The character criterion: full cones and dimension drop

Source and verification

This article explains family 299 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: characters and central sequences

A character of a complex $C^{\ast }$-algebra is a nonzero $\ast$-homomorphism to $\mathbb{C}$. On a unital algebra it preserves the unit. Characters are precisely one-dimensional representations, so the absence of characters says that the algebra has no scalar representation.

For a nonzero unital algebra $A$ and a free ultrafilter $\omega$ on $\mathbb{N}$, its norm ultrapower is

$$\begin{array}{c}{A}_{\omega }={\ell }^{\infty }(\mathbb{N},A)/I_{\omega },\\ I_{\omega }=\{(a_n):\underset{n\to \omega }{\lim }\Vert a_n\Vert =0\}.\end{array}$$

Elements are bounded sequences, with sequences identified when their difference tends to zero in operator norm along $\omega$. The norm central-sequence algebra is

$$F_{\omega }(A)=A_{\omega }\cap A^{\prime },$$

where $A$ is embedded by constant sequences. Thus a representative $(x_n)$ is central when $\Vert x_{n}a-a{x}_n\Vert {\to }_{\omega }0$ for every fixed $a\in A$. These are norm conditions; a tracial ultrapower using the trace $2$-norm would be a different object.

The Jiang–Su algebra $\mathcal{Z}$ is a standard unital infinite-dimensional $C^{\ast }$-algebra used in regularity and classification. An algebra is $\mathcal{Z}$-stable if

$$A\cong A{\otimes }_{\min }\mathcal{Z}.$$

This tensor product uses representations on separate Hilbert-space factors. The maximal tensor product, used in the finite construction below, instead has the universal property for pairs of commuting representations on the same Hilbert space. For general algebras these tensor products can differ.

🏷️ The character criterion

Kirchberg–Rørdam criterion

For every nonzero unital separable complex $C^{\ast }$-algebra $A$ and every free ultrafilter $\omega$, the manuscript states

$$\begin{array}{c}{F}_{\omega }(A)\text{ has no characters}\\ ⟺A\cong A{\otimes }_{\min }\mathcal{Z}.\end{array}$$

No nuclearity, simplicity, trace, dimension or comparison hypothesis is imposed. (OpenAI, 2026)

The criterion turns a representation-theoretic obstruction in the central-sequence algebra into an absorption property of $A$. The finite ingredient is a separate theorem:

A finite tensor construction

If $D$ is any nonzero unital characterless complex $C^{\ast }$-algebra, some finite maximal tensor power admits a unital $\ast$-homomorphism

$$I(2,3)⟶D^{{\otimes }_{\max }m}.$$

Separability is not required, and $m$ may depend on $D$. The map need not be injective.

Here the dimension-drop algebra consists of continuous functions $f:[0,1]\to M_2\otimes M_3$ satisfying

$$\begin{array}{c}f(0)\in M_2\otimes 1_3,\\ f(1)\in 1_2\otimes M_3.\end{array}$$

Its irreducible representations have dimensions two, three or six, so it has no characters and is subhomogeneous: irreducible dimensions have a common finite bound. Any nonzero unital image retains these two properties.

The finite theorem does not assert that a finite tensor power is $\mathcal{Z}$-stable. For separable $D$, the manuscript additionally deduces that the infinite minimal tensor power is $\mathcal{Z}$-stable and contains a unital copy of $\mathcal{Z}$.

🏷️ Positive comparison and the missing defect

A positive element is full if the closed two-sided ideal it generates is the entire algebra. Equivalently, it has nonzero image in every irreducible representation.

For positive elements in matrix algebras over $B$, Cuntz subequivalence $a\precsim b$ means that, after adding zero matrix corners when needed, there are matrices $x_j$ with $x_j^{\ast }b{x}_j\to a$ in norm. Write $[a]\le [b]$ for this comparison. Addition of classes is induced by direct sum; orthogonal positive elements satisfy $[a+b]=[a]+[b]$.

