A Counterexample to Kaplansky's Direct-Finiteness Conjecture in Characteristic Two

1. Supply geometry, algebra and size separately

Enumerate a field $Q=\{a_0,\ldots,a_{q-1}\}$ of order $q=2^h$ and define the index simplex $S_d=\{(a_{k_1},\ldots,a_{k_n}):k_i\ge0,\ \sum_i k_i\le d\}$. Set $A=Q^n$, take $O\subset S_{(q-2)/2}$, and write $m=|O|$. Interpolate the indicator values of each hole $o$ by $f_o$, and set $x_v=(f_o(v))_{o\in O}$. Every entry of $x_vx_v^T$ has polynomial degree at most $q-2$, so its sum vanishes both on full lines and on $A$. At holes the values are coordinate vectors; restricting to $V=A\setminus O$ yields $$\sum_{v\in f}x_vx_v^T=0,\qquad\sum_{v\in V}x_vx_v^T=-I_m.$$ The two bounds on the number of holes imply $0<m<t=|V|\le m\ell^2$. Incidence girth protects local subgroup embeddings, the upper bound enables character packing, and $m<t$ detects the reverse defect.

Proposition 2.3 · pp. 5–6 Theorem 3.1 · p. 6

2. Turn existence into a terminating prescription

For $h=2,3,\ldots$, choose the least prime $\ell\equiv-1\pmod{q^{100}}$, the field $K=\mathbb F_{2^{h(\ell-1)}}$ defined by the first irreducible polynomial in the specified order, its subfield $Q$, and the first primitive $\ell$th root. For each $h$, enumerate the finitely many pairs $(O,\mathcal F)$ and test size, full-line containment, regularity and girth. Theorem 5.1 guarantees success for all sufficiently large $h$ and every field enumeration, so the outer search terminates too.

Finite-field linear algebra, finite local multiplication tables, prescribed conjugation relations and finite-group matrix units then determine the group and two finite expressions. There is no need to normalize those expressions by deciding equality of group words. This is a theoretical terminating prescription, not a claim of practical complexity or already computed output.

Section 6; Equation (6.2) · pp. 22–24 [Xyl11, Theorem 1.1, print p. 66 / PDF p. 2]

1. Local embeddings justify averages and the rectangular identity

Put $K_v^0=(\mathbb F_\ell^2,+)$ at each point and $L_f^0=(\mathbb F_\ell,+)$ at each member. Glue incident $L_f^0$ into distinct one-dimensional subspaces of $K_v^0$ to form $N$. Lemma 3.2 protects these local groups by reducing a hypothetical collapse to a weighted planar graph with one exceptional vertex. After minimization, black degrees are at least three, ordinary white degrees at least two, and all but possibly one face have length at least twelve, contradicting Euler's formula. Merely writing a presentation would not supply this embedding guarantee.

Use averages $P_v=\kappa^{-1}\sum_{u\in K_v}u$ in the embedded local groups and put $P=\operatorname{diag}(P_v)$, $U_{-,v}=-\kappa x_vP_v$, $T_{v,-}=P_vx_v^T$. The denominator is invertible because $\ell$ is odd. Partitioning $K_v\setminus\{1\}$ by incident one-dimensional subgroups and using the two pairings gives $$UP=U,\qquad PT=T,\qquad UT=I_m.$$ The factors are still rectangular and do not yet constitute a direct-finiteness counterexample.

Lemma 3.2 · pp. 7–9 Section 4.1; Equation (4.4) · p. 10

2. After character packing, return to the subgroup algebra

Set $H=N\times\langle z\rangle$, $|z|=\ell$, and adjoin stable letters implementing coordinate isomorphisms and character twists to form $L$. The manuscript directly proves $H\hookrightarrow L$ by extending the regular permutation action of $H$. With the central idempotent $e=\ell^{-1}\sum_r\zeta^{-r}z^r$, orthogonal character idempotents pack the $t\le m\kappa$ coordinates into $m$ slots, yielding $J_0C=eP$. Then $$a_{\mathrm{mat}}=UJ_0+(1-e)I_m,\qquad b_{\mathrm{mat}}=CT+(1-e)I_m$$ satisfy $a_{\mathrm{mat}}b_{\mathrm{mat}}=I_m$.

The reverse defect $\Delta=b_{\mathrm{mat}}a_{\mathrm{mat}}-I_m$ is detected through $$J_0\Delta C=e(TU-P)\in\operatorname{Mat}_t(K[H]).$$ Apply $\epsilon:K[H]\to K$, $nz^r\mapsto\zeta^r$. The result is $X^TX-I_t$, where $X=(x_v)$. Since $m<t$, the kernel of $X$ is nonzero, so this matrix is nonzero. The embedding $K[H]\hookrightarrow K[L]$ preserves that fact. No extension of $\epsilon$ to all of $K[L]$ is asserted.

