d214c92 3ff9228 7e25769 b6037d2 bfc8b79
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KAPLANSKY'S QUASITRACE CONJECTURE: A COUNTEREXAMPLE
and a failure of tensor-product stable finiteness

A guided tour, written so that a motivated high-school student can follow
the whole argument. You need vectors, complex numbers, and a little
calculus; everything else is defined as we go.

Author: OpenAI (September 23, 2026)
Rewritten for a general audience (v2).
================================================================================


0. HOW TO READ THIS DOCUMENT
----------------------------

The whole document is climbing toward one object: a separable unital complex
C*-algebra A that admits a "normalized 2-quasitrace" but has no "tracial
state". A is not built out of matrices; it is built out of certain infinite
shifts. The point of the construction is that at every finite test -- at
every matrix size and every word length -- a topological obstruction (a
Chern class) blocks the shift from being an isometry. So the shifts carry a
permanent, uniform defect; and then a general existence theorem harvests a
2-quasitrace that must carry the same defect.

Sections 1-5 give the headline and the basic vocabulary. Sections 6-8
state the three external theorems we need and lay out the game plan.
Section 9 proves a dimension-counting lemma (Lemma 1). Sections 10-11 give
the two bundle lemmas. Section 12 constructs the algebra B. Section 13
proves no matrix algebra over B is properly infinite. Section 14 proves
the defect-pair lemma (Lemma 3); Section 15 assembles the proof of Theorem
1. Section 16 proves Corollary 1. Section 17 is a glossary; Section 18
collects the references.


1. THE HEADLINE RESULT
----------------------

Theorem 1 (main). There exist a separable unital complex C*-algebra A, a
normalized 2-quasitrace on A, and positive contractions a, b in A such
that every normalized 2-quasitrace t on A satisfies

      t(a+b) - t(a) - t(b)  >=  1/144.

Read the inequality as a defect test. The left side is the additivity defect
of t at the pair (a, b); for an honest trace it is always zero. So the
existence of a 2-quasitrace with nonzero defect, at a fixed pair, means
that no normalized 2-quasitrace on A can be a trace. Therefore A has at
least one normalized 2-quasitrace and has no tracial state.

That refutes the quasitrace conjecture: the claim, open since the 1950s,
that every normalized 2-quasitrace on every unital complex C*-algebra is a
trace. It also answers Kaplansky's 1951 AW*-factor question negatively, by
a chain of equivalences recalled in Section 4.

Corollary 1 (tensor products and stable finiteness).

 (i) There exist two unital simple stably finite complex C*-algebras whose
     minimal C*-tensor product is properly infinite. One factor can be
     taken to be C_r*(F_2), the reduced C*-algebra of the free group on
     two generators.

(ii) There exists a separable unital complex C*-algebra B with a normalized
     2-quasitrace such that B (x)_min C_r*(F_2) is properly infinite and
     admits no normalized 2-quasitrace.

(iii) There exists a separable unital stably finite complex C*-algebra
      with no tracial state.

The three parts use related but different algebras; Section 16 says which.
All tensor products are minimal (spatial) C*-tensor products.


2. FROM NUMBERS TO OPERATORS: THE VOCABULARY
--------------------------------------------

2.1 Complex numbers. C = {x + iy : x, y in R}, with i^2 = -1. Write z-bar
for the complex conjugate of z.

2.2 Hilbert space. A Hilbert space H is a complex vector space with a
sesquilinear inner product <u, v> (linear in v, conjugate-linear in u),
symmetric-conjugate <u, v> = <v, u>-bar, satisfying <u, u> > 0 for u != 0,
and complete in the norm ||u|| = sqrt(<u, u>). The finite example we need
is C^n with <u, v> = sum_{j=1}^n (u_j)-bar v_j. The infinite example is
l^2(N) = square-summable sequences.

2.3 Bounded operators. A linear map T : H -> H is bounded if
sup { ||Tv|| : ||v|| <= 1 } is finite; that supremum is the operator norm
||T||. The set B(H) of all bounded operators is an algebra under pointwise
addition and composition. Every T in B(H) has a unique adjoint T* in B(H)
with <u, Tv> = <T*u, v> for all u, v; this pairing defines T*.

2.4 Self-adjoint, positive, projections, isometries. T is self-adjoint when
T = T*; positive when T = S*S for some S; a projection is a self-adjoint
idempotent P = P* = P^2. An isometry is a V in B(H) with V*V = 1 (but
possibly V V* != 1: the range can be a proper subspace of H). Write 1 or
1_H for the identity operator.

2.5 C*-algebra. A C*-algebra is a subalgebra C of some B(H) that is closed
under adjoints and closed in operator norm. It is unital when it contains
1_H; separable when it has a countable norm-dense subset. M_m(C) is the
C*-algebra of m-by-m matrices over C, identified with operators on the
m-fold direct sum of H.


3. TRACES AND TRACIAL STATES
---------------------------

3.1 Linear functionals. A linear functional on a C*-algebra C is a linear
map f : C -> C. It is positive when f(x*x) >= 0 for all x. It is normalized
when f(1) = 1.

3.2 Traces. A trace on C is a linear functional tau with
tau(xy) = tau(yx) for all x, y in C. A tracial state is a trace that is
positive and normalized.

3.3 Why a trace is nontrivial in infinite dimensions. The matrix analogue
tau(A) = (1/n) Tr(A) has the cyclic property tau(AB) = tau(BA) because the
ordinary trace does. In general C*-algebras traces can fail to exist, and
whether they exist is closely tied to the "dimensions" of projections.

3.4 Two relations that break the trace. Suppose x_1, ..., x_N in a unital
C*-algebra C satisfy

      (R1)   sum_{i=1}^N x_i* x_i   =   1,
      (R2)   sum_{i=1}^N x_i  x_i*  <=   c 1,

for some 0 <= c < 1. If tau is a tracial state, applying tau to (R1) and
(R2) and using tau(x_i* x_i) = tau(x_i x_i*) gives

      1  =  sum_i tau(x_i* x_i)  =  sum_i tau(x_i x_i*)
         =  tau( sum_i x_i x_i* )  <=  c tau(1)  =  c,

so 1 <= c, contradicting c < 1. The whole point of the construction is to
produce (R1), (R2) with c = 2/3 in an algebra where no matrix algebra is
properly infinite (Section 5). Milhoj and Rordam posed exactly this: a
uniformly bounded tuple whose matrix matrix amplification is arbitrarily
delayed proper infiniteness [MR, Remark 4.7].


4. QUASITRACES
--------------

4.1 From trace to quasitrace: drop linearity. A 1-quasitrace on a unital
complex C*-algebra C is a function tau : C -> C such that

  (i)   tau(x*x) = tau(xx*) >= 0 for every x in C;
  (ii)  tau(h + ik) = tau(h) + i tau(k) for every pair of self-adjoint
        h, k in C;
  (iii) the restriction of tau to any commutative C*-subalgebra of C is
        complex linear.

Condition (iii) is the weakening: we require linearity only on commuting
elements. The axioms are due to Kaplansky's program; the history is in
[Gow, Introduction]. A 1-quasitrace is normalized when tau(1) = 1.

4.2 2-quasitraces. A 2-quasitrace on C is a 1-quasitrace T on M_2(C)
whose restriction to the upper-left corner matches tau: if iota : C ->
M_2(C) sends z to diag(z, 0), then T(iota(z)) = tau(z) for every z in C.
Requiring the matrix extension is real structure, not a technicality.
Kirchberg (unpublished, 2006) constructed a 1-quasitrace that is not a
2-quasitrace; that example does not answer Kaplansky's question. Our
Theorem 1 produces a genuine 2-quasitrace.

