---
abstract: 'Let $R$ be the free $\CC$-algebra on $x$ and $y$ modulo the relations $x^5=yxy$ and $y^2=xyx$ endowed with the $\ZZ$-grading $\deg x=1$ and $\deg y=2$. Let $\BB_3$ denote the blow up of $\CC\PP^2$ at three non-colliear points. The main result in this paper is that the category of quasi-coherent $\cO_{\BB_3}$-modules is equivalent to the quotient of the category of $\ZZ$-graded $R$-modules modulo the full subcategory of modules $M$ such that for each $m \in M$, $(x,y)^nm=0$ for $n \gg 0$. This reduces almost all representation-theoretic questions about $R$ to algebraic geometric questions about the del Pezzo surface $\BB_3$. For example, the generic simple $R$-module has dimension six. Furthermore, the main result combined with results of Artin, Tate, and Van den Bergh, imply that $R$ is a noetherian domain of global dimension three'
address: ' Department of Mathematics, Box 354350, Univ. Washington, Seattle, WA 98195'
author:
- 'S. Paul Smith'
title: 'A non-commutative homogeneous coordinate ring for the third del Pezzo surface'
---

[^1]

Introduction
============

We will work over the field of complex numbers.

The surface obtained by blowing up $\PP^2$ at three non-colinear points is, up to isomorphism, independent of the points. It is called the [third del Pezzo surface]{} and we will denote it by $\BB_3$.

Let $R$ be the free $\CC$-algebra on $x$ and $y$ modulo the relations $$\label{relations}
x^5=yxy \qquad \hbox{and } \qquad y^2=xyx.$$ Give $R$ a $\ZZ$-grading by declaring that $$\deg x=1 \qquad \hbox{and } \qquad \deg y=2.$$

In this paper we prove there is a surprisingly close relationship between the non-commutative algebra $R$ and the third del Pezzo surface. This relationship can be exploited to obtain a deep understanding of the representation theory of $R$.

\[thm.main1\] Let $R$ be the non-commutative algebra $\CC[x,y]$ with relations (\[relations\]). Let $\Gr R$ be the category of $\ZZ$-graded left $R$-modules. Then there is an equivalence of categories $$\Qcoh \BB_3 \equiv {{\Gr R}\over{\sT}}$$ where the left-hand side is the category of quasi-coherent $\cO_{\BB_3}$-modules and the right-hand side is the quotient category modulo the full subcategory $\sT$ consisting of those modules $M$ such that for each $m \in M$, $(x,y)^nm=0$ for $n \gg 0$.

Theorem \[thm.main1\] is a consequence of the following result.

\[thm.main2\] Let $R$ be the non-commutative algebra $\CC[x,y]$ with relations (\[relations\]). There is an automorphism $\s$ of $\BB_3$ having order six, and a line bundle $\cL$ on $\BB_3$ such that $R$ is isomorphic to the twisted homogeneous coordinate ring $$B(\BB_3,\cL,\s) : = \bigoplus_{n \ge 0} H^0(\BB_3,\cL_n)$$ where $$\cL_n:= \cL \otimes (\s^*)\cL \otimes \cdots \otimes (\s^*)^{n-1}\cL.$$

Results of Artin, Tate, and Van den Bergh now imply that $R$ is a 3-dimensional Artin-Schelter regular algebra and therefore has the following properties.

Let $R$ be the algebra $\CC[x,y]$ with relations (\[relations\]). Then

1.  $R$ is a left and right noetherian domain;

2.  $R$ has global homological dimension 3;

3.  $R$ is Auslander-Gorenstein and Cohen-Macaulay in the non-commutative sense;

4.  the Hilbert series of $R$ is the same as that of the weighted polynomial ring on three variables of weights 1, 2, and 3;

5.  $R$ is a finitely generated module over its center [@ST94 Cor. 2.3];

6.  $R^{(6)}:=\oplus_{n=0}^\infty R_{6n}$ is isomorphic to $\bigoplus_{n=0}^\infty H^0(\BB_3\cO(-nK_{\BB_3})$;

7.  $\Spec R^{(6)}$ is the anti-canonical cone over $\BB_3$, i.e., the cone obtained by collapsing the zero section of the total space of the anti-canonical bundle over $\BB_3$.

This close connection between $R$ and $\BB_3$ means that almost all aspects of the representation theory of $R$ can be expressed in terms of the geometry of $\BB_3$. We plan to address this question in another paper.

The justification for calling $R$ a non-commutative homogeneous coordinate ring for $\BB_3$ is the similarity between the equivalence of categories in Theorem \[thm.main1\] and following theorem of Serre:

> if $X \subset \PP^n$ is the scheme-theoretic zero locus of a graded ideal $I$ in the polynomial ring $S=\CC[x_0,\ldots,x_n]$ with its standard grading, and $A=S/I$, then there is an equivalence of categories $$\label{eq.Serre}
> \Qcoh X \equiv {{\Gr A}\over{\sT}}$$ where the right-hand side is the quotient category of $\Gr A$, the category of graded $A$-modules, by the full subcategory $\sT$ of modules supported at the origin.