The relevant dimension-drop criterion requires more than two full orthogonal elements. It asks for positive contractions $a_1,a_2$ with exact equivalence

$$\begin{array}{c}{a}_1=v^{\ast }v,a_2=v{v}^{\ast },\\ a_1{a}_2=0,\end{array}$$

and, for some $\epsilon >0$, a comparison of the unused part of the unit:

$$1-a_1-a_2\precsim (a_1-\epsilon {)}_{+}.$$

Here $(a-\epsilon {)}_{+}$ is obtained by applying $t\mapsto \max (t-\epsilon ,0)$ to a positive operator. The Rørdam–Winter implication used in the manuscript then gives a unital map from $I(2,3)$. It requires no stable-rank-one assumption. Mere equality of Cuntz classes is not a substitute for the displayed exact equivalence.

The proof therefore has two analytic tasks: construct a full matrix cone in a finite tensor power, then arrange that its unit defect fits into a positive cutoff of one corner.

🏷️ Finite tests that survive every representation

Absence of characters is witnessed by finitely many commutators. The closed ideal generated by commutators of selfadjoint elements is the whole algebra; otherwise its quotient would be nonzero, unital and commutative, and hence would have a character.

Choose a finite expression in those commutators within distance less than one of the unit. The real span $V$ of their selfadjoint entries is finite-dimensional, and its image contains a noncommuting pair under every unital representation of $D$.

The manuscript then constructs tensor labels that remain independent in all relevant representations. If ${\dim }_{\mathbb{R}}V=q$ and $r$ labels are needed, choose $n$ with

$$2^n>2qn+r-1.$$

There are tensors $x_1,\dots ,x_r\in V^{{\otimes }_{\mathbb{R}}n}$ whose images remain linearly independent after any tensor product of linear maps of rank at least two on the individual legs.

The dimension count is useful to understand. It suffices to consider surjections from $V$ onto ${\mathbb{R}}^2$. Their $n$-tuples have parameter dimension $2qn$; a normalized potential dependence adds $r-1$ parameters. Vanishing of that dependence imposes $2^n$ independent equations on the label choices. The incidence set therefore projects to a measure-zero subset of all labels, by Sard’s theorem. One choice outside it works simultaneously for every tuple of surjections.

Applying this to represented copies of $V$ requires care for maximal tensor products: an irreducible representation need not be a spatial tensor product. The weak closures of its commuting tensor legs are factors, and multiplication is injective on their algebraic tensor product. Real independence of selfadjoint labels also implies complex independence. Together these facts let the labels survive arbitrary irreducible representations, including after an additional tensor factor is attached.

🏷️ A full square-zero element and the matrix cone

Choose a real basis $v_1,\dots ,v_q$ of $V$ and suitable independent labels $x_i$. In a finite tensor power define

$$\begin{array}{c}s=\sum _i{v}_i\otimes x_i\otimes 1,\\ t=\sum _i{v}_i\otimes 1\otimes x_i.\end{array}$$

If their images commuted in an irreducible representation, independence of both label families would force every represented $[v_i,v_j]$ to vanish, contradicting the finite commutator witness.

Set $z=[s,[s,t]]$, which is selfadjoint. Both its positive and negative parts are full. For example, if its image were positive, then for every vector $\xi$ the function

$$f_{\xi }(\lambda )=⟨e^{i\lambda s}t{e}^{-i\lambda s}\xi ,\xi ⟩$$

in that representation would be bounded and concave on all of $\mathbb{R}$, since its second derivative is the negative of the conjugated double commutator. It must be constant, forcing $[s,t]=0$. Negativity gives the same contradiction using convexity.

Fullness of $z_{+}$ and $z_{-}$ supplies finitely many elements $c_i=z_{-}{d}_i{z}_{+}$ whose generated ideal is the whole algebra. Orthogonality gives $c_i{c}_j=0$. Attach another independent label family and set

$$w_0=\sum _i{c}_i\otimes x_i^{\prime }.$$

Then $w_0^2=0$, while independence prevents its image from vanishing in any irreducible representation. Normalize it to a full square-zero contraction $w$ of norm one.

Take its polar decomposition $w=u|w|$ in the bidual. Its initial and final support projections are orthogonal, and the polar partial isometry supplies matrix units $E_{ij}$ on their sum. The positive element

$$h=|w|+|w^{\ast }|$$

commutes with these matrix units. Thus $f\otimes e_{ij}\mapsto f(h)E_{ij}$ defines a homomorphism from $C_0((0,1],M_2)$.

