Amplification and strong 1-boundedness

SinceN≤r are an increasing sequence of subalgebras of N and^WN≤r=N, we have that

h(N:M) =sup

r

h(N≤r:M)≥

∞

∑

j=1

λ²_jh(N_j:M_j).

(ii) If (M,τ) is Connes embeddable, and N≤M is hyperfinite, then the inclusion N≤M has the microstates extension property.

(iii) If(M_j,τ_j),j=1,2are Connes-embeddable, thenM₁∗1≤M₁∗M₂has the microstates extension property.

We remark that (i) makes sense from our intuitive description of1-bounded entropy. The quantity h(N:Q) (resp. h(N:M)) is supposed to be a measurement of “how many" embeddings there are of N into an ultraproduct of matrices which have an extension to Q(resp. M). If Q≤M has the microstates extension property, then every embedding of Qextends to M and thus the quantities h(N:Q),h(N:M)should be the same.

Proof. (i): This is an exercise from Lemma 3.0.7.

(ii): Without loss of generality we may, and will, assume thatN,M have separable predual. Fix a free ultrafilter ω ∈βN\N. Let Θ: N→∏k→ωMk(C) be a trace-preserving embedding. Since M is Connes-embeddable, there exists a trace-preserving embedding Ψ:M →∏k→ωMk(C). By [Con76, Jun07a] there exists a unitaryu∈∏k→ωMk(C)so thatΨ

N=ad(u)◦Θ. SetΘe=ad(u^∗)◦Ψ.

ThenΘe is the desired extension.

(iii): Without loss of generality we may, and will, assume that M1,M2 have separable predual.

Our desired result is now a consequence of [Voi98, Theorem 2.4], [Pop14, Corollary 0.2]. Namely, fix a free ultrafilterω∈βN\N, and letM =∏k→ωMk(C). Then, by assumption, there exists trace- preserving embeddingsΘj:M_j→M,j=1,2. By [Voi98, Theorem 2.4], [Pop14, Corollary 0.2] there exists a unitaryu∈M which is Haar distributed and freely independent ofΘ1(M₁)∨Θ2(M₂). Then Θ₁(M₁),uΘ₂(M₂)u^∗are freely independent, and this produces an extensionΘ:e M₁∗M₂→M ofΘ.

Now we are ready to prove Proposition 1.0.3, and in fact we also give the analogous result forh.

We remark that the point h(M^t)≤t⁻²h(M) fort∈(0,1)was already shown in [Hay18, Propostion A.13(ii)].

Proposition 3.0.9. Suppose that M is a II₁-factor and τ is its canonical trace. Fort∈(0,∞), let M^t be thet^th compression ofM. Then

h(M^t) = 1

t²h(M) and h(M^t) = 1 t²h(M).

Proof. We first handle the case whent∈(0,1). Fix a hyperfinite II₁-subfactor R of M. Let p∈R be a projection withτ(p) =t. Observe thatM^t∼=pM p. LetN=pM p+ (1−p)R(1−p)SinceRis a

factor, we may find partial isometriesv₁,· · ·,v_n∈Rso thatv^∗_jv_j≤pand∑jv_jv^∗_j=1. Thus

N∨R⊇W^∗(v₁,v₂,· · ·,v_n)∨(pM p+ (1−p)) =M.

So

N∨R=M. (3.0.2)

Step 1. We show thath(M^t)≥ ¹

t²h(M). By (3.0.2) and the fact thatN∩R=pRp+ (1−p)R(1−p)is diffuse, we have

h(M) =h(N∨R)≤h(N) +h(R) =h(N)≤t²h(pM p) + (1−t)²h((1−p)R(1−p)) =t²h(pM p) =t²h(M^t).

Here we use the analogues of [Hay18, Lemma A.12 and Proposition A.13 (i)] for h.

Step 2. We prove thath(M^t)≤ ¹

t²h(M). We start with the following claim.

Claim. N≤Mhas the microstates extension property. Suppose thatω is a free ultrafilter on the natural numbers, and letM=∏k→ωMk(C)be the tracial ultraproduct ofMk(C). LetΘ: N→M be a trace-preserving embedding. SinceN∩Ris hyperfinite, we know thatN∩R≤Rhas the microstates extension property by Lemma 3.0.8 (ii). So there exists a trace-preserving embedding Ψ:R→M withΨ

_N∩R=Θ

_N∩R. Letv₁,· · ·,v_nbe as before the proof of Step 1. DefineΘe:M→M by

Θ(x) =e

∑

i,j

Ψ(v_i)Θ(v^∗_ixv_j)Ψ(v_j)^∗.

