Hyperinvariant subspaces: weighted shifts and backward transport

Source and verification

This article explains family 293 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: three forms of invariance

Let $T$ be a bounded operator on a complex Hilbert space $H$. A closed subspace $L$ is invariant when $TL\subseteq L$. If $p$ is its orthogonal projection, this is equivalent to $(1-p)Tp=0$; it does not require $pT=Tp$.

The commutant $\{T{\}}^{\prime }$ consists of all bounded operators commuting with $T$. A subspace is hyperinvariant for $T$ when every member of $\{T{\}}^{\prime }$ leaves it invariant. An operator algebra is transitive when its only common closed invariant subspaces are zero and $H$. Thus a transitive commutant is exactly the absence of nontrivial hyperinvariant subspaces.

A von Neumann algebra $M\subseteq \mathcal{B}(H)$ is a unital self-adjoint operator algebra closed in the strong operator topology, which means pointwise norm convergence on vectors. A factor has scalar center. A factor of type ${\mathrm{II}}_1$ has a faithful normal tracial state $\tau$, normalized by $\tau (1)=1$, and no nonzero minimal projections. A subspace is affiliated with $M$ when its orthogonal projection belongs to $M$.

These definitions give a useful bridge. If $T\in M$, the projection onto every hyperinvariant subspace of $T$ belongs to $M$. Indeed, each unitary in $M^{\prime }$ and its inverse commute with $T$. Hyperinvariance makes the subspace reducing for these unitaries, so its projection commutes with $M^{\prime }$ and lies in $M^{\prime \prime }=M$.

Finally, $T$ is norm-quasinilpotent if

$$\underset{n\to \infty }{\lim }\Vert T^n{\Vert }^{1/n}=0.$$

Equivalently, its spectrum is $\{0\}$. This allows $T\ne 0$ and does not by itself decide whether invariant or hyperinvariant subspaces exist.

🏷️ Two constructions in one family

The first manuscript works with a prescribed irrational angle $\theta \in (0,1)$. In the tracial representation, unitaries $U,V$ satisfy

$$VU=e^{2\pi i\theta }UV.$$

They generate the hyperfinite ${\mathrm{II}}_1$ factor $R_{\theta }$; hyperfiniteness means that finite-dimensional subalgebras approximate the factor in the strong topology.

Continuous weighted irrational rotations

For every prescribed irrational $\theta$, the manuscript constructs a continuous $f:\mathbb{T}\to [0,1]$ with exactly one zero, at $1$, and ${\int }_{\mathbb{T}}\log fdm=-\infty$, such that $T_f=Uf(V)$ is nonzero and norm-quasinilpotent. Every projection $p\in R_{\theta }$ satisfying $(1-p)T_{f}p=0$ is zero or one. No Diophantine restriction on $\theta$ is imposed. (OpenAI, 2026)

Here $m$ is normalized arc length. The conclusion excludes invariant projections in the specified containing factor. Arbitrary invariant subspaces of its Hilbert-space representation need not have their projections in that factor. The bridge above nevertheless excludes all nontrivial hyperinvariant subspaces.

A direct operator with transitive commutant

On every separable infinite-dimensional complex Hilbert space, the companion manuscript constructs a nonzero norm-quasinilpotent operator $S$ whose commutant is transitive, proper and strongly closed. Its concrete model has many ordinary invariant subspaces and satisfies

$$\begin{array}{c}\Vert S\Vert =e^{-10},\\ \Vert S^n\Vert \le e^{-10\lfloor (n+1{)}^2/4\rfloor }.\end{array}$$

(OpenAI, 2026a)

The direct model exposes a particularly concrete mechanism: each fiber has only coordinate-tail invariant subspaces, while a commuting operator pushes those tails backward. A probability-preserving transformation cannot sustain the resulting strict descent of tail indices.

🏷️ Weighted shifts over a probability space

Let $K={\ell }^2(\mathbb{Z})$ with basis $e_j$, and write

$$K_{\ge h}=\overline{\mathrm{span}}\{e_j:j\ge h\}.$$

The conventions $K_{\ge -\infty }=K$ and $K_{\ge +\infty }=0$ include the two trivial subspaces. A bilateral weighted shift sends $e_j$ to ${\beta }_j{e}_{j+1}$. Every coordinate tail is invariant when the weights are positive.