Although the matrix units may only belong to the bidual, the image belongs to the original algebra: each $h{E}_{ij}$ is one of $|w|,|w^{\ast }|,w,w^{\ast }$, and approximation by polynomials vanishing at zero handles every $f$. Unitization gives

$$\eta :C⟶P,$$

where $C$ consists of the continuous $M_2$-valued functions on $[0,1]$ whose value at zero is scalar, and $P$ is a finite maximal tensor power of $D$. The element $\eta (t{e}_{11})=|w|$ is full. No polar partial isometry or support projection is asserted to lie in $P$.

🏷️ A cube construction for the unit defect

Write $b=\eta (t{e}_{11})$. Fullness gives $0<\delta <1$, a cutoff $c=(b-\delta {)}_{+}$, and a finite positive integer $N$ with

$$[1_P]\le N[c].$$

This can be obtained from an exact matrix witness after approximating the unit by a finite sum through $c$. The other cone corner gives an orthogonal, exactly equivalent element $c_2$ with $c+c_2\le 1$.

Choose an integer $L\ge 1$ with $2^L\ge N$, and put

$$\begin{array}{c}R=P^{{\otimes }_{\max }L},\\ e=c^{\otimes L},k=N^L.\end{array}$$

The $2^L$ tensor products choosing $c$ or $c_2$ in each leg are orthogonal and equivalent. Tensoring the fullness comparison gives the two-sided bounds

$$N[e]\le [1_R]\le k[e].$$

The integers $N$ and $k$ have different roles and need not agree.

The nuclear cone algebra $C$ has tensor powers represented as continuous matrix functions on a cube, with the matrix action in a coordinate becoming scalar when that coordinate is zero. A continuous family of embeddings of $M_2$ indexed by the simplex interpolates among the available tensor legs. At a simplex vertex it uses exactly that coordinate; on a face it uses only the positive-weight coordinates. The family is built using a continuous path from the identity to the matrix flip.

In the $k$-cube define

$$\begin{array}{c}g(s)=(1-2s/\delta {)}_{+},\\ H(t)=1-\prod _{j}g(t_j).\end{array}$$

For $t\ne 0$, put $d(t)={\max }_j{t}_j$ and choose weights

$$r_i(t)=\frac{(t_i/d(t)-1/2{)}_{+}}{{\sum }_j(t_j/d(t)-1/2{)}_{+}}.$$

Using the simplex embeddings ${\theta }_r$, set

$$a_i(t)=H(t){\theta }_{r(t)}(e_{ii}),i=1,2.$$

They extend continuously by zero at the origin. They are orthogonal and exactly equivalent through $\sqrt{H(t)}{\theta }_{r(t)}(e_{21})$, and their sum is $H(t)1$. Their defect is the scalar function $X=g^{\otimes k}$.

Let $q(s)=(s-\delta {)}_{+}{e}_{11}$ and place it in one tensor coordinate at a time:

$$y_j=g^{\otimes (j-1)}\otimes q\otimes g^{\otimes (k-j)}.$$

These elements are pairwise orthogonal because $qg=0$. Wherever $y_j$ is nonzero, the $j$th coordinate exceeds $\delta$ and all others are below $\delta /2$. Thus $r(t)$ is exactly the $j$th vertex and $H(t)=1$. It follows that

$$a_1{y}_j=y_j.$$

After applying ${\eta }^{\otimes k}$, retain the same symbols and put $Y={\sum }_j{y}_j$. The comparison $[\eta (g)]\le [1_P]\le N[c]$, tensorized in each coordinate, gives

$$k[X]\le N[Y].$$

The auxiliary algebra $R$ removes these unequal multiplicities without cancellation in Cuntz order:

$$\begin{array}{rl}[X\otimes 1_R]& \le k[X\otimes e]\\ & \le N[Y\otimes e]\\ & \le [Y\otimes 1_R].\end{array}$$

Set $b_i=a_i\otimes 1_R$. Since $a_{1}Y=Y$, functional calculus gives an exact comparison witness:

$$\begin{array}{c}{z}_0=\sqrt{2Y}\otimes 1_R,\\ z_0^{\ast }(b_1-1/2{)}_{+}{z}_0=Y\otimes 1_R.\end{array}$$

Consequently $1-b_1-b_2\precsim (b_1-1/2{)}_{+}$. The pair remains exactly equivalent and orthogonal. This is the required defect comparison.

