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A singular two-dimensional affine toric variety in the normal fan construction must come from a full-dimensional pointed cone; cones of smaller dimension give smooth charts with a torus factor. Let its primitive boundary generators be $v_0,v_{t+1}$. Choose an oriented lattice basis with $v_0=(1,0)$, and write the other generator as $(a,r)$, $r>0$. Primitivity gives $\gcd (a,r)=1$. Changing the second basis vector by a multiple of the first lets us write the cone asHere $r>1$ because determinant one would give a smooth chart. The sublattice $N^{\prime }=\mathbb{Z}(1,0)+\mathbb{Z}(-q,r)$ has index $r$, and its affine chart is ${\mathbb{C}}^2$. The quotient $N/N^{\prime }$ is cyclic, generated by $(0,1)$, whose coordinates in the two boundary generators are $(q/r,1/r)$. Thus the kernel of the corresponding finite map of tori acts on the two affine coordinates as $(x,y)\mapsto ({\zeta }^{q}x,\zeta y)$. More explicitly, invariant monomials are exactly those $x^i{y}^j$ with $qi+j\equiv 0(\mathrm{mod}r)$, which are the elements of the original dual lattice in the first quadrant. Their ring is $\mathbb{C}[{\sigma }^{\vee }\cap M]$. Hence the singularity is the cyclic quotientThe choice of the two boundary axes is unordered; exchanging them replaces $q$ by its inverse modulo $r$.

The Hirzebruch–Jung resolution is computed by the negative Euclidean algorithm. Put $r_0=r$, $r_1=q$ and recursively setThe nonzero remainders strictly decrease, $0\le r_{i+1}<r_i$, and their greatest common divisor stays one; the process terminates with $r_t=1$, $r_{t+1}=0$. All $b_i\ge 2$, and unwinding the recursion gives the negative continued fractionStart with $v_0=(1,0)$ and $v_1=(0,1)$ and define $v_{i+1}=b_i{v}_i-v_{i-1}$. The determinant of every consecutive pair is one, by induction. Moreoveragain by the same recursion and the initial values $r,q$. Each intermediate ray is therefore inside $\sigma$. To check the ordering explicitly, the first coordinates after $v_1$ are negative, the second coordinates positive, and successive determinants are positive. At the last step the determinant with the endpoint is zero. The vector $v_{t+1}$ is primitive, points in the same direction as $(-q,r)$, and hence equals it. Subdivide by all the intermediate rays $v_1,\dots ,v_t$. Every new cone is regular, and the subdivision has the same support, so it gives a proper smooth resolution.

Each internal ray has a compact invariant curve $E_i\cong {\mathbb{P}}^1$, because the quotient fan at it has the two opposite primitive rays. The relation $v_{i-1}+v_{i+1}=b_i{v}_i$ gives its self-intersection $E_i^2=-b_i$. This can also be checked directly: on the chart for $(v_{i-1},v_i)$ use coordinates $(x,y)$ with $E_i=\{y=0\}$; on the next chart the coordinates are $(u,v)=(x^{b_i}y,x^{-1})$, with $E_i=\{u=0\}$. The normal-coordinate transition is that of ${\mathcal{O}}_{{\mathbb{P}}^1}(-b_i)$, proving the self-intersection. Neighboring exceptional curves meet transversely once and nonneighbors do not meet.

Since $b_i\ge 2$, there is no exceptional minus-one curve. Combinatorially, deleting $v_i$ gives a cone of determinant $b_i>1$. Deleting a longer consecutive block cannot give a regular cone either: determinants of successively more distant rays satisfy the same recursion with coefficients at least two and grow from $0,1$ to integers at least two. Thus no proper smooth coarsening of this subdivision exists. This proves the minimality of the minimal resolution of a cyclic quotient surface singularity; equivalently its exceptional chain has no curve that can be blown down to a smooth point. The number of exceptional curves is the length $t$ of the negative continued fraction.

