Algebraic surface derived McKay in characteristic zero

The two-dimensional diagonal algebra, proved directly. Let (R,m,k0) be an integral two-dimensional local ring essentially of finite type over a field. Let F2→F1→F0 be a finite free complex with all homology finite length and H0=k0. Normalize R in its fraction field, obtaining a finite semilocal normal domain S. Finiteness is the actual theorem thm-integral-closure-finite-finite-type-domain-over-field, followed by localization. Every maximal localization of S above m has dimension two and is S2, hence Cohen–Macaulay. Any parameter sequence of R is therefore regular on S (check at these finitely many maximal localizations). In particular Hom_R(k0,S)=Ext_R^1(k0,S)=0; the usual two short exact sequences for the regular parameters prove this Ext vanishing even without appealing to a finite-module depth convention.

The finite free complex is split exact at every nonmaximal prime. On principal opens covering the punctured spectrum, clear denominators in its contracting homotopies. This produces finitely many identities d $h_j+h_{j}d=a_j^N$ id, for generators $a_j$ of m. They survive tensoring with S, so the homology of F⊗S is m-power torsion. H2 is a torsion submodule of a finite free S-module and is zero. Its boundary B1 is therefore a finite sum of S and has Ext depth at least two. If H1 were nonzero it would have a nonzero m-socle; in 0→B1→Z1→H1→0 the vanishings $\mathrm{Hom}(k0,B1)={\mathrm{Ext}}^1(k0,B1)=0$ would lift this socle to Hom(k0,Z1). But Z1 embeds in a sum of S and has no such socle. Thus H1=0 as well.

For a nonnegative complex U and a nonnegative flat resolution V, the truncation U→H0(U) gives a surjection H1(U⊗V)→H1(H0(U)⊗V): its kernel complex has homology zero in degrees≤0, which remains so after tensoring with the nonnegative flat complex. Apply this to $U=F$ and a free resolution V of S. The left group is H1(F⊗S)=0, so Tor_1^R(k0,S)=0. Since S is FINITE over the local R, this makes S free: lift a basis of S/mS to $R^r\to S$, apply the Tor exact sequence to its kernel, and use finite Nakayama. It has positive rank and is faithfully flat. Exactness of F⊗S therefore descends to F. Now F is a finite free resolution of k0, so the actual Auslander–Buchsbaum–Serre criterion makes R regular. This proves the entire surface diagonal criterion without a big-CM theorem, Frobenius, spreading, or analytic field embedding.

Construct the projective G-Hilbert component over k. Use thm-hilbert-scheme-represents-projective-flat-families over k for length |G| subschemes of M. Intersect the finitely many closed equalizers defining G-invariance. The universal finite-flat section bundle has canonical G-action. Maschke's averaging and its finitely many isotypic projectors split this bundle; the locus where every irreducible representation ρ has multiplicity dim ρ is open and closed. It is the projective G-cluster scheme. A cluster's support is one orbit, because units of distinct orbit components would give more than one invariant section. Its invariant section bundle is the unit line.

Faithfulness gives the nonempty free locus U⊂M. Its quotient U/G is the open of free-orbit clusters, as a functor on families. Let Y be its reduced closure. It is integral projective of dimension two. Invariant functions act on the invariant unit line by scalars; these ring homomorphisms glue to the Hilbert–Chow map τ:Y→X, birational and an isomorphism over U/G. The universal G-cluster is denoted C⊂Y×M.

The quotient's canonical algebra. On a quotient affine chart write $R=S^G$, with S the semilocal smooth cover. Invariant Noether-normalization parameters form a regular local ring A of dimension two over which S is finite. The A-parameters form a regular sequence in the regular localizations of S; Auslander–Buchsbaum makes S finite free over A. Reynolds averaging splits R as an A-direct summand, so R is Cohen–Macaulay. Finite regular-base trace duality identifies $\omega R=(\omega S{)}^G$: apply Hom_A(−,A) to the averaging summand and tensor with ωA. The given invariant volume identifies ωS with S equivariantly, so $\omega R=R$. These identifications glue to $\omega X=OX$. Normality follows directly: an invariant fraction integral over R is integral over normal S and therefore lies in $S^G$.

