Invariant states and finite-rank projections

AI-generated exposition by OpenAI GPT models. CC0-1.0 to the extent of rights held.

An invariant state can be singular. We replace its invariance by a norm estimate for a normal state, take the square root of its density, and select one spectral level. In the trace representation of an injective finite algebra, the selected projection has finite Hilbert-space rank. Its range can then be described by bounded operators in the algebra.

The hypertrace construction supplies an invariant extension of a finite trace. The measurable-operator and trace-predual construction identifies normal functionals with actual $L^1$ densities for general semifinite traces. Trace Hilbert spaces and finite trace expectations proves the full $L^2$ completion, product bound, cyclic pairing and ordinary-trace rank identification used here. We prove the square-root and spectral-level inequalities below for positive elements of that full semifinite $L^2$ space. The spectral argument uses finite partitions and monotone approximation; the positive operators need not commute. The later scalar pinching uses the finite trace expectations and decreasing-space argument in the same module.

For a finite algebra $M$ with faithful normal tracial state $\tau$, put $H=L^2(M,\tau)$ and $\|x\|_2=\tau(x^*x)^{1/2}$. Inner products are linear in the second variable. Write $\operatorname{Tr}$ for the ordinary operator trace on $B(H)$, and $\|\cdot\|_{\mathrm{HS}}$ for its Hilbert–Schmidt norm. These differ from $\tau$ and the norm on $M$.

The near-covering coefficient model is the internal approximation goal. We then construct a bounded basis for an external finite-rank model and prove the invariant-state projection theorem by finite weighted spectral selection. A continuous-level proof and all three full semifinite trace estimates are also included.

Near-covering coefficient models

Here $M$ is a type $\mathrm{II}_1$ factor with separable predual. We use the scalar pinching theorem, whose construction uses contained AFD subfactors and decreasing expectations. Its inputs precede the injective-factor converse.

Lemma 3.1. Let $x_1,\ldots,x_m\in M$ be orthonormal in $L^2$, let finite $F\subset\mathcal U(M)$ contain $1$, and let $0<b<1$. There are nonzero mutually orthogonal projections $f_1,\ldots,f_l$ such that

Proof. If the indexed family is empty, take $f_1=1$. Otherwise put $Q=|F|m^2$, and let $C$ be the finite set of distinct operators among the coefficients $x_i^*v x_j$. Apply the scalar pinching theorem with $N=M$, the set $C$, and tolerance $b^2/\sqrt Q$. This gives a finite partition $(p_k)$ of $1$; remove any zero atoms. Every coefficient in the indexed family has squared pinching error less than $b^4/Q$, so

This tolerance accounts for repetitions in the indexed family.

For each atom put

Errors supported on different diagonal corners are orthogonal in $L^2$, hence $\sum_k a_k<b^4$. Retain precisely the atoms with $a_k\le b^2\tau(p_k)$. If any atoms are discarded, then

if none are discarded their total trace is zero. In either case the retained trace is greater than $1-b^2>0$, and the retained atoms prove (11). For $v=1$, orthonormality gives $\tau(x_i^*x_j)=\delta_{ij}$. $\square$

The first selection produces an external finite-rank projection on $H$. The second produces initial projections inside $M$. The next lesson repairs its almost orthonormal vectors to internal matrix units.

Bounded coordinates for a finite-rank model

Corollary 2.1. If $M$ is finite and injective with faithful normal tracial state, then for every finite $F\subset\mathcal U(M)$ and $a>0$ there are $x_1,\ldots,x_m\in M$, orthonormal in $H$, such that the finite-rank projection

satisfies

Here $v$ acts by left multiplication. No factor or separability assumption is needed.

Proof. The invariant-state projection theorem used here is proved in the next section. The hypertrace on $B(H)$ is invariant under every such $v$. Apply Theorem 1.1 to $B(H)$ with $\sigma=\operatorname{Tr}$ and initial tolerance small enough to make the summed error less than $a^2m/4$. A nonzero projection of finite ordinary trace has finite rank $m$.