The base space $X={\mathbb{Z}}_2$ can be viewed as infinite binary digit strings, equipped with independent fair digits. Addition of one with carries gives an invertible probability-preserving map $\sigma x=x+1$. Its finite residue $T_m(x)=x\mathrm{mod}{2}^m$ is uniform on $\{0,\dots ,2^m-1\}$. Addition cycles these residue classes, which proves ergodicity: an invariant event is independent of every finite residue partition and hence of itself.

The Hilbert space is $H_0=L^2(X;K)$, whose vectors are square-integrable fields $x\mapsto f(x)\in K$. The operator acts fiberwise:

$$\begin{array}{c}(Sf)(x)=S_{x}f(x),\\ S_x{e}_j={\beta }_j(x)e_{j+1}.\end{array}$$

The weights must accomplish two tasks: classify invariant subspaces within a fiber and permit transport between fibers.

For completeness, the source’s logarithmic weights are explicit. Put $c=10$, $D=100$, $N_m=2^m$, and let block $I_m$ consist of $8^{m-1}\le k<8^m$. Set $J_m=D(N_m-2)$ and, for $k\in I_m$, define

$$\begin{array}{rl}{a}_k^{+}(x)& =c(k+1)+J_m+D{T}_m(x),\\ a_k^{-}(x)& =c(k+1)+J_m\\ & +D(N_m-1-T_m(x)).\end{array}$$

Take $a_0^{+}=a_0^{-}=c$, use $a_j=a_j^{+}$ for $j\ge 0$ and $a_j=a_{-j-1}^{-}$ for $j<0$, and set ${\beta }_j=e^{-a_j}$. The two central weights equal $e^{-10}$. The linear part of $a_j$ controls decay; the opposite residue corrections at the two ends control the backward transport.

🏷️ Why every fiber subspace is a tail

Define $w_0=1$ and $w_{j+1}/w_j={\beta }_j$. Multiplication by $w_j$ identifies the weighted space with norm ${\sum }_j|w_j{z}_j{|}^2$ with $K$, and carries right translation to $S_x$.

The companion proves a quadratic weighted-translation criterion. It requires $w$ positive and decreasing, $\log w_j$ convex for $j\le 0$ and concave for $j\ge 0$, and

$$\begin{array}{c}\sum _{j\ne 0}{w}_j^{1/j}<\infty ,\\ \frac{\log w_n}{n^2}⟶-a,\\ \frac{\log w_{-n}}{n^2}⟶a,\\ a>\log 3.\end{array}$$

For a negative index the summand is $w_{-n}^{-1/n}$, not $w_{-n}^{1/n}$. Under these hypotheses, all closed invariant subspaces of right translation are coordinate tails.

The proof converts invariance into vanishing correlations. If $z$ belongs to an invariant subspace and $b$ is a continuous linear functional annihilating that subspace, then

$$\sum _j{b}_j{z}_{j-r}=0(r\ge 0).$$

Convex envelopes of the logarithms of the coefficients locate two coefficients whose product dominates the middle of one correlation. With $\log 3<\lambda <a$, the other middle terms have total relative size at most

$$C_{\lambda }=2\sum _{n\ge 1}{e}^{-\lambda n}=\frac{2}{e^{\lambda }-1}<1.$$

Both infinite tails, divided by that same distinguished product, tend to zero. Summable adjacent-weight ratios provide the dominated-convergence bounds. Controlling unnormalized tails would be insufficient because the distinguished product can itself become small.

This forces the support of each annihilator to have a finite maximum and the support of each nonzero subspace vector to have a finite minimum, with the former no larger than the latter. The zero-shift correlation then separates the endpoint coordinates. Hahn–Banach identifies the subspace with exactly one coordinate tail. (OpenAI, 2026a)

For the constructed weights, the residue corrections are uniformly $O(k^{1/3})$, so cumulative logarithms satisfy

$$\begin{array}{c}\log w_n=-5n(n+1)+O(n^{4/3}),\\ \log w_{-n}=5n(n+1)+O(n^{4/3}).\end{array}$$

The criterion applies with $a=5$. Monotonicity across block boundaries follows from the increments of $J_m$ and the nesting of residues.