4.3 Normalization on M_2. Suppose tau is a normalized 2-quasitrace on C,
with T the 1-quasitrace on M_2(C). By convention T(diag(1,0)) = 1. By the
square axiom inside M_2(C), T(diag(0,1)) = T(diag(1,0)) = 1. The two
diagonal corner units commute, so by abelian linearity
T(1_{M_2(C)}) = T(diag(1,0)) + T(diag(0,1)) = 2. This is the standard
accounting: a 2-quasitrace is "twice" the 1-quasitrace normalization.

4.4 Kaplansky's question, in two equivalent forms.

Conjecture (Kaplansky, 1951; equivalent form from [MR, Section 2]).
Every normalized 2-quasitrace on every unital complex C*-algebra is a
tracial state.

Equivalently, every AW*-factor of type II_1 is an honest von Neumann
algebra (i.e. has a Banach-space predual). Here an AW*-algebra is a
C*-algebra in which every set of orthogonal projections has a supremum
that is itself a projection; the type classification follows the von
Neumann-algebra pattern. The recent [Gow] paper (arXiv:2601.04431v2)
places these equivalences alongside versions using maximal abelian
subalgebras and bidual retractions, [Gow, Theorem 4.11].


5. PROPERLY INFINITE AND STABLE FINITE
--------------------------------------

5.1 Properly infinite. A unital C*-algebra C is properly infinite when it
contains two isometries s_1, s_2 with orthogonal ranges:

      s_i* s_j  =  delta_ij 1_C,   i, j in {1, 2}.

Think of it as a Hilbert-hotel statement: two isometric copies of the whole
algebra sit inside it, pointing in orthogonal directions. Every matrix
algebra over a properly infinite C*-algebra is itself properly infinite.
A C*-algebra C is stably properly infinite when M_m(C) is properly infinite
for some m.

5.2 Stable finiteness. A projection P in C is finite if P = V*V implies
P = VV* (no proper "shrink" through an isometry). C is finite when 1 is
finite; stably finite when every matrix algebra M_m(C) is finite. In a
unital simple C*-algebra, absence of stable proper infiniteness is
equivalent to stable finiteness [MR, Section 2]. We use only the absence-
of-proper-infiniteness, which is the condition Section 7 needs.

5.3 Why anybody cares. In a tracial state, the value tau(P) on projections
acts like a "dimension function", and dimension functions in turn classify
projections in finite C*-algebras. Stably finite algebras were thought to
be exactly the ones with dimension functions, hence with traces. Corollary
1 (iii) separates those: it produces a stably finite, unital, separable
C*-algebra with no tracial state at all.


6. THREE FACTS FROM THE LITERATURE
----------------------------------

We quote the three external results we need, in forms that fit this
document.

Theorem 2 (Haagerup, exact case). Every 2-quasitrace on a unital exact
C*-algebra is a tracial state [Haagerup, Theorem 5.11]. First draft 1991,
published 2014.

"Exact" is a regularity property of C*-algebras, stable under natural
constructions; nuclear C*-algebras are exact. Theorem 2 is why the algebra
A of Theorem 1 must be non-exact. It also explains why a finite-dimensional
or matrix-based construction cannot possibly work: every finite-dimensional
C*-algebra is exact, so a 2-quasitrace there is always a trace.

Theorem 3 (Haagerup, trace obstructions). For a unital C*-algebra D, the
following are equivalent:

   (i)   D has no tracial state;
  (ii)   there exist finitely many y_1, ..., y_n in D with
           sum_i y_i* y_i = 1  and  || sum_i y_i y_i* || < 1;
 (iii)   D (x)_min C_r*(F_infinity) is properly infinite.

The equivalence (i) <=> (ii) is Haagerup, Lemma 2.1. The equivalence
(i) <=> (iii) is the tensor-product formulation recorded in
[MR, Theorem 3.1]. We use (ii) in Section 16 to deduce Corollary 1(iii),
and (iii) to deduce Corollary 1(i) and (ii).

6.4 From F_infinity to F_2. The group F_infinity (finitely supported words
in infinitely many free generators) embeds in F_2 (free on two generators)
by a_n -> a^n b a^{-n}. The induced *-hom C_r*(F_infinity) -> C_r*(F_2)
is unital and injective [Haagerup, Remark 2.5(a)]. The minimal tensor
product preserves injective *-homs, so for any unital D there is a unital
inclusion

      D (x)_min C_r*(F_infinity)  ->  D (x)_min C_r*(F_2).

The two isometries whose existence (iii) asserts survive this inclusion.
Consequently, for every unital C*-algebra D,

      D has no tracial state   ==>   D (x)_min C_r*(F_2) is properly
                                      infinite.

The upshot: any unital C*-algebra with no tracial state immediately produces
a properly infinite minimal tensor product, with one factor equal to the
fixed algebra C_r*(F_2).


7. THE CUNTZ-BLACKADAR-HANDELMAN EXISTENCE THEOREM
--------------------------------------------------

Theorem 4 (Cuntz-Blackadar-Handelman; [MR, Theorem 2.3]). A unital
C*-algebra C admits a normalized 2-quasitrace if and only if no matrix
algebra M_m(C), m >= 1, is properly infinite.

We use Theorem 4 in the "if" direction: absence of proper infiniteness at
every matrix size forces a normalized 2-quasitrace to exist. The
dimension-function formulation is due to Blackadar and Handelman [BH];
the original projection-theoretic viewpoint is Cuntz [Cuntz1978].
Predecessors are discussed in [BR].

The theorem has no exactness, simplicity, or representation hypothesis.
That is exactly what makes an unexact counterexample possible: Theorem 2
forbids a defect if C is exact, but Theorem 4 permits a 2-quasitrace to
exist even in a badly behaved algebra, as long as no M_m(C) is properly
infinite.


8. THE GAME PLAN
----------------

We want a unital C*-algebra B = C*(1, x_1, ..., x_24) with

      (R1)   sum_{i=1}^{24} x_i* x_i   =   1,
      (R2)   sum_{i=1}^{24} x_i  x_i*  <=  (2/3) 1,
      (H)    M_m(B) is not properly infinite for every m >= 1.

The game plan:

  * From (H) and Theorem 4, B has a normalized 2-quasitrace; so does
    A = M_48(B), since M_k(A) = M_{48k}(B) is not properly infinite either.

  * From (R1), (R2) and the "defect lemma" in Section 14, we get positive
    contractions a, b in A such that every normalized 1-quasitrace on A
    satisfies
         tau(a+b) - tau(a) - tau(b)  >=  (1 - 2/3) / (2 * 24)  =  1/144.
    Every normalized 2-quasitrace on A is in particular a normalized
    1-quasitrace on A, so the defect bound covers all of them.

Thus Theorem 1 follows once B with (R1), (R2), (H) exists.

Why not take B to be matrices? Because, by Theorem 2, matrices are exact,
so any 2-quasitrace there becomes a trace. We need a non-exact B.

Why not have B be any old algebra with (R1), (R2)? Because then B might
easily be properly infinite -- the x_i can themselves be isometries into
orthogonal parts of the identity. Condition (H) is the hard part: no matrix
algebra over B is properly infinite, so the x_i never succeed in being
isometries, at any matrix size.

The construction: for each "test" (m, R) we build a finite-dimensional
Hilbert space and 24 bounded shifts acting on it, satisfying (R1), (R2).
The point is: any m-by-2m matrix W over B with W*W = I_{2m} would, after
approximating by a bounded polynomial P, contradict a Chern class of the
bundle that the shifts generate at (m, R). The bound on R is chosen from
the degree of P; larger R only makes the test stronger. B is then the
C*-algebra generated by all segments of all the shifts, taken in one
bounded product.

Sections 9-11 give the bundle machinery. Section 12 gives B (Theorem 5);
Section 13 proves (H); Sections 14-15 finish Theorem 1.