If $t=1$, then $r/q=b_1$ is an integer, so coprimality gives $q=1$. Conversely $q=1$ gives the single coefficient $[r{]}^{-}$ and one curve. If $t=2$, thenThe numerator and denominator on the right are coprime, so $r=b_1{b}_2-1$ and $q=b_2\ge 2$. Therefore $q>1$ and $q\mid r+1$. Conversely, if these conditions hold, then $b_1=(r+1)/q\ge 2$, $b_2=q\ge 2$ give exactly this two-term expansion. Hence the requested equivalences are

Now consider a complete fan in dimension two with three rays and at most one singularity. Its three two-dimensional cones are the consecutive ray pairs. At least two cones are regular, and they share a ray. After an oriented lattice change, take their consecutive generators to be $v_1=(1,0)$, $v_2=(0,1)$ and $v_3=(-1,b)$: the first cone is a basis, and $\det (v_2,v_3)=1$ forces the first coordinate $-1$. Completeness forces $b<0$, because the third pair must span the missing sector with positive determinant $\det (v_3,v_1)=-b$. Write $b=-r$, $r\ge 1$. The unique positive primitive relation is $v_1+r{v}_2+v_3=0$, so the fan is that of the weighted projective plane $\mathbb{P}(1,r,1)$, or $\mathbb{P}(1,1,r)$ after permuting weights. For $r=1$ this is ${\mathbb{P}}^2$; for $r>1$ its remaining cone has type $X_{r,1}$. This proves the full three-ray classificationThese are the complete three-ray toric surfaces with at most one singular point, including the smooth case.

For four rays and exactly the specified singularity, place the three regular cones consecutively. Normalize the middle one to have generators $(1,0),(0,1)$. Its neighboring regular cones then force the cyclic ray listThe remaining cone has determinant $\det (v_4,v_1)=1-ab=r$, so $ab=1-r$. Thus $a,b$ have opposite signs, and $s=|a|$ is a positive divisor of $r-1$. In the oriented lattice basis $v_4,(0,-1)$,so the singular cone has quotient parameter $a\mathrm{mod}r$. It follows that its parameter is congruent to $\pm s$. If identification with the specified singularity exchanges the two boundary rays, the parameter is inverted; this does not change the criterion, since for $st=r-1$ we have $s^{-1}\equiv -t(\mathrm{mod}r)$, and $t$ is the complementary positive factor. The same identity with a minus sign handles $(-s{)}^{-1}$.

Conversely suppose $s\mid r-1$ and $s\equiv \epsilon q(\mathrm{mod}r)$, $\epsilon \in \{1,-1\}$. Choose $a=\epsilon s$ and $b=-\epsilon (r-1)/s$ in the displayed four rays. They are primitive and cyclically ordered, with all four gaps less than a half-plane; the opposite signs of $a,b$ ensure completeness. Three consecutive determinants are one, and the fourth is $r$. Its quotient parameter is $a\equiv q(\mathrm{mod}r)$, so the only singular point has the required type. We have proved the four-ray completion criterion for a cyclic quotient surface singularity:

Let $u,v,w$ be the consecutive primitive toric ray vectors around the compact invariant curve $F=F_v$. Smoothness gives oriented determinants $\det (u,v)=\det (v,w)=1$, hence $u+w=bv$ for an integer $b$. Choose an algebraic torus character $m$ with $⟨m,u⟩=0$ and $⟨m,v⟩=1$. The principal divisor on a toric variety has coefficient one at $F$, coefficient $b$ at $F_w$, and zero at $F_u$. The other invariant curves do not meet $F$, while each adjacent one meets it transversely once. Intersecting this principal divisor with the proper curve $F$ gives $0=F^2+b$. This proves the self-intersection formula for a toric surface divisor even when the ambient surface is not complete.