The same weights give a uniform power bound. If $d(j)=j+1$ for $j\ge 0$ and $d(j)=-j$ for $j<0$, then $a_j(x)\ge 10d(j)$. The smallest sum of $n$ consecutive values of $d$ is $\lfloor (n+1{)}^2/4\rfloor$, obtained by centering the interval around $-1,0$. This proves the stated bound on $\Vert S^n\Vert$ uniformly over $x$, rather than merely proving quasinilpotence separately in each fiber.

🏷️ A commuting operator that moves tails backward

Seek a map $V_x{e}_j=b_j(x)e_{j-1}$ from the fiber at $x$ to the fiber at $\sigma x$. The associated global operator is defined by $(Vf)(\sigma x)=V_{x}f(x)$. Commutation requires

$$b_{j+1}(x){\beta }_j(x)=b_j(x){\beta }_{j-1}(\sigma x).$$

This recurrence has a positive measurable solution ${\tilde{b}}_j$ once ${\tilde{b}}_0=1$ is fixed. The issue is boundedness in both directions.

Except when $T_m(x)=2^m-1$, adding one increases the positive-end logarithmic correction by $D$ and decreases the negative-end correction by $D$. The wrap events have probabilities $2^{-m}$, whose sum is finite. Borel–Cantelli therefore leaves only finitely many exceptional blocks almost surely; independence of those events is unnecessary.

Summing the logarithmic recurrence forward and backward gives, for a pointwise finite constant $C(x)$,

$$\begin{array}{c}\log {\tilde{b}}_{\pm n}(x)\\ \le -90n+C(x)+O(n^{1/3}).\end{array}$$

Thus ${\tilde{b}}_j(x)$ tends to zero at both ends. Its countable supremum $L(x)$ is measurable and finite almost everywhere. Dividing every coefficient at the same $x$ by this common $L(x)$ preserves the recurrence and makes $0<b_j(x)\le 1$. Neither integrability of $C(x)$ nor a uniform bound on the unnormalized coefficients is required.

Probability preservation makes $V$ a contraction. Positivity of every coefficient gives

$$\overline{V_x{K}_{\ge h}}=K_{\ge h-1},\overline{V_{x}K}=K.$$

These are dense-image statements; no bounded inverse is asserted.

Now let a closed subspace be invariant under $S,V$ and every scalar multiplier $M_g$, where $(M_{g}f)(x)=g(x)f(x)$. Since adjoints of the multipliers are included, its projection is a measurable field $P_x$. Fiber classification gives $\mathrm{ran}{P}_x=K_{\ge s(x)}$ for a measurable extended-integer label $s$. The backward map forces

$$s(\sigma x)\le s(x)-1whens(x)\in \mathbb{Z}.$$

A full fiber also maps densely to a full fiber. Consequently the set $E=\{s=-\infty \}$ is forward invariant modulo null sets, hence invariant by probability preservation. Ergodicity makes its measure zero or one.

If its measure is zero, put $F_j=\{s\le j\}$. The inclusions $\sigma F_j\subseteq F_{j-1}\subseteq F_j$ force equal measures. Decreasing $j$ gives $\mu (F_j)=\mu (E)=0$. Thus $s=+\infty$ almost everywhere. The only common invariant subspaces are therefore zero and $H_0$.

All these generators commute with $S$, so $\{S{\}}^{\prime }$ is transitive. It is strongly closed because the equation $AS=SA$ passes to strong limits. It is proper because $S$ is nonscalar. The ordinary invariant subspaces $L^2(X;K_{\ge h})$ remain present, but $V$ moves each of them out of itself.

🏷️ The circle construction: a finite convolution obstruction

The other manuscript excludes invariant projections directly inside the rotation factor. In its concrete representation on $L^2(\mathbb{R}/\mathbb{Z};{\ell }^2(\mathbb{Z}))$,

$$\begin{array}{c}(U\xi )(x)=S_0\xi (x),\\ S_0{e}_k=e_{k+1},\\ (\pi (a)\xi )(x{)}_k=a(x+k\theta )\xi (x{)}_k.\end{array}$$

Elements of the factor have measurable operator fields with covariance

$$\begin{array}{c}p(x+\theta )=S_0^{-1}p(x)S_0,\\ \tau (p)=\int p(x{)}_{00}dx.\end{array}$$

For a positive-almost-everywhere weight, covariance and ergodicity imply that a nontrivial invariant projection has $0<p(x{)}_{00}<1$ almost everywhere. Thus both $p(x)e_0$ and $(1-p(x))e_0$ have nonzero central coordinates.