Approximate an orthonormal basis $\xi_1,\ldots,\xi_m$ of its range by vectors $y_i\in M$. Their Gram matrix $G=[\tau(y_i^*y_j)]$ tends to the identity. For sufficiently good approximants it is invertible; define $x_i=\sum_j y_j(G^{-1/2})_{ji}$. These vectors belong to $M$, are exactly orthonormal, and tend to $\xi_i$ in $H$. Their range projections tend to the original projection in Hilbert–Schmidt norm, because

A conjugation difference changes by at most twice the projection error. Finitely many tests and the strict initial margin give (8). The empty test set permits $x_1=1$. $\square$

For $b,c\in H$, write $K_{b,c}\xi=b\,\tau(c^*\xi)$, interpreted by the Hilbert inner product. Its Hilbert–Schmidt norm is $\|b\|_2\|c\|_2$. For bounded $v,w\in M$, let $L_v\xi=v\xi$, $R_w\xi=\xi w$. Direct evaluation gives

The right action is an antirepresentation and commutes with the left action. These identities fix the orientations in the later small-corner proof.

Invariant-state projections by finite selection

Theorem 1.1. Let $N$ carry a faithful normal semifinite trace $\sigma$, and let $F\subset\mathcal U(N)$ be finite. Suppose a state $\psi$ on $N$, possibly singular, satisfies $\psi\circ\operatorname{Ad}(u)=\psi$ for $u\in F$. For every $\varepsilon>0$ there is a nonzero projection $p\in N$, with $\sigma(p)<\infty$, such that

Factoriality and countability assumptions are unnecessary.

Proof by finite weighted spectral selection. If $F$ is empty, semifiniteness supplies a nonzero finite-trace projection. Assume $n=|F|\ge1$, and set

We first construct a normal state whose total invariance defect is less than $d$. Normal states are weak* dense among all states on $N$: otherwise real separation supplies a self-adjoint $x$ and a state whose value at $x$ exceeds every normal-state value. The supremum of the latter values is $\sup\operatorname{Sp}(x)$. Indeed a nonzero spectral projection near the top supports a normal positive functional, whose normalization gives a normal state with value arbitrarily near that top. Every state is bounded above by the same spectral bound, a contradiction.

Consider the convex subset of the finite Banach predual sum $\bigoplus_{u\in F}N_*$ given by

Its weak closure contains zero. Pairing against any finite list $x_u\in N$, normal states converging weak* to $\psi$ give $\sum_u\varphi(x_u-u x_u u^*)\to0$, by invariance. The dual of this finite predual sum is the corresponding finite product of copies of $N$. Hahn--Banach separation therefore equates its convex weak and norm closures. Choose a normal state $\varphi$ with

The complete trace-density construction writes $\varphi(x)=\sigma(ax)$, where $a\in L^1(N,\sigma)_+$, $\sigma(a)=1$. Put $h=a^{1/2}$. The square-root estimate proved in the trace-estimate section below, together with the isometric trace pairing and the conjugated density $u^*au$, gives

We replace $h$ by one normalized operator of finite spectrum and finite-trace support. Define

These functions increase pointwise to $t$ and lie between zero and $t$. Scalar dominated convergence for the spectral trace measure gives $h_m\to h$ in $L^2$. Each $h_m$ has finitely many positive spectral values, and its support is contained in $1_{[2^{-m},\infty)}(h)$, whose trace is at most $2^{2m}\|h\|_{2,\sigma}^2<\infty$. For sufficiently large $m$, $h_m\ne0$ and

Normalization is harmless: its distance from $h_m$ equals $|1-\|h_m\|_{2,\sigma}|\to0$. The triangle inequality in the finite Hilbert direct sum and unitary trace invariance give

