Kasparov's technical theorem

Written by GPT-6.1 Sol (OpenAI). Public domain (CC0).

Constructing a Kasparov product requires combining two operators whose errors are compact in different directions. A partition of the identity in a multiplier algebra lets us assign one error to one part and the other error to the complementary part. The partition must also commute modulo the ideal with the operators already present. Kasparov's technical theorem supplies exactly this partition.

We prove the theorem, including its graded form, by constructing a strict sum of positive operators. The proof includes the required convexity argument, the reduction from $\sigma$-unital to separable data, and the estimates for each error. We also prove the multiplier quotient lifting used in a shorter separation argument when no commutator control is requested. Finally we use the theorem to extend an operator homotopy from an ideal; this is the technical input to the comparison of cycle relations.

We use the earlier continuous functional-calculus proofs. Theorem 1.1 proves its bounded-functional and strong-support inputs locally, using the GNS construction of Lesson 05, Lemma 3.0a. The ideal version of quasicentral approximate identities belongs to Lifting completely positive maps; Section 1 supplies the additional derivating-subspace version needed here. For a countably generated Hilbert module, the identification $\mathcal L(E)=M(\mathcal K(E))$ and $\sigma$-unitality of $\mathcal K(E)$ are proved in Lesson 05, Lemma 2.0.

1. Approximate identities with commutator control

An approximate identity means positive contractions $u_\lambda$ such that $u_\lambda a\to a$ and $au_\lambda\to a$ in norm for every $a$ in the algebra. A subspace $\Delta$ of an ambient C*-algebra derives a subalgebra $A$ if $[d,a]\in A\qquad(d\in\Delta,\ a\in A). \tag{1.1}$ Here the commutator is ordinary until grading is explicitly introduced. Deriving is weaker than preserving $A$ by left and right multiplication separately.

Theorem 1.1 (convex quasicentrality). Suppose $\Delta$ derives $A$. From any positive contractive approximate identity of $A$, finite convex combinations can be chosen to form an approximate identity quasicentral for $\Delta$: $\|[u_\lambda,d]\|\longrightarrow0\qquad(d\in\Delta). \tag{1.2}$ More precisely every tail of the original approximate identity has convex combinations meeting any finite collection of approximate-identity and commutator tests. If $A$ is $\sigma$-unital and $\Delta$ is separable, the combinations may be chosen as a sequence.

Proof. Write $C$ for the ambient C*-algebra. First every bounded complex linear functional $f$ on $C$ is a vector coefficient in some Hilbert-space representation of the forced unitization $C^+$. Here is a proof that needs only Hahn–Banach and the GNS construction proved in Lesson 05 Lemma 3.0a. The zero functional is immediate; normalize a nonzero one to $\|f\|=1$. In $M_2(C^+)$ use the unital self-adjoint subspace consisting of matrices $\begin{pmatrix}\lambda1&a\\b&\mu1\end{pmatrix}, \qquad a,b\in C,\quad\lambda,\mu\in\mathbb C.$ On it define $F\left(\begin{pmatrix}\lambda1&a\\b&\mu1\end{pmatrix}\right) =\tfrac12(\lambda+\mu+f(a)+\overline{f(b^*)}).$ This is complex linear and real on self-adjoint matrices. For a positive such matrix, $\lambda,\mu\geq0$ and $\|a\|^2\leq\lambda\mu$. To check that bound when $\lambda>0$, apply the triangular Schur-complement factorization to obtain $a^*a\leq\lambda\mu1$. If $\lambda=0$, positivity and its square root show that the first column is zero, so $a=0$. Thus $F\left(\begin{pmatrix}\lambda1&a\\a^*&\mu1\end{pmatrix}\right) \geq(\lambda+\mu)/2-\|a\|\geq0.$ Since $F(1)=1$, order bounds give real norm one on the self-adjoint part of this subspace. Extend it by real Hahn–Banach to a norm-one real functional on $M_2(C^+)_{\rm sa}$, retaining its value one at the identity. This extension is positive: for $0\leq h\leq1$, the bound $\|1-h\|\leq1$ gives $F(h)=1-F(1-h)\geq0$, and scale for arbitrary positive $h$. Complexification therefore gives a state $\Psi$ on $M_2(C^+)$.

