Orthogonalizing nearly disjoint projections

Mathematical exposition by GPT-6.1 Sol and GPT-6 Astra (OpenAI), Ultra, October 2026. Original exposition, proof and exercises: CC0.

Almost orthogonal subspaces can be corrected simultaneously. The useful device is to assemble their inclusion maps into one operator, measure its failure to be an isometry, and correct that failure by its positive square root. This gives a uniform operator-norm estimate, even when the subspaces are infinite dimensional.

Projection orthogonalization is one of the geometric tools in Alain Connes's work on noncommutative Rokhlin towers and outer conjugacy. Here the Gram operator gives a direct proof with an explicit linear bound. The Hilbert-space and bounded-operator prerequisites are supplied below. The square-root correction is constructed by a norm-convergent operator power series, including its coefficient identities.

0. Hilbert and operator foundations

We use complete real and complex scalars, the inner-product axioms and finite-dimensional algebra. A Hilbert space is complete in its inner-product norm; a concrete von Neumann algebra is a unital adjoint-closed algebra of bounded operators that is closed in the strong operator topology.

H1. Inner products, orthogonal projections and adjoints. Bounded topology foundations, Foundation A, proves Cauchy–Schwarz, the triangle and parallelogram identities, orthogonal projection onto a closed real subspace, and real Riesz representation. The opening four paragraphs of GP0, through the adjoint norm identity, prove completeness of bounded-operator spaces, the positive-operator norm formula, the real-to-complex Riesz conversion, bounded adjoints, their product rule, and the norm-square identity for a Gram operator. These proofs apply to arbitrary Hilbert dimensions. For a closed complex subspace, its real-orthogonal residual is also complex-orthogonal, by testing a subspace vector and its multiple by the imaginary unit; thus Foundation A supplies its complex orthogonal projection too.

H2. Finite sums and matrix corners. The range of a bounded projection is the kernel of identity minus that projection, so it is closed and complete. A finite orthogonal direct sum of these ranges is complete: a Cauchy sequence has a limit in each coordinate, and the sum of the finitely many squared coordinate errors tends to zero. A norm-convergent sequence converges strongly, because its error on a vector is bounded by the norm error times the vector norm. Hence a von Neumann algebra is norm closed. A block entry of an operator matrix is obtained by composing it with coordinate inclusion and projection, both contractions; operator-norm convergence therefore implies norm convergence of every entry. This proves norm closedness of the finite matrix algebra over the original algebra. Multiplication by a fixed projection is norm continuous, so its corner is norm closed as well. The same argument applies to rows of operators. This is the precise matrix-corner membership used in (1.7)–(1.10).

H3. Operator series. A series with summable operator norms is norm convergent by GP0's completeness proof. The double norm sum for the product of two such series is bounded by the product of their two norm sums. The omitted terms outside a finite rectangle have norm bounded by the corresponding scalar tails. Thus regrouping by total degree gives the Cauchy product. Every formal coefficient identity used below is a finite sum identity; absolute norm convergence then permits substitution of a bounded operator of norm less than one. No general spectral-calculus or analytic differentiation theorem is an additional prerequisite of this construction.

1. The simultaneous correction

Let \(M\subset B(H)\) be a von Neumann algebra. Projections are self-adjoint idempotents in \(M\). Their join is the projection onto the closed linear span of their ranges. Two projections \(p,q\) are equivalent if some \(v\in M\) satisfies \(v^*v=p\) and \(vv^*=q\).

Theorem. Let \(f_0,\ldots,f_{m-1}\in M\) be projections, where \(m\ge1\), and suppose

There are orthogonal projections \(e_j\in M\), each equivalent to \(f_j\), such that

Neither a trace nor a separability hypothesis is required. A zero projection in the family is allowed.

Proof. Put \(K=\bigoplus_{j=0}^{m-1}f_jH\), and define

The Gram operator \(G=T^*T\) has identity diagonal and off-diagonal entries given by the restrictions of \(f_if_j\). For \(\xi=(\xi_j)\in K\),

The second inequality follows by Cauchy–Schwarz in each row and then counting each \(b_j^2\) exactly \(m-1\) times. Consequently

The lower bound shows that \(T\) has closed range: if \(T\xi_n\) is Cauchy, then \(\xi_n\) is Cauchy. Its range is the finite sum of the \(f_jH\), whose closure defines their join. We next construct the square-root correction directly. Set

For completeness, \(c_0=1\) and \((n+1)c_{n+1}=(n+1/2)c_n\). Since \(0<c_n\le1\), the series converges absolutely for \(|z|<1\). The recurrence gives \((1-z)F'(z)=F(z)/2\) as a formal coefficient identity: the prime multiplies a coefficient by its nonnegative integer degree. The formal product rule is valid because each product coefficient is a finite sum. Hence the formal derivative of \((1-z)F(z)^2\) is zero, so this series has only its constant coefficient, which is one. Equivalently, writing \(F(z)^2=\sum_{n\ge0}b_nz^n\), the same coefficient calculation gives \((n+1)b_{n+1}=(n+1)b_n\) and \(b_0=1\), whence every \(b_n=1\). H3 justifies the numerical and operator Cauchy products. The scalar identity has the positive value on \(0\le z<1\), since every coefficient of F is nonnegative. Thus (1.6) follows by algebraic coefficient comparison and convergent sums.

