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Consider the stack $Ell$ (of groupoids) of elliptic curves. I'm interested in the autoequivalence 2-group of $Ell$, the objects of which consists of transformations $Ell \Rightarrow Ell: Ring \to Gpd$ valued in equivalences of groupoids. The arrows are isomorphisms of such transformations.

In a chat opinions ranged from optimistic that it would be large to hunches it would be small (in fact $\mathbb{Z}/2 \rightrightarrows \ast$)

Even if this 2-group is very small, I would also be interested in knowing if it's possible to calculate much of the endomorphism monoidal groupoid of $Ell$, given that we know a very explicit presentation of it (using a Hopf algebroid built from finitely generated polynomial rings). Here we would take all transformations, not just those valued in equivalences.

EDIT, Will Jagy: to get fuller context, you can scroll arbitrarily, and conveniently, back in the transcript where David's announcement of this question occurs pretty late in this segment. I will check later today, the terminal hour marker '16' may change, that is how the system works.

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My hunch is that it is $\pm 1$. You can see that its DM-compactification has automorphism group $\mathbb C^\ast$ using that it is a weighted projective space $\mathbb P(4,6)$ (in the stack sense). But only $\pm 1$ fix the point at infinity. Of course this is not a proof because I see no reason for an automorphism of $Ell$ to extend to the boundary. – Dan Petersen Jun 28 '13 at 7:23
But I'm not interested in the automorphism group... – David Roberts Jun 28 '13 at 10:06
@DanPetersen: Isn't your suggestion that the automorphism 2-group is $B \mathbb{Z}/2$? – Akhil Mathew Jun 28 '13 at 12:31
@AkhilMathew: That is what I thought when I wrote it, but now I think I may have been hasty. For instance, I do not see how what I wrote rules out the existence of a horribly non-geometric natural transformation from the identity map $M_{1,1} \to M_{1,1}$ to itself. – Dan Petersen Jun 28 '13 at 15:00
Maybe what I said earlier can be made to work by using the results of Noohi in : he determines the automorphism 2-group of an arbitrary weighted projective stack. – Dan Petersen Sep 6 '13 at 7:29

1 Answer 1

up vote 3 down vote accepted

The 2-group is $B\mathbb Z/2$. In other words, the automorphism $1$-group of $M_{1,1}$ is trivial, and the identity functor $M_{1,1} \to M_{1,1}$ has exactly one non-identity invertible natural transformation to itself: the one which sends a family of elliptic curves $\xi \colon E \to S$ to $\xi \circ i \colon E \to S$, where $i$ is inversion in the group structure of $E$.

The claim about the $1$-group was explained in the comments: the automorphism $1$-group of $\overline M_{1,1}$ is $\mathbb G_m$, since $\overline M_{1,1} \cong \mathbb P(4,6)$. None of these automorphisms fix the point at infinity. So we need only to determine the natural equivalences from the identity to itself.

Such a natural equivalence would assign to any $\newcommand{\id}{\mathrm{id}}\xi \colon E \to S$ an automorphism $a_\xi \colon E \to E$ over $S$, and it should satisfy the conditions of a natural transformation. For any $E \to S$ the automorphism group of $E$ over $S$ has two distinguished elements, $\id$ and $-\id$, and these are stable under pullback. In particular given $S' \to S$ and $\xi' \colon E' \to S'$ the pullback of $\xi$, if $a_\xi$ is trivial or inversion in the group, then the same holds for $a_{\xi'}$. The converse holds by descent if $S' \to S$ is étale.

Now let $X \to M_{1,1}$ be an étale cover by a scheme, let $\eta \colon C \to X$ be the pullback of the universal family. There is an open dense $U \subset X$ such that the only automorphism of $C$ over $U$ is inversion in the group. Then the same holds globally on $X$, so $a_\eta = \pm \id$, because the isomorphism scheme of $C$ over $X$ is separated. Let $\xi \colon E \to S$ be arbitrary. There is an étale cover $S' \to S$ such that $\xi'$ is pulled back from $\eta$. Then $a_{\xi'}$ is $\pm \id$. Then the same is true for $a_\xi$.

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Ah, I see - I didn't realise that $Ell$ was of the form that Noohi calculate Aut of. Thanks for getting back to this (I did see the update of the paper on the arXiv, but didn't connect the two things) – David Roberts Sep 6 '13 at 15:29
No problem. Minor quibble: strictly speaking it's not $\mathit{Ell}$ but its Deligne--Mumford compactification which is a stack of that form. – Dan Petersen Sep 6 '13 at 20:14
Yep, I think I got that. – David Roberts Sep 8 '13 at 2:37

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