The initial weight is measurable. It places very deep logarithmic drops on rare sets, using divisible frequencies $q_i$:

$$\begin{array}{c}{g}_{\ast }(x)=\sum _i{c}_i\lfloor \{q_{i}x\}+\{q_i\theta \}\rfloor ,\\ f_{\ast }(x)=e^{-g_{\ast }(x)}.\end{array}$$

Only finitely many drops occur almost everywhere, but their expected total cost is infinite. The angle stays fixed throughout the construction.

Invariance gives weighted orthogonality identities along positive orbit pairs. Sampling and compact-set approximation extend them to carefully coupled endpoints $x,y$. A mixed-radix expansion supplies independent digits for these couplings. The characters $\{q_{i}x\}$ themselves are dependent, since they use overlapping digit tails. Uniform $L^2$ bounds on both marginal densities prevent concentration on the exceptional sets where measurable projection fields lack continuity. (OpenAI, 2026b)

To specify the weights in the identities, let $F_0=1$ and, for $n\ge 1$, set

$$\begin{array}{c}{F}_n(x)=\prod _{r=0}^{n-1}{f}_{\ast }(x+r\theta ),\\ F_{-n}(x)=1/F_n(x-n\theta ).\end{array}$$

These are positive finite functions almost everywhere. Put $\xi (x)=p(x)e_0$ and $\gamma (y)=(1-p(y))e_0$, and define

$$\begin{array}{c}{A}_v(x)={\xi }_v(x)/F_v(x),\\ B_w(y)=F_{-w}(y)\overline{{\gamma }_{-w}(y)}.\end{array}$$

The resulting exact identities have the form

$$\begin{array}{c}\sum _{v\in \mathbb{Z}}{A}_v(x)B_{-n-v}(y)=0\\ (n\ge 0).\end{array}$$

On an event of fixed positive probability, the coefficients in a finite window are bounded and the two central coefficients are bounded away from zero. Two Laurent polynomials with these properties have at least one negative-or-zero product coefficient bounded away from zero, uniformly over the compact set of admissible pairs. To see nonvanishing, take the first nonzero exponent of each polynomial: their sum is nonpositive and its product coefficient cannot cancel.

The infinite tail must cancel this finite contribution. Put $R(k)=F_k(y)/F_k(x)$. Coordinate order and monotonicity of $F_k$ give

$$\begin{array}{c}\frac{F_{n+v}(y)}{F_v(x)}\\ \le \min \{R(v),R(n+v)\}\le 1.\end{array}$$

The last bound and the vector norm bounds ensure absolute convergence of the convolutions. If either $v$ or $n+v$ lies outside the window, its ratio can control the omitted term. For each such index $k$, split the exceptional event into two parts. On the good event $E$ containing the central-coefficient bounds, the construction gives

$$\begin{array}{c}R(k)\le \eta +1_{H_k^{(1)}}+1_{H_k^{(2)}}\text{on }E,\\ \mathbb{P}(H_k^{(1)}\mid x)\le \delta ,\\ \mathbb{P}(H_k^{(2)}\mid y)\le \delta .\end{array}$$

These conditional estimates hold under the full coupling law, before restricting to the event with good central coefficients. For the first exceptional sum, place its indicator beside the squared coordinates of the vector depending on $x$ and condition on $x$. Cauchy–Schwarz on probability times counting measure bounds the expected sum by $\sqrt{\delta }$. For the second part, place the indicator beside the squared coordinates depending on $y$. Each vector has squared coordinate sum at most one.

Thus no union bound over infinitely many exceptional indices is needed. If the good event has probability at least $\rho$ and the finite coefficient bound is $\epsilon$, a window of radius $K$ yields

$$\rho \epsilon \le (2K+1)(\eta +2\sqrt{\delta }).$$

The parameters are chosen so that the right side is at most $\epsilon /j$ at stage $j$, while $\rho$ stays positive. This contradicts the exact convolution identities for sufficiently large $j$.

🏷️ Continuity through an unbounded intertwiner

The measurable weight must still be replaced by a continuous weight with exactly one zero. The construction smooths each logarithmic drop, moves its support into an arc shrinking to zero, and adds a nonnegative background forcing divergence there. It obtains

$$\begin{array}{c}g=g_{\ast }+1+b\circ \sigma -b,\\ \sigma x=x+\theta ,\end{array}$$

where $b$ is finite almost everywhere but may be unbounded and nonintegrable. The function $g$ is finite and continuous away from zero and tends to infinity there. Hence $f(e^{2\pi ix})=e^{-g(x)}$ has the required continuity and zero set.