Actual kernels and adjunction over k. Let πY,πM be the projections from Y×M. Define Φ(A)=RπM*(OC⊗LπYA) with trivial G action on A, and Ψ(B)=[RπY(OC∨⊗πMωM[2]⊗LπMB)]^G. The universal OC is perfect on Y×M: it is finite-flat over Y, and the relative smooth diagonal Koszul resolution of M gives finite locally free amplitude two. Thus these are bounded coherent functors. The normalized projective trace over k, extended by tensoring finite Cech/projective-resolution complexes with an arbitrary affine k-algebra, proves Ψ⊣Φ even while Y is singular. This uses no regularity of the Y base; every k-algebra is flat over k.

We give the finite-complex construction over the possibly singular base that underlies step 7.1 and its convolution. First the universal cluster sheaf $P={\mathcal{O}}_C$ is perfect on $Y\times M$, with a local finite-projective resolution of length two. Indeed, around any closed point of $Y$, shrink to an affine base $\mathrm{Spec}A$ on which the family is contained in an affine open $U=\mathrm{Spec}B$ of $M$: a finite subscheme of a projective scheme is contained in a hyperplane-complement affine open, and properness of $C\to Y$ permits shrinking the base to retain this containment. Its module $N=\mathrm{\Gamma }(C,{\mathcal{O}}_C)$ is finite locally free over $A$. The diagonal $B$-module over $B{\otimes }_{k}B$ has a length-two finite-projective resolution $D_{\bullet }$. To justify that bound, the diagonal of the smooth surface $U$ is a regular immersion of codimension two: on étale coordinates its ideal is generated by the two coordinate differences, whose Koszul complex is exact. Start a finite-free resolution of its finitely presented module; after two steps the last syzygy is locally projective by those local Koszul resolutions, hence is a finite-projective module. Regard $D_{\bullet }\to B$ as a resolution of right $B$-modules via its second factor. Every term is projective over that $B$, since $B{\otimes }_{k}B$ is free as a right $B$-module, and its augmentation module $B$ is projective. Consequently its short exact sequences split as right $B$-modules. Tensoring this resolution on the right with $N$ is therefore still exact and resolves $N$, with the first $B$ acting as the functions on $U$. Its terms are direct summands of finite sums of $B{\otimes }_{k}N$, which are finite-projective over $A{\otimes }_{k}B$ because $N$ is finite-projective over $A$. This proves perfection on the indicated chart; off $C$ the sheaf is zero. These charts cover $C$ and give the assertion globally. This argument uses flatness of the cluster family, not regularity of $A$. Next let $R$ be any Noetherian coordinate algebra of an affine open in $Y$ or $Y\times Y$, and let $L$ be a perfect complex on $M_R=M{\times }_k\mathrm{Spec}R$. We will use the following explicit properties of $T(L)=R\mathrm{\Gamma }(M_R,L)$. Resolve $L$ by finite sums of ample negative twists; the ample global-generation criterion [F4] supplies finite generators at every stage, since the coherent syzygies are on the projective Noetherian scheme $M_R$. The local finite-projective bounds of $L$, bounded uniformly on a finite cover, make a sufficiently late syzygy locally free, so this gives a bounded complex of finite-rank vector bundles representing $L$. Cover $M$ by finitely many affine opens; their finite intersections are affine since $M$ is separated. The total Čech complex of these vector bundles is a bounded complex of flat $R$-modules: on each intersection its section module is finite-projective over $R{\otimes }_k{B}_I$, and $R{\otimes }_k{B}_I$ is free over $R$. It has coherent, bounded cohomology by proper coherence [F4]. These facts imply that $T(L)$ is perfect, with no assumption on regularity of $R$. Here is the local algebra argument: a bounded coherent complex over a Noetherian local ring has a bounded-above resolution by degreewise finite free modules, obtained by choosing finite generators successively downwards; cancel every invertible entry of its differentials to make this resolution minimal. A bounded flat representative gives a uniform finite interval for its derived fibers. In the minimal resolution, tensoring with the residue field makes all differentials zero, so all free terms outside that interval vanish. Thus the resolution is finite; these local finite-free models prove perfection. For every $R$-algebra $R^{\mathrm{\prime }}$, the same bounded flat Čech complex computes derived base change, because its terms tensor to precisely the vector-bundle sections on the base-changed affine intersections. It follows that $T(L){\otimes }_R^{\mathbb{L}}{R}^{\mathrm{\prime }}\cong R\mathrm{\Gamma }(M_{R^{\mathrm{\prime }}},L{\otimes }_R^{\mathbb{L}}{R}^{\mathrm{\prime }})\mathrm{.