🏷️ Converting the comparison to dimension drop

The manuscript includes the stable-rank-free Rørdam–Winter argument. First, a cutoff of a Cuntz comparison is realized by an element $v$ with an exact initial square and a final square in the hereditary subalgebra generated by the target. A hereditary subalgebra generated by $b\ge 0$ is $\overline{bBb}$.

This moves the defect into one corner. Two orthogonal copies of the defect are then combined with the original matrix-cone structure. A weighted matrix flip arranges commuting completely positive contractive order-zero maps from $M_3$ and $M_2$, whose units sum to one. Order zero means preservation of orthogonality of positive elements.

Such maps give the usual presentation of $I(2,3)$: at a scalar parameter $r\in (0,1)$ they rescale to commuting copies of $M_3$ and $M_2$, while at the endpoints only the appropriate one remains. Polynomial evaluation at these fibers proves the resulting map is bounded and extends to a unital homomorphism.

The source checks membership in the original algebra whenever bidual support projections or polar decompositions are used. It also retains the scalar fiber at zero in the section algebra, and weights the flip on the region where the two pieces overlap. These checks prevent an unjustified use of a bidual element as an element of the original algebra. The construction also covers zero defect.

🏷️ Applying the finite result to central sequences

If $F_{\omega }(A)$ has no characters, its finite commutator witness lies in a separable subalgebra $D$ containing the same unit and still having no characters.

For each finite $m$, reindexing central sequences produces $m$ commuting unital images of this $D$ inside $F_{\omega }(A)$. At each output index, choose an old coordinate satisfying finitely many algebraic-relation tests and commutator bounds. The latter are legitimate because, at that fixed output index, the other lifted coordinates are fixed elements of $A$. A diagonal choice makes all errors tend to zero in the ordinary limit, and hence along any free ultrafilter.

Injectivity of the reindexed maps is unnecessary. The universal property of the maximal tensor product gives the composite

$$I(2,3)⟶D^{{\otimes }_{\max }m}⟶F_{\omega }(A).$$

Its image is a separable subhomogeneous characterless algebra with the ambient unit. The imported Kirchberg–Rørdam absorption theorem for such a central subalgebra gives $A\cong A{\otimes }_{\min }\mathcal{Z}$. Conversely, absorption supplies a central subalgebra of this kind; a character of $F_{\omega }(A)$ would restrict to a character of it, a contradiction.

For the infinite-tensor consequence, compose the finite map with the canonical unital quotient from maximal to minimal tensor powers and then include that finite stage in $D^{{\otimes }_{\min }\infty }$. Its image remains characterless and subhomogeneous even if the map has a kernel. The imported Dadarlat–Toms theorem gives absorption of this infinite tensor power. Transporting $z\mapsto 1\otimes z$ through the absorption isomorphism yields a unital copy of $\mathcal{Z}$. (OpenAI, 2026)

🏷️ Key steps and proof outlines

Fullness supplies a finite, exact comparison witness

The cone-to-dimension-drop argument starts with a full positive corner $b$. “Full” means the closed two-sided ideal it generates is the entire algebra. Consequently there are finitely many $h_i,{\ell }_i$ such that ${\sum }_i{h}_{i}b{\ell }_i$ is close to the unit. Choose a small cutoff $\delta >0$ so that replacing $b$ by $c=(b-\delta {)}_{+}$ leaves

$$S=\sum _{i=1}^N{h}_{i}c{\ell }_i$$

invertible. The cutoff is essential: later geometric supports need a genuine gap. The source’s “From a cone map with a full corner” section turns this approximation into an exact comparison. Set

$$T=(h_i{c}^{1/2}{)}_{i=1}^N,Z=(c^{1/2}{\ell }_i{)}_{i=1}^N{}^{\mathrm{T}},d=Z^{\ast }Z.$$

