# Resolution of singularities of this cubic surface?

Let $A = \mathcal O(Y)^{SL_2(\mathbb C)}$ be the ring of invariant functions on $Y := \mathrm{Hom}(\mathbb Z^2, SL_2(\mathbb C))$. We can identify $A$ with the quotient of $\mathbb C[x,y,z]$ by the ideal generated by $f(x,y,z) = x^2+y^2+z^2-xyz-4$. (If $a$ and $b$ are the generators for $\mathbb Z_2$, then under this identification, $x$, $y$, and $z$ map to the functions $\rho \mapsto \mathrm{Tr}(\rho(g))$, for $g$ being $a$, $b$, and $ab$, respectively.) This realizes $\mathrm{Spec}(A)$ as a (singular) cubic surface in $\mathbb C^3$.

We can also realize $A$ as the $\mathbb Z_2$-invariants of $\mathbb C[X^{\pm 1}, Y^{\pm 1}]$, where $\mathbb Z_2$ acts by inverting $X$ and $Y$. This isomorphism is given by sending $x \mapsto X+X^{-1}$, $\quad y \mapsto Y+Y^{-1}$, and $z \mapsto XY+X^{-1}Y^{-1}$.

Q: Does $\mathrm{Spec}(A)$ have a "nice" resolution of singularities that has appeared in the literature?

[edit: relation corrected, comment below added] In particular, I'd be interested in a resolution that has an interpretation as some kind of moduli space, analogous to the way that the Hilbert scheme of $n$ points in $\mathbb C^2$ resolves $(\mathbb C^2)^n / S_n$.

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A blow-up on the origin (in your first presentation) resolves it, no? –  Mariano Suárez-Alvarez Feb 14 '13 at 3:11
(I compute $f(x,y,z)$ with your trace functions and I get $4$, not zero. Are you sure?) –  Mariano Suárez-Alvarez Feb 14 '13 at 3:53
That's what I get for writing the relation from memory - there is indeed a $-4$ term. –  Peter Samuelson Feb 14 '13 at 4:10
It seems that your surface the Cayley cubic: there are a lot of interpretations of the Cayley cubic! –  M P Feb 14 '13 at 10:18
I think you're right - I'd also be interested in hearing about other interpretations of this surface (or references). –  Peter Samuelson Feb 14 '13 at 17:50