The author is grateful to the following people: Amer Iqbal for directing the author to some papers in high energy physics where the algebra $R$ appears in a hidden form; Paul Hacking and Sándor Kovács for passing on some standard results about del Pezzo surfaces and algebraic geometry; and Darin Stephenson for telling the author that the algebra $R$ is an iterated Ore extension, and for providing some background and guidance regarding his two papers cited in the bibliography.

The non-commutative algebra $R=\CC[x,y]$ with $x^5=yxy$ and $y^2=xyx$ {#sect.nc.ring}
=====================================================================

The following result is a straightforward calculation. The main point of it is to show that $R$ has the same Hilbert series as the weighted polynomial ring on three variables of weights 1, 2, and 3.

\[Stephenson\] \[prop.steph\] The ring $R:=\CC[x,y]$ with defining relations $$x^5=yxy \qquad \hbox{and} \qquad  y^2=xyx$$ is an iterated Ore extension of the polynomial ring $\CC[w]$. Explicitly, if $\zeta$ is a fixed primitive $6^{\th}$ root of unity, then:

1.  $R= \CC[w][z;\s][x;\tau,\d]$ where $\s \in \Aut \CC[z]$, $\tau \in \Aut \CC[w,z]$, and $\d$ is the $\tau$-derivation defined by $$\begin{aligned}
       \s(w)& =\zeta w,
       \\
       \tau(w) &=  - \zeta^2 w, \quad    \tau(z)  = \zeta z,
       \\
       \d(w) & = z, \quad    \d(z)  =- w^2;
       \end{aligned}$$

2.  A set of defining relations of $R=\CC[z,w,x]$ is given by $$\begin{aligned}
    zw = & \zeta wz,
    \\
    xw = & -\zeta^2wx+z ,
    \\
    xz  = & \zeta zx- w^2;\end{aligned}$$

3.  $R$ has basis $\{ w^iz^jx^k \; | \; i,j,k \ge 0\}$;

4.  $R$ is a noetherian domain;

5.  The Hilbert series of $R$ is $(1-t)^{-1}(1-t^2)^{-1}(1-t^3)^{-1}$.

Define the elements $$\begin{aligned}
w: & = \,  y-x^2
\\
z: & = \, xw + \zeta^2 wx 
\\
&= \, xy +\zeta^2 yx - \zeta x^3\end{aligned}$$ of $R$. Since $y$ belongs to the ring generated by $x$ and $w$, $\CC[x,y]=\CC[x,w]=\CC[x,w,z]$. It is easy to check that $$\label{steph.relns}
zw=\zeta wz, \quad xw= z- \zeta^2 wx, \quad xz=\zeta zx - w^2.$$

Let $R'$ be the free algebra $\CC\langle w,x,z \rangle$ modulo the relations in (\[steph.relns\]). We want to show $R'$ is isomorphic to $R$. We already know there is a homomorphism $R' \to R$ and we will now exhibit a homomorphism $R \to R'$ by showing there are elements $x$ and $Y$ in $R'$ that satisfy the defining relations for $R$. Define the element $Y:=w+x^2$ in $R'$. A straightforward computation in $R'$ gives $$xwx-x^2w = w^2+wx^2$$ so $$Y^2= w^2+x^2w + wx^2 + x^4 =  xwx+x^4=xYx.$$ In the next calculation we make frequent use of the fact that $1-\xi+\xi^2=0$. Deep breath... $$\begin{aligned}
YxY& = (w+x^2)xw  +wx^3 + x^5
 \\
 & = (w+x^2) (z-\zeta^2wx)+ \big[ wx^3 + x^5 \big]
\\
& = x^2 z -\zeta^2 x^2wx + \big[  wz -\zeta^2 w^2x +wx^3 + x^5 \big]
\\
& = x(\zeta zx -w^2)  -\zeta^2 x(z-\zeta^2wx)x +  \big[  wz -\zeta^2 w^2x +wx^3 + x^5 \big]
\\
& =(\zeta-\zeta^2)xzx - xw^2 - \zeta xwx^2 +  \big[  wz -\zeta^2 w^2x +wx^3 + x^5 \big]
\\
& =(\zeta-\zeta^2)(\zeta zx -w^2)x  -   (z-\zeta^2wx) w    -\zeta(z- \zeta^2 wx)x^2 
\\
& \qquad  \qquad + \big[  wz -\zeta^2 w^2x +wx^3 + x^5 \big]
\\
& =(\zeta^2-\zeta^3)zx^2 -(\zeta-\zeta^2)w^2x - zw +\zeta^2 wxw -\zeta zx^2 - wx^3  
\\
& \qquad  \qquad + \big[  wz -\zeta^2 w^2x +wx^3 + x^5 \big]
\\
& =(\zeta^2-\zeta^3-\zeta)zx^2   +\zeta^2 wxw  + \big[ (1-\zeta) wz -\zeta w^2x + x^5 \big]
\\
& = \zeta^2 wxw   +   \big[ (1-\zeta) wz -\zeta w^2x + x^5 \big]
 \\
& = \zeta^2 w(z-\zeta^2wx)   + \big[ -\zeta^2  wz -\zeta w^2x + x^5 \big]
\\
& = x^5.
 \end{aligned}$$ Since $YxY=x^5$, $R$ is isomorphic to $R'$. Hence $R$ is an iterated Ore extension as claimed. The other parts of the proposition follow easily.