9. LEMMA 1: DIMENSION COUNTING AVOIDS SMALL COLLECTIVE IMAGES
-------------------------------------------------------------

Lemma 1 (incidence). Let E, F be smooth Hermitian complex vector bundles
of ranks d, r over a compact smooth manifold Y of real dimension 2h. Let
q >= 1 be an integer and let A be a finite set of pairs (s, t) with
1 <= s <= d and 0 <= t <= r, in which each pair satisfies

      (qs - t) (r - t) - s (d - s)   >   h.       (*)

Then there exist smooth bundle maps T_1, ..., T_q : E -> F such that, for
every y in Y, every (s, t) in A, and every complex s-plane S in E_y, the
span

      sum_{i=1}^q T_i(y) S

has complex dimension strictly greater than t. The statement also holds if
Y is a single point; then h = 0.

Idea of the proof: each tuple (T_1, ..., T_q) is a point of a big vector
space V. Fix one pair (s, t) in A. The "bad" tuples are those for which
some s-plane S in some fiber E_y has its image inside a t-plane of F_y.
Parametrize bad tuples by triples (y, S, U, T), where S is an s-plane in
E_y and U is a t-plane in F_y; that family has real dimension

      2h     +  2 s (d - s)  +  2 t (r - t)    [ choices of y, S, U ]
    + 2 M     -  2 q s (r - t),                [ choices of bad tuple T ]

where M = dim_C V is the complex dimension of parameter space. The
condition (*) is precisely the statement that this real dimension is
strictly less than 2M. So the image of the bad set under the projection
to V is a measure-zero subset of V (the same argument as in Sard's
theorem, with a Lipschitz cover of V by small boxes), and cannot exhaust
V. The complement is open and dense, so a generic tuple works for one
pair (s, t). Take a finite intersection over the pairs in A and use that
a finite union of measure-zero sets has measure zero.

The dimension count is the whole content: with more maps q, or with F of
larger rank r, or with a smaller base Y (smaller h), the inequality (*)
is easier to satisfy. All constant choices in Section 11 are made so that
(*) holds in the two regimes we need.

The technical smoothness step, that the "set of bad tuples" is an embedded
manifold and its image under a smooth map is measure zero, is a standard
statement about images of manifolds of strictly smaller real dimension;
see [Sard] for the original reference.


10. LEMMA 2: NORMALIZATION TO COLUMN SUM 1 AND ROW SUM 2/3
---------------------------------------------------------

Lemma 2 (continuous normalization). Let E, F be Hermitian complex vector
bundles of positive ranks d, r over a compact manifold Y. Suppose continuous
bundle maps

      T_1, ..., T_n : E -> F

satisfy the collective expansion

      dim_C  sum_{i=1}^n T_i(y) S   >=  2 dim_C S     (**)

for every y in Y and every complex linear subspace S of E_y. Then there
exist continuous bundle maps X_1, ..., X_n : E -> F such that

      column sum:   sum_{i=1}^n X_i* X_i   =   I_E,
      row sum:      sum_{i=1}^n X_i X_i*   <=  (2/3) I_F.

The constant 2 in (**) is what allows target eigenvalues to be paired in
twos (indices 2s-1, 2s) so that the log det inequality (***) is 2 per
source index; combined with the (3/2) log det P penalty, this gives the
row bound (2/3) I_F. A larger growth factor theta in (**) would also work;
the choice 2 is minimal.

Idea of the proof. In each fiber of E, write P for a positive-definite
endomorphism of E_y; put

      D_y(P) = sum_{i=1}^n T_i(y) P T_i(y)^*     (positive on F_y),
      g_y(P) = log det(I_{F_y} + D_y(P))  -  (3/2) log det P.

The scalar (3/2) is the "penalty" that will enforce row sum 2/3. The
argument has four steps.

(A) A common compact sublevel set. For s in {1, ..., d}, choose a rank-s
projection p_s onto the top s eigenvectors of P. Then P >= lambda_s p_s,
where lambda_s is the s-th eigenvalue of P. Positivity and eigenvalue
monotonicity give D_y(P) >= lambda_s D_y(p_s) in the eigenvalue sense.
Because (**) holds, the 2s-th eigenvalue of D_y(p_s) is bounded below by a
single epsilon > 0 independent of y, s, p_s (the Grassmann bundle of
rank-s projections is compact, and eigenvalues are continuous). Pairing
target eigenvalues 2s-1 and 2s yields the uniform bound

      log det(I + D_y(P))  >=  2 sum_{s=1}^d log(1 + epsilon lambda_s).    (***)

Set f(t) = 2 log(1 + epsilon t) - (3/2) log t on t > 0. Then f is bounded
below, and f(t) -> +infinity as t -> 0+ or t -> +infinity. Combining f on
each eigenvalue with (***) shows that every P with g_y(P) <= C has all
eigenvalues in a common compact interval [a, A] subset (0, infinity). Since
g_y(I_E) <= C trivially for C = max_y g_y(I_E), any minimizer of g_y lies
in that compact set, so a global minimizer P_y exists.

(B) Strict convexity. Fix P_0 > 0 and a nonzero Hermitian H in one fiber.
Along the positive-definite geodesic P(u) = P_0^{1/2} e^{uH} P_0^{1/2}, with
eigenvalues h_j of H and vectors v_{ij} = T_i P_0^{1/2} e_j,

      D_y(P(u))  =  sum_{i,j}  e^{u h_j} v_{ij} v_{ij}^*.

Expanding det(I + D) as a sum over row subsets gives
det(I + D_y(P(u))) = sum_alpha c_alpha e^{u alpha} with c_alpha >= 0. For
each j the total coefficient of e^{u h_j} is positive (because P_0^{1/2} e_j
is nonzero and (**) makes the common kernel of the T_i trivial). Since some
h_j != 0, there are positive exponentials at two distinct exponents
(0 and some h_j). Hence log det(I + D_y(P(u))) is the log of a positive sum
of exponentials at two distinct exponents, whose second derivative is the
variance of those exponents under positive weights and is strictly positive.
Meanwhile log det P(u) = log det P_0 + u tr H is affine in u. Therefore
g_y(P(u)) is strictly convex.

(C) Uniqueness and continuity. Strict convexity forces uniqueness of the
minimizer P_y. For continuity, observe that the minimizers lie in the common
compact matrix set aI <= P <= AI from step (A). If y_k -> y and P_{y_k} is
the minimizer at y_k, any subsequential limit P_infinity satisfies
g_y(P_infinity) <= g_y(P') for every positive-definite comparison P'
(simply pass to the limit in the minimizing inequality evaluated against a
fixed P'). Uniqueness forces P_infinity = P_y, and therefore every
subsequence converges to P_y. Working in local unitary frames and using
uniqueness on the overlap, the local minimizers glue to a continuous positive
definite endomorphism P of E.

(D) Normalization. Let Q = I_F + D(P). Differentiating g_y at the minimizer
in an arbitrary Hermitian source direction K gives

      0  =  tr[ ( sum_i T_i* Q^{-1} T_i  -  (3/2) P^{-1} ) K ].

The Hermitian pairing tr(MK) is nondegenerate, so

      sum_i T_i* Q^{-1} T_i   =   (3/2) P^{-1}.

Define

      X_i  =  sqrt(2/3)  Q^{-1/2} T_i P^{1/2}.

That is a continuous bundle map E -> F by the continuous functional
calculus on Q and P. The order of the factors gives

      sum_i X_i* X_i  =  (2/3) P^{1/2}  (sum_i T_i* Q^{-1} T_i)  P^{1/2}
                      =  (2/3) P^{1/2}  (3/2) P^{-1}  P^{1/2}
                      =  I_E,
      sum_i X_i X_i*  =  (2/3) Q^{-1/2} D(P) Q^{-1/2}
                      =  (2/3) ( I_F - Q^{-1} )
                      <= (2/3) I_F.

