9
$\begingroup$

I can't find any information about the canonical ring of Klein's quartic curve (the one with 168 automorphisms). I would imagine there is a lot known about the structure of this ring. Can anybody help me please? Thank you!

$\endgroup$
5
  • 2
    $\begingroup$ What kind of answer are you looking for? The syzygies in the minimal free resolution of the canonical ring, a la Green's conjecture? $\endgroup$ Commented Oct 10, 2016 at 6:23
  • $\begingroup$ I added the AG tag. I hope that's OK. $\endgroup$ Commented Oct 10, 2016 at 14:56
  • $\begingroup$ your question is tautological $\endgroup$ Commented Oct 10, 2016 at 15:10
  • $\begingroup$ YES! thank you folks (really lots of interesting info), what I meant if "a la Green's conjecture" (I had notes 100 years ago but I can't find them; however, not about this specific curve; I vaguely recall there were remarks about a rank-2 bundle over the curve, must have been an extension of K_X). I wanted to know this ring as precisely as possible, to construct explicit bases of H^0(nK_X) (yep, I should have used additive notation to go with the word "divisor"); a related issue are of course higher Weierstrass points. $\endgroup$ Commented Oct 12, 2016 at 4:13
  • $\begingroup$ GOT IT! It's in David Eisenbud's Green's Conjecture: An Orientation (1991). The structure of the canonical ring of curves is known for genus less that or equal to 8 (at that time...) and also for smooth plane curves of any genus, so... it's overkill! John Mangual: You may be happy to see the abstract "tries to explain the attractiveness of canonical rings"... but the article is a little technical. Listen everybody, I had such a great experience here, it was my first time daring to post on MO and I enjoyed hearing back and learning the things that you found beautiful, thank you! $\endgroup$ Commented Oct 17, 2016 at 5:02

3 Answers 3

11
$\begingroup$

Klein quartic, $X$, is a smooth degree $4$ plane curve, given by the equation $F(x,y,z) = x^3 y + y^3 z + z^3 x = 0$. Doesn't this mean that the canonical ring is $k[x,y,z]/(x^3 y + y^3 z + z^3 x)$?

Details: This embedding is the canonical embedding, since we compute by adjunction that $\mathcal{O}(1)$ is the canonical bundle. Thus, we have an injection $k[x,y,z]/F \hookrightarrow \bigoplus_{n=0}^{\infty} H^0(X, K^{\otimes n})$. The Hilbert series of the left hand side is $(1-t^4)/(1-t)^3 = 1+3t + \sum_{n \geq 2} (4n-2) t^n$. The Hilbert series of the canonical ring is $1+gt + \sum_{n \geq 2} ((2g-2)n-(g-1)) t^n$. These match, so the map is an isomorphism. (Actually, I think I recall that the canonical ring of a nonhyperelliptic curve is always generated in degree $1$, but it seemed easier to check the Hilbert series in this case.)

$\endgroup$
1
  • 3
    $\begingroup$ You recall correctly that the canonical ring of a nonhyperelliptic curve is always generated in degree 1. That's a theorem of Max Noether. $\endgroup$ Commented Oct 10, 2016 at 15:11
2
$\begingroup$

There are many resources for this, one of easy treatments can be found in

https://www.math.hmc.edu/~ursula/teaching/math189/finalpapers/julia.pdf

$\endgroup$
1
  • 2
    $\begingroup$ Thank you, I had not seen this gorgeous reference; but, my question was about the ring structure of \oplus H^0(X,K^n), where X is the curve and K the canonical bundle $\endgroup$ Commented Oct 10, 2016 at 5:22
-6
$\begingroup$

Not sure what "canonical" means in this context. Half of algebraic geometry is called "canonical ring" or "fundamental class" or some other vacuous term.


Please consider Elkies The Klein Quartic in Number Theory and in general the book The Eightfold Way is online. In the translation of Klein's original work we find this futuristic image:

enter image description here

Also in German. The translated version has improved images


I have no idea what the above really means, but it sounds like we get really lucky. These results are due to Riemann -- long before Serre, Grothendieck, Hartshorne or Shafarevich. How could he have anything more complex in mind?

The only ring mention in Klein's paper is $\mathbb{C}[x,y,z]/(x^3y + y^3z+z^3x)$ -- the one David Speyer wrote -- as well as $SL(2, \mathbb{F}_7)$ and $\mathbb{Z}(\frac{1 + \sqrt{-7}}{2})$. Which one is the canonical ring?