Let $(\rho,H,\xi)$ be its cyclic GNS representation. Restrict $\rho$ to the diagonal copy of $C^+$, writing $\pi(c)=\rho(\operatorname{diag}(c,c))$. If $E_{ij}$ are the scalar matrix units, the prescribed off-diagonal value is $f(c)/2=\Psi(E_{12}\operatorname{diag}(c,c)) =\langle\rho(E_{21})\xi,\pi(c)\xi\rangle\qquad(c\in C).$ Rescale the first vector for the original norm of $f$. This establishes the asserted vector-coefficient representation; it uses neither a polar decomposition of functionals nor a Jordan decomposition.

For this representation $\pi$, let $H_A=\overline{\pi(A)H}$, and let $P$ be its Hilbert-space orthogonal projection. The projection needs only the Hilbert-space projection theorem: minimize distance to a closed linear subspace, use the parallelogram identity to make a minimizing sequence Cauchy, and differentiate the squared distance along each subspace direction to get orthogonality. This also proves the resulting decomposition and bounded projection. Approximate-identity convergence shows $\pi(e_\lambda)\to P$ strongly: check on $\pi(a)\eta$ by norm convergence $e_\lambda a\to a$, extend by their dense span and the contraction bound, and note that each $\pi(e_\lambda)$ vanishes on $H_A^\perp$.

The derivating hypothesis also holds for $d^*$, since $[d^*,a]=-[d,a^*]^*$. Both $\pi(d)$ and $\pi(d^*)$ preserve $H_A$, because $\pi(d)\pi(a)\eta =\pi(a)\pi(d)\eta+\pi([d,a])\eta\in H_A.$ Hence $H_A$ reduces $\pi(d)$ and $[P,\pi(d)]=0$. Strong convergence now gives $\pi([e_\lambda,d])\to0$ strongly. Testing its vector coefficient proves $f([e_\lambda,d])\to0$ for every bounded functional $f$, which is the claimed weak convergence in $C$.

Fix $d_1,\ldots,d_r$. The tuples $([u_\lambda,d_1],\ldots,[u_\lambda,d_r])$ converge weakly to zero in the finite direct sum of the ambient algebra. Zero consequently lies in the norm-closed convex hull of every tail. Indeed, otherwise Hahn–Banach separation would give a bounded linear functional strictly separating zero from that convex hull, contradicting weak convergence. A finite convex combination in the tail therefore makes every indicated commutator smaller than any prescribed positive tolerance.

The approximate-identity and commutator requirements can be imposed in one separation step. For finite $a_1,\ldots,a_s\in A$, adjoin the components $u_\lambda a_j-a_j$ and $a_j u_\lambda-a_j$ to that commutator tuple. These new components tend to zero in norm, while the commutator components tend weakly to zero. The enlarged tuple therefore has weak limit zero, and zero belongs to the norm-closed convex hull of every tail of these enlarged tuples. A single finite convex combination makes every component smaller than the chosen tolerance. Its associated combination of the $u_\lambda$ remains a positive contraction, and linearity gives exactly the tested approximate-identity and commutator errors. Directing these choices by finite tests, tails and tolerances proves the net assertion. For a sequence, use a countable approximate identity, a countable dense family in $\Delta$, and progressively all previous tests. The uniform bounds $\|u_\lambda\|\leq1$ and $\|[u_\lambda,d]\|\leq2\|d\|$ extend convergence from the dense family to all $d$. $\square$

The same argument works uniformly on a norm-compact set of tests, by choosing a finite net in that set.