}$ This also proves the projection formula with a bounded coherent $R$-complex: choose its bounded-above finite-free resolution and use the same bounded flat Čech totalization. The bound on the Čech direction makes the comparison in each total degree finite. We now construct the relative trace rather than apply a regular-base theorem to a singular ring. Tensor the normalized field trace of [F4] with $R$ to obtain $T({\omega }_{M_R\mathrm{/}R})[2]\to R$; the flat Čech description identifies this with base change of the trace on $M$. Evaluation, cup product and that trace define a canonical map $T(L^{\vee }\otimes {\omega }_{M_R\mathrm{/}R})[2]⟶T(L{)}^{\vee }\mathrm{.}$ Both sides are perfect and both commute with derived base change by the preceding construction. It suffices to test the cone at closed points: $R$ is of finite type over the algebraically closed $k$, so their residue fields equal $k$. There the base-changed trace is exactly the normalized surface trace of [F4], and the displayed map is its Serre-duality map. It is an isomorphism for a vector bundle by that supplied pairing, and for a bounded vector-bundle complex by its finite filtration and exact triangles. Hence the cone has zero derived fibers at every closed point; its finite minimal free models at maximal ideals vanish by the same cancellation argument. Its coherent cohomology is therefore zero, so the displayed trace map is an isomorphism over $R$. Its naturality is the actual evaluation-and-trace naturality. Since $\mathrm{\mid }G\mathrm{\mid }$ is invertible, the averaging idempotent takes the equivariant invariant summand in every one of these constructions; invariants are exact, commute with base change, preserve perfect complexes as direct summands, and commute with the dual perfect pairing. Applied on affine opens of $Y$, this gives the actual adjunction $\mathrm{\Psi }⊣\mathrm{\Phi }$ of step 7.1. Explicitly, for $B\in D^b({\mathrm{Coh}}_{G}M)$ choose a bounded equivariant vector-bundle resolution. An ample bundle becomes $G$-linearized after tensoring its finitely many translates; summing the translates of generating sections gives equivariant bundle surjections, and smoothness bounds the last syzygy. Put $L=P\otimes {\pi }_M^{\ast }{B}^{\vee }$ and $H=[R{\pi }_{Y\ast }L{]}^G$, a perfect complex on $Y$. The trace and the dual averaging projector identify $\mathrm{\Psi }(B)=H^{\vee }$. For a bounded coherent complex $A$ on $Y$, evaluation and the projection formula give $R\mathcal{H}o{m}_Y(H^{\vee },A)=H{\otimes }^{\mathbb{L}}A=[R{\pi }_{Y\ast }(L{\otimes }^{\mathbb{L}}{\pi }_Y^{\ast }A){]}^G$. These are sheaf identities, checked by the preceding affine calculation and glued by their canonical evaluation maps. Applying $R\mathrm{\Gamma }(Y,-)$ and the ordinary pullback/pushforward adjunction for ${\pi }_M$ identifies the two sides with $R{\mathrm{Hom}}_Y(\mathrm{\Psi }(B),A)$ and $R{\mathrm{Hom}}_G(B,\mathrm{\Phi }(A))$, respectively. This proves the global left adjunction with its actual invariant summand. Here $\mathrm{\Phi }$ is bounded coherent because $P$ is flat over $Y$ and all projections are projective, and $\mathrm{\Psi }(B)$ is perfect by the finite-complex calculation just given. In particular, no perfection of a residue sheaf on $Y$ has been assumed. Finally put $V=Y\times Y\times M$, let $p:V\to Y\times Y$, and let $P_i$ be the pullback of $P$ by $(y_1,y_2,m)\mapsto (y_i,m)$. Define $S=[R{p}_{\ast }(P_1^{\vee }\otimes P_2){]}^G,Q=[R{p}_{\ast }(P_2^{\vee }\otimes P_1\otimes {\omega }_M)[2]{]}^G\mathrm{.}$ The preceding perfect-pushforward calculation proves that both are perfect on the singular $Y\times Y$, and its relative trace identifies $Q=S^{\vee }$. Bounded Čech totalization and the projection formula identify $Q$, with the first variable the input and the second the output, as the actual kernel of $\mathrm{\Psi }\mathrm{\Phi }$; this is the convolution, not a postulated kernel with only the correct pointwise dimensions. All tensor products in these formulas are derived. For closed points $y_1,y_2$, universal-family flatness makes the pullbacks of $P_i$ the actual cluster sheaves $K_i={\mathcal{O}}_{C_{y_i}}$. Derived base change therefore gives $R{\mathrm{Hom}}_{Y\times Y}(Q,k_{(y_1,y_2)})\cong S{\otimes }^{\mathbb{L}}{k}_{(y_1,y_2)}\cong R{\mathrm{Hom}}_G(K_1,K_2)\mathrm{.}$ The first equality uses perfection and $Q=S^{\vee }$; it is not an underived restriction of a singular-base dual. Likewise for fixed $y$ the derived first-variable pullback $E_y=L{i}_{\{y\}\times Y}^{\ast }Q$ is perfect on $Y$ and equals $\mathrm{\Psi }\mathrm{\Phi }(k_y)=\mathrm{\Psi }(K_y)$, since $P$ is flat over $Y$. Adjunction, or the same perfect base-change calculation, identifies $R{\mathrm{Hom}}_Y(E_y,k_z)$ with $R{\mathrm{Hom}}_G(K_y,K_z)$. This proves precisely the perfection, base-change, and point-Hom assertions needed before $Y$ is smooth.

