Nf$ and $Mf$ are the same (up to units) mod $p$. So once you fix a $p$ that you're interested in, "most" ways to clear denominators give the same answer! Moreover, this answer will be the same as just clearing in the most obvious way (multiplying by $N$ the lcm), so you might as well just do that.

Chris Grossack (they/them) (Apr 13 2024 at 13:45):

Am I making some easy mistake? It's possible

John Baez (Apr 13 2024 at 15:20):

If we think of an affine algebraic variety as presented by 'generators' (variables $x_1, x_2, \dots$) and 'relations' (polynomial equations $P_j(x_1, \dots, x_n) = 0$), what worries me is that an affine algebraic variety doesn't come with a god-given presentation.

Once we choose a presentation, you're right that there's a clear strategy to get a presentation using only polynomials with integer coefficients: take each polynomial $P_j$ and multiply it by the least common multiple of all the divisors of all the coefficients. But suppose we have two quite different-looking presentations....

John Baez (Apr 13 2024 at 15:22):

I should cook up a concrete example, and study this generators and relations business in more detail - especially since it's fundamental in algebraic geometry: there can be 'relations between relations', called 'syzygies', and thinking about this led people to invent the concept of a 'resolution' of a module, and cohomology for modules of rings, and 'cofibrant replacement', and ultimately n-categories (since there can be relations between relations between relations...).

John Baez (Apr 13 2024 at 15:30):

I've dived reasonably deep into some of this stuff, yet I'll have to work a bit to come up with a finitely generated $\mathbb{Q}$-algebra $A$ that has different nonisomorphic but equally plausible '$\mathbb{Z}$-forms', i.e. algebras over $\mathbb{Z}$ which when tensored with $\mathbb{Q}$ give $A$.

John Baez (Apr 13 2024 at 15:31):

If you asked me for a $\mathbb{C}$-algebra that had different $\mathbb{R}$-forms I'd instantly say

$\langle x , y \vert x^2 + y^2 = 1 \rangle$

and

$\langle x, y \vert x^2 - y^2 = 1 \rangle$

The circle and the hyperbola become the same over $\mathbb{C}$!

John Baez (Apr 13 2024 at 15:32):

And I could talk about this in the fancy language of 'Galois descent', at least rather slowly and haltingly. But descending from $\mathbb{Q}$ to $\mathbb{Z}$ is not something I'm comfortable with.

Chris Grossack (they/them) (Apr 13 2024 at 15:49):

What about this, then? It's more intrinsic, but it's also more likely that I'll make a mistake, since I'm using topological intuition that I'm not sure how to make precise in the AG world.

Let $X$ be smooth and proper over $\mathbb{Q}$, if we can show it's still proper over $\mathbb{Z}$ (composing with the unique map $\mathbb{Q} \to \mathbb{Z}$) then the image of the non-smooth locus "should be" closed in $\mathbb{Z}$, which means it corresponds to a single ideal $(N)$. Then $X$ should be everywhere smooth over $\mathbb{Z}[N^{-1}]$. Now you can reduce mod $p$ (by base change) for any $p \in \mathbb{Z}[N^{-1}]$. That is, for all but finitely many primes.

Chris Grossack (they/them) (Apr 13 2024 at 15:50):

Also sorry for not typing "Spec" everywhere. I'm on my phone, haha

John Baez (Apr 14 2024 at 09:01):

Your approach is too abstract for my feeble brain. Luckily, I realized that my cute little "circle versus hyperbola" example can be modified to give a bunch of nonisomorphic affine schemes over $\mathbb{Z}$ that become isomorphic over $\mathbb{Q}$. Namely, ellipses!

For example, the circle

$\langle u,v \vert u^2 + v^2 = 1 \rangle$

is isomorphic to the ellipse

$\langle x, y \vert x^2 + 4 y^2 = 1 \rangle$

as schemes over $\mathbb{Q}$, using the coordinate change

$u \mapsto x, v \mapsto 2 y$

This map is invertible as a map of schemes over $\mathbb{Q}$ because we can define the inverse

$x \mapsto u, y \mapsto v/2$

But as a map of schemes over $\mathbb{Z}$ it's not invertible because we can't divide by 2!

Indeed, there are infinitely many nonisomorphic schemes over $\mathbb{Z}$ - different shapes of ellipses - that become isomorphic over $\mathbb{Q}$.

John Baez (Apr 14 2024 at 09:03):

(And then there are infinitely many nonisomorphic ellipses over $\mathbb{Q}$ that become isomorphic over $\mathbb{R}$, because in $\mathbb{R}$ but not in $\mathbb{Q}$ every positive number is a square. But this is captured by Galois descent in its classic form - i.e., Galois descent for field extensions!)

John Baez (Apr 14 2024 at 09:15):

To sound like an algebraic geometer, instead of saying "ellipse" I should say "quadric". So, we've seen that a quadric over $\mathbb{Q}$ has infinitely many different $\mathbb{Z}$-forms, also known as "models". The latter can then have different numbers of points over $\mathbb{F}_p$, though apparently only for finitely many primes $p$.

John Baez (Apr 14 2024 at 09:17):

For example,

$\langle u,v \vert u^2 + v^2 = 1 \rangle$

and

$\langle x, y \vert x^2 + 4 y^2 = 1 \rangle$

are dramatically different over $\mathbb{F}_2$, but they're isomorphic over $\mathbb{F}_p$ for all primes $\ne 2$, since in those fields $2$ is invertible.

John Baez (Apr 14 2024 at 09:20):

I should compute the zeta functions of these two quadrics and see if they're the same or not!

John Baez (Apr 14 2024 at 10:07):

They should be a product of Euler factors, one for each prime, and these factors should agree except for the prime $2$.

Chris Grossack (they/them) (Apr 14 2024 at 12:02):

are dramatically different over $\mathbb{F}_2$, but they're isomorphic over $\mathbb{F}_p$ for all primes $\neq 2$, since in those fields $2$ is invertible.

This was kind of my point earlier! Here's a proof theoretic angle, which might convince you, though I'm still not sure how to make it totally precise:

If $X$ has two presentations over $\mathbb{Q}$, then there should be a proof of this fact (by which I mean something like a sequence of tietze transformations but for ring presentations). But any such proof is finitely long, so one only has to divide finitely many times. But then the same proof will work over any $\mathbb{F}_p$ with $p$ large enough for all those divisions to be allowed!

Chris Grossack (they/them) (Apr 14 2024 at 12:04):

As an aside, I can't actually find anything on "tietze transformations for algebra presentations"... Surely someone has studied that, though! Does it have a different name?

Morgan Rogers (he/him) (Apr 14 2024 at 14:03):

This is also what I meant when I talked about '"most" characteristics of field'. It's nice to see how this issue is more subtle than I had imagined, though!!