If $F^2=-1$, then $b=1$ and $v=u+w$. The vectors $u,w$ form a lattice basis since $\det (u,w)=1$, and their toric cone is the union of the two toric cones adjacent to $v$. Deleting $v$ is consequently a smooth toric fan coarsening. Its inverse is precisely the toric blowup at a torus-fixed point, obtained by inserting the sum of the two primitive basis vectors. Conversely, that insertion gives $u+w=v$, and hence algebraic self-intersection $-1$. Thus the toric contraction of a minus-one curve satisfiesFor the cyclic quotient description, first take a strongly convex toric cone of dimension two. Write its primitive boundary vectors as $a,b$ and put $N_0=\mathbb{Z}a+\mathbb{Z}b$, $r=[N:N_0]$. Since $a$ is primitive, extend it to a basis of $N$ and write $b=ua+rc$ with $\gcd (u,r)=1$. Thus $N/N_0$ is cyclic of order $r$. In the lattice $N_0$, the toric cone is smooth and its toric variety is ${\mathbb{A}}^2$. Passing to the larger lattice $N$ takes the finite quotient whose invariant monomials are precisely those in the smaller dual lattice $M=N^{\ast }\subset N_0^{\ast }$. Relative to the two boundary coordinates, $c=(b-ua)/r$, so a generator acts with weights $(-u,1)$ modulo $r$. Replacing that generator by a suitable power makes the first weight one and the second some $q$ with $0<q<r$ and $\gcd (q,r)=1$. Therefore, for the singular case,The cyclic quotient surface singularity can equivalently be described by the first-quadrant toric cone inIndeed its coordinate ring is $k[x^i{y}^j:i,j\ge 0,i+qj\equiv 0(\mathrm{mod}r)]=k[x,y{]}^{{\mu }_r}$. Characteristic zero ensures the stated ordinary finite-group quotient interpretation.

The restriction to singular toric cones of dimension two is necessary for $r>1$. A smooth toric cone has $r=1$ and gives ${\mathbb{A}}^2$ with the trivial quotient. A toric cone of dimension one gives ${\mathbb{A}}^1\times k^{\ast }$, and the zero toric cone gives $(k^{\ast }{)}^2$. These latter surfaces cannot be a quotient of the displayed kind, since they have nonconstant invertible regular functions, whereas $k[x,y{]}^{{\mu }_r}$ has only constant units. Thus the blanket affine-surface wording requires these separate cases.

For the singular case, compute the negative continued fraction by the ceiling form of the Euclidean algorithm. Start with $a_0=r$, $a_1=q$, and, while $a_i>0$, setThen $0\le a_{i+1}<a_i$, so the process terminates at $a_{s+1}=0$. The gcd is unchanged at each step, hence $a_s=1$. Since $a_{i-1}>a_i>0$, every $b_i\ge 2$. Rearranging the recurrence givesNow define lattice vectorsTo verify this Hirzebruch–Jung resolution algorithm, write $v_i=(p_i,a_i)/r$ with $p_0=0$, $p_1=1$ and $p_{i+1}=b_i{p}_i-p_{i-1}$. Induction gives $a_i\equiv q{p}_i(\mathrm{mod}r)$, so $v_i\in L$. It also gives $p_i{a}_{i+1}-a_i{p}_{i+1}=-r$. Thus every adjacent pair has absolute determinant $1/r$, the covolume of $L$, and is a basis of $L$. The $p_i$ increase strictly, while the $a_i$ decrease to zero, so these toric rays occur in order inside the first quadrant. At termination, $a_s=1$ and the determinant identity gives $p_{s+1}=r$, hence $v_{s+1}=(1,0)$. All boundary and inserted vectors are primitive because they occur in lattice bases.

Subdivide the quadrant by $v_1,\dots ,v_s$. Every resulting toric cone is smooth by the smoothness criterion for a toric variety, and the fan subdivision gives a proper birational toric morphism, hence a resolution. Each interior toric ray has a complete one-dimensional star, so its exceptional curve of a surface resolution is ${\mathbb{P}}^1$. The recurrence yieldsThis proves minimality: no exceptional curve of a surface resolution can be contracted to a smooth point by the Castelnuovo contraction criterion. There is also a direct toric fan check. For fixed $i$, the determinants of $v_i$ with $v_{i+h}$, measured in units of the lattice covolume, satisfy the same recurrence with coefficients at least two and initial values zero and one. They are at least $h$. Omitting any intervening toric ray therefore leaves a toric cone of determinant at least two, so no nontrivial coarsening is smooth. Thus this is the minimal resolution of a cyclic quotient surface singularity.