Sections 4.2–4.3; Equations (4.5)–(4.13) · pp. 10–12 Section 4.4; Equations (4.14)–(4.17) · p. 12

3. Complete the identity in a finite-group matrix corner

Character idempotents of $F=(\mathbb Z/\ell)^m\rtimes S_m$ give linearly independent elements $p_{ij}$ with $p_{ij}p_{kr}=\delta_{jk}p_{ir}$. For $G=L\times F$, the map $$\Phi:\operatorname{Mat}_m(K[L])\hookrightarrow K[G],\qquad \Phi(Z)=\sum_{i,j}Z_{ij}p_{ij}$$ is injective and multiplicative, but $\Phi(I_m)=f_{\mathrm{blk}}$ is the corner identity. Set $$a_{\mathrm{out}}=1-f_{\mathrm{blk}}+\Phi(a_{\mathrm{mat}}),\qquad b_{\mathrm{out}}=1-f_{\mathrm{blk}}+\Phi(b_{\mathrm{mat}}).$$ The complement annihilates the image on both sides, so the forward product is $1$ and the reverse defect is $\Phi(\Delta)\ne0$. The construction inverts only the odd number $|(\mathbb Z/\ell)^m|$, never $|S_m|$.

Local zero-sum relators of length at most three give a finite presentation for $N$. Finitely many stable letters and relations, followed by the finite factor $F$, give a finite presentation for $G$. The embeddings $K_v\le N\le H\le L\le G$ retain elements of order $\ell$. Finite presentation is not inferred merely from finite generation of the support subgroup.

Section 4.5; Equations (4.18)–(4.25) · pp. 13–14 Section 4.6 · p. 14

1. Choose uniform parameters and compatible holes

Apply Xylouris to the coprime residue $q^{100}-1$ and modulus $q^{100}$ to obtain $\ell\le Cq^{100L}$ for absolute constants $C>0,L$; Xylouris gives $L=5.18$. Thus $s$, and consequently $n$, have bounds independent of $h$, making the asymptotic estimates uniform. Partition the $D$ directions into bundles of size $q$ and use a Vandermonde construction of spaces $W_r,E_j$ such that any at most ten have direct sum.

Choose holes with small independent probability and delete collisions. The retained holes meet each translate of $W_r$ at most once, have bounded occupancy in relevant cosets, and avoid four simultaneous witnesses for an exceptional rectangle of replacement lines. These three conditions control compatibility of the later trades and the event counts.

[Xyl11, Theorem 1.1, print p. 66 / PDF p. 2] Lemmas 5.2–5.4 · pp. 15–18

2. Exclude short slab polygons, then exchange full lines

For each coset of $U_j=d_j\oplus E_j$, independently choose a two-dimensional space $P_{j,C}\subset E_j$ and partition into three-dimensional slabs. Directness of the direction spaces restricts candidate triangles, quadrilaterals and pentagons to a few types. Each bad event depends on at most four variables, and the sum of event probabilities at any variable is $O(q^{-7})$. The weighted local lemma proved in the manuscript simultaneously eliminates collisions of holes within slabs and the short polygons.

Partition each slab into full lines. At every hole $o$ and bundle $r$, remove the fan of $q$ lines through $o$ and insert $q-1$ parallel lines in $o+W_r$. Every non-hole point either loses and gains exactly one incidence or changes neither; the hole loses all its incidences. Thus the degree on $V$ stays exactly $D$. A short incidence cycle would yield either an excluded generic polygon or a replacement rectangle already forbidden by the hole choice.

Lemmas 5.5–5.7 · pp. 19–21 Lemmas 5.8–5.9; proof of Theorem 5.1 · pp. 21–22

1. The left inverse acts on every configuration

The coefficient supports are finite, so the double sum remains finite even for infinitely supported $x\in K^G$: $T_c(T_dx)(g)=\sum_{u,v}c_ud_vx(guv)=T_{cd}x(g)$. The order is $cd$, not $dc$. With $a=a_{\mathrm{out}}$ and $b=b_{\mathrm{out}}$, we have $T_aT_b=T_{ab}=\mathrm{id}$, hence $T_b$ is injective. The finite field supplies a finite alphabet, and the finite support of $b$ supplies finite memory.

Lemma 7.1 · p. 25

2. A defect coefficient selects a missing point indicator

If $y=T_bx$, then $T_{ba}y=y$; conversely, this equality gives $y=T_b(T_ay)$. Hence $$\operatorname{im}T_b=\{y:T_{ba}y=y\}.$$ Choose a coefficient $(ba-1)_h\ne0$. The point indicator $\delta_h$ satisfies $$(T_{ba}\delta_h-\delta_h)(1_G)=(ba-1)_h\ne0,$$ so it is outside the image. The same image is described by the finite local equations $\sum_u(ba-1)_uy(gu)=0$. This proves nonsurjectivity and non-surjunctivity. It does not claim that a particular nonzero coefficient has actually been computed by executing the search.

Proposition 7.2 · p. 26 Corollary 7.3 · p. 26