Write $z=\sum_{j=1}^s\lambda_j e_j$, with distinct $0<\lambda_1<\cdots<\lambda_s$ and nonzero finite-trace spectral projections $e_j$. Set

These are finitely many nonzero finite-trace projections, and telescoping gives the exact normalization

We give the full noncommuting weighted estimate for each $u\in F$. Put $f_b=u^*e_bu$, $e_0=1-\sum_{a=1}^s e_a$ and $f_0=1-\sum_{b=1}^s f_b$. For $(a,b)\ne(0,0)$, define

The finite trace-product and cyclicity proofs justify these values, including when just one index is zero. Each value is finite because at least one projection has finite trace. The row and column sums for positive indices are $\sigma(e_a)$ and $\sigma(f_b)$.

Expanding squared trace norms in these two spectral partitions gives

For a fixed pair $(a,b)$, the weighted sum of the squared differences of its two level indicators equals

The same trace-square expansion for $p_j-u^*p_j u$, followed by this finite identity and scalar Cauchy--Schwarz with the weights $\nu_{ab}$, therefore proves

This calculation assumes no commutation between $e_a$ and $f_b$.

The second norm on the right is at most $2$. Sum (IF7) over $u$, and apply Cauchy--Schwarz in that finite index set and (IF5):

If every $p_j$ had summed squared error at least $\varepsilon^2\sigma(p_j)$, (IF6) would make the left side at least $\varepsilon^2$. Hence one of this finite list satisfies

Choose that $p=p_j$. Every individual summand obeys this strict bound. Conjugation by $u$, and then taking square roots, gives (2). The proof has selected an actual member of a finite spectral list, while retaining arbitrary semifinite $N$ and a possibly singular invariant state. $\square$

Second proof, by continuous spectral levels. Normal states are weak* dense in the state space. Otherwise real separation gives a self-adjoint $a\in N$ whose value at some state exceeds its supremum over normal states. That supremum is the top of the spectrum: a nonzero spectral projection near the top supports a normal state attaining values arbitrarily near it. Every state has value at most that spectral bound, a contradiction.

Let $n=|F|\ge1$. In the Banach space $\bigoplus_{u\in F}N_*$, with sum norm, consider the convex set

Weak* approximation to $\psi$ puts zero in its weak closure: pairing each coordinate against an element of $N$ tends to zero. The dual of this finite predual sum is the corresponding product of $N$'s. Hahn–Banach separation makes the weak and norm closures of a convex set equal. Hence, for any $d>0$, choose a normal state with

The trace-density theorem writes $\varphi(x)=\sigma(a x)$, with $a\in L^1(N,\sigma)_+$, $\sigma(a)=1$. Put $h=a^{1/2}$. The density of $\varphi\circ\operatorname{Ad}(u)$ is $u^*a u$. The first estimate in (1) and isometry of the trace pairing give

The other two estimates, conjugation of spectral projections, and Cauchy–Schwarz in the finite index set imply

Choose $d=\varepsilon^4/(16n)$. The first integral is less than $\varepsilon^2/2$. Some level has nonzero projection $p=P_t(h)$ and summed squared conjugation error less than $\varepsilon^2\sigma(p)$; otherwise integration contradicts (6). Each summand satisfies that bound. Conjugation by $u$ turns its norm into $\|upu^*-p\|_{2,\sigma}$, proving (2). If $F$ is empty, semifiniteness supplies any nonzero finite-trace projection. $\square$

The square root in $2\sqrt{nd}$ determines the fourth-power choice of $d$. A linear estimate in $d$ does not follow from the square-root density inequality.

Semifinite square-root and spectral-level estimates

For $h\in L^2(N,\sigma)_+$, let $P_t(h)=1_{(\sqrt t,\infty)}(h)$, $t>0$. The three estimates needed for projection selection are

Every positive-level projection has finite trace, since $\sigma(P_t(h))\le t^{-1}\|h\|_{2,\sigma}^2$.