Since $x, y, z$ are the coordinates of an equation $x^3 y + y^3 z + z^3 x = 0$, then $x, y, z$ can be thought of as parameterizing solutions to the set $$\{x^3 y + y^3 z + z^3 x = 0\}$$

So the coordinate ring might be the coordinate ring of $\mathbb{P}^N$ mod a single relation: $$ \mathbb{C}[x,y,z]/(x^3 y + y^3 z + z^3 x ) $$ In fact, $x, y, z$ are sections of the line bundle $\mathcal{O}_X$ and if $i+j+k = n$ then maybe we can multiply them to get sections of the other sheafs: $$ x^i y^j z^k \in H^0(X, n \mathcal{O}_C)$$ Looks like the coordinate ring is the direct sum of all of these possibilities excepting for the possibility that $x^3 y + y^3z + z^3 x = 0$.


If we then re-read Klein, there is even more to say:

\begin{eqnarray} x &=& \sum_{\beta \equiv 1 \mod \sqrt{-7}} \mathrm{Re}(\beta) \, q^{\beta \overline{\beta}/7} \\ y &=& \sum_{\beta \equiv 2 \mod \sqrt{-7}} \mathrm{Re}(\beta) \, q^{\beta \overline{\beta}/7} \\ z &=& \sum_{\beta \equiv 4 \mod \sqrt{-7}} \mathrm{Re}(\beta) \, q^{\beta \overline{\beta}/7} \end{eqnarray}

and these theta functions explicitly solve the equation $x^3 y + y^3 z + z^3 x = 0$ in terms of modular functions over $\mathbb{H}/SL(2, \mathbb{F}_7)= \langle z \mapsto z + 7, z \mapsto - \frac{1}{z} \rangle $. These sections $x,y,z \in \mathbb{H}$ are modular cusp forms of weight 2 over $X(7)$ -- one must verify that $x, y, z, \neq 0$ exceept at the "cusps".

The fact that these theta functions parameterize a curve both on $\mathbb{H}$ and in projective space $\mathbb{P}^3$ does not seem trivial at all. A lot of effort may go into showing the rather strange-looking coordinates we chose do not have exceptional behavior.

And these are generators of your canonical ring!


Birational Geometry Old and New (Antonella Grassi) has this much to say:

Example-Theorem 29: If $C$ is a curve of genus $g \geq 2$ the divisor $3 K_C$ is very ample. There is an embedding $\phi: C \to \mathbb{P}^N$ such that if $H$ is a hyperplane $H \cdot C = 3K_C$. Then the canonical ring $R(C, K_C)$ can be reconstructed from the coordinate ring of $\mathbb{P}^N$. In particular, $R$ is finitely generated. In fact $3K_C$ determines a pluricanonical embedding.

Antonella Grassi explain that nothing too exceptional occurs, so that for example $K_C = \mathcal{O}(-1)$, and the coordinates $x,y,z$ can be identified with hyperplanes $\{ x=0\}, \{ y = 0\}, \{z=0\}$ and then $[x],[y],[z]$ are hyperplane divisors and $H\cdot C$ is just setting $x$, $y$, or $z = 0$.

For much more please turn to the standard texts:

  • Hartshorne Algebraic Geometry
  • Shafarevich Basic Algebraic Geometry
$\endgroup$
12
  • 5
    $\begingroup$ John: I have been called many names in my (long) life, but this is the first time I'm being called "vacuous": that is BRILLIANT! that's exactly what I am. Thank you, I had read the book but many readers will benefit from your reference! My answer to your puzzlement (in the slight chance you care to know) is in the comment to Amdeberhan, but my vacuousness (now that I know) is eternal! :-) $\endgroup$ Commented Oct 10, 2016 at 5:26
  • 16
    $\begingroup$ How can you say that the canonical ring $$R(X, \, K_X) = \bigoplus _{n \in \mathbb{N}} H^0(X, \, nK_X)$$ is a vacuous term? $\endgroup$ Commented Oct 10, 2016 at 6:16
  • 16
    $\begingroup$ If you're not sure what "canonical ring" means in this context, maybe this is not a question for you to answer? Any algebraic geometer can understand the question perfectly well. $\endgroup$ Commented Oct 10, 2016 at 6:40
  • 6
    $\begingroup$ I occasionally turn to MO in the mornings as a respite from reading the news. I thought that at least here we would all understand what words actually mean. $\endgroup$ Commented Oct 10, 2016 at 14:41
  • 3
    $\begingroup$ OK, but what does that have to do with addressing a comment to me in which you write "There's no need to take it personally" – what were you referring to? $\endgroup$ Commented Oct 11, 2016 at 5:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .