Quasitraces: a uniform additivity defect and tensor finiteness

Source and verification

This article explains family 294 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Preliminaries: traces, quasitraces and finiteness

A tracial state on a unital $C^{\ast }$-algebra $C$ is a positive linear functional $\tau$ with $\tau (1)=1$ and $\tau (xy)=\tau (yx)$. It assigns a consistent normalized size to positive operators, including sums of operators that do not commute.

A $1$-quasitrace retains some of these properties while weakening additivity. It is a function $\tau :C\to \mathbb{C}$ satisfying

$$\tau (x^{\ast }x)=\tau (x{x}^{\ast })\ge 0,$$

which is complex linear on every commutative $C^{\ast }$-subalgebra and satisfies $\tau (h+ik)=\tau (h)+i\tau (k)$ for selfadjoint $h,k$. A $2$-quasitrace also has a $1$-quasitrace extension to $M_2(C)$ agreeing with $\tau$ on the upper-left corner.

The normalization is important. If $\tau (1_C)=1$, its corner-preserving matrix extension has value two on the identity of $M_2(C)$. Equivalent diagonal corners have equal value, and distinct diagonal corners commute. A normalized quasitrace on the matrix algebra itself is a different normalization.

A unital algebra is stably finite if every isometry in every matrix algebra over it is unitary. It is properly infinite if it contains two isometries with orthogonal ranges:

$$s_i^{\ast }{s}_j={\delta }_{ij}1(i,j\in \{1,2\}).$$

The statement that no matrix algebra over $C$ is properly infinite is weaker than stable finiteness in general. The manuscript uses the following imported existence theorem: $C$ admits a normalized $2$-quasitrace if and only if none of its matrix algebras is properly infinite.

Kaplansky’s quasitrace question asks whether every $2$-quasitrace is a trace. The matrix-extension condition is part of the question. For exact $C^{\ast }$-algebras, Haagerup’s theorem gives an affirmative answer; exactness means that minimal tensoring preserves short exact sequences.

🏷️ A fixed positive pair with a uniform defect

The manuscript's quasitrace counterexample

There is a separable unital complex $C^{\ast }$-algebra $A$ admitting normalized $2$-quasitraces, and fixed positive contractions $a,b\in A$, such that every normalized $2$-quasitrace satisfies

$$\tau (a+b)-\tau (a)-\tau (b)\ge \frac{1}{144}.$$

In particular, $A$ has no tracial state and is necessarily nonexact. (OpenAI, 2026)

The pair is fixed independently of the quasitrace. The result includes existence of quasitraces, so the universal inequality is not vacuous.

The construction first produces a separable algebra

$$B=C^{\ast }(1,x_1,\dots ,x_{24})$$

with the incompatible column and row identities

$$\begin{array}{c}\sum _{i=1}^{24}{x}_i^{\ast }{x}_i=1,\\ \sum _{i=1}^{24}{x}_i{x}_i^{\ast }\le \frac{2}{3}1.\end{array}$$

A tracial state would assign the same value to the two sums and give $1\le 2/3$. At the same time, a family of finite topological tests prevents every matrix algebra over $B$ from being properly infinite. The existence theorem then supplies quasitraces. Finally, taking $A=M_{48}(B)$ converts the column–row discrepancy into the displayed fixed-pair defect.

There are three distinct tasks: normalize maps to obtain the column–row discrepancy, exclude all finite polynomial candidates for proper infiniteness, and telescope the discrepancy without assuming the additivity being disproved.

🏷️ Continuous matrix scaling from subspace expansion

A finite-rank complex vector bundle is a continuous family of finite-dimensional complex vector spaces, locally identified with a product. A Hermitian metric provides fiberwise adjoints and positive operators. The construction uses maps between bundles over a compact manifold, so their normalization must vary continuously even when no global frame exists.

Suppose $T_1,\dots ,T_N:E\to F$ are continuous bundle maps such that every source subspace $S$ expands collectively:

$$\begin{array}{c}\dim \sum _i{T}_i(y)S\ge 2\dim S\\ (y\in Y).\end{array}$$

The sum means linear span, not a direct sum. The scaling lemma constructs continuous maps $X_i$ with

$$\begin{array}{c}\sum _i{X}_i^{\ast }{X}_i=I_E,\\ \sum _i{X}_i{X}_i^{\ast }\le \frac{2}{3}{I}_F.\end{array}$$

For a positive definite source matrix $P$, write

$$\begin{array}{c}{D}_y(P)=\sum _i{T}_i(y)P{T}_i(y{)}^{\ast },\\ g_y(P)=\\ \log \det (I+D_y(P))-\frac{3}{2}\log \det P.\end{array}$$

The coefficient $3/2$ lies strictly between the source scale one and the expansion factor two. This margin prevents the minimizer from approaching either boundary of the positive cone.

To see this quantitatively, let ${\lambda }_1\ge \cdots \ge {\lambda }_d>0$ be the eigenvalues of $P$. Compactness of the Grassmann bundles and subspace expansion give a uniform $ϵ>0$ such that the $2s$th eigenvalue of $D_y(p)$ is at least $ϵ$ for every rank-$s$ orthogonal projection $p$. Since $P\ge {\lambda }_s{p}_s$ for a top-eigenspace projection $p_s$,

$$\begin{array}{c}\log \det (I+D_y(P))\\ \ge 2\sum _{s=1}^d\log (1+ϵ{\lambda }_s).\end{array}$$

The scalar function $2\log (1+ϵt)-\frac{3}{2}\log t$ tends to $+\infty$ both as $t\downarrow 0$ and as $t\to \infty$. Thus minimizers lie in a common compact interval of positive definite matrices.

Uniqueness comes from strict convexity along curves $P(u)=P_0^{1/2}{e}^{uH}{P}_0^{1/2}$ with nonzero Hermitian $H$. A Gram determinant expansion writes $\det (I+D(P(u)))$ as a positive sum of exponentials. It contains the constant term and a nonzero singleton exponent. The second derivative of the logarithm of this sum is the variance of its exponents under the corresponding normalized positive weights, hence is strictly positive. The term $\log \det P(u)$ is affine.

Compactness and uniqueness make the minimizer continuous in $y$. Because the determinant potential is unchanged by unitary changes of frame, the local minimizers agree on overlaps and define a global bundle endomorphism.

At the minimizer, put $Q=I+D(P)$. Differentiation gives

$$\sum _i{T}_i^{\ast }{Q}^{-1}{T}_i=\frac{3}{2}{P}^{-1}.$$

Therefore

$$X_i=\sqrt{\frac{2}{3}}{Q}^{-1/2}{T}_i{P}^{1/2}$$

has column sum $I_E$ and row sum $\frac{2}{3}(I_F-Q^{-1})\le \frac{2}{3}{I}_F$. The order of factors matters; no commutation of $P$ with the maps $T_i$ is assumed.

🏷️ A finite topological test for each degree

The relevant topological invariant is the Chern class of a complex vector bundle. The $k$th class vanishes when $k$ exceeds the bundle rank. The total class multiplies under direct sums, and a trivial bundle contributes the factor one.

For the hyperplane line bundle on complex projective space ${\mathbb{CP}}^K$, its first class $z$ has $z^K\ne 0$ but $z^{K+1}=0$. Consequently, on a product of projective spaces, the product of these top powers is nonzero. These facts supply a concrete obstruction to a large trivial subbundle. For example, a line bundle with nonzero first Chern class cannot itself be trivial.

Fix integers $m,R\ge 1$, where $m$ is the matrix size and $R$ bounds the polynomial degree. The manuscript constructs bundles $E_0,\dots ,E_R$, with $E_0$ trivial, and

$$d_0=64m{4}^R,\mathrm{rank}{E}_j=4^j{d}_0.$$

At each step, old line-bundle multiplicities are divided by four, and a fresh line-bundle block of rank $\frac{15}{4}\mathrm{rank}{E}_j$ is added. Thus total rank multiplies by four while the cumulative trivial part stays small.

Choose the dimension $K_i$ of the $i$th projective-space factor to equal the total multiplicity of its line bundle in

$$G={\left(\underset{j=0}{\overset{R}{⨁}}{E}_j\right)}^{\oplus m}.$$

Then $G$ has a trivial part of rank

$$h_0=\frac{4}{3}m{d}_0(1-4^{-(R+1)})<2m{d}_0$$

and nontrivial line-bundle blocks of multiplicities $K_i$. With $p={\sum }_i{K}_i$,

$$c_p(G)=\prod _i{z}_i^{K_i}\ne 0.$$

If $G$ contained a trivial subbundle of rank $k>h_0$, its Hermitian complement would have rank $h_0+p-k<p$, but would retain the same nonzero $p$th Chern class. This contradicts vanishing above rank. In particular, $G$ cannot contain a trivial subbundle of rank $2m{d}_0$, regardless of its position relative to the displayed line summands.

The difficult compatibility requirement is to construct expanding maps between successive bundles while keeping their number fixed at 24. A dimension-counting lemma handles this. For source rank $d$, target rank $r$, base real dimension $2h$, and $q$ candidate maps, the inequality

$$(qs-t)(r-t)-s(d-s)>h$$

ensures that a generic tuple sends every $s$-plane to a collective image of dimension greater than $t$. The bad tuples are parametrized by a source $s$-plane and a target $t$-plane; the displayed inequality makes that incidence space lower-dimensional than the space of tuples.

Sixteen block-diagonal maps into the reduced old blocks handle subspaces of dimension at most $m$. A flag argument is needed because a source subspace may have correlated components in several blocks: intersect it with successive block sums, apply expansion on the quotients, and add the quotient dimensions.

Eight additional maps into the fresh block handle larger subspaces. They are constructed over the smaller product of projective spaces used by the current source and target, whose complex dimension is less than $\frac{20}{3}m\mathrm{rank}{E}_j$. Applying the incidence inequality with $t=2s$ gives the required expansion. The scaling lemma then produces the normalized maps. After level $R$, simple doubled bundles continue the construction without changing the finite topological test. (OpenAI, 2026)

🏷️ Excluding every polynomial isometry candidate

At each point $y$ of each test base, form the Hilbert space ${⨁}_{j\ge 0}{E}_j(y)$. The normalized maps define 24 forward shifts. Their norms are at most one, their column sum is the identity, and their row sum is at most $2/3$; on level zero the row sum is zero.

Take these shifts simultaneously in the bounded product over all tests $(m,R)$ and all base points. The algebra generated by the resulting 24 elements is separable, even though the ambient product may be enormous: rational-coefficient $\ast$-polynomials form a countable dense subset.

Suppose $M_m(B)$ contained two isometries with orthogonal ranges. Their concatenation would give a rectangular matrix $W=[s_1{s}_2]$ with $W^{\ast }W=I_{2m}$. Approximate it by a matrix $P$ of $\ast$-polynomials with $\Vert P-W\Vert <1/5$. Then

$$\begin{array}{c}\Vert P^{\ast }P-I\Vert \\ \le 2\Vert P-W\Vert +\Vert P-W{\Vert }^2<\frac{1}{2},\\ P^{\ast }P\ge \frac{1}{2}I.\end{array}$$

Choose $R$ at least the length of every word occurring in $P$. The test for this $(m,R)$ is already present in the product. Applied to level zero, a word of length at most $R$ reaches only levels zero through $R$: a generator raises the level, an adjoint lowers it, and an adjoint at level zero kills the vector.

Thus restricting the same polynomial matrix at every base point yields a continuous bundle map

$$P_0:E_0^{\oplus 2m}⟶G.$$

Compression preserves the lower bound $P_0^{\ast }{P}_0\ge \frac{1}{2}I$. Polar normalization gives a continuous isometric embedding of the trivial bundle $E_0^{\oplus 2m}$ into $G$, contradicting the Chern-class test.

This excludes every matrix size. It also explains the order of quantifiers: all tests are assembled first; a proposed isometry pair determines a finite polynomial approximation; only then does its degree select the test that defeats it.

🏷️ Telescoping without noncommutative additivity

The final quantitative step works for a general tuple of length $N$ with column sum one and row sum at most $c1$, where $c<1$. Set

$$\begin{array}{c}{u}_i=x_i^{\ast }{x}_i,v_i=x_i{x}_i^{\ast },\\ H_i=\sum _{j\le i}{u}_j,J_i=\sum _{j\le i}{v}_j,\\ H_0=J_0=0.\end{array}$$

Define positive contractions in $M_{2N}(B)$ by

$$\begin{array}{rl}a& =\mathrm{diag}(H_0,\dots ,H_{N-1},\\ & 1-J_0,\dots ,1-J_{N-1}),\\ b& =\mathrm{diag}(u_1,\dots ,u_N,\\ & 1-v_1,\dots ,1-v_N).\end{array}$$

Let $\tau$ be any normalized $1$-quasitrace on this matrix algebra, and let $\phi (z)$ be its value on $z$ placed in the first diagonal corner. The square identity gives equal values on equivalent corners and $\phi (u_i)=\phi (v_i)$. Linearity on commuting diagonal corners gives

$$\phi (1)=\frac{1}{2N}.$$

The first $N$ blocks contribute the scalar defects $\phi (H_i)-\phi (H_{i-1})-\phi (u_i)$. The remaining blocks contribute

$$\begin{array}{rl}& \phi (2-J_i)-\phi (1-J_{i-1})\\ & -\phi (1-v_i)\\ & =-\phi (J_i)+\phi (J_{i-1})\\ & +\phi (v_i).\end{array}$$

Each scalar-translation identity takes place in the commutative algebra generated by the identity and one selfadjoint element. It does not assert additivity of $\phi$ on noncommuting positive elements.

Adding the scalar defects telescopes the partial sums and cancels $\phi (u_i)$ against $\phi (v_i)$:

$$\begin{array}{c}\tau (a+b)-\tau (a)-\tau (b)\\ =\phi (1)-\phi (J_N)\\ \ge \frac{1-c}{2N}.\end{array}$$

The last inequality uses positivity and linearity on $C^{\ast }(1,J_N)$, where $0\le J_N\le c1$. Taking $N=24$ and $c=2/3$ gives $1/144$.

Apply the existence theorem directly to $A=M_{48}(B)$: no matrix algebra over $A$ is properly infinite because each is a matrix algebra over $B$. Thus normalized $2$-quasitraces on $A$ exist, and every one satisfies this already-proved $1$-quasitrace inequality.

🏷️ Tensor products and the distinct finiteness conclusions

The minimal, or spatial, tensor product uses representations on separate Hilbert spaces. The manuscript imports a trace-obstruction theorem: if a unital algebra $D$ has no tracial state, then

$$D{\otimes }_{\min }{C}_r^{\ast }({\mathbb{F}}_{\infty })$$

is properly infinite. The subgroup inclusion of the countably generated free group into the two-generator free group gives a unital inclusion of reduced group algebras. Minimal tensoring preserves this inclusion, so the conclusion holds with $C_r^{\ast }({\mathbb{F}}_2)$ as well. A reduced group algebra is the norm closure of the group’s left regular representation.

The resulting consequences involve different algebras:

• The explicit separable algebra $B$ admits normalized $2$-quasitraces, but its tensor product with $C_r^{\ast }({\mathbb{F}}_2)$ is properly infinite and admits none. The construction proves absence of proper infiniteness in every matrix algebra over $B$; it does not assert that $B$ is stably finite.
• An imported equivalence with Kaplansky’s finite-factor question yields a simple stably finite algebra $M$ with no tracial state. There is no separability assertion for $M$. Since $C_r^{\ast }({\mathbb{F}}_2)$ is also simple and stably finite, this gives two unital simple stably finite algebras whose minimal tensor product is properly infinite.
• A finitely generated unital subalgebra of $M$ retains a finite column–row trace obstruction. It is separable, inherits stable finiteness, and has no tracial state. Simplicity is not asserted for this subalgebra.

For the second branch, the imported equivalence produces a type ${\mathrm{II}}_1$ $A{W}^{\ast }$-factor that is not a $W^{\ast }$-algebra. An $A{W}^{\ast }$-algebra has projection-generated annihilators of subsets; the $W^{\ast }$ condition means that it has a Banach-space predual. The proof uses established structural results about these factors, rather than identifying the explicit 24-generator algebra with that factor.

🏷️ Key steps and proof outlines

Put every finite obstruction into one bounded product

For each pair $(m,R)$ of positive integers, the source constructs a bundle test for matrix size $m$ and polynomial word length $R$. Its maps $X_i^{(j)}:E_j\to E_{j+1}$ satisfy a column identity and a row bound. Make them the blocks of forward shifts $S_i$ on ${⨁}_{j\ge 0}{E}_j$.

The column identity implies $X_i^{(j)\ast }{X}_i^{(j)}\le 1$, so every shift has norm at most one, uniformly over all tests and base points. They therefore define one 24-tuple in the bounded product of the corresponding operator algebras. Let $B$ be the unital $C^{\ast }$-algebra this tuple generates. Rational-coefficient $\ast$-polynomials give a countable dense subset, even though the ambient product has many coordinates. (OpenAI, 2026)

The identities hold coordinatewise and hence in $B$:

$$\sum _{i=1}^{24}{x}_i^{\ast }{x}_i=1,\sum _{i=1}^{24}{x}_i{x}_i^{\ast }\le \frac{2}{3}1.$$

A trace would immediately contradict these equations by additivity and $\tau (x_i^{\ast }{x}_i)=\tau (x_i{x}_i^{\ast })$. A quasitrace does not permit that sum, so its existence remains compatible with the relations.

Turn a hypothetical isometry pair into one forbidden bundle map

If $M_m(B)$ were properly infinite, it would contain two isometries $s_1,s_2$ with orthogonal ranges. Their rectangular concatenation $W=[s_1{s}_2]$ satisfies $W^{\ast }W=I_{2m}$. Choose a polynomial matrix $P$ with $\Vert P-W\Vert <1/5$. Then

$$\Vert P^{\ast }P-I\Vert \le 2\Vert P-W\Vert +\Vert P-W{\Vert }^2<\frac{1}{2},P^{\ast }P\ge \frac{1}{2}I.$$

Only after this approximation is fixed, choose $R$ at least its largest word length and evaluate the already constructed $(m,R)$ coordinate. A length-$R$ word applied to level zero reaches only levels zero through $R$: generators raise the level, adjoints lower it, and an adjoint at level zero annihilates it. Thus $P$ restricts to a continuous bundle map

$$P_0:E_0^{\oplus 2m}⟶{\left(\underset{j=0}{\overset{R}{⨁}}{E}_j\right)}^{\oplus m}$$

with $P_0^{\ast }{P}_0\ge I/2$ on every fiber. Polar normalization makes it a continuous isometric bundle embedding, contradicting the finite topological test for that pair $(m,R)$.

This explains why countably many tests suffice for every matrix size and every possible isometry pair. No common polynomial approximation degree is assumed. Nor is it necessary to trivialize the infinitely many Hilbert bundles simultaneously: each tested map uses only finitely many continuous bundle blocks.

Package all failures of additivity into one fixed pair

For a general column-row tuple of length $N$, write $u_i=x_i^{\ast }{x}_i$, $v_i=x_i{x}_i^{\ast }$ and partial sums $H_i,J_i$. The two diagonal matrices specified above place the column increments in their first $N$ corners and complementary row increments in their last $N$ corners.

A normalized quasitrace on $M_{2N}(B)$ gives each equivalent diagonal unit the value $1/(2N)$. Its square identity identifies the values of $u_i$ and $v_i$. Within each diagonal corner, scalar translation is legitimate because it occurs in the commutative algebra generated by one selfadjoint element. After computing these scalar quantities, ordinary addition of real numbers telescopes them:

$$\tau (a+b)-\tau (a)-\tau (b)=\phi (1)-\phi (J_N).$$

Only now use $0\le J_N\le c1$ in the commutative algebra $C^{\ast }(1,J_N)$ to obtain $(1-c)/(2N)$. Thus one pair of positive contractions works for every normalized quasitrace, with a uniform defect $1/144$ when $N=24,c=2/3$. The matrices are chosen from the tuple before choosing the quasitrace.

The algebra construction proves absence of proper infiniteness in every matrix algebra, which is the hypothesis used by the imported existence theorem. It does not prove stable finiteness of this particular explicit algebra. Keeping that distinction is necessary when transferring the result to the separate tensor-product and finite-factor consequences.

🏷️ Connections and verification boundaries

The Jiang–Su stability family uses traces and Cuntz comparison to establish regularity; this family concerns how quasitraces can survive outside the exact setting while full trace additivity fails. The nuclear-ultrapower family also encodes all finite tests in a large operator construction and then uses a proposed approximation to select a contradictory test.

All principal manuscript sections were read, including the scaling proof, bundle tests, polynomial obstruction and fixed-pair calculation. The quasitrace existence criterion, characteristic-class identities, finite-factor equivalence and tensor-product theorem remain imported inputs. The exposition is a source and selected-argument review, not an independent certification of the complete research claim.

🏷️ Manuscripts and verification scope

• A counterexample to Kaplansky’s quasitrace conjecture and failure of tensor-product stable finiteness — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The documented selections include a separable algebra with normalized 2-quasitraces and a fixed positive pair whose real additivity defect is at least 1/144 for every such quasitrace. The linked scope also includes the distinct simple stably finite tensor-product example, separable stably finite traceless example, and loss of normalized 2-quasitraces under a minimal tensor product. These branches have different simplicity and separability hypotheses. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. A counterexample to Kaplansky’s quasitrace conjecture and failure of tensor-product stable finiteness.