Multiplication by $e^{b(x+k\theta )}$ on coordinate $k$ defines an injective positive affiliated operator $H_b$. The coboundary identity gives the domain-preserving intertwining relation

$$\begin{array}{c}{H}_b{T}_f\xi =e^{-1}{T}_{\ast }{H}_b\xi \\ (\xi \in \mathrm{Dom}{H}_b).\end{array}$$

This is not a bounded similarity. The source proves that transporting an invariant projection’s range through this affiliated operator preserves its trace. For spectral cuts $E_N=1_{[1/N,N]}(H_b)$, polar decomposition on $p\wedge E_N$ gives the lower bound for the transported trace. Applying the inverse operator to $(1-p)\wedge E_N$ gives the complementary bound. Letting $N$ increase proves equality of traces and transfers the projection obstruction.

The infinite logarithmic integral follows directly from the nonnegative costs; integrating the coboundary would be invalid without integrability of $b$. Quasinilpotence follows from the continuous truncated logarithms $\log \max (f,e^{-M})$. Their integrals decrease to $-\infty$, while irrational-rotation averages converge uniformly. Choosing $M$ first and then the orbit length bounds $\Vert T_f^n\Vert$ by $e^{-an}$ for any prescribed $a>0$.

🏷️ Product realization and spectral distinctions

The circle manuscript includes a separate product-of-circles model with genuinely independent coordinates. It has a positive-almost-everywhere measurable weight, infinite logarithmic cost, zero determinant, norm-quasinilpotence and no nontrivial invariant projection in its containing finite factor. The Fuglede–Kadison determinant here is $\exp (\tau (\log |T|))$, with $\exp (-\infty )=0$.

For this selected product model, $|T|$ has pure-point spectral type, and every finite family of polar conjugates generates an atomic abelian algebra. The full family generates a diffuse algebra, with no minimal projections. Finite joint tuples have countable range; an atom for the entire orbit would have a finite orbit of equal-trace atoms, contradicting total ergodicity unless the algebra were scalar. The weight is nonconstant, excluding that case.

The product model’s Brown measure, the tracial spectral-distribution measure defined from the logarithmic determinant potential, is ${\delta }_0$, since its spectrum is $\{0\}$. A point-mass Brown measure alone does not exclude invariant projections. Likewise, pure-point spectral type does not require every spectral point to be an eigenvalue.

The direct companion also has finite atomic and full diffuse polar-conjugate algebras. Its fibers are compact weighted shifts, but the global operator is not compact: an orthonormal sequence of scalar functions tensored with $e_0$ has images of the same nonzero norm. Its scalar multiplier algebra has infinite multiplicity and is not maximal abelian in $\mathcal{B}(H_0)$, so the classical density theorem for transitive algebras containing a maximal abelian algebra does not apply.

🏷️ Key steps and proof outlines

Prove trace preservation for an unbounded transport

The rotation construction replaces a measurable weight by a continuous one through an affiliated positive operator $H$. Suppose $H$ is injective, $A$ preserves its domain, and $HA\zeta =BH\zeta$ there. For an $A$-invariant projection $p$, let $P$ project onto the closure of $H(\mathrm{ran}p\cap \mathrm{Dom}H)$.

Every commutant unitary preserves the domain and the range of $p$, and commutes with $H$ on that domain. It therefore preserves the transported closed space in both directions, giving $P\in M$. The intertwining identity makes that space $B$-invariant. The remaining issue is whether this transport could collapse a nontrivial projection to zero or inflate it to one. (OpenAI, 2026b)

Put $E_N=1_{[1/N,N]}(H)$. On $q_N=p\wedge E_N$, the operator $H{q}_N$ is bounded and bounded below. Its polar decomposition identifies $q_N$ with a projection below $P$, so

$$\tau (P)\ge \tau (p\wedge E_N)\ge \tau (p)+\tau (E_N)-1.$$

The meet inequality follows by polar decomposition of $(1-E_N)p$: its initial support is $p-(p\wedge E_N)$ and its final support is at most $1-E_N$. Since injectivity gives $E_N\uparrow 1$, this proves $\tau (P)\ge \tau (p)$.