It is an immediate consequence of the relations that $x^6=y^3$. Hence $x^6$ is in the center of $R$.

The del Pezzo surface $\BB_3$
=============================

Let $\BB_3$ be the surface obtained by blowing up the complex projective plane $\PP^2$ at three non-collinear points. We will write $E_1$, $E_2$, and $E_3$ for the exceptional curves associated to the blowup.

The $-1$-curves on $\BB_3$ lie in the following configuration $$\label{six.lines}
 \qquad
\UseComputerModernTips
\xymatrix{
&& &  \save []+<0cm,0.1cm>*\txt<4pc>{$\scriptstyle{E_2}$} \restore \ar@{-}[dddlll] 
& \ar@{-}[dddrrr]   \save []+<0cm,0.1cm>*\txt<4pc>{$\scriptstyle{E_3}$} \restore 
\\ 
& \save []+<-0.3cm,0cm>*\txt<4pc>{$\scriptstyle{L_1}$} \restore 
\ar@{-}[rrrrr] &&&&& \save []+<0.5cm,0cm>*\txt<4pc>{$\scriptstyle{X=0}$} \restore  &
\\  
\save []+<-0.2cm,0.1cm>*\txt<4pc>{$\scriptstyle{Z=0}$} \restore  &&&&&&&
\save []+<0.3cm,0.1cm>*\txt<4pc>{$\scriptstyle{Y=0}$} \restore 
&&&&& 
\\ 
\save []+<-0.2cm,-0.1cm>*\txt<4pc>{$\scriptstyle{s=0}$} \restore   &&&&&&&  \save []+<0.3cm,-0.1cm>*\txt<4pc>{$\scriptstyle{u=0}$} \restore    &
\\
&
 \save []+<-0.3cm,0cm>*\txt<4pc>{$\scriptstyle{E_1}$} \restore 
 \ar@{-}[rrrrr] &&&&&  \save []+<0.5cm,0cm>*\txt<4pc>{$\scriptstyle{t=0}$} \restore 
\\
&&& \ar@{-}[uuulll]   \save []+<0cm,-0.1cm>*\txt<4pc>{$\scriptstyle{L_3}$} \restore  
&  \save []+<0cm,-0.1cm>*\txt<4pc>{$\scriptstyle{L_2}$} \restore  \ar@{-}[uuurrr]  }$$ where $L_1$, $L_2$, and $L_3$ are the strict transforms of the lines in $\PP^2$ spanned by the points that are blown up. (The labeling of the equations for the $-1$-curves will be justified shortly.)

The union of the $-1$-curves is an anti-canonical divisor and is, of course, ample.

The Picard group of $\BB_3$ {#sect.Pic}
---------------------------

The Picard group of $\BB_3$ is free abelian of rank four. We will identify it with $\ZZ^4$ by using the ordered basis $$H,\; -E_2, \; -E_1, \; -E_3$$ where the $E_i$s are the exceptional curves over the points blown up and $H$ is the strict transform of a line in $\PP^2$ in general position, i.e., missing the points being blown up. Thus $$H=(1,0,0,0), \quad E_1=(0,0,-1,0), \quad E_2=(0,-1,0,0), \quad E_3=(0,0,0,-1).$$ The canonical divisor $K$ is $-3H+E_1+E_2+E_3$ so the anti-canonical divisor is $$-K=(3,1,1,1).$$

Cox’s homogeneous coordinate ring
---------------------------------

By definition, Cox’s homogeneous coordinate ring [@Co] for a complete smooth toric variety is $$S:= \bigoplus_{[\cL] \in \Pic X} H^0(X,\cL).$$ For the remainder of this paper $S$ will denote Cox’s homogeneous coordinate ring for $\BB_3$.

Let $X,Y,Z,s,t,u$ be coordinate functions on $\CC^6$. One can present $\BB_3$ as a toric variety by defining it as the orbit space $$\BB_3:={{\CC^6 -W}\over{(\CC^\times)^4}}$$ where the irrelevant locus, $W$, is the union of nine codimension two subspaces, namely $$\label{irrel.locus}
\begin{array}{ccc}
\begin{array}{c}
X=t=0\\
Y=s=0\\
Z=u=0
\end{array}
\qquad
\begin{array}{c}
X=Y=0\\
Y=Z=0\\
Z=X=0
\end{array}
\qquad
\begin{array}{c}
s=t=0\\
u=t=0\\
s=u=0
\end{array}
\end{array}$$ and $(\CC^\times)^4$ acts with weights $$\begin{array}{rrrrrr}
X & \phantom{x} & 1 & 1 & 0 & 1 \\
Y &  \phantom{x} & 1 & 0 & 1 & 1 \\
Z & \phantom{x} & 1 & 1 & 1& 0 \\
s & \phantom{x} & 0 & -1 & 0 & 0\\
t & \phantom{x} & 0 & 0 & -1 & 0 \\
u & \phantom{x} & 0 & 0 & 0 & -1
\end{array}.$$ Therefore $S$ is the $\ZZ^4$-graded polynomial ring $$S=\CC[X,Y,Z,s,t,u]$$ with the degrees of the generators given by their weights under the $(\CC^\times)^4$ action. It follows from Cox’s results [@Co Sect. 3] that $$\Qcoh \BB_3 \equiv {{\Gr(S,\ZZ^4)}\over{\sT}}$$ where $\Gr(S,\ZZ^4)$ is the category of $\ZZ^4$-graded $S$-modules and $\sT$ is the full subcategory consisting of all direct limits of modules supported on $W$.