The last equality uses Q = I_F + D(P), and the order is
Q^{-1/2} D(P) Q^{-1/2} = Q^{-1/2}(Q - I)Q^{-1/2} = I - Q^{-1}.

Hence the X_i satisfy the lemma.

Remarks on the proof. The existence of a unique minimizer is what makes
the construction canonical. If the minimizer depended on the trivialization
of E, or on a partition of unity, the continuity assertion (C) would fail
across overlaps and the construction of the X_i as bundle maps would break.
As it stands, the normalization X_i is determined by the T_i alone, so the
X_i are genuine bundle maps over Y.

Number-theoretic note: the conclusion "row sum <= (2/3) I_F" is exactly
the bound required by (R2) once the shifts are read off the X_i. The
expansion factor 2 in (**) is what we produce in Proposition 1 from the
16 + 8 map count; the resulting (2/3) row bound is the constant used in
the algebra of Section 12.


11. PROPOSITION 1: BUNDLE TESTS AT MATRIX SIZE m AND DEGREE R
-------------------------------------------------------------

Fix integers m >= 1 and R >= 1. Proposition 1 builds a compact base Y and
Hermitian vector bundles E_0, ..., E_R over Y, together with 24 continuous
maps between successive bundles, such that:

  (T1) each bundle is a sum of powers of line bundles;
  (T2) the 24 maps satisfy the scaled identities of Lemma 2 at each level;
  (T3) the sum  E_0 + E_1 + ... + E_R  retains a nonzero top Chern class
       after allowing trivially-many  E_0  summands.

The Chern class will be what obstructs  m  by  2m  matrices over the
generated algebra.

Proposition 1. For every pair (m, R) of positive integers there are a
compact smooth manifold Y = Y_{m,R} and Hermitian vector bundles
E_0, E_1, ..., E_{R+1}, ... over Y, with E_0 trivial of rank d_0 and
rank E_j = d_j for 0 <= j <= R, and continuous bundle maps

      X_i^{(j)}  :  E_j  ->  E_{j+1},        i = 1, ..., 24,  j >= 0,

such that:

   (a)  sum_{i=1}^{24} (X_i^{(j)})* X_i^{(j)}  =  I_{E_j}             (column)
        sum_{i=1}^{24} X_i^{(j)} (X_i^{(j)})*  <= (2/3) I_{E_{j+1}}   (row)

   (b)  ( E_0 + E_1 + ... + E_R )^{m}  contains no trivial subbundle of
        rank  2 m  d_0.

Here  d_0 = 64 m 4^R  and  d_j = 4^j d_0.

Construction.

Let us define everything in advance. For 1 <= i <= R put

      b_i  =  (15/4)  d_{i-1}                        (a positive integer).

For 0 <= i <= j <= R put

      a_{0,j}  =  d_0 4^{-j}  =  64 m  4^{R-j},
      a_{i,j}  =  b_i 4^{-(j-i)}  =  960 m  4^{R + 2i - j - 2}    (i >= 1).

Interpret  a_{i,j}  as "the multiplicity of the line bundle L_i in bundle
E_j". These are positive integers, and a_{i,j} is divisible by 4 whenever
i <= j <= R-1 (needed for the map construction below); they are also at
least 256 m in that range.

The base. Put

      K_i  =  m  sum_{j=i}^R  a_{i,j}  =  320 m^2  4^{2i-2} (4^{R-i+1} - 1).
      Y    =  product_{i=1}^R  CP^{K_i}.

These are positive integers, and K_i is exactly the total multiplicity of
the line bundle L_i in the bundle G to be tested. The complex dimension of
Y is eta = sum_i K_i.

The line bundles. Let L_0 be the trivial line bundle on Y. For i >= 1, let
L_i be the pullback of the hyperplane line bundle under the projection Y ->
CP^{K_i} onto the i-th factor. The integral cohomology ring is

      H^*(Y; Z)  =  Z[z_1, ..., z_R]  /  ( z_1^{K_1+1}, ..., z_R^{K_R+1} ),

with  z_i = c_1(L_i)  of degree 2. In particular, the monomial
product_{i=1}^R  z_i^{K_i}  is nonzero: it is the top-degree monomial
in the polynomial ring modulo ideal, and no factorization can zero it out
because top degree forces every exponent to be at its bound K_i.

The bundles. Put

      E_j  =  direct sum_{i=0}^j  L_i^{a_{i,j}},       0 <= j <= R.

Step rank recursion. By definition,

      rank E_{j+1}  =  sum_{i=0}^{j+1} a_{i,j+1}
                    =  a_{0,j+1} + sum_{i=1}^j a_{i,j+1} + a_{j+1,j+1}.

Compute each piece:  a_{0,j+1} = (1/4) d_0 4^{-j}? Not exactly; use the
multiplicities:

      a_{0,j+1}     =  (1/4) a_{0,j},                ( = (1/4) d_0 4^{-j} )
      a_{i,j+1}     =  (1/4) a_{i,j}      (1 <= i <= j),
      a_{j+1,j+1}   =  b_{j+1} = (15/4) d_j.

Therefore

      rank E_{j+1}  =  (1/4) (a_{0,j} + sum_{i=1}^j a_{i,j}) + (15/4) d_j
                    =  (1/4) d_j + (15/4) d_j
                    =  4 d_j.

Hence  d_{j+1} = 4 d_j  as claimed, by induction.

The retained Chern class. Form

      G  =  ( E_0 + E_1 + ... + E_R )^{m}.

Since E_0 is trivial of rank d_0, G contains a trivial summand of rank

      h_0  =  m  sum_{j=0}^R a_{0,j}
           =  m  sum_{j=0}^R 64 m  4^{R-j}
           =  (4/3) m  d_0 ( 1 - 4^{-(R+1)} )
           <   2 m  d_0.

The complementary summand is  direct sum_{i=1}^R  L_i^{K_i},  with  total
"trivial-normalized" rank  p = sum_i K_i. Whitney's formula then gives

      c_p(G)  =  product_{i=1}^R  z_i^{K_i}   !=  0.

Now suppose, for contradiction, that G had a trivial subbundle T of rank k
with k > h_0. Then by the exact sequence 0 -> T -> G -> G/T -> 0,  the
quotient G/T has rank  h_0 + p - k < p.  Its top Chern class in degree 2p
vanishes because it is the top Chern class of a bundle of rank < p. By
Whitney,  c_p(G) = c_p(G/T) = 0,  contradicting  c_p(G) = product z_i^{K_i}
nonzero. So G has no trivial subbundle of rank > h_0, and in particular none
of rank 2 m d_0.

The 24 maps on the test base. Let us build, for each  j in {0, ..., R-1},
24 continuous maps  T_1^{(j)}, ..., T_{24}^{(j)} : E_j -> E_{j+1}  over Y,
such that for every subspace of every fiber of E_j, the collective image
spans at least twice as much. Then Lemma 2 will convert these to
normalized maps X_i^{(j)}.

The construction is by cases.

Case (a): small subspaces. For the source block  L_i^{D}  in E_j whose
replacement in E_{j+1} is  L_i^{D/4}  (D = a_{i,j}),  take  q = 16  maps
between multiplicity spaces underlying L_i. For 1 <= s <= m, set  t = 2s,
r = D/4  (rank of the target block). Then (see arithmetic below) the
incidence inequality (*) of Lemma 1 holds with h = 0 (base is a point, since
we build blockwise matrices, not bundle maps). Apply Lemma 1 for each
1 <= s <= m  and assemble the 16 maps diagonally across the line-bundle
blocks (they act as the identity on L_i and as a 16-tuple of matrices on
the multiplicity space). This doubles every s-plane, s <= m, in each block.