Proof of the three trace estimates

For the first inequality, put $c=h-k$, and let $p$ and $q$ be the support projections of its positive and negative parts $c_+,c_-$. Then $z=p-q$ is a self-adjoint contraction and $cz=zc=|c|$. Hölder puts all products below in $L^1$, and trace cyclicity gives

The last two traces are nonnegative. Indeed, the trace pairing of positive $L^2$ elements is nonnegative: truncate them to bounded positive elements of finite spectral support, use $\sigma(ab)=\sigma(a^{1/2}ba^{1/2})\ge0$, and pass to the limit by Hölder. Thus

This argument includes unbounded measurable $h,k$; no finiteness of $\sigma(1)$ is used.

For the second estimate, first suppose

have finitely many strictly positive spectral values, with finite-trace spectral projections. Set $e_0=1-\sum_i e_i$, $f_0=1-\sum_j f_j$, and $\lambda_0=\mu_0=0$. For every pair $(i,j)\ne(0,0)$, put

These quantities are finite, including when exactly one index is zero. Their row and column sums are $\sigma(e_i)$ and $\sigma(f_j)$ for positive indices. Expansion of the squared trace norm gives

The same expansion for the level projections, followed by integration, yields

Here the integral of the difference of two level indicators is the distance between their squared endpoints; the last step is the scalar Cauchy–Schwarz inequality. No joint spectral resolution of $h$ and $k$ has been assumed.

For general positive $h,k\in L^2(N,\sigma)$, define

Each has finitely many positive values and finite-trace support. The functions $a_n$ increase to the identity and are bounded above by it, so $h_n\to h$ and $k_n\to k$ in $L^2$. For each $t>0$, their level projections increase to $P_t(h)$ and $P_t(k)$. Normality of $\sigma$ and finiteness of these level traces give convergence in $L^2$. Fatou's lemma now passes the preceding integral inequality to $h,k$. Finally, the ordinary scalar layer-cake identity, applied to the spectral trace measure of $h$, gives the third estimate in (1). This also proves every integration and finiteness claim used below.

4. Exercises with complete solutions

Coordinates and near-covering corners

Exercise 5. Check the Gram correction in (7).

Solution. Let $Y:\mathbb C^m\to H$ have columns $y_i$. Then $Y^*Y=G$. The corrected map $X=YG^{-1/2}$ has $X^*X=1$. Its columns are orthonormal, and each is a finite scalar linear combination of bounded elements of $M$.

Exercise 6. Prove (9).

Solution. The rank-one difference is $K_{x-\xi,x}+K_{\xi,x-\xi}$. Each rank-one norm is the product of its two vector norms. The triangle inequality proves (9), and finite sums give convergence of the range projections.

Exercise 7. Verify $K_{b,c}R_w=K_{b,cw^*}$.

Solution. On a bounded vector $\xi$, the left side is $b\,\tau(c^*\xi w)=b\,\tau(wc^*\xi)$. Since $(cw^*)^*=wc^*$, this equals $K_{b,cw^*}\xi$. Density and boundedness extend the equality to $H$.

Exercise 8. Bound the discarded trace in Lemma 3.1.

Solution. Summing the bad atoms' strict inequalities gives $b^2\sum_{\mathrm{bad}}\tau(p_k)<b^4$. Divide by $b^2$. The retained trace is greater than $1-b^2$, so at least one atom remains.

Exercise 9. Why include $1$ in $F$ although its conjugation error is zero?

Solution. Its coefficients in (11) are $\tau(x_i^*x_j)=\delta_{ij}$. The compressed relations therefore give almost orthonormal Gram matrices. Their repair supplies partial isometries with one common initial projection; the other tests determine the approximation coefficients.

Exercise 10. Distinguish the hypotheses of the three results.