Since $S=TZ$, one has $S^{\ast }S\le \Vert T{\Vert }^{2}d$. Invertibility of $S$ forces $d$ to be invertible. The column $V=({\ell }_i{d}^{-1/2}{)}_i$ therefore satisfies

$$V^{\ast }\mathrm{diag}(c,\dots ,c)V=1.$$

This is the concrete witness behind $[1]\le N[c]$ in Cuntz comparison; no cancellation property is being assumed. (OpenAI, 2026)

Continuously move a matrix corner without violating boundary conditions

On the cube model of the tensor cone, a coordinate equal to zero requires its matrix factor to be scalar. A matrix embedding chosen arbitrarily at each point need not respect this condition. The simplex family ${\theta }_r$ solves that compatibility problem explicitly.

For two factors, let $F$ be the flip unitary and use

$$U(s)=\frac{1+F}{2}+e^{i\pi s}\frac{1-F}{2}.$$

Its endpoints are the identity and the flip. For several factors, recursively combine the embedding on the first factors with the last one, using $s=r_k$ and normalized earlier weights $r_i/(1-r_k)$. Although those normalized weights may oscillate when $r_k\to 1$, the estimate

$$\Vert {\theta }_r(a)-1^{\otimes (k-1)}\otimes a\Vert \le 2\Vert a\Vert \Vert U(r_k)-F\Vert$$

is uniform in the earlier embedding and tends to zero. Thus the construction is continuous at the problematic vertex. When a weight is zero, its factor stays scalar by induction. This is why the geometric construction defines an element of the original tensor cone, rather than merely a pointwise matrix formula.

The same membership issue can be checked coefficientwise. Expand a matrix-valued cube section in a basis containing the identity. A coefficient involving nonidentity factors in coordinates $S$ must vanish whenever some $t_i=0$ for $i\in S$. With $p(t)={\prod }_{i\in S}{t}_i$, the functions $f(t)p(t)/(p(t)+\epsilon )$ converge uniformly to that coefficient. Each approximant belongs to the required tensor algebra. This proves the cube model has exactly the stated boundary conditions.

Absorb multiplicities by amplification, rather than cancellation

The two equivalent orthogonal corners provide $2^L$ orthogonal copies of $e=c^{\otimes L}$. Taking $2^L\ge N$ and $k=N^L$ yields

$$N[e]\le [1_R]\le k[e],R=P^{\otimes L}.$$

Meanwhile the cube construction produces a defect $X$ and a positive element $Y$ with $k[X]\le N[Y]$. Tensoring combines the inequalities in the useful direction:

$$[X\otimes 1_R]\le k[X\otimes e]\le N[Y\otimes e]\le [Y\otimes 1_R].$$

Geometric support separation gives $a_{1}Y=Y$, so $Y$ sits entirely where the first corner has value one. Multiplication by $\sqrt{2Y}$ then realizes its comparison to $(a_1-1/2{)}_{+}$ exactly. The resulting defect comparison, together with exact equivalence of the two corners, is the input to the manuscript’s dimension-drop criterion. The later central-sequence and absorption steps use this finite construction as an interface; no finite tensor power is claimed to absorb the Jiang–Su algebra merely from this comparison.

🏷️ Connections

Jiang–Su stability and comparison explains other regularity mechanisms involving comparison and central splitting. Here the initial hypothesis is solely absence of characters in the norm central-sequence algebra. The finite tensor construction supplies the additional dimension-drop structure rather than assuming it.

🏷️ Manuscripts and verification scope

• The Kirchberg–Rørdam character criterion — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The documented selection is the equivalence between absence of nonzero characters on the norm central-sequence algebra and Jiang–Su absorption for every nonzero unital separable complex C*-algebra and every free ultrafilter. No nuclearity, simplicity, trace or comparison assumption is imposed. The finite maximal-tensor dimension-drop construction and infinite minimal-tensor absorption are additional manuscript results; a finite power is not asserted to absorb the Jiang–Su algebra. 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. The Kirchberg–Rørdam character criterion.