For S in the whole E_j with dim S <= m, telescope: put S_k = S intersect
(V_1+...+V_k), S_0=0, W = sum_{i=1}^{16} T_i S, W_k = W intersect (F_1+...+F_k),
W_0=0. Since T_i is block diagonal, S_k/S_{k-1} maps into W_k/W_{k-1} with the
k-th block inside F_k, and dim(S_k/S_{k-1}) <= m, so the block estimate gives
dim(W_k/W_{k-1}) >= 2 dim(S_k/S_{k-1}); summing over k telescopes to
dim W >= 2 dim S.

Case (b): large subspaces. Take q = 8  maps, now on the base Y_{j+1} (the
"prefix"  product_{i=1}^{j+1} CP^{K_i}, whose complex dimension we call
eta_j). The source is E_j  (pulled back to Y_{j+1}); the target is the
fresh block  L_{j+1}^{b_{j+1}}  inside E_{j+1}  (also pulled back). For
s > m, set  t = 2s;  apply Lemma 1 with  h = eta_j  (complex dimension of
the base). This handles every subspace of rank s > m in E_j.

Combining cases (a) and (b) gives 24 maps  T_i^{(j)} : E_j -> E_{j+1}
satisfying

      dim_C  sum_{i=1}^{24} T_i^{(j)}(y)  S  >=  2 dim_C S

for every subspace S of E_j(y). Apply Lemma 2 to obtain X_i^{(j)} satisfying
(a).

The arithmetic.

Case (a). The incidence expression with q = 16, t = 2s is

      (16s - 2s)(D/4 - 2s) - s (D - s)   =   (5/2) s D  -  27 s^2.

Since  D >= 64 m  and  1 <= s <= m,  this is

      >=  s ( (5/2)(64 m) - 27 s )  >=  s ( 160 m - 27 m )  =  133 m s,
      >  0  (strict, since m, s >= 1).

The strict positivity is the incidence inequality (*) with h = 0.

Case (b). The incidence expression with q = 8, t = 2s is

      (8s - 2s)(15d/4 - 2s) - s (d - s)   =   (43/2) s d  -  11 s^2.

Since  s <= d  and  s >= m + 1,  this is

      >=  s ( (43/2) d - 11 d )  =  (21/2) s d
      >  (20/3) m d
      >  eta_j                                (from the dimension estimate).

The dimension estimate:  eta_j = sum_{i=1}^{j+1} K_i  and Stirling-style
bound

      eta_j  <=  (4/3) m sum_{i=1}^{j+1} b_i  =  (5/3) m d_0 (4^{j+1} - 1)
             <  (20/3) m d_j.

The inequality  eta_j < (20/3) m d_j  is exactly the "h" in Lemma 1 for
case (b). Combined with (21/2) s d > (20/3) m d, the incidence inequality
strictly holds.

Continuation beyond level R. For j >= R, define E_{j+1} = E_j + E_j
(direct sum of two copies) and set

      X_1^{(R)}  =  2^{-1/2} iota_1,        X_2^{(R)}  =  2^{-1/2} iota_2,
      X_i^{(R)}  =  0    ( 3 <= i <= 24 ),

where  iota_1, iota_2 : E_R -> E_R + E_R  are the two summand inclusions. Then:

The column sum:  (1/2) iota_1* iota_1 + (1/2) iota_2* iota_2 = (1/2) I_E
+ (1/2) I_E = I_E.  Good.

The row sum:  (1/2) iota_1 iota_1* + (1/2) iota_2 iota_2* = (1/2) P_1 +
(1/2) P_2  where P_i are projections onto the i-th summand of the target.
Since P_1 + P_2 = I_{E_{j+1}},  the row sum equals (1/2) I_{E_{j+1}}, which
is <= (2/3) I_{E_{j+1}}.  So (a) holds.

Nothing changes the retained Chern class of Proposition 1: E_0 + ... + E_R
over Y is the same bundle as before; the continuation lives only at levels
above R. So (b) also holds. This completes the construction.


12. THEOREM 5: THE CONSTRUCTION OF B
-----------------------------------

We now assemble B out of all the finite tests of Proposition 1 at once.

For each pair (m, R) of positive integers, let Y = Y_{m,R} and
E_0, E_1, E_2, ...  be the bundles of Proposition 1, together with the
24 maps  X_i^{(j)} : E_j -> E_{j+1}  over Y. For each y in Y, form the
Hilbert-space direct sum

      H_{m,R,y}  =  E_0(y) (+) E_1(y) (+) E_2(y) (+) ...

of the fibers. This Hilbert space is infinite-dimensional; it is a direct
sum of finite-dimensional spaces, complete in the direct-sum norm
||(v_j)|| = sqrt( sum_j ||v_j||^2 ).

For i = 1, ..., 24, define S_i^{m,R}(y) in B(H_{m,R,y}) by the block rule:
on the j-th summand E_j(y), S_i^{m,R}(y) acts as the single block

      S_i^{m,R}(y) v  =  X_i^{(j)}(y) v   in   E_{j+1}(y),
      S_i^{m,R}(y) v  =  0                in   every other summand.

This is the forward shift along level j -> j+1, with the specific block
X_i^{(j)}(y). Boundedness: each (X_i^{(j)})* X_i^{(j)} <= I_{E_j}, so
||X_i^{(j)}|| <= 1; the direct sum of blocks of norm <= 1 has norm <= 1.
Thus  ||S_i^{m,R}(y)|| <= 1.

Form the bounded C*-product algebra

      D  =  prod^{(bounded)}_{m,R,y}  B(H_{m,R,y}),

the C*-algebra whose elements are the uniformly-bounded families
( T_{m,R,y} )_{ (m,R,y) }  with pointwise operations and norm
sup_{m,R,y} ||T_{m,R,y}||. Each coordinate is one of the B(H_{m,R,y}).

For each i in {1, ..., 24}, define

      x_i  =  ( S_i^{m,R}(y) )_{m,R,y}   in  D.

Because  ||S_i^{m,R}(y)|| <= 1  uniformly over (m, R, y), the family x_i is
bounded, so x_i lies in D.

The two relations in D.

Column sum. Fix (m, R, y). On a vector in the j-th summand,

      sum_i (S_i^{m,R}(y))* (S_i^{m,R}(y)) v_j
      =  sum_i (X_i^{(j)}(y))* X_i^{(j)}(y) v_j
      =  v_j                          (Proposition 1(a), column identity).

So  sum_i x_i* x_i = 1  in D.

Row sum. Fix (m, R, y). On a vector v in the j-th summand with j >= 1, the
only nontrivial contributions to  sum_i S_i S_i*  come from the shifts on
level j-1, i.e. from  sum_i X_i^{(j-1)} (X_i^{(j-1)})*  on E_j(y). By
Proposition 1(a) (row identity),

      0  <=  sum_i S_i^{m,R}(y) (S_i^{m,R}(y))*  <=  (2/3) I_{H_{m,R,y}}.

On level 0 there are no incoming shifts (no level -1), so the row sum
vanishes there. Consequently,

      sum_{i=1}^{24} x_i x_i*   <=   (2/3) 1  in  D.

Define

      B  =  C*(1, x_1, ..., x_24)  subset  D.

Then B is a unital sub-C*-algebra of D. It is separable: the *-polynomials
in the 24 generators with rational complex coefficients form a countable
norm-dense set. And it satisfies

      (R1)  sum_{i=1}^{24} x_i* x_i   =   1,
      (R2)  sum_{i=1}^{24} x_i x_i*  <=  (2/3) 1,

inherited from D.