Off-diagonal orthogonality and diagonal regularity. A G-equivariant map between cluster sheaves sends their invariant cyclic generator 1 to a scalar. If nonzero it is surjective and equal lengths make it an isomorphism, hence the clusters are equal. Thus distinct clusters have equivariant Hom zero. Equivariant surface Serre duality and the invariant form also give Ext_G^2=0 for distinct clusters. Negative and above-two Ext vanish because M is smooth. Hence off the diagonal, point-Hom for Q occurs only in degree one. A minimal finite free model there is therefore one term, a shifted vector bundle. Its support is proper: pairs with different τ-image have disjoint clusters and zero Ext. Indeed support lies in Y×_X Y, whose exceptional product has dimension at most two inside the four-dimensional Y×Y. A vector bundle cannot have proper support on an integral open. Therefore Q vanishes off the diagonal.

Fix a closed point $y\in Y$, put $K_y={\mathcal{O}}_{C_y}$ and $E_y=\mathrm{\Psi }\mathrm{\Phi }(k_y)$. Step 8.1 proves that $E_y$ is perfect before regularity of $Y$ is known. At any closed point $z$, ${\mathrm{Hom}}_Y^i(E_y,k_z)={\mathrm{Ext}}_G^i(K_y,K_z)$ vanishes outside $0\le i\le 2$ since $M$ is a smooth surface and averaging is exact. Over a local ring, cancelling invertible differential entries in a finite-free representative of a perfect complex leaves a finite minimal free complex $F$. The differentials of $\mathrm{Hom}(F,k_z)$ are zero, so the stated point-Hom vanishing forces $F^j=0$ outside $[-2,0]$. If $z\mathrm{\ne }y$, step 9.1 makes all the point-Hom groups zero, so that minimal model is zero. Consequently the coherent cohomology of $E_y$ is supported only at $y$: a nonempty support away from $y$ in a finite-type $k$-scheme would have a closed point away from $y$. The stalks of its cohomology sheaves at $y$ therefore have finite length. Let $ϵ:E_y\to k_y$ be the counit of the actual left adjunction. Its adjoint is ${\mathrm{id}}_{K_y}$ by the triangular identity. An equivariant endomorphism of $K_y$ sends the cyclic invariant generator $1$ to an element of $H^0(K_y{)}^G=k$, so is a scalar. Thus precomposition with $ϵ$ gives an isomorphism $\mathrm{Hom}(k_y,k_y)\to \mathrm{Hom}(E_y,k_y)$, carrying $1$ to ${\mathrm{id}}_{K_y}$; this fixes its normalization. The induced degree-one map is injective, as follows. Naturality and the triangular identity identify the map ${\mathrm{Ext}}_Y^1(k_y,k_y)⟶{\mathrm{Ext}}_Y^1(E_y,k_y)\cong {\mathrm{Ext}}_G^1(K_y,K_y)$ with applying $\mathrm{\Phi }$ to extensions. The first group is the Zariski tangent space ${\mathrm{Hom}}_k({\mathfrak{m}}_y\mathrm{/}{\mathfrak{m}}_y^2,k)$. Its vector is represented by $\mathrm{Spec}D\to Y$, where $D=k[\eta ]\mathrm{/}({\eta }^2)$, and its associated extension of $k_y$ by itself is the pushforward of ${\mathcal{O}}_{\mathrm{Spec}D}$. Since the universal cluster is flat over $Y$, applying $\mathrm{\Phi }$ gives exactly the universal embedded cluster deformation $0\to K_y\to K_D\to K_y\to 0$. If its class is zero, an equivariant ${\mathcal{O}}_M$-linear splitting identifies this extension with the trivial