If all exceptional algebraic self-intersections are $-2$, all $b_i=2$ and induction gives $[2,\dots ,2{]}^{-}=(s+1)/s$. Since $r,q$ are coprime, this forces $r=s+1$, $q=s$. Conversely $r/(r-1)$ has exactly $r-1$ coefficients equal to two. ThereforeFor the final toric fan, the minimal smooth fan subdivision of the toric cone from $(0,1)$ to $(r,1)$ inserts the toric rays $(j,1)$ for $1\le j<r$. Adjacent determinants have absolute value one, and $(j-1,1)+(j+1,1)=2(j,1)$, so its inserted curves have algebraic self-intersection $-2$ and it is the minimal fan subdivision just described. The second original toric cone, from $(r,1)$ to $(1,0)$, is already smooth.

Write $v_j=(j,1)$ and $e=(1,0)$. In the resolved toric fan, the rightmost interior toric ray obeys $v_{r-1}+e=v_r$, so its curve has algebraic self-intersection $-1$ and can be blown down to a smooth point. After deleting it, the same relation $v_{r-2}+e=v_{r-1}$ applies to the new rightmost interior toric ray. Continue deleting $v_r,v_{r-1},\dots ,v_1$ in that order. Each deletion is the inverse of a toric blowup at a torus-fixed point. The last toric fan has only the quadrant toric cone and defines ${\mathbb{A}}^2$. ConsequentlyAll maps induce the identity on the dense algebraic torus, so this composite is the same morphism as $X_{{\Sigma }^{\prime }}\to X_{\Sigma }\to {\mathbb{A}}^2$. The intermediate smooth toric fans need not include the singular toric fan $\Sigma$.

Use the boundary-orientation convention in which $-p/q$ surgery on the unknot gives $L(p,q)$. The negative continued fraction isAttach three two-handles to a zero-handle along a chain of unknots, with adjacent linking numbers $+1$ and framings $-3,-2,-2$. Equivalently, take the plumbing of oriented disk bundles over three spheres with Euler numbers $-3,-2,-2$. Successive slam-dunk moves change the boundary surgery coefficient to $-3-1/(-2-1/(-2))=-7/3$, proving that its oriented boundary is $L(7,3)$ in this convention.

The zero sections are a basis of second homology, and the intersection form isIts successive leading principal minors are $-3,5,-7$. Equivalently those of $-Q$ are $3,5,7>0$, so Sylvester's criterion proves that $Q$ is negative definite. This explicitly constructs the required filling.

If lens spaces are instead labeled by positive surgery with its boundary orientation, use the chain with framings $-2,-4$, whose coefficient is $-7/4$; its boundary is orientation-preservingly the positive-surgery $L(7,3)$, since orientation reversal changes $q=4$ to $q=3$. Its negative-definite form is $\left(\begin{array}{cc}-2& 1\\ 1& -4\end{array}\right)$. This is only a convention change in the named boundary, and specifies a valid filling in either labeling system.

Plumbing identifies a base disk times a fiber disk in one oriented rank-two disk bundle with such a neighborhood in another, exchanging base and fiber coordinates. Their zero sections then intersect transversely once. For disk bundles over $S^2$ arranged along a chain with Euler numbers $-a_i$, the intersection form has diagonal entries $-a_i$, adjacent entries one and other entries zero. The same manifold has a Kirby diagram consisting of adjacent Hopf-linked unknots with framings $-a_i$. Its boundary is described by the corresponding negative continued fraction, with the orientation fixed by the surgery convention.