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\begin{document}

Speaker: Wei Ho (Columbia)

Title: Coregular representations and genus 1 curves (joint with Manjul Bhargava)

Date: January 26, 2012

\section{Philosophy.}

We want to get explicit constructions for moduli spaces. One possible way to do this is to construct it as $V/G$ where $V$ is a linear representation of $G$. The motivations involve finding explicit descriptions, computation, finding compactifications, and getting unirationality of the moduli space.

\begin{example}[Plane curves]
We think of plane curves as degree $n$ polynomials in $\bP^2$, but only up to change of basis. This is the quotient $\Sym^n(3) / \GL_3$.

If we restrict to smooth plane curves, this corresponds to some nondegeneracy condition on $\Sym^n(3) / \GL_3$, which gives an open subset. When the curve is smooth, it has genus $(n-1)(n-2)/2$. Also, it naturally has a degree $n$ line bundle obtained by pulling back $\cO_{\bP^2}(1)$.

When $n=3$, we get $\Sym^3(3) / \GL_3$ which correspond to genus 1 curves $C$ together with a degree 3 line bundle $L$ modulo a natural notion of equivalence. Over $\bC$, the line bundle isn't really extra data since they differ by some automorphism of $C$ (we lose it when we look at the coarse moduli space). Furthermore, over $\bC$, both moduli spaces are $\bP^1$ ($j$-invariant). The ring of (semi-)invariants is a polynomial ring generated in degrees 4 and 6 (call them $S$ and $T$).

What about an arbitrary field $K$ (avoid characteristic 2 and 3)? First, degree 3 line bundles need not be equivalent under some $K$-automorphism of $C$. In general, the invariants $S$ and $T$ give an equation for the Jacobian of $C$: $y^2 = x^3 + Sx + T$ (Weierstrass form).

We have $\cM_{1,1} = \{y^2 = x^3 + Ax + B\}$, but note that $(A,B)$ and $(\lambda^4 A, \lambda^6 B)$ give the isomorphic curves. So we can think of $\cM_{1,1}$ as $(A,B)$ modulo this scaling action.
\end{example}

If you want to classify some objects such that the $\bC$-points of the coarse moduli space is, for example, a weighted projective space, then you might look for linear representations with a polynomial ring of invariants. Such representations are called {\bf coregular} and have been classified by Littelmann when the representation is irreducible.

\section{$3 \otimes 3 \otimes 3$.}

Consider the representation $V = V_1 \otimes V_2 \otimes V_3$ where $\dim V_i = 3$. The group is $G = \GL(V_1) \times \GL(V_2) \times \GL(V_3)$. If we pick bases, we can picture these 3-tensors as a $3 \times 3 \times 3$ arrays of scalars. Taking flattenings in different ways, we get 3 different $3 \times 3$ matrices of linear forms in 3 variables $x,y,z$. Taking the determinant of these matrices gives a cubic equation in $\bP^2$ and hence a smooth genus 1 curve (generically). It turns out these curves are all isomorphic, so we get 3 different line bundles $L_1, L_2, L_3$. The relation among the line bundles is that $L_1^2 = L_2 \otimes L_3$.

To see the isomorphisms, we consider the restriction of the $3 \times 3$ matrix to the plane cubic. Its kernel is 1-dimensional and gives a point in a different $V_i$, and such correspondences are isomorphisms. This gives us maps $C_1 \to C_2 \to C_3 \to C_1$, but the composition is not the identity. But note that automorphisms of an elliptic curves are either translations by a point in the Jacobian or a certain kind of flip. In particular, we also get the data of a special kind of point in the Jacobian of $C$.

Conversely, given all of this information, one can recover the tensor, so we get an isomorphism of moduli spaces.

\begin{theorem} For any field $K$ (of characteristic not $2$ or $3$), we have an isomorphism
\begin{align*}
  \{ \text{non-degenerate orbits of $V(K) / G(K)$}\} &\cong \{(C,L_1,L_2,L_3) \mid L_1^2 = L_2 \otimes L_3\}\\
  &\cong \{(C,L,P) \mid P \text{ is a special point of $\Jac(C)$} \}.
\end{align*}
\end{theorem}

The ring of invariants is generated by polynomials of degrees $6, 9, 12$. They give an equation for $\Jac(C)$: 
\[
y^2 + d_9 y = x^3 + d_6 x^2 + d_{12} x
\]
The marked point corresponds to $x=y=0$.

Without changing the flattenings, we can see all 3 line bundles from a single perspective (on $C_1$). Namely, we interpret the tensor as a linear map $\bP(V_1^*) \to \bP(V_2 \otimes V_3)$. Let $Y$ be the determinantal hypersurface and let $X$ be the Segre variety $\bP(V_2) \times \bP(V_3)$. The adjugate matrix gives a birational involution $\beta \colon \bP(V_2 \otimes V_3) \dasharrow \bP(V_2 \otimes V_3)$ with base locus $X$ (and $Y$ is blown down to $X$). This map gives a way to construct the other two line bundles.

~

The construction in the last paragraph generalizes if we replace $3 \times 3$ matrices by $3 \times 3$ Hermitian matrices (over a composition algebra). The corresponding ``Severi varieties'' are
\begin{align*}
\bP^2 &\xrightarrow{\cO(2)} \bP^5\\
\bP^2 \times \bP^2 &\xrightarrow{\cO(1,1)} \bP^8\\
\Gr(2,6) &\xrightarrow{\cO(1)} \bP^{14}\\
E^{16} &\to \bP^{26}
\end{align*}
(the last space is a homogeneous space for ${\rm E}_6$). The corresponding split forms of the representations are 
\begin{align*}
(\GL_3 \times \GL_3, 3 \otimes \Sym^2(3)),\\
(\GL_3^3, 3 \otimes 3 \otimes 3),\\
(\GL_3 \times \GL_6, 3 \otimes \bigwedge^2(6)),\\
(\GL_3 \times {\rm E}_6, 3 \otimes 27).
\end{align*}

There is another family $2 \otimes V$ where $V$ is a special kind of representation (the simplest example being $V = 2 \otimes 2 \otimes 2$.

%(Are there automorphisms of the Abelian surface torsor for $\wedge^3 C^9$?)

%(Ask about enhanced nilpotent cones)

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\end{document}