one, respecting its inclusion and quotient. Multiplication by $\eta$ is the composite of quotient and inclusion, so that identification also respects the $D$-action. Thus $K_D\cong K_y{\otimes }_{k}D$ as $G$-equivariant ${\mathcal{O}}_M{\otimes }_{k}D$-modules, with identity reduction. The quotient generator $1$ maps to an invariant global section, hence to a scalar in $D$ reducing to $1$, which is a unit. Multiplication by this unit does not change the annihilator of the generator, so the two embedded quotient ideals are identical. The induced tangent vector of the actual $G$-cluster Hilbert scheme is zero. Since $Y$ is a closed reduced component, its cotangent map from that Hilbert scheme is surjective and its tangent map is injective. The original tangent vector of $Y$ is therefore zero. This proves the claimed degree-one injection without supposing $Y$ smooth. Form the triangle $C\to E_y\stackrel{ϵ}{\to }{k}_y\to C[1]$. Its contravariant long Hom sequence is ${\mathrm{Hom}}^i(k_y,k_y)⟶{\mathrm{Hom}}^i(E_y,k_y)⟶{\mathrm{Hom}}^i(C,k_y)⟶{\mathrm{Hom}}^{i+1}(k_y,k_y)\mathrm{.}$ The groups of negative degree for $k_y,k_y$ vanish, and those for $E_y,k_y$ vanish by the amplitude just established. The degree-zero isomorphism and degree-one injection imply ${\mathrm{Hom}}^i(C,k_y)=0$ for every $i\le 0$. We do not assume $C$ perfect, since the residue-field projective dimension is precisely what is being proved. Over $R={\mathcal{O}}_{Y,y}$ take a bounded-above, degreewise finite free resolution of this bounded coherent cone; finite generators of successive kernels construct it over the Noetherian ring, and it may continue infinitely to the left. Cancel unit entries starting at the top and proceeding downwards. Only finitely many terms occur in each degree, so this gives a bounded-above minimal free complex $L$ without introducing any term to its right. Such a bounded-above projective complex computes derived Hom, and its minimality gives ${\mathrm{Hom}}_R^i(C,k)={\mathrm{Hom}}_R(L^{-i},k)$. The vanishing for $i\le 0$ forces every $L^j$ with $j\ge 0$ to be zero. Hence $H^j(C)=0$ for $j\ge 0$. The cohomology sequence of the triangle now gives $H^0(E_y)=k_y$. The local ring $R$ is integral, two-dimensional and essentially of finite type over the field, because $Y$ is the integral surface of step 5.1 and $y$ is closed. The three-term free complex representing $(E_y{)}_y$ has finite-length homology and degree-zero homology the residue field. Apply exactly the proved criterion in steps 1.1–3.1 to it. That criterion makes $R$ regular and the complex a free resolution of its residue field, so its negative cohomology is zero. Its canonical counit $E_y\to k_y$ is therefore an isomorphism globally, since both have support at $y$. This applies at every closed point. All remaining local rings are localizations of regular closed-point local rings, and finite-type regular schemes over the perfect field $k$ are smooth, so $Y$ is smooth. The argument has supplied a finite-free point complex over each singular local ring before invoking regularity; it has never assumed a finite-free resolution of $k_y$ or of the cone.