Theorem 5. Let B = C*(1, x_1, ..., x_24) sub D be as above. Then B is a
separable unital complex C*-algebra satisfying (R1), (R2), and for every
m >= 1 the matrix algebra M_m(B) is not properly infinite.

The rest of this section proves "not properly infinite"; Section 13 is that
proof.


13. EVERY MATRIX ALGEBRA OVER B FAILS TO BE PROPERLY INFINITE
--------------------------------------------------------------

Proof of the second assertion of Theorem 5. Suppose, towards a contradiction,
that M_m(B) is properly infinite for some m >= 1. Then there are isometries
s_1, s_2 in M_m(B) with orthogonal ranges:

      s_i* s_j  =  delta_ij  1_{M_m(B)},    i, j in {1, 2}.

Their rectangular concatenation

      W  =  [ s_1   s_2 ]  in  M_{m, 2m}(B)

is an isometry in the rectangular sense:

      W* W  =  I_{2m}     in  M_{2m}(B).

Note that W W* need not be  I_m; that is the point.

Step 1: approximation by a polynomial. The generators x_i and their
adjoints generate B as a norm-closed *-algebra, so *-polynomials in the
x_i and x_i* are norm-dense in B. Consequently, *-polynomials in the x_i
(entry by entry) are norm-dense in M_{m, 2m}(B). Choose such a
polynomial-entry matrix P with

      || P - W ||  <  1/5.

Because ||W|| = 1 (it is an isometry), standard C*-algebra estimates give

      || P* P - I_{2m} ||  <=  2 ||P - W||  +  || P - W ||^2
                           <   2/5  +  1/25  =  11/25  <  1/2.

Hence

      (1/2) I_{2m}  <=  P* P   in  M_{2m}(B).    (lower bound)

Step 2: choose R from the word length of P. Let R >= 1 be at least the
length of every word in the entries of P. (A word here is a product of
letters x_i and x_i*; its length is the number of letters.)

Step 3: evaluate in the coordinate (m, R). Let the pair (m, R) be the one
of the test bundle of Proposition 1. The coordinate projection

      pi_{m,R,y} : D -> B(H_{m,R,y})

restricts to a *-hom  pi_{m,R,y} | B : B -> B(H_{m,R,y}), because B is a
subalgebra of D. Applying this entry by entry to P yields

      P(y)  =  pi_{m,R,y}(P)  in  M_{m, 2m}( B(H_{m,R,y}) ),

and the operator-inequality evaluation of the lower bound gives

      P(y)* P(y)  >=  (1/2) I_{2m}  in  M_{2m}( B(H_{m,R,y}) ).

Step 4: restriction to level zero. Let

      iota_0  :  E_0(y)^{2m}  ->  H_{m,R,y}^{2m}

include the trivial rank-(2 m d_0) block E_0(y)^{2m} of the source, and

      iota_F  :  F_{m,R}(y)   ->  H_{m,R,y}^m

include the level-0-through-R block

      F_{m,R}  =  ( E_0 + E_1 + ... + E_R )^m

of the target. Both are isometric inclusions, and iota_F* iota_F = I.

Because every word in the entries of P has length <= R, and because each
generator x_i raises level by 1 while x_i* lowers level by 1 (and vanishes
on level 0), every block of P(y) sends iota_0 of the level-zero source
input into iota_F of the level-0-through-R target. Precisely:

      P(y) iota_0  =  iota_F P_0(y)

for a unique continuous bundle map P_0(y) : E_0(y)^{2m} -> F_{m,R}(y).

Step 5: P_0 is injective. Compressing P* P >= (1/2) I between iota_0 and
iota_0*:

      P_0(y)* P_0(y)
      =  iota_0* P(y)* iota_F iota_F* P(y) iota_0
      =  iota_0* P(y)* P(y) iota_0
      >= (1/2) iota_0* iota_0
      =  (1/2) I_{E_0(y)^{2m}}.

So P_0(y)* P_0(y) is invertible, and P_0(y) has full rank 2 m d_0. It is a
continuous bundle map from E_0^{2m} to F_{m,R}.

Step 6: the isometric image. Define

      Q(y)  =  P_0(y) ( P_0(y)* P_0(y) )^{-1/2}  :  E_0(y)^{2m} -> F_{m,R}(y).

The positive operator P_0(y)* P_0(y) has spectrum in [1/2, infinity), so
the inverse square root is defined by continuous functional calculus and
depends continuously on y; Q(y) is continuous in y. And

      Q(y)* Q(y)  =  I_{2 m d_0}  on  E_0(y)^{2m},

so Q(y) is an isometric embedding of the trivial bundle E_0^{2m} (rank
2 m d_0) into F_{m,R}(y). Therefore the image Q(Y_{m,R}) is a subbundle of
F_{m,R} isomorphic to the trivial bundle of rank 2 m d_0:

      F_{m,R}  contains a trivial subbundle of rank  2 m d_0.

Step 7: contradiction. By Proposition 1(b) and the definition of
F_{m,R}, the bundle F_{m,R} = (E_0 + ... + E_R)^m contains no trivial
subbundle of rank 2 m d_0. That contradicts Step 6. Hence our assumption
fails: for every m >= 1,  M_m(B) is not properly infinite.

This completes the proof of Theorem 5.


14. LEMMA 3: FROM THE RELATIONS (R1), (R2) TO A UNIFORM DEFECT PAIR
-------------------------------------------------------------------

We now turn the abstract relations (R1), (R2) into a specific pair of
positive contractions with a uniform additivity defect. The construction is
elementary: it is a telescope along the partial sums of the x_i* x_i and
x_i x_i*, with complements placed in the "upper half" of a block-diagonal
matrix to force positivity of every entry.

Lemma 3 (defect). Let B be a unital complex C*-algebra, let N >= 1, and let
x_1, ..., x_N in B satisfy

      sum_{i=1}^N x_i* x_i    =   1,
      sum_{i=1}^N x_i  x_i*  <=   c  1

for some 0 <= c < 1. Then the block-diagonal matrices

      a  =  diag( H_0, H_1, ..., H_{N-1},  1-J_0, 1-J_1, ..., 1-J_{N-1} ),
      b  =  diag( u_1, u_2, ..., u_N,      1-v_1, 1-v_2, ..., 1-v_N ),

where  u_i = x_i* x_i,  v_i = x_i x_i*,  H_i = sum_{j=1}^i u_j,
J_i = sum_{j=1}^i v_j,  and  H_0 = J_0 = 0,  are positive contractions in
M_{2N}(B) satisfying, for every normalized 1-quasitrace tau on M_{2N}(B),

      tau(a+b) - tau(a) - tau(b)  >=  (1-c) / (2 N).

Positivity check. Each entry is positive: 0 <= H_i <= H_N = 1,
0 <= u_i <= 1, and 0 <= J_i, v_i <= J_N <= c 1, so 1 - J_i and 1 - v_i are
positive too. Norm check: each entry has norm <= 1, so a and b, as
block-diagonal matrices, have norm <= 1.

Proof of the defect bound. Fix a normalized 1-quasitrace tau on M_{2N}(B).
Let  iota_k : B -> M_{2N}(B)  be the inclusion into the k-th diagonal
corner.

Step 1: corners have common value. Define

      phi(z)  =  tau( iota_1(z) )      for  z in B.

For any z >= 0 in B and any two corner indices k, l, let t be the matrix
with z^{1/2} in its (k, l) entry and zeros elsewhere. Then

      t* t  =  iota_l(z),     t t*  =  iota_k(z).