Observe thatv^∗_iMv_j⊆pM p⊆N for alli,j, so the formula above makes sense. For all 1≤i,j,k,l≤n and allx,y∈M we have

Ψ(v_i)Θ(v^∗_ixv_j)Ψ(v_j)^∗Ψ(v_k)Θ(v^∗_kyv_l)Ψ(v_l)^∗=Ψ(v_i)Θ(v^∗_ixv_j)Ψ(v^∗_jv_k)Θ(v^∗_kyv_l)Ψ(v_l)^∗

=δj=kΨ(v_i)Θ(v^∗_ixv_j)Θ(v^∗_jv_j)Θ(v^∗_jyv_l)Ψ(v_l)^∗

=δj=kΨ(v_i)Θ(v^∗_ixvjv^∗_jvjv^∗_jyv_l)Ψ(v_l)^∗

=δ_j=kΨ(v_i)Θ(v^∗_ixv_jv^∗_jyv_l)Ψ(v_l)^∗.

From here, it is direct to show that Θ(xy) =e Θ(x)ee Θ(y)for allx,y∈M. It is also direct to show that Θe preserves adjoints. Finally, for allx∈M:

τω(Θ(x)) =e

∑

i,j

τω(Θ(v^∗_ixv_j)Ψ(v^∗_jv_i)) =

∑

i

τω(Θ(v^∗_ixv_iv^∗_iv_i)) =

∑

i

τ(xv_iv^∗_i) =τ(x).

This proves the claim.

By (3.0.2) and the fact that N∩Ris diffuse, we have that

h(M) =h(N∨R:M) =h(N:M) =h(N),

the last step following by the claim and Lemma 3.0.8 (i). By Lemma 3.0.5,

h(M) =h(N)≥t²h(pM p).

This completes the proof of Step 2 and hence also the proof forh whent∈(0,1). The caset=1 is trivial, whereas ift>1, then ¹_t ∈(0,1)and so

h(M) =h((M^t)^1t) =t²h(M^t),

For the case of h, it was already shown in [Hay18, Proposition A.13(ii)] that h(M^t)≤ ¹

t²h(M) for t∈(0,1), and the proof of the opposite inequality proceeds in a similar manner to Step 2 above.

With Lemma 3.0.5 on direct sums and Proposition 3.0.9 on amplifications in hand, we are ready to finish the proof of Proposition 1.0.2 showing that strong1-boundedness of all tracial von Neumann algebras with Property (T) is equivalent toh(M)≤0 for allII₁factors with Property (T).

Proof of Proposition 1.0.2. (iii) =⇒ (ii). Assume that everyII₁Property (T) factor has nonpositive 1-bounded entropy. LetM be a tracial von Neumann algebra with Property (T), and we will show that h(M)≤0. Decomposing the center ofM into diffuse and atomic parts, we see that there is a countable index setJ (potentially empty) so that

M=M0⊕^M

j∈J

M_j,

where M0 either has diffuse center or is{0}, and each M_j is a factor. Since M0 either has diffuse center or is {0}, we know h(M₀)≤0. Since M has Property (T), each Mj has Property (T) by [Pop06a, Proposition 4.7.2]. Thus, by (3.0.1), we haveh(M)≤0.

(ii) =⇒ (i). If every tracial von Neumann algebra with Property (T) satisfiesh(M)≤0, then it is strongly1-bounded since strong1-boundedness is equivalent toh(M)<∞by [Hay18, Proposition A.16].

(i) =⇒ (iii). Proceeding by contraposition, assume thatNis a Property (T) factor withh(N)>0,

and we will show that there is a tracial von Neumann algebra(M,τ)with Property (T) such that h(M,τ) =∞. Let(N^4−k,τk)be the compression ofM, and let

(M,τ) =^M

k∈N

2^−k(N^4−k,τ_k).

By [Pop06a, Proposition 4.7.1] we know thatM has Property (T). By Lemma 3.0.5 and Proposition 3.0.9, for each j∈N, we have

h(M,τ)≥4^−jh(N^4−j) +

∑

k̸=j

4^−kh(N^4−j) =4^jh(N) +

∑

k̸=j

4^kh(N).

Sinceh(N)>0, we know thatN is Connes-embeddable, and thush(N)≥0. So

h(M)≥4^jh(N)for all j∈N.

Letting j→∞we see thath(M) =∞, i.e. M is not strongly1-bounded.

Finally, we show that for each finite von Neumann algebra M with Property (T), there exists a faithful normal tracial state τ such that(M,τ)is strongly 1-bounded. As in (iii) =⇒ (ii), write M=^L∞_j=0M_j such that M0 is zero or has diffuse center, andM_j is a factor for j≥1. Let τ_j be the unique tracial state on Mj. Since h(M₀)≤0 and since h(M_j)<∞ for j≥1 by Theorem 1.0.1, we may choose nonnegative constants(λ_j)^∞_j=0 such that∑^∞_j=0λj=1, ∑j∈N0λ²_jh(M_j)<∞, andλj>0 if and only ifM_j̸=0. Letτbe the faithful normal tracial state onMgiven by^L∞_j=0λjτj. It follows by (3.0.1) thath(M,τ)<∞.