Simultaneous shifted representations remain critical

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Statement and scope

A possible way to disprove cofiniteness was to demand representations of many consecutive targets at once and force a higher-genus or general-type obstruction. The following calculation rules out that particular geometric argument. It is not a point-counting theorem.

Fix $r\ge3$, $L\ge2$, distinct rational numbers $h_1,\ldots,h_L$, and forms in disjoint blocks of $r$ variables

$$F_j(\mathbf x_j)=\sum_{i=1}^r a_{ji}x_{ji}^r, \qquad a_{ji}\in\mathbb Q^\times.$$

Let

$$V:\quad F_j(\mathbf x_j)=n+h_j\quad(1\le j\le L).$$

Eliminating $n$ and homogenizing gives

$$X\subset\mathbb P^{rL}:\quad F_j(\mathbf x_j)-F_1(\mathbf x_1) =(h_j-h_1)z^r\quad(2\le j\le L). \tag{1}$$

Theorem. The projective variety $X$ is geometrically integral, normal, and a complete intersection, with

$$\dim X=L(r-1)+1,\qquad \omega_X\simeq\mathcal O_X(-r-1). \tag{2}$$

The affine chart $V=\{z\ne0\}$ is smooth. Set-theoretically, the singular locus of $X$ consists exactly of points with $z=0$ and at least two entire root blocks zero. It is empty for $L=2$ and has codimension $2r$ for $L\ge3$.

In particular $X$ is normal Gorenstein with ample anticanonical bundle, for every $L$. Positivity of the coefficients is not needed for these geometric statements.

Complete intersection and integrality

Work over an algebraically closed characteristic-zero field $k$. Order the equations in (1) by increasing $j$. Each is monic, up to a nonzero scalar, in the previously unused variable $x_{j,r}$. They form a regular sequence, giving the complete-intersection dimension in (2) and Cohen--Macaulayness.

For integrality introduce $T$. The affine cone ring can equivalently be written

$$R=k[z,T,\mathbf x_1,\ldots,\mathbf x_L]/ (F_j-T-h_jz^r:1\le j\le L).$$

Eliminating $T=F_1-h_1z^r$ recovers the original cone ring. The monic equations in the $x_{j,r}$ make $R$ finite free over

$$k[z,T,(x_{ji})_{i<r}].$$

It is therefore torsion-free over $A=k[z,T]$ and injects into its generic fiber over $K=k(z,T)$. That generic fiber is the tensor product of

$$K[\mathbf x_j]/(F_j-(T+h_jz^r)).$$

Each factor is geometrically integral: over an algebraic closure of $K$, its degree-$r$ homogenization is a smooth diagonal hypersurface in $\mathbb P^r$. Such a hypersurface is irreducible, since distinct positive-degree components would meet and create singular points. The tensor product is thus integral. Its subring $R$ is integral, proving geometric integrality of $X$.

The singular locus and normality

Put $d_j=h_j-h_1$. A dependency among the differentials of (1), with coefficients $\lambda_j$ for $j\ge2$, satisfies

$$\lambda_j\nabla F_j=0,\qquad \left(\sum_{j\ge2}\lambda_j\right)\nabla F_1=0,\qquad z^{r-1}\sum_{j\ge2}\lambda_jd_j=0. \tag{3}$$

The gradient of a diagonal $F_j$ vanishes exactly when its whole block is zero. If at most one block is zero, (3) forces every $\lambda_j=0$. On $z\ne0$, two zero blocks are impossible: they would give $0=(h_j-h_k)z^r$. Thus $V$ is smooth.

When $z=0$, the last condition in (3) disappears. If precisely $t$ blocks are zero, the dependency space has dimension $\max(t-1,0)$. This proves the asserted set-theoretic description. For a prescribed zero pair, all remaining blocks satisfy $F_j=0$. For $L\ge3$ the resulting projective stratum has dimension

$$(L-2)(r-1)-1.$$

Subtracting this from $\dim X$ gives codimension $2r$. For $L=2$ both zero blocks would leave no projective point. The statement is about the reduced singular set; the Jacobian singular subscheme may have nilpotents.

Cohen--Macaulayness gives Serre's condition $S_2$, and the singular codimension gives regularity in codimension one. Serre's criterion therefore proves normality. Complete-intersection adjunction gives

$$\omega_X =\mathcal O_X\bigl((L-1)r-(rL+1)\bigr) =\mathcal O_X(-r-1),$$

as claimed.

What cofiniteness would imply, and what was not proved

Assume now the coefficients are positive, the shifts are integers, and $S_r$ is cofinite. The canonical-core tail bound allows a finite cutoff $Y$ such that targets with a representation using a larger core have upper density less than $1/(4L)$. Targets with fewer than $r$ summands have density zero. It follows that a positive lower proportion of $n\le X$ have, for every $j$, an exactly-$r$-term representation of $n+h_j$ using only cores at most $Y$.

There are only finitely many tuples of forms from this cutoff. Consequently, along an unbounded sequence of $X$, one fixed tuple $(F_1,\ldots,F_L)$ accounts for $\gg X$ distinct such targets. Choosing one representation of each gives $\gg B^r$ positive integer points of $V$ with roots $O(B)$, where $B=X^{1/r}$.

No estimate contradicting this consequence was established. In particular, dimension and adjunction do not yield $o(B^r)$. Each additional shifted equation introduces exactly $r$ new root variables. The archimedean critical counting scale remains

$$B^{rL-r(L-1)}=B^r.$$

For the unit quintic form, even the finite joint system is locally soluble for every local target $n$, by the established local universality.

The argument does not give a gap theorem. It excludes the proposed general-type explanation, while leaving the distribution of distinct integer targets, and hence complement infinitude for $r\ge5$, open.