2
$\begingroup$

Let $Y(3)$ be the fine moduli space (say, over $\mathbb{C}$) representing elliptic curves equipped with a full level 3 structure. Abstractly, there are 24 such structures for any elliptic curve, but thanks to every elliptic curve having $[-1]$ as an automorphism, there are generically only 12 equivalence classes. Thus, the natural map $Y(3)\rightarrow\text{Spec }\mathbb{C}[j]$ to the $j$-line is generically 12-to-1.

On the other hand, the "forget structure" map $Y(3)\rightarrow\mathcal{M}_{1,1}$ is finite etale and has geometric fibers of size 24, so I would think of it as a 24-to-1 map.

And yet, I can only imagine the composition $Y(3)\rightarrow \mathcal{M}_{1,1}\rightarrow \text{Spec }\mathbb{C}[j]$ to be the same as the 12-to-1 map described above.

Please, so that my christmas isn't ruined trying to figure out if I've been operating on a huge misunderstanding for the last year, can someone tell me that this apparent contradiction is because the morphism from $\mathcal{M}_{1,1}$ to the $j$-line isn't representable and hence it doesn't make much sense to talk about degree? (I've unravelled the definition and I think I've convinced myself that the map isn't representable, but I would still like to confirm that this is the cause of the apparent contradiction)

$\endgroup$
  • 5
    $\begingroup$ The map $j:M_{1,1} \rightarrow \mathbf{A}^1$ is indeed proper and etale but not representable, just like the map to the coarse moduli space from any separated DM stack (of finite type over a noetherian ring) when the stack isn't an algebraic space. (The map $Y(n) \rightarrow M_{1,1}$ is the stack quotient by the non-free action of ${\rm{GL}}_2(\mathbf{Z}/n\mathbf{Z})$ whereas $Y(n) \rightarrow \mathbf{A}^1$ is the scheme quotient by that same non-free action.) Ho-ho-ho, merry Christmas. $\endgroup$ – user74230 Dec 25 '14 at 6:11
9
$\begingroup$

If the map from $M_{1, 1}$ to the $j$-line can be said to have a degree, that degree should be $\frac{1}{2}$, which makes everything work out. The reason is that its fibers are generically not a finite set but a finite groupoid, namely $\text{pt}/\mathbb{Z}_2$ (corresponding to the $-1$ automorphism), which has groupoid cardinality $\frac{1}{2}$.

$\endgroup$
  • $\begingroup$ Presumably one can make this precise by defining a stacky fundamental class living in rational cohomology, but I don't know the details. $\endgroup$ – Qiaochu Yuan Jan 3 '15 at 7:49
5
$\begingroup$

This is to record User74230's answer to your question.

The map $j:M_{1,1}→\mathbb A^1$ is indeed proper and etale but not representable, just like the map to the coarse moduli space from any separated DM stack (of finite type over a noetherian ring) when the stack isn't an algebraic space. (The map $Y(n) \to M_{1,1}$ is the stack quotient by the non-free action of $GL_2(\mathbb Z/n\mathbb Z)$ whereas $Y(n)\to\mathbb A^1$ is the scheme quotient by that same non-free action.) Ho-ho-ho, merry Christmas.

$\endgroup$
  • $\begingroup$ I don't think that $j$ is etale. $\endgroup$ – Tom Graber Dec 28 '14 at 9:45
  • $\begingroup$ @TomGraber Thank you for your comment. I think User74230 meant to write "quasi-finite" instead of "etale". This makes me wonder: aren't all etale morphisms representable? $\endgroup$ – Ariyan Javanpeykar Dec 29 '14 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.