The quasitrace axiom  tau(t* t) = tau(t t*)  gives

      tau( iota_k(z) )  =  phi(z)     for all k and all z >= 0.        (14.1)

Step 2: diagonal matrices. If z_1, ..., z_{2N} are positive elements of B,
the diagonal matrix  diag(z_1, ..., z_{2N})  is a sum of pairwise-commuting
corner embeddings iota_k(z_k). By abelian linearity of tau (axiom (iii) in
Definition 4.1), plus (14.1),

      tau( diag(z_1, ..., z_{2N}) )  =  sum_{k=1}^{2N} phi(z_k).       (14.2)

Step 3: corner normalization. Take every z_k = 1. Then
diag(1, ..., 1) = 1_{M_{2N}(B)}, so by normalization
tau(1_{M_{2N}(B)}) = 1. Equation (14.2) then gives

      2 N phi(1)  =  1,    i.e.   phi(1)  =  1 / (2 N).                 (14.3)

Step 4: square equality within a corner. Apply the square axiom to the single
element t = iota_1(x_i) in M_{2N}(B): t*t = iota_1(x_i* x_i) and
t t* = iota_1(x_i x_i*), so

      phi(u_i)  =  tau(iota_1(x_i* x_i))                          (by 14.1)
                =  tau(iota_1(x_i x_i*))                          (by axiom (i))
                =  phi(v_i).                                       (14.4)

Step 5: the actual defect.  Because a and b are generally noncommuting
diagonal matrices over the (generally noncommuting) algebra B, the defect
of tau is NOT entrywise zero.  Apply (14.2) to a+b, a, b separately:

      D  :=  tau(a+b) - tau(a) - tau(b)
          =  sum_{k=1}^{2N} [ phi( (a+b)_{k,k} ) - phi( a_{k,k} )
                              - phi( b_{k,k} ) ].                (14.5)

First-half contribution.  For 1 <= k <= N:  a_{k,k} = H_{k-1},
b_{k,k} = u_k,  and  (a+b)_{k,k} = H_{k-1} + u_k = H_k.  The k-th
contribution is  phi(H_k) - phi(H_{k-1}) - phi(u_k).  Summing over
k = 1, ..., N telescopes (the H_k are partial sums; H_0 = 0, H_N = 1):

      sum_{k=1}^N [ phi(H_k) - phi(H_{k-1}) - phi(u_k) ]
      =  phi(H_N) - phi(H_0) - sum_{k=1}^N phi(u_k)
      =  phi(1) - sum_{k=1}^N phi(u_k).                             (14.6)

Second-half contribution.  For 1 <= j <= N (block index k = N+j):
a_{N+j, N+j} = 1 - J_{j-1},  b_{N+j, N+j} = 1 - v_j,  and
(a+b)_{N+j, N+j} = (1 - J_{j-1}) + (1 - v_j) = 2 - J_j.  The contribution
is

      phi(2 - J_j) - phi(1 - J_{j-1}) - phi(1 - v_j).

Use abelian linearity within the commutative subalgebras C*(1, J_j),
C*(1, J_{j-1}), C*(1, v_j) separately.  The elements J_{j-1}, J_j, v_j
do not commute with one another in general; each individually commutes
only with 1.  That is exactly what the quasitrace axioms allow.

      phi(2 - J_j)
      = phi( 1 + (1 - J_j) )                 (in C*(1, J_j))
      = phi(1) + phi(1 - J_j)                (1 and 1 - J_j commute)
      = phi(1) + phi(1) - phi(J_j)           (in C*(1, J_j))
      = 2 phi(1) - phi(J_j),

      phi(1 - J_{j-1})  =  phi(1) - phi(J_{j-1}),    (in C*(1, J_{j-1}))
      phi(1 - v_j)      =  phi(1) - phi(v_j).        (in C*(1, v_j))

Therefore the j-th second-half contribution equals

      [2 phi(1) - phi(J_j)] - [phi(1) - phi(J_{j-1})] - [phi(1) - phi(v_j)]
      =  2 phi(1) - phi(1) - phi(1)  -  phi(J_j) + phi(J_{j-1}) + phi(v_j)
      =  - phi(J_j) + phi(J_{j-1}) + phi(v_j).

Summing over j = 1, ..., N telescopes to

      sum_{j=1}^N [ -phi(J_j) + phi(J_{j-1}) + phi(v_j) ]
      =  - phi(J_N) + phi(J_0) + sum_{j=1}^N phi(v_j)
      =  - phi(J_N) + sum_{j=1}^N phi(v_j).                          (14.7)

The sum does NOT collapse further: phi(J_N) is phi evaluated at the sum of
the v_j, and is generally not equal to the sum of phi(v_j), because the v_j
do not commute with one another.

Combining (14.6) and (14.7):

      D  =  phi(1) - sum_{k=1}^N phi(u_k) - phi(J_N) + sum_{j=1}^N phi(v_j).

By the corner square equality (14.4), phi(u_k) = phi(v_k) for every k, so
the two sums cancel exactly:

      D  =  phi(1) - phi(J_N).                                       (14.8)

Step 6: bound on phi(J_N).  The self-adjoint element J_N = sum_{j=1}^N v_j
lies in B, and by the row relation (R2),

      0  <=  J_N  <=  c 1        in B       (c = 2/3 in the application).

Since J_N and 1 commute, they generate a commutative C*-subalgebra
C*(1, J_N) of B, on which phi is positive and complex linear.  Therefore

      phi(J_N)  <=  c phi(1).                                        (14.9)

Combine (14.8), (14.9), and the corner normalization (14.3):

      D  =  phi(1) - phi(J_N)
        >=  phi(1) - c phi(1)
        =  (1 - c) phi(1)
        =  (1 - c) / (2 N).

This is exactly the claimed defect bound.

Remarks.

(1) Where the positivity comes from.  The positivity of the defect is
carried by phi(J_N) <= c phi(1), i.e. by the row relation (R2).  Without
(R2), the defect could be zero; with (R2), the complement 1 - J_N is
strictly positive under any positive linear functional phi, and phi(J_N)
is strictly less than phi(1).

(2) Where the quasitrace axioms enter.  Two places:  the corner argument
(Step 1) uses  tau(t* t) = tau(t t*);  the square equality (14.4) uses the
same axiom inside a single corner.  Everywhere else, the argument uses only
corner additivity (Section 14, Step 2), abelian linearity for the
complement identities, and the normalization phi(1) = 1/(2N).

End of Lemma 3.


15. PROOF OF THEOREM 1 (MAIN THEOREM)
-------------------------------------

We now assemble the ingredients. Recall from Theorem 5: B is a separable
unital complex C*-algebra with generators x_1, ..., x_24 satisfying

      (R1)  sum_i x_i* x_i  =  1,
      (R2)  sum_i x_i x_i* <=  (2/3) 1,
      (H)   M_m(B) is not properly infinite for every m >= 1,

and A = M_48(B) is the algebra with which we work.

Step A: A has a normalized 2-quasitrace. For every m >= 1,
M_m(A) = M_{48 m}(B) is not properly infinite, by (H). Theorem 4
(Cuntz-Blackadar-Handelman) therefore supplies a normalized 2-quasitrace
on A.

Step B: A has a uniform defect pair. Apply Lemma 3 to B with N = 24 and
c = 2/3. It yields positive contractions

      a, b  in  M_{2 N}(B)  =  M_48(B)  =  A

such that, for every normalized 1-quasitrace tau on A,

      tau(a+b) - tau(a) - tau(b)  >=  (1 - 2/3) / (2 * 24)  =  1/144.

Every normalized 2-quasitrace on A is, by definition, a normalized
1-quasitrace on A; hence the same bound holds for every normalized
2-quasitrace on A.

Combine Steps A and B: A is a separable unital complex C*-algebra (as
M_48(B), with B separable unital), A has a normalized 2-quasitrace, and
there are positive contractions a, b in A such that every normalized
2-quasitrace on A has defect at least 1/144 at (a, b). This is Theorem 1.