For the opposite bound, transport $(1-p)\wedge E_N$ by $H^{-1}$. Its image is orthogonal to the range of $P$, because

$$⟨H^{-1}\eta ,H\zeta ⟩=⟨\eta ,\zeta ⟩=0.$$

The same polar argument gives $\tau (1-P)\ge \tau (1-p)$. Thus equality holds. Both directions are needed: injectivity alone would not justify preservation of trace by an unbounded map.

Separate infinite logarithmic cost from pointwise divergence

Smoothing and relocating logarithmic drops gives nonnegative continuous terms $h_i^{\prime }$ supported in arcs shrinking to zero. Their sum is locally finite away from zero, and Tonelli gives infinite integral. Neither fact alone says the sum tends to infinity as the point approaches zero.

The source adds a background built from bumps $v_j$ with $v_j(0)=1$ and short supports $I_j$. Choose $W_j=2^{j+1}$ so the first $W_j$ translates of each support are disjoint and ${\sum }_j{W}_j\mu (I_j)<\infty$. Then

$$g_0=1+\sum _j\left(v_j-\frac{1}{W_j}\sum _{n=1}^{W_j}{v}_j\circ {\sigma }^{-n}\right).$$

Each negative averaged bump has norm at most $1/W_j$, so their sum is uniformly bounded by $1/2$. Every finite collection of positive bumps tends to its number of terms near zero. Hence $g_0\ge 1/2$ and $g_0(x)\to \infty$ as $x\to 0$. The transfer functions have summable supports, making their sums finite almost everywhere by Borel–Cantelli. This proves the coboundary identity without needing integrability of the transfer. Exponentiation now gives continuity and exactly one zero, while the separate nonnegative integral calculation gives infinite logarithmic cost.

A stationary probability law cannot support a descending finite tail index

In the direct companion, fix one candidate common invariant subspace. Scalar multipliers and their adjoints force its projection to be a measurable field $P_x$. Apply global invariance to the countable basis vectors, then intersect the resulting conull set over integer iterates of the base transformation. This produces one invariant conull set for this fixed subspace; no uncountable intersection over all candidate subspaces is used. (OpenAI, 2026a)

Fiber classification gives $\mathrm{ran}{P}_x=K_{\ge s(x)}$, with measurable label because $\{s\le j\}=\{⟨P_x{e}_j,e_j⟩=1\}$. The backward intertwiner gives $s(\sigma x)\le s(x)-1$ at finite labels and preserves full fibers.

Let $E=\{s=-\infty \}$. Forward invariance and probability preservation make it invariant modulo null sets, so ergodicity makes its measure zero or one. If zero, set $F_j=\{s\le j\}$. The inclusions $\sigma F_j\subseteq F_{j-1}\subseteq F_j$ force all these probabilities equal. Continuity from above sends them to $\mu (E)=0$. Their countable union excludes every finite label, leaving only the zero fiber almost everywhere. This closes transitivity even though the direct operator retains many ordinary invariant subspaces.

🏷️ Connections and verification boundaries

Bicentralizer spectral recovery also depends on controlling representatives before passing to operator limits. Here the corresponding issues are normalization of backward intertwiners and domains of affiliated operators. The nuclear-ultrapower obstruction similarly turns an analytic construction into a final finite or measure-theoretic contradiction.

The direct companion’s principal proof sections and the circle paper’s projection-field, coupling, convolution and continuity arguments were read. The full prescribed-angle parameter construction and the product appendix’s separate sampling and obstruction proofs were not independently audited. The conclusions above remain attributed to the manuscripts; ordinary invariance, factor affiliation and hyperinvariance retain their distinct meanings.

🏷️ Manuscripts and verification scope

• Invariant-projection counterexamples for every irrational rotation — pinned manuscript and companion source.
• Backward intertwiners and a transitive commutant — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The documented selections include the prescribed-angle continuous single-zero weight and finite-factor invariant-projection obstruction, a product realization and its selected Brown measure δ₀, and a nonzero norm-quasinilpotent operator with a proper strongly closed transitive commutant on every separable infinite-dimensional complex Hilbert space. These statements distinguish affiliated invariant projections from arbitrary ordinary invariant subspaces. 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 2026a. Backward intertwiners and a transitive commutant.

🐻  OpenAI 2026b. Invariant-projection counterexamples for every irrational rotation.