The labelling of the $-1$-curves the diagram (\[six.lines\]) is explained by the existence of the morphisms $$\UseComputerModernTips
\xymatrix{
& \BB_3 \ar[dl] \ar[dr]  
\\
\PP^2 && \PP^2 
}
\qquad
\UseComputerModernTips
\xymatrix{
&  \save []+<-0.2cm,0.2cm>*\txt<4pc>{$\text{ $(X,Y,Z,s,t,u) $}$} \restore     \ar[dl] \ar[dr] 
\\
 \save []+<-0.6cm,-0.2cm>*\txt<4pc>{$\text{$(Xsu,Ytu,Zst)$}$} \restore&&
 \save []+<+0.3cm,-0.2cm>*\txt<4pc>{$\text{$(YZt,XZs,XYu)$}$} \restore
}$$ that collapse the $-1$-curves.

An order six automorphism $\s$ of $\BB_3$
-----------------------------------------

The cyclic permutation of the six $-1$-curves on $\BB_3$ extends to a global automorphism of $\BB_3$ of order six. We now make this explicit.

The category of graded rings consists of pairs $(A,\G)$ consisting of an abelian group $\G$ and a $\G$-graded ring $A$. A morphism $(f,\theta):(A,\G) \to (B,\Upsilon)$ consists of a ring homomorphism $f:A \to B$ and a group homomorphism $\theta:\G \to \Upsilon$ such that $f(A_i) \subset B_{\theta(i)}$ for all $i \in \G$.

Let $\tau:S \to S$ be the automorphism induced by the cyclic permutation $$\label{defn.tau}
\UseComputerModernTips
\xymatrix{
X \ar[r] & u \ar[r] & Y \ar[r]_\tau & t \ar[r] & Z \ar[r] & s \ar@{}@/_1pc/[lllll] | {<}
}$$ and let $\theta:\ZZ^4 \to \ZZ^4$ be left multiplication by the matrix $$\theta= \begin{pmatrix}
2 & -1 & -1 & -1   \\
1 & -1 & -1 & 0 \\
1 & 0 & -1 & -1 \\
1 & -1 & 0 & -1
\end{pmatrix}.$$ Then $(\tau,\theta):(S,\ZZ^4) \to  (S,\ZZ^4)$ is an automorphism in the category of graded rings. Because the irrelevant locus (\[irrel.locus\]) is stable under the action of $\tau$, $\tau$ induces an automorphism $\s$ of $\BB_3$. It follows from the definition of $\tau$ that $\s$ cyclically permutes the six $-1$-curves

Since $(\tau,\theta)^6=\id_{(S,\ZZ^4)}$ the order of $\s$ divides six. But the action of $\s$ on the set of $-1$-curves has order six, so $\s$ has order six as an automorphism of $\BB_3$.

Fix a primitive cube root of unity $\omega$. The left action of $\theta$ on $\ZZ^4=\Pic \BB^3$ has eigenvectors $$v_1=\begin{pmatrix} 1    \\  1 \\ 1 \\ 1 \\ \end{pmatrix}, \quad
v_2=\begin{pmatrix} 3    \\  1 \\ 1 \\ 1 \\ \end{pmatrix}, \quad
v_3=\begin{pmatrix} 0    \\  1 \\ \omega \\ \omega^2 \\ \end{pmatrix}, \quad
v_4=\begin{pmatrix} 0    \\  1 \\ \omega^2 \\ \omega \\ \end{pmatrix},$$ with corresponding eigenvalues $-1$, $+1$, $\omega^2$, $\omega$.

A twisted hcr for $\BB_3$
=========================

A sequence of line bundles on $\BB_3$ {#sect.Dn}
-------------------------------------

We will blur the distinction between a divisor $D$ and the class of the line bundle $\cO(D)$ in $\Pic\BB_3$.

We define a sequence of divisors: $D_0$ is zero; $D_1$ is the line $L_1$; for $n \ge 1$ $$D_n:=(1+\theta + \cdots + \theta^{n-1})(D_1).$$ We will write $\cL_n:=\cO(D_n)$. Therefore $$\cL_n  = \cL_1 \otimes \s^* \cL_1 \otimes \ldots \otimes (\s^*)^{n-1} \cL_1.$$ For example, $$\begin{array}{cc}
\begin{array}{l}
\cO(D_1)=\cL_1 = \cO(1,1,0,1) \\
\cO(D_2)=\cL_2 = \cO(1,1,0,0) \\
\cO(D_3)=\cL_3 = \cO(2,1,1,1) \\
\cO(D_4)=\cL_4 = \cO(2,1,0,1) \\
\cO(D_5)=\cL_5 =  \cO(3,2,1,1) \\
\cO(D_6)=\cL_6 = \cO(3,1,1,1) \\
\phantom{xyz} \\
\phantom{xyz}
\end{array}
\begin{array}{l} 
=\cO(L_1 )\\
=\cO(L_1 - E_3 )\\
=\cO(L_1 + L_2 - E_3)  \\
=\cO(L_1 + L_2 -E_1 - E_3)   \\
=\cO(L_1 + L_2 + L_3  -E_1 - E_3)  \\
=\cO(L_1 + L_2 + L_3  -E_1 -E_2 - E_3) \\
=\cO( 3H  -E_1 -E_2 - E_3)\\
=\cO(-K).
\end{array}
\end{array}$$

\[lem.Dn\] Suppose that $m \ge 0$ and $0 \le r \le 5$. Then $$D_{6m+r} = D_{r}-mK.$$

