For an infinite graph, positivity of the Markov trace does not by itself prove factoriality. The path model gives a concrete way to check the center: a central element cannot distinguish two prefixes that end at the same vertex. Its value can depend only on the tail. We will prove that this tail carries no nonconstant bounded invariant function.
We assume Paths, local projections and a faithful trace, conditional expectation in a finite tracial von Neumann algebra, and elementary probability on countable products. We use only explicit conditional probabilities, finite sums, and the fact that summable event probabilities imply that only finitely many of those events occur almost surely. The last fact follows by bounding the probability of any event after time by the tail of the convergent sum. References for the broader trace picture are [Jones], [Wassermann] and [Moore].
Throughout, the graph is , , is given by (9.3), and , and are the path algebras and trace from the preceding lesson. Put .
The full path diagonal
In , let be the diagonal algebra spanned by , and let . Write for the space of infinite paths with , , and . The diagonal projection of a prefix is its cylinder set . Its probability is
These probabilities are compatible by the eigenvector equation (9.1). They give a probability measure on , and identify , with its trace, with . One can construct the measure by the transition probabilities
Multiplying these probabilities along a prefix gives (10.1).
Lemma 10.1. The full path diagonal is maximal abelian in . If , its function on is invariant under replacement of a finite prefix by another prefix of the same length and endpoint. In particular, for every , it is measurable with respect to
Proof. The trace-preserving expectations converge in to the identity: their Hilbert-space projections have increasing ranges with dense union. If commutes with , bimodularity implies that commutes with . The commutant of a full diagonal in each matrix block of is that diagonal. Hence , and its limit lies in . Since is bounded, this says . Thus .
For a central , commute it with the matrix unit for two prefixes of the same length and endpoint. This partial isometry replaces by and preserves the later path. Conjugating a later diagonal cylinder verifies this description directly using (9.5). The equality therefore makes the function of equal on these two prefixes with any common later continuation, almost everywhere. There are only countably many pairs of prefixes, so their exceptional null sets can be combined into one.
For fixed , the finitely many prefixes reaching any given endpoint consequently give the same function of the continuation. The function is therefore measurable with respect to the endpoint and all later vertices, namely .
The diagonal in this lemma includes every path projection. It differs from the algebra generated by the even-numbered , whose maximal-abelian claim was refuted in Commuting projections need not be maximal abelian.
Counting bridges without crossing the boundary
Let be the number of length- nearest-neighbor paths from to that stay nonnegative. Use when is not an integer in .
Lemma 10.2. The bridge count is
If is a fixed length- prefix ending at , then, for ,
Proof. Without the boundary, a path from to has downward steps, giving the first binomial coefficient. Reflect the part of a boundary-crossing path up to its first visit to . This is a bijection with unrestricted paths starting at and ending at , counted by the second coefficient. Subtracting proves (10.3).
Given the endpoint , every length- prefix has the same probability . The Markov rule (10.2) makes the subsequent conditional path law depend only on . Thus conditioning on the whole tail still gives the uniform law on the prefixes. Exactly of them extend the specified prefix . This proves (10.4), first tested against finite tail cylinders and then against the sigma algebra they generate. The denominator is positive at every endpoint actually visited.
How far the Markov path goes
Lemma 10.3. Almost surely,
If , then . If , then .
Proof for . Put
Here and . For an ordinary random walk with upward probability , the probability, starting at , of never reaching is . To verify it, stop the martingale on first reaching or a level . The two boundary values determine the probability of the lower exit. Let ; the probability of hitting is . Every path that hits before a finite time is counted for all sufficiently large , which justifies this limit.
Condition this biased walk, started at zero, on never reaching . Its upward transition from is , and the downward transition is . The identities
show that the conditioned walk is exactly (10.2).
For the unconditioned walk, almost surely. An elementary proof uses its independent increments : for each , exponential Markov bounds give
for some . Indeed, the logarithm of has value and first derivative zero at , so for sufficiently small positive it is less than ; apply this to both signs. The bounds are summable. Conditioning on an event of positive probability preserves an almost-sure assertion. The resulting positive speed also implies .