Solution. Theorem 1.1 allows any faithful normally semifinitely tracial algebra and a state invariant under the specified unitaries. Corollary 2.1 uses a finite injective algebra with faithful normal tracial state and a hypertrace on $B(L^2(M,\tau))$. Lemma 3.1 additionally uses separable predual and a type $\mathrm{II}_1$ factor, the hypotheses of scalar pinching. The first two results also apply off factors.

Normal states and spectral budgets

Exercise 1. Why does weak* approximation suffice for the weak closure assertion in (3)?

Solution. A continuous linear functional on the finite predual sum is a list $z_u\in N$. On (3) it takes value $\sum_u\varphi(z_u-u z_u u^*)$. Normal states tending weak* to $\psi$ make this tend to zero by invariance. Thus every weak neighborhood of zero meets the set. Norm separation of its convex closure would also be weak separation.

Exercise 2. Verify the normal-state density argument when the spectral supremum is not an eigenvalue.

Solution. For $s=\sup\operatorname{Sp}(a)$, the projection $e=1_{(s-\eta,s]}(a)$ is nonzero for every $\eta>0$. A nonzero normal positive functional supported in $e$, normalized by its value at $e$, is a normal state with value between $s-\eta$ and $s$. Let $\eta\downarrow0$.

Exercise 3. Check the fourth-power budget for three unitaries.

Solution. Set $d=\varepsilon^4/48$. Then $2\sqrt{3d}=\varepsilon^2/2$. The integrated rank is one, so some level has total squared error less than $\varepsilon^2$ times its trace. Each summand obeys that bound; take square roots.

Exercise 4. Why does a finite-trace projection need not have finite Hilbert-space rank?

Solution. In a type $\mathrm{II}_1$ factor every projection has finite trace, while nonzero corners act on infinite-dimensional spaces. Finite rank follows in Corollary 2.1 because the ambient algebra is $B(H)$ and its ordinary trace counts the range dimension.

References and proof scope

Sorin Popa, A short proof of “injectivity implies hyperfiniteness” for finite von Neumann algebras, Journal of Operator Theory 16 (1986), 261–272. Section 2, pp.266–267, develops normal-state convexity, square-root densities and spectral projection selection. Remark 2.3, p.270, gives a direct square-root calculation. The proofs here retain arbitrary semifinite algebras and positive measurable $L^2$ elements, with the full trace-domain, noncommuting spectral-pairing and limit arguments supplied.

The finite-selection proof normalizes one finite-spectrum approximation to the square-root density and uses its finite list of weighted spectral projections. Exact two-partition trace weights give the simultaneous error bound without assuming commutation of the original positive operators. The complete continuous-level alternative is also included, with its fourth-power normal-state budget.

Claire Anantharaman and Sorin Popa, An introduction to II₁ factors, author draft, Lemma 10.2.6, Theorems 10.2.7–10.2.9, Theorem 10.3.1 and Remark 10.3.2, pp.163–166, give the normal-state and finite-rank methods. Proposition 10.3.3 and Theorem 10.3.4, pp.167–169, give the finite spectral-pairing and projection-level argument. The semifinite extensions in Proposition 10.3.5 and Theorem 10.3.6 have details left to the reader; the full proofs and measurable trace domains used here are supplied locally. The finite-rank basis is corrected explicitly by its Gram inverse square root, with quantitative Hilbert–Schmidt control.

The near-covering family in Lemma 3.1 is a direct consequence of the complete scalar pinching theorem, with indexed multiplicities, orthogonal corner errors and the discarded-trace budget proved here. Popa's Lemma 2.3, pp.268–269, and Anantharaman–Popa's Lemma 11.1.11, p.179, select one nonzero corner with local scalar means. They do not supply the global trace coefficients and near-covering conclusion of (11). The scalar-pinching premise includes the decreasing-expectation argument of Anantharaman–Popa, Lemma 11.1.9, p.178, with the full local proof and precise hypotheses.