Since $\theta^6=1$, $$\begin{aligned}
\sum_{i=0}^{6m+r-1}\theta^i & =(1+\theta+\cdots+\theta^5)\sum_{j=0}^{m-1}\theta^{6j} + \theta^{6m}(1+\theta + \cdots+\theta^{r-1})
\\
&=(1+\theta + \cdots+\theta^{r-1})  +  m(1+\theta+\cdots+\theta^5)\end{aligned}$$ where the sum $1+\theta+\cdots +\theta^{r-1}$ is empty and therefore equal to zero when $r=0$. Therefore $D_{6m+r}= D_{r}  + mD_6   = D_{r}-mK$, as claimed.

Vanishing results
-----------------

For a divisor $D$ on a smooth surface $X$, we write $$h^i(D):=\dim H^i(X,\cO_{X}(D)).$$ We need to know that $h^1(D)=h^2(D)=0$ for various divisors $D$ on $\BB_3$.

If $D-K$ is ample, then the Kodaira Vanishing Theorem implies that $h^0(K-D)=h^1(K-D)=0$ and Serre duality then gives $h^2(D)=h^1(D)=0$.

The notational conventions in section \[sect.Pic\] identify $\Pic \BB_3$ with $\ZZ^4$ via $$aH-cE_1-bE_2-dE_3 \equiv (a,b,c,d)$$ where $H$ is the strict transform of a line in $\PP^2$. The intersection form on $\BB_3$ is given by $$H^2=1, \qquad E_i.E_j=-\d_{ij}, \qquad H.E_i=0,$$ so the induced intersection form on $\ZZ^4$ is $$(a,b,c,d) \cdot(a',b',c',d')=aa'-bb'-cc'-dd'.$$

\[lem.van\] Let $D=(a,b,c,d) \in \Pic\BB_3\equiv \ZZ^4$. Suppose that $$\label{eq.ample2}
(a+3)^2> (b+1)^2+(c+1)^2+(d+1)^2$$ and $$\label{eq.ample3}
b,\, c,\, d >-1, \qquad \hbox{and} \qquad  a+1>b+c,\, b+d, \, c+d.$$ Then $D-K$ is ample, whence $h^1(D)=h^2(D)=0$.

The effective cone is generated by $L_1$, $L_2$, $L_3$, $E_1$, $E_2$, and $E_3$ so, by the Nakai-Moishezon criterion, $D-K$ is ample if and only if $(D-K)^2>0$ and $D.L_i>0$ and $D.E_i>0$ for all $i$. Now $D-K=(a+3,b+1,c+1,d+1)$, so $(D-K)^2>0$ if and only if (\[eq.ample2\]) holds and $(D-K).D'>0$ for all effective $D'$ if and only if (\[eq.ample3\]) holds.

Hence the hypothesis that (\[eq.ample2\]) and (\[eq.ample3\]) hold implies that $D-K$ is ample. The Kodaira Vanishing Theorem now implies that $h^0(K-D)=h^1(K-D)=0$. Serre duality now implies that $h^2(D)=h^1(D)=0$.

\[lem.h1D\] For all $n \ge 0$, $h^1(D_n)=h^2(D_n)=0$.

The value of $D_n$ for $0 \le n \le 6$ is given explicitly in section \[sect.Dn\]. We also note that $D_7 =
D_1+D_6=(4,2,1,2)$. It is routine to check that conditions (\[eq.ample2\]) and (\[eq.ample3\]) hold for $D=D_n$ when $n=0,2,3,4,5,6,7$. Hence $h^1(D_n)=h^2(D_n)=0$ when $n=0,2,3,4,5,6,7$.

We now consider $D_1$ which is the $-1$-curve $X=0$. (Since $(D_1-K).D_1 = 0$, $D_1-K$ is not ample so we can’t use Kodaira Vanishing as we did for the other small values of $n$.) It follows from the exact sequence $0 \to \cO_{\BB_3} \to \cO_{\BB_3}(D_1) \to \cO_{D_1}(D_1) \to 0$ that $H^p(\BB_3,\cO_{\BB_3}(D_1)) \cong H^p(\BB_3,\cO_{D_1}(D_1))$ for $p=1,2$. However, $\cO_{D_1}(D_1)$ is the normal sheaf on $D_1$ and as $D_1$ can be contracted to a smooth point on the second del Pezzo surface, $\cO_{D_1}(D_1) \cong \cO_{D_1}(-1)$. But $D_1 \cong \PP^1$ so $H^p(\BB_3, \cO_{D_1}(D_1))
\cong H^p(\PP^1,\cO(-1))$ which is zero for $p=1,2$. It follows that $h^1(D_1)=h^2(D_1)=0$.