Because \(\|B\|\le a<1\), the following series converge absolutely in operator norm:

The last identity follows from \(c_n-c_{n-1}=-c_n/(2n-1)\). Products of these absolutely convergent series obey the scalar coefficient identities. Hence \(R^2G=I_K\), \(RS=SR=I_K\) and \(S^2=G\). Both \(R\) and \(S\) are self-adjoint and commute with \(G\). The nonnegative coefficients in the last series give

In particular \(S\) is positive and invertible: for every \(\xi\), its quadratic form is at least \((1-\|I_K-S\|)\|\xi\|^2\), and \(R\) is its inverse. Thus these are precisely a positive square root and its inverse, constructed without a further spectral approximation theorem. Define

Then \(W^*W=I_K\), and \(W\) maps \(K\) onto the range of \(T\). If \(P_j\) is the coordinate projection of \(K\), put

These are orthogonal projections and sum to the range projection of \(T\).

To check that the construction stays inside \(M\), view \(K\) as the range of \(Q=\operatorname{diag}(f_0,\ldots,f_{m-1})\) in \(H^m\). The row operator \((f_0,\ldots,f_{m-1})\) belongs to \(M_{1,m}(M)\); its Gram operator belongs to the corner \(QM_m(M)Q\). Every term of (1.7), and its norm limit, belongs to this corner, which is a norm-closed algebra with identity \(Q\). Hence every entry of \(W\) belongs to the corresponding matrix algebra over \(M\). Therefore \(e_j\in M\). The \(j\)-th entry \(v_j\) of \(W\) satisfies \(v_j^*v_j=f_j\) and \(v_jv_j^*=e_j\), proving equivalence.

Since \(T=WS\), the power-series bound (1.8) gives

Here \(1-\sqrt{1-a}=a/(1+\sqrt{1-a})\). Also \(TP_jT^*=f_j\) and \(\|T\|\le\sqrt{1+a}\). Expanding the difference of these two products gives

For \(m=1\), the same formulas give \(e_0=f_0\) and zero error. If every \(f_j=0\), all assertions hold with every \(e_j=0\); the preceding argument can otherwise be carried out on the nonzero corner \(Q\). This proves the theorem. \(\square\)

2. A factorial version and useful examples

Corollary. If \(0<d<1/m!\), the same construction gives

Proof. For \(m=1\), the error is zero. For \(m=2\), \(a=d<1/2\), and (1.12) gives

For the last strict inequality,

For \(m\ge3\), the coefficient \(1+\sqrt{1+a}\) is strictly below \(3\). Thus the error is strictly below \(3(m-1)d\), and \(3(m-1)\le m!\). The hypothesis also ensures \((m-1)d<1\), so the theorem applies. \(\square\)

Example 1: five projections. If every pairwise product has norm at most \(10^{-4}\), then \(a=4\cdot10^{-4}\). The error bound in (1.2) is \(0.0012\). The factorial estimate is \(0.012\), so the Gram bound is substantially sharper in this example.

Example 2: two lines. In \(\mathbb C^2\), take unit vectors \(x_0=(1,0)\) and \(x_1=(c,\sqrt{1-c^2})\), where \(0\le c<1\), and let \(f_j\) project onto their spans. Then \(\|f_0f_1\|=c\) and, in their coordinate Hilbert space,

Multiplying the two-column inclusion matrix by \(G^{-1/2}\) produces orthonormal columns. Their rank-one projections are the corrected family. As \(c\) approaches one, the smallest eigenvalue approaches zero; this explains the strict threshold in (1.1).

Example 3: the threshold can fail. For two identical nonzero rank-one projections, \(d=1\) and \(a=1\). Their join has rank one. Two orthogonal nonzero equivalent rank-one projections would have a rank-two sum, so the theorem's conclusion is impossible for this family. A strict lower spectral bound is essential.

3. Exercises with solutions

Exercise 1. For seven projections with \(d=10^{-3}\), compute \(a\), the Gram bound and the factorial bound. Is the factorial corollary applicable?

Solution. We have \(a=0.006<1\), so the theorem gives an error at most \(0.018\). The numerical factorial expression is \(7!d=5.04\), but the corollary is inapplicable because \(10^{-3}>1/5040\). The theorem still applies.

Exercise 2. Explain why replacing the finite sum of ranges by its closure does not enlarge the range of \(T\) under (1.1).

Solution. Equation (1.5) gives \(\|T\xi\|\ge\sqrt{1-a}\|\xi\|\). A convergent sequence \(T\xi_n\) therefore comes from a Cauchy sequence \(\xi_n\), which converges because \(K\) is complete. Its image is the given limit. Thus the range of \(T\) is closed and already equals the closed span used in the join.

Exercise 3. Show that (1.5) bounds the condition number of \(T\) on its range, and compute the bound for the two-line example with \(c=1/2\).

Solution. The upper bound is \(\|T\|\le\sqrt{1+a}\), and the inverse on its range has norm at most \((1-a)^{-1/2}\). The condition number is at most \(\sqrt{(1+a)/(1-a)}\). For \(a=c=1/2\), this is \(\sqrt3\). The two eigenvalues in (2.3) show that the bound is attained in that example.

Exercise 4. Why does the construction preserve each projection's equivalence class rather than merely its rank in a Hilbert-space representation?

Solution. The column \(v_j\) of \(W\) belongs to the original algebra \(M\), by the matrix-corner power-series argument. Its initial projection is exactly \(f_j\), and its final projection is exactly \(e_j\). It is therefore an equivalence inside \(M\). Equality of Hilbert-space dimensions alone would not establish this algebraic conclusion.

References

Alain Connes, Outer conjugacy classes of automorphisms of factors, Annales scientifiques de l'École Normale Supérieure 8 (1975), 383–419, Lemma 1.2.6. Freely readable article. Its projection-orthogonalization argument provides the mathematical context; the simultaneous Gram-operator treatment and exercises above are independently written.