Exact target

Does Sylvester incidence force a linear presentation?

Fouli, Grate, Kara, LaClair, McCullough, Nguyen, Sobieska, and Udanshive conjecture that, for a simple matroid $M$ with quadratic graded Möbius algebra $A(M)=S/I_M$,

$$M\text{ is Sylvester}\quad\Longleftrightarrow\quad I_M\text{ is linearly presented}.$$

Here Sylvester means that every pair lies in at least one 3-element circuit. The paper proves the forward direction. The open reverse direction is the exact vanishing statement

$$\beta^S_{2,d}(A(M))=0\qquad(d\ge4).$$

Exact inverse-system theorem

Sylvester incidence is precisely avoidance of rank two

Let $R_M=(I_M)_2$ and pair quadratic coefficients. Its orthogonal space has one canonical generator for every rank-two flat:

$$R_M^\perp=\operatorname{span}\{Q_L\},\qquad Q_L=\sum_{\{e,f\}\subset L}z_ez_f.$$

If a pair has a two-point closure, its $Q_L$ has rank two. Conversely, any nonzero coefficient on a Sylvester line gives a three-point principal minor with determinant $2a_L^3\neq0$. Therefore

$$M\text{ is Sylvester}\quad\Longleftrightarrow\quad \mathbb P(R_M^\perp)\cap\{\operatorname{rank}\le2\}=\varnothing.$$

Sharp boundary of the shortcut

The rank gap does not control first syzygies by itself

There is a three-plane of zero-diagonal quadrics in four variables that avoids every rank-one and rank-two quadric, yet its orthogonal quadratic ideal defines an Artinian algebra with

$$\operatorname{HF}_A=(1,4,3),\qquad \boxed{\beta^S_{2,4}(A)=3}.$$

Seven exact cut determinants certify the rank gap, and exact Koszul reduction gives $\dim K_{2,4}=18$ and $\operatorname{rank}d_3=15$. This countermodel is not a graded Möbius algebra: it lacks the partition of pair supports into matroid lines. It rules out a generic Veronese/secant proof, not the conjecture.

The full $PG(3,2)$ supplies the structural positive control. It has the same quadratic line ideal as its rank-three truncation, but exact rational computation gives $\beta_{2,4}=\beta_{2,5}=\beta_{2,6}=0$. The full quadratic ideal removes the truncation's nonlinear spanning block.

Canonical carrier

The Koszul complex splits by the flat spanned by a basis term

The algebra has a basis $y_F$ indexed by flats. A Koszul basis vector $y_F\otimes e_J$ carries the intrinsic label

$$X(F,J)=\operatorname{cl}_M(F\cup J).$$

Every nonzero differential term moves one element of $J$ into $F$, so this closure is unchanged. Thus every internal-degree complex and its second homology split canonically:

$$K_{\bullet,d}(A(M))=\bigoplus_{X\in L(M)}K_{\bullet,d}(M;X),\qquad \beta_{2,d}=\sum_X\dim H_2(K_{\bullet,d}(M;X)).$$

The $X$-block is supported entirely inside the restriction $M|X$. This turns the global conjecture into one functorial spanning-block problem, without choosing coordinates, circuit generators, or an ordering.

The attack closes the functoriality gap as well: both Sylvester incidence and quadraticity pass to every flat restriction. If $X$ has rank $s$, its second-homology block can occur only on the three internal diagonals

$$d\in\{s,s+1,s+2\}.$$

Exact positive result

The symmetric line average really does contract rank two

For a line $L$, two points $a,b\in L$, and any third point $c\in L\setminus\{a,b\}$, the Koszul differential gives

$$d\bigl((y_a+y_b-y_c)\otimes e_a\wedge e_b\wedge e_c\bigr)=2y_L\otimes e_a\wedge e_b.$$

Averaging this identity over every third point produces a canonical primitive, with no chosen point and no regularity assumption on the line size. Thus every rank-two nonlinear block vanishes in characteristic zero.

Exact obstruction carrier

The top diagonal is a pointed-hyperplane triangle complex

For a rank-$r$ restriction $N$, form a graph whose vertices are hyperplane non-incidences $(H,a),\ a\notin H$. A rank-$(r-2)$ flat $F$ with $F\vee a\vee b=E$ gives an edge between $(F\vee a,b)$ and $(F\vee b,a)$. Rank-three extensions of rank-$(r-3)$ flats attach triangular cells.

These are exactly the degree-$r$ Koszul bases and differentials, so

$$H_2(K_{\bullet,r}(N;E))\cong H_1(\mathcal T(N);\Bbbk).$$

This identity isolates the unresolved global invariant. It does not assert that the displayed homology is nonzero under the conjecture's hypotheses.

Stopping boundary

Third points miss the spanning block by one rank

A line-average insertion makes $\{a,b,c\}$ a rank-two triple. But the coefficient flat in $K_{3,r}$ has rank $r-3$. Therefore

$$\operatorname{rk}(G\vee a\vee b\vee c)\le(r-3)+2=r-1<r.$$

Every direct third-point insertion projects to zero on the top diagonal, regardless of rank- or line-size coefficients. The correct triangular cells use independent rank-three extensions, hence global hyperplane and corank-two incidence.

Nor can the proposed T-chord propagation restore the lost information: every Sylvester matroid is already T-chordal. On the paper's designated Sylvester-but-nonquadratic control, the exact rational audit finds

$$\dim H_{2,4}\bigl(T(PG(3,2));E\bigr)=209,$$

entirely in the spanning block. This is not a counterexample: that algebra is not quadratic. It proves that line incidence plus T-chordality cannot substitute for the full algebraic coherence of quadratic generation.

Source audit

No preferred third point is available

The source introduction once says “unique circuit of size 3,” while its formal Section 4 definition, examples, and Murty's Sylvester property require only existence. The distinction matters: every pair of $U_{2,6}$ lies in four 3-circuits.

The proposed attack therefore uses no chosen third point. Its only rank-two operator is the symmetric average over every third point on the line:

$$\eta_{ab}=\frac{1}{|\operatorname{cl}(a,b)|-2}\sum_{c\in\operatorname{cl}(a,b)\setminus\{a,b\}}e_c.$$

Exact calibration over Q

The flat blocks recover both positive and negative controls

A dependency-free exact script constructs the flat lattice and Koszul matrices directly, verifies that the differential preserves the flat label, and checks that block homology sums to global homology.

The last value agrees with the published Betti table. These are carrier checks, not an extrapolation from examples.

Attack result

The generic geometry stops; the line tensor remains

If the graded Möbius algebra is quadratic, its defining ideal is exactly the quadratic line ideal. The next question is therefore whether every finite linear space with lines of size at least three has a linearly presented line ideal, or whether quadratic GMA structure forces one additional incidence identity.

The allowed route is a uniform presentation built from the line-partition tensor. Chosen third points, elimination-order searches, and matroid enumeration do not address this remaining structure. The quadraticity theorem of LaClair et al. remains a formal dependency.

Scope and credit

Exact reductions, one stopped architecture, and no conjecture claim

The conjecture and its proved direction belong to the authors of the 2026 source paper. The algebra presentation and T-chordality theorem belong to LaClair, Mastroeni, McCullough, and Peeva. The flat restriction proof, line primitive, pointed-hyperplane reformulation, rank-gap characterization, and countermodel have not received external review. They neither prove nor refute the source conjecture.