Thus $h^1(D_n)=h^2(D_n)=0$ when $0 \le n \le 7$. We have also shown that $D_n-K$ is ample when $2 \le n \le 7$. We now argue by induction. Suppose $n \ge 8$ and $D_{n-6}-K$ is ample. Now $D_n-K=D_{n-6}-K -K$. Since a sum of ample divisors is ample, $D_n-K$ is ample. It follows that $h^1(D_n)=h^2(D_n)=0$.

The twisted homogeneous coordinate ring $B(\BB_3,\cL,\s)$
---------------------------------------------------------

We assume the reader is somewhat familiar with the notion of twisted homogeneous coordinate rings. Standard references for that material are [@AV], [@ATV1], [@ATV2], and [@AZ].

The notion of a $\s$-ample line bundle [@AV] plays a key role in the study of twisted homogeneous coordinate rings. Because $\cL_6$ is the anti-canonical bundle and therefore ample, $\cL_1$ is $\s$-ample. This allows us to use the results of Artin and Van den Bergh in [@AV] to conclude that the twisted homogeneous coordinate ring $$B(\BB_3,\cL,\s) = \bigoplus_{n=0}^\infty B_n = \bigoplus_{n=0}^\infty H^0(\BB_3,\cL_n)$$ is such that $$\Qcoh \BB_3 \equiv {{\Gr B}\over{\sT}}$$ where $\sT$ is the full subcategory of $\Gr B$ consisting of those modules $M$ such that for each $m \in M$, $B_nm=0$ for $n \gg 0$. It then follows that $B$ has a host of good properties—see [@AV] for details.

We will now compute the dimensions $h^0(D_n)$ of the homogeneous $B_n$ of $B$. We will show that $B$ has the same Hilbert series as the non-commutative ring $R$, i.e., the same Hilbert series as the weighted polynomial ring with weights 1, 2, and 3. As usual we write $\chi(D) =h^0(D)-h^1(D)+h^2(D)$. The Riemann-Roch formula is $$\chi(\cO(D))=\chi(\cO)+\hbox{${{1}\over{2}}$} D\cdot (D-K)$$ where $K$ denotes the canonical divisor. We have $\chi(\cO_{\BB_3})=1$ and $K_{\BB_3}^2=  6$.

\[lem.h0D\] Suppose $0 \le r \le 5$. Then $$h^0(D_{6m+r})=\begin{cases} (m+1)(3m+r) & \text{if $r \ne 0$,}
\\
3m^2+3m+1 & \text{if $r=0$}
\end{cases}$$ and $$\sum_{n=0}^\infty h^0(D_n) t^n ={{1}\over{(1-t)(1-t^2)(1-t^3)}} .$$

Computations for $1 \le r \le 5$ give $D_{r}^2=r-2$ and $D_{r}\cdot K= -r $. Hence $$\begin{aligned}
\label{T.series}
\chi(D_{6m+r}) &= 1 + \hbox{${{1}\over{2}}$} (D_{r}-mK)\cdot (D_{r}-(m+1) K)
\\
 &= 1 + \hbox{${{1}\over{2}}$} \big(D_r^2-(2m+1)D_r.K +6m(m+1)^2\big)
 \\
 & = (3m+r)(m+1)\end{aligned}$$ for $m \ge 0$ and $1 \le r \le 5$. When $r=0$, $D_r=0$ so $$\chi(D_{6m}) = 3m^2+3m+1.$$ By Lemma \[lem.h1D\], $\chi(D_n)=h^0(D_n)$ for all $n \ge 0$ so it follows from the formula for $\chi(D_n)$ that $$\label{eq.chi.diff}
h^0(D_{n+6})-h^0(D_n)=n+6$$ for all $n \ge 0$.

To complete the proof of the lemma, it suffices to show that $h^0(D_n)$ is the coefficient of $t^n$ in the Taylor series expansion $$f(t):={{1}\over{(1-t)(1-t^2)(1-t^3)}} = \sum_{n=0}^\infty  a_n t^n.$$ Because $$(1-t^6)f(t) = (1-t+t^2)(1-t)^{-2} = 1 + \sum_{n=1}^\infty nt^n,$$ it follows that $$\begin{aligned}
(1-t^6)f(t)  & =  a_0+a_1t+\cdots+a_5t^5 + \sum_{n=0}^\infty (a_{n+6}-a_n)t^{n+6} 
\\
& = 1+t+2t+\cdots+5t^5 + \sum_{n=6}^\infty nt^n
\\
& = 1+t+2t+\cdots+5t^5 + \sum_{n=0}^\infty (n+6)t^{n+6}.\end{aligned}$$ In particular, if $0 \le r \le 5$, $a_r=h^0(D_r)$. We now complete the proof by induction. Suppose we have proved that $a_i=h^0(D_i)$ for $i \le n+5$. By comparing the expressions in the Taylor series we see that $$a_{n+6}= a_n + (n +6)= h^0(D_n)+n+6= h^0(D_{n+6})$$ where the last equality is given by (\[eq.chi.diff\]).