16. PROOF OF COROLLARY 1 (TENSOR PRODUCTS AND STABLE FINITENESS)
----------------------------------------------------------------

Three separate constructions go into the three parts of Corollary 1. We
state the ingredients first, then take each part.

16.1 Ingredients.

(a) The factor input (Kaplansky, from [MR, Section 2] and
    [Gow, Theorem 4.11]). The assertion

        "every normalized 2-quasitrace on every unital complex
         C*-algebra is a tracial state"

    is equivalent to the assertion

        "every AW*-factor of type II_1 is a W*-algebra
         (i.e. has a Banach-space predual)."

    Theorem 1 negates the first assertion, so it negates the second. Hence
    there exists an AW*-factor M of type II_1 that is not a W*-algebra.

(b) Every II_1 AW*-factor admits a normalized 2-quasitrace; it admits a
    tracial state exactly when it is a W*-algebra [MR, Section 2].
    Consequently M has a normalized 2-quasitrace and no tracial state.

(c) Every finite AW*-factor is simple as a C*-algebra
    [Gow, Remark 2.7]. Since M is of type II_1 (finite by type), M is
    simple.

(d) For a unital simple C*-algebra, absence of stable proper infiniteness
    is equivalent to stable finiteness [MR, Section 2].

(e) The reduced C*-algebra R = C_r*(F_2) of the free group on two
    generators is simple [Powers], unital, and stably finite. Stable
    finiteness comes from the faithful canonical trace delta_e (with
    delta_e evaluated at the coefficient of the identity): extensions to
    matrix algebras are also faithful, so a nonunitary isometry in some
    M_n(R) is impossible.

(f) The Haagerup trace-obstruction in tensor form (Theorem 3): a unital
    C*-algebra D has no tracial state if and only if
    D (x)_min C_r*(F_infinity) is properly infinite. By 6.4, F_infinity
    embeds in F_2, so D (x)_min C_r*(F_2) is properly infinite whenever D
    has no tracial state.

16.2 Proof of (i): find two unital simple stably finite algebras whose
tensor product is properly infinite.

Take M as in (a). By (b)-(d): M is unital, simple, has a normalized
2-quasitrace (so no M_k(M) is properly infinite by Theorem 4), and
therefore is stably finite. It has no tracial state.

Take R = C_r*(F_2). By (e), R is unital, simple, and stably finite.

By (f), the absence of a tracial state on M makes
M (x)_min C_r*(F_2) properly infinite. Both factors are unital simple
stably finite. That is (i). One factor is R = C_r*(F_2) as claimed.

16.3 Proof of (ii): an explicit separable B with a normalized 2-quasitrace
whose tensor with C_r*(F_2) loses the 2-quasitrace.

Take the algebra B of Theorem 5.

Tracelessness. By (R1) and (R2) together with the argument of Section 3.4
(c = 2/3 < 1), B has no tracial state: applying a tracial state to (R1)
and (R2) gives 1 <= 2/3, a contradiction. Note this is tracelessness, not
defect: the pair (a,b) of Lemma 3 is not needed here.

2-quasitrace on B. By (H) and Theorem 4, B has a normalized 2-quasitrace.

Proper infiniteness. By (f), B has no tracial state, so
B (x)_min C_r*(F_2) is properly infinite.

Loss of the 2-quasitrace. Since B (x)_min C_r*(F_2) is properly infinite
(the unit is properly infinite), the criterion (Theorem 4) excludes a
normalized 2-quasitrace on it. This is (ii).

16.4 Proof of (iii): a separable stably finite C*-algebra with no tracial
state.

Take M as in (a). It has no tracial state. By Haagerup's criterion
(Theorem 3 (ii)), there are finitely many y_1, ..., y_n in M with

      sum_{i=1}^n y_i* y_i  =  1,
      || sum_{i=1}^n y_i y_i* ||  <  1.

Let

      C  =  C*(1, y_1, ..., y_n)   subset  M.

Then C is a separable unital complex C*-algebra.

C is stably finite. Suppose V in M_k(C) is an isometry (V* V = 1). Then V
is an isometry in M_k(M). Since M is stably finite (16.2), V is unitary in
M_k(M), so V V* = 1. Then VV* computed within M_k(C) equals 1 too, so V is
unitary in M_k(C). No such V is proper, so C is stably finite.

C has no tracial state. Suppose tau were a tracial state on C. On the
positive self-adjoint element  sum_i y_i y_i*  of norm < 1, we have

      tau( sum_i y_i y_i* )  <=  || sum_i y_i y_i* || * tau(1)  <  1.

On the other hand, by the trace property and on the y-tuple relation

      tau( sum_i y_i y_i* )  =  tau( sum_i y_i* y_i )  =  tau(1)  =  1.

Contradiction. Hence C has no tracial state. This is (iii).

The three algebras in (i), (ii), (iii) are different: M (for (i) and
(iii)), R = C_r*(F_2) (for (i)), B of Section 12 (for (ii)), and C of
Section 16.4 (for (iii)). None of the three parts implies the others.

End of Corollary 1.


17. GLOSSARY OF KEY TERMS
-------------------------

A short glossary for the reader who has gotten this far. Each entry is one
or two sentences.

Adjoint. For a bounded operator T on a Hilbert space H, the unique operator
T* with <u, Tv> = <T*u, v> for all u, v.

AW*-algebra. A C*-algebra in which every set of orthogonal projections
has a supremum that is itself a projection. Named after von Neumann's
algebras but characterized algebraically, without a chosen Hilbert-space
representation.

AW*-factor. An AW*-algebra whose center is C 1. Type classification (I,
II, III, II_1, II_infinity, ...) follows the von Neumann-algebra pattern.

C*-algebra. A norm-closed *-subalgebra of some B(H). Separable if it has
a countable norm-dense subset.

C_r*(F_2). The reduced C*-algebra of the free group on two generators. It
carries a canonical faithful trace delta_e. It is simple [Powers], unital,
and stably finite; these are the only properties we use.

Exact. A regularity property of unital C*-algebras; nuclear algebras are
exact. Closed under various standard constructions.

Hilbert space. A complete complex inner product space.

Minimal tensor product. The spatial C*-tensor product, unique on the class
of C*-algebras. It preserves injective *-homomorphisms.

Nuclear. A C*-algebra for which every *-hom into B(H) is automatically
well-approximated by finite-rank maps. Nuclear => exact.

Positive contraction. A self-adjoint T with <v, Tv> >= 0 for all v and
with ||T|| <= 1.

Projection. A self-adjoint idempotent P = P* = P^2.

Properly infinite. A unital C*-algebra C is properly infinite when it
contains two isometries s_1, s_2 with orthogonal ranges; equivalently,
W*W = I for some rectangular isometry W. Stably properly infinite means some
M_m(C) is properly infinite. Stable finiteness forbids a nonunitary isometry in
any M_m(C); for unital simple C this is exactly the opposite condition.

Quasitrace (1-, 2-). A 1-quasitrace is a function tau with tau(x*x) = tau(xx*) >= 0,
tau(h + ik) = tau(h) + i tau(k) for self-adjoint h, k, and tau complex-linear on
each commutative subalgebra (Definition 4.1). A 2-quasitrace also extends to a
1-quasitrace on M_2.

Stable finiteness. A unital C*-algebra C is stably finite when no matrix
algebra M_k(C) contains a nonunitary isometry. In the simple unital case,
equivalent to absence of stable proper infiniteness [MR, Section 2].

Trace. A linear functional tau with tau(xy) = tau(yx). A tracial state is
a positive normalized trace.

W*-algebra. A C*-algebra with a Banach-space predual; equivalently, a von
Neumann algebra (up to abstract isomorphism).


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--------------

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End of document.