Proof for . The transition probabilities are
with the downward probability zero at . For , the function is harmonic for this chain. Stop it between a lower level and an upper level . Solving for the two boundary probabilities, then letting , gives
The stopping time in the finite interval is almost surely finite: there is a fixed positive probability of an exit within a fixed number of steps from every state of the interval. Repeated blocks therefore have geometrically decreasing survival probability. The limit in (10.6) is justified as in the preceding boundary-exit argument.
After any visit to , the chance of stepping up and never returning is
The Markov property bounds the probability of further returns by a geometric sequence. Each fixed state is visited finitely often. A countable intersection of these full-measure assertions shows that .
Finally, let be ordinary symmetric random walk from zero. By (10.1) and (10.3),
The right side is summable by the same exponential bound for independent increments. Thus , first for each rational , and then altogether.
A bridge forgets its fixed prefix
Lemma 10.4. For every fixed and endpoint reachable at time , almost surely
Proof. Write , , and . All binomial arguments below are integers because and .
First suppose , so . Put and . For any fixed integer , cancellation of factorials gives
For example, the ratio is
;
each finite product has the corresponding limit because and . The formula remains valid for negative by taking reciprocal finite products.
The two numerator arguments in (10.3) correspond to and . Since and , their respective limits in (10.8) are
Meanwhile,
.
The resulting quotient is
At , both this numerator difference and the denominator's limiting value vanish. We must retain their exact difference rather than divide the two zero limits. Put . Telescoping gives
For integers ,
By Lemma 10.3, and . The fixed shifts used below therefore lie in this range for all sufficiently large . For each fixed , the same finite-product calculation as (10.8) gives
. Dividing its difference term by introduces the two additional factors
Every term of (10.9) therefore has limiting ratio . There are terms, giving . This completes both cases.
Factoriality of the Markov trace
Theorem 10.5. For every , the von Neumann algebra is an approximately finite-dimensional II₁ factor. The Markov trace on the path algebra is therefore a factorial tracial state.
Proof. Let . By Lemma 10.1 it is a bounded function on paths, measurable with respect to every . For a fixed prefix ending at at time , conditional expectation gives
The ratio is a conditional probability, hence lies in . Lemma 10.4 and bounded convergence show that the right side tends to
.
It follows that has zero integral over every cylinder. Finite cylinder functions are dense in , so is constant. Thus the center of is scalar.
The trace is faithful and normalized. The block at vertex zero at level has at least paths: concatenate, independently, the two length-four loops and . Faithfulness embeds their matrix blocks in , whose matrix sizes are consequently unbounded. Thus this finite factor is infinite dimensional, hence II₁. Its increasing union of finite-dimensional path algebras is weakly dense, proving the approximation assertion.
The proof supplies the needed conclusion at the critical value as well as above it. It does not assert that the even-generator algebra is maximal abelian. The full path diagonal gives the correct intermediate algebra.
Exercises
Exercise 10.1 — introductory. Count the nonnegative length-five paths from zero to one. Use the answer to find for .
Solution. Formula (10.3) gives . Each path has probability , so the total is .
Exercise 10.2 — intermediate. For the critical chain, find the probability, starting at four, of ever visiting one. Show why transience does not imply a positive limiting speed.
Solution. Equation (10.6) gives . Every finite state is visited finitely often, so the chain tends to infinity, but Lemma 10.3 gives . These statements describe different scales and are compatible.
Exercise 10.3 — intermediate. Let . Determine , the limiting speed, and the probability that the unconditioned biased walk from zero never hits .
Solution. The equation gives . Hence , , speed , and survival probability . Also .
Exercise 10.4 — advanced. Explain why agreement of a bounded function's integrals with a constant on all path cylinders implies almost-everywhere equality to that constant.
Solution. Finite linear combinations of cylinder indicators form the union of the finite-coordinate measurable function spaces. Their closure is all of , since the coordinate sigma algebras generate the full sigma algebra. If is the difference from the constant, then and is orthogonal to this dense union. Therefore in , hence almost everywhere. This also explains why the factoriality proof needs no claim about uniqueness of all traces on the AF algebra.
References
Vaughan F. R. Jones, Index for subfactors, Inventiones Mathematicae 72 (1983), 1–25.