### Remark.

It wasn’t necessary to compute $\chi(D_n)$ in the previous proof. The proof only used the fact that $\chi(D_{n+6})-\chi(D_n)=n+6$ which can be proved directly as follows: $$\begin{aligned}
\chi(D_{n+6})-\chi(D_n)  &=  \hbox{${{1}\over{2}}$} D_{n+6}\cdot(D_{n+6}-K) -  \hbox{${{1}\over{2}}$} D_n\cdot(D_{n}-K) 
\\
& =  \hbox{${{1}\over{2}}$} (D_{n+6}-D_n)\cdot(D_{n+6}+D_n -K)
\\
&=-K\cdot(D_r-(m+1)K)
\\
& = 6(m+1) - K\cdot D_r
\\
&=n+6.\end{aligned}$$

The isomorphism $R \to B(\BB_3,\cL,\s)$
---------------------------------------

The ring $B$ has the following basis elements in the following degrees: $$\begin{array}{ccccccc}
\begin{array}{l}
\cO(1,1,0,1) \\
\cO(1,1,0,0) \\
\cO(2,1,1,1) \\
\cO(2,1,0,1) \\
\cO(3,2,1,1) \\
\cO(3,1,1,1)
\end{array}
\begin{array}{l} 
X \\ 
Xu \\
XYu\\
XYtu\\
XYZtu\\
XYZstu
\end{array}
\begin{array}{l} 
\\ 
Zt \\
YZt\\
YZt^2\\
YZ^2t^2\\
YZ^2st^2
\end{array}
\begin{array}{l} 
\\ 
\\
XZs\\
XZst\\
XZ^2st\\
XZ^2s^2t
\end{array}
\begin{array}{l} 
\\ 
\\
\\
X^2su\\
X^2Zsu\\
X^2Zs^2u
\end{array}
\begin{array}{l} 
\\ 
\\
\\
\\
X^2Yu^2\\
X^2Ysu^2
\end{array}
\begin{array}{l} 
\\ 
\\
\\
\\
\\
XY^2tu^2 \, \; Y^2Zt^2u
\end{array}
\end{array}$$ Although $B$ is a graded subspace of Cox’s homogeneous coordinate ring $S$,

> [ *the multiplication in $B$ is not that in $S$.* ]{}

The multiplication in $B$ is Zhang’s twisted multiplication [@Ztwist] with respect to the automorphism $\tau$ defined in (\[defn.tau\]): the product in $B$ of $a \in B_m$ and $b \in B_n$ is $$\label{eq.Z.tw.mult}
a *_B b := a\tau^m(b).$$ To make it clear whether a product is being calculated in $B$ or $S$ we will write $x$ for $X$ considered as an element of $B$ and $y$ for $Zt$ considered as an element of $B$. Therefore, for example, $$\begin{aligned}
x^5 & =X\tau(X)\tau^2(X)\tau^3(X)\tau^4(X)\tau^5(X) 
\\
&
= XuYtZ
\\
&= (Zt) Y(uX) 
\\
&
=  Zt\tau^2(X)\tau^3(Zt) 
 \\
 & = yxy\end{aligned}$$ and $$y^2=Zt\tau^2(Zt)=Zt(sX) = X(sZ)t =  X\tau(zt)\tau^3(X)=xyx.$$ The following proposition is an immediate consequence of these two calculations.

\[prop.R-&gt;B\] Let $R$ be the free algebra $\CC\langle x,y\rangle$ modulo the relations $x^5=yxy$ and $y^2=xyx$. Then there is a $\CC$-algebra homomorphism $$R=\CC[x,y] \to B(\BB_3,\cL,\s), \qquad x \mapsto X, \; y \mapsto Zt.$$

\[lem.low.degs\] The homomorphism in Proposition \[prop.R-&gt;B\] is an isomorphism in degrees $\le 6$.[^2]

By Proposition \[prop.steph\], $R$ has Hilbert series $(1-t)^{-1}(1-t^2)^{-1}(1-t^3)^{-1}$, so the dimension of $R_n$ in degrees $1,2,3,4,5,6$ is $1,2,3,4,5,7$.

The $n^{\th}$ row in the following table gives a basis for $B_n$, $1 \le n \le 6$. One proceeds down each column by multiplying on the right by $x$. There wasn’t enough room on a single line for $B_6$ so we put the last two entries for $B_6$ on a new line. $$\begin{array}{cccccc}
\begin{array}{l} 
x=X \\ 
x^2=Xu \\
x^3=XYu\\
x^4=XYtu\\
x^5=XYZtu\\
x^6=XYZstu\\
\phantom{x}
\end{array}
\!\!\!\!
\begin{array}{l} 
\\ 
y=Zt \\
yx=YZt\\
yx^2=YZt^2\\
yx^3=YZ^2t^2\\
yx^4=YZ^2st^2\\
\phantom{x}
\end{array}
\!\!
\begin{array}{l} 
\\ 
\\
xy=XZs\\
y^2=XZst\\
y^2x=XZ^2st\\
y^2x^2=XZ^2s^2t\\
\phantom{x}
\end{array}
\!\!
\begin{array}{l} 
\\ 
\\
\\
x^2y=X^2su\\
xy^2=X^2Zsu\\
xy^2x=X^2Zs^2u\\
 yx^2y=Y^2Zt^2u
\end{array}
\!\!
\begin{array}{l} 
\\ 
\\
\\
\\
x^3y=X^2Yu^2\\
x^2y^2=X^2Ysu^2\\
x^4y=XY^2tu^2
\end{array}
\end{array}$$ The products involving $x$ and $y$ were made by using the formula (\[eq.Z.tw.mult\]) in the same way that it was used to show that $x^5=yxy$.

\[lem.gend.gl.sects\] $\cL_2$ is generated by its global sections.