The kernel counit Q→OΔ now has a perfect cone whose restriction to every point is zero. Derived Nakayama makes the cone zero, so ΨΦ=id and Φ is fully faithful.

Essential surjectivity and crepancy. The equivariant target has Serre functor [2], by the projective surface trace pairing and exact averaging. After Y is smooth projective the Fourier–Mukai image is admissible. The [2] Serre duality makes its left and right orthogonals equal, so it is an orthogonal direct summand. Db_G(M) is indecomposable: a free-orbit sheaf is simple, every indecomposable $O_M\otimes \rho$ maps nontrivially to it, every finite-length equivariant sheaf receives a map from some $O_M\otimes \rho$, and every indecomposable coherent equivariant sheaf maps nontrivially to a point-orbit quotient. All coherent sheaves therefore lie in one orthogonal summand, and they detect any nonzero bounded complex. Hence the nonzero fully faithful image is the whole target.

The equivalence identifies Serre functors and gives $\omega Y=OY$. X is canonical because the given Z is smooth and crepant: further divisorial discrepancies are nonnegative over smooth Z. The pullback to Y of the invariant quotient volume therefore has effective exceptional divisor. It is a nonzero holomorphic section of $\omega Y=OY$ on connected projective Y and hence is nowhere zero. Thus τ is crepant.

Identify Y with the GIVEN crepant surface resolution, without a factorization theorem. Normalize the graph of Y⇢Z to obtain a projective normal surface T with birational projections p:T→Y and q:T→Z. The shared rational canonical form has strictly positive order at every p-exceptional curve D, and order zero on every nonexceptional divisor. At the generic point of D, normality of T and [F6] make its local ring a DVR, so its order is a divisorial valuation. If u,v are parameters at its center on smooth Y, write $u=t^a$ u0, $v=t^b$ v0 in that valuation, with a,b≥1. Expanding du∧dv shows order at least a+b−1≥1. In characteristic zero a separating residue parameter together with t computes this order of the top differential. If D is nonexceptional, p is an isomorphism at the generic divisor and the order is zero. The identical assertion for q uses the same canonical form, since both resolutions are crepant. Therefore p and q contract exactly the same curves.

The normalization map from $T$ to the graph in $Y\times Z$ is finite by [F1]. If $p$ contracted an integral curve, step 14.1 would make $q$ contract that same curve; its image in the graph would then be a single point. A finite morphism has zero-dimensional fibres, so this is impossible. Thus $p$ contracts no curve. Every positive-dimensional projective fibre contains an integral curve (intersect an irreducible positive-dimensional component by sufficiently many hyperplanes); consequently $p$ is quasi-finite. The identical argument applies to $q$. Both are proper, so [F7] makes them finite. They are birational onto the normal integral surfaces $Y$ and $Z$, so the finite birational normality argument in [F7] makes each an isomorphism. Hence $Y\cong Z$ over $X$, and composing this isomorphism with $\mathrm{\Phi }$ proves the claimed equivalence over $k$.