A line bundle $\cL$ on a variety is generated by its global sections if and only if for each point $p$ on the variety there is a section of $\cL$ that does not vanish at $p$. In this case, $H^0(\BB_3,\cL_2)$ is spanned by $Xu$ and $Zt$. One can see from the diagram (\[six.lines\]) that the zero locus of $Xu$ does not meet the zero locus of $Zt$, so the common zero locus of $Xu$ and $Zt$ is empty.

As a $\CC$-algebra, $B$ is generated by $B_1$ and $B_2$.

It follows from the explicit calculations in Lemma \[lem.low.degs\] that the subalgebra of $B$ generated by $B_1$ and $B_2$ contains $B_m$ for all $m \le 6$. It therefore suffices to prove that the twisted multiplication map $B_2 \otimes B_n \to B_{n+2}$ is surjective for all $n \ge 5$.

By definition, $B_2=H^0(\cL_2)$ and this has dimension two so, by Lemma \[lem.gend.gl.sects\], there is an exact sequence $0 \to \cN \to B_2 \otimes \cO_{\BB_3} \to \cL_2 \to 0$ for some line bundle $\cN$. In fact, $\cN \cong \cL_2^{-1}$.

By definition, $\cL_{n+2}=\cL_2 \otimes \cM$ where $\cM \cong \cO(D_{n+2}-D_2)$, and the twisted multiplication map $B_2 \otimes B_n \to B_{n+2}$ is the ordinary multiplication map $$B_2 \otimes H^0(\cM) =  H^0(\cL_2) \otimes H^0(\cM) \to H^0(\cL_2 \otimes \cM).$$

Applying $-\otimes \cM$ to the exact sequence $0 \to \cL_2^{-1} \to B_2 \otimes \cO_{\BB_3} \to \cL_2 \to 0$ and taking cohomology gives an exact sequence $$0 \to H^0(\cL_2^{-1} \otimes \cM) \to B_2 \otimes H^0(\cM) \to H^0(\cL_2 \otimes \cM) \to H^1(\cL_2^{-1} \otimes \cM).$$ Hence, if $h^1(\cL_2^{-1} \otimes \cM)=0$, then $B_2B_n=B_{n+2}$. But $$\cL_2^{-1} \otimes \cM \cong \cO(-D_2 +D_{n+2}-D_2)$$ so we need to show that $D_{6m+r}-2D_2-K$ is ample whenever $6m+r \ge 7$ and $0 \le r \le 5$. By Lemma \[lem.Dn\], $D_{6m+r}=D_r-mK$. We therefore need to check that conditions (\[eq.ample2\]) and (\[eq.ample3\]) in Lemma \[lem.van\] hold for the divisors $D$ in the following table: $$\begin{array}{lll}
 &&  D:=D_r-2D_2 - (m+1)K\\
 r=0   \quad & m \ge 2 \quad & (3m+1,m-1,m+1,m+1) \\
  r=1 \quad & m \ge 1  \quad& (3m+2,m,m+1,m+2) \\
   r=2  \quad & m \ge 1 \quad & (3m+2,m,m+1,m+1) \\
    r=3  \quad & m \ge 1  \quad& (3m+3,m,m+2,m+2) \\
     r=4  \quad & m \ge 1  \quad & (3m+3,m,m+1,m+2) \\
      r=5  \quad & m \ge 1 \quad & (3m+4,m+1,m+2,m+2).
      \end{array}$$ This is a routine though tedious task.

\[thm.isom\] Let $R$ be the free algebra $\CC\langle x,y\rangle$ modulo the relations $x^5=yxy$ and $y^2=xyx$. Then the $\CC$-algebra homomorphism $$\Phi: R=\CC[x,y] \to B(\BB_3,\cL,\s), \qquad x \mapsto X, \; y \mapsto Zt,$$ is an isomorphism of graded algebras.

By Lemma \[lem.low.degs\], $B_1$ and $B_2$ are in the image of $\Phi$. It follows that $\Phi$ is surjective because $B$ is generated by $B_1$ and $B_2$. But $\Phi(R_n) \subset B_n$, and $R$ and $B$ have the same Hilbert series, so $\Phi$ is also surjective.

Consider $R^{(3)}\supset \CC[x^3,xy,yx]$. Since $\dim R_6=7=(\dim R_3)^2 -2$ there is a 2-dimensional space of quadratic relations among the elements $x^3$, $xy$, and $yx$. Hence $R^{(3)}$ is not a 3-dimensional Artin-Schelter regular algebra. The relations in the degree two component of $R^{(3)}$ are generated by $$(x^3)^2=(xy)^2=(yx)^2.$$

[12]{}

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[^1]: The author was supported by the National Science Foundation, Award No. 0602347

[^2]: We will eventually prove that the homomorphism is an isomorphism in all degrees but the low degree cases need to be handled separately.