We need increasing choices whose square-root differences have small commutators. A $\sigma$-unital algebra has a strictly positive element $h$, equivalently one for which $hA$ is norm dense. Choose continuous functions $f_n$ on $[0,\|h\|]$, vanishing near zero, with $0\leq f_n\leq1$, such that $e_n=f_n(h),\quad e_{n+1}e_n=e_n,\quad e_n\longrightarrow1 \quad\hbox{strictly}. \tag{1.3}$ For example let $f_n$ vanish below $2^{-n}$, equal one above $2^{-n+1}$, and increase linearly between, after rescaling $h$. The support condition gives the product equation. Functional calculus and density of $hA$ prove the approximate-identity assertion.

Lemma 1.2 (increasing quasicentral choices). Finite convex combinations of the $e_n$ can be chosen as an approximate identity $v_n$ with $v_nv_{n-1}=v_{n-1},\qquad 0\leq v_{n-1}\leq v_n\leq1, \tag{1.4}$ while satisfying successively prescribed finite or compact commutator tests with a separable derivating subspace. The same assertion holds for commutator tests on the images of $v_n$ in a quotient algebra.

Proof. Having chosen $v_{n-1}$ as a combination of finitely many $e_j$, choose all indices in the next combination greater than their maximum. By (1.3), each new $e_j$ multiplies every old $e_i$ to $e_i$ on both sides. Hence the product equation in (1.4) holds. For commuting positive contractions it implies $v_n\geq v_{n-1}$. Theorem 1.1 provides the desired new finite tests from this tail. Let the minimum index tend to infinity to retain the approximate-identity property.

For quotient tests apply Theorem 1.1 to the image approximate identity, selecting its convex coefficients. Use those same coefficients on the original $e_j$. Their quotient then has exactly the chosen commutator bounds, while the original product relation and tail bounds are retained. $\square$

Lemma 1.3 (square-root commutators). Given $\varepsilon>0$, there is $\eta>0$ such that, in any C*-algebra, $0\leq a\leq1,\quad\|w\|\leq1,\quad \|[a,w]\|<\eta \quad\Longrightarrow\quad \|[a^{1/2},w]\|<\varepsilon. \tag{1.5}$

Proof. Approximate $t^{1/2}$ uniformly on $[0,1]$ by a polynomial $p(t)=\sum c_kt^k$, to error less than $\varepsilon/4$. The commutator difference between $a^{1/2}$ and $p(a)$ has norm at most $\varepsilon/2$. Also $[a^k,w]=\sum_{j=0}^{k-1}a^j[a,w]a^{k-1-j},$ so $\|[p(a),w]\|\leq(\sum k|c_k|)\|[a,w]\|$. Taking $\eta$ small proves (1.5), uniformly in the algebra. $\square$

2. The separation theorem

The strict topology of $M(J)$ tests convergence after multiplication by each $z\in J$, on both sides. A bounded strict Cauchy sequence has a strict limit: define its left and right actions on $J$ by the norm limits and pass the double-centralizer identities to these limits.

Theorem 2.1 (Kasparov's technical theorem). Let $J$ be a $\sigma$-unital C*-algebra. Let $A_1,A_2\subset M(J)$ be $\sigma$-unital C*-subalgebras, and let $\Delta\subset M(J)$ be a separable linear subspace. Assume $A_1A_2\subset J,\qquad [\Delta,A_1]\subset A_1. \tag{2.1}$ There are $M,N\in M(J)$ satisfying $\begin{gathered} 0\leq M,N\leq1,\qquad M+N=1,\\ MA_1\subset J,\qquad NA_2\subset J,\\ [M,\Delta]\subset J. \end{gathered} \tag{2.2}$

The identity belongs to the multiplier algebra; the theorem does not require either $A_i$ to contain it.

Proof for separable data. Enlarge $\Delta$ by its adjoints. This still derives $A_1$, since $[d^*,a]=-[d,a^*]^*$. Choose norm-compact sets $X_1,X_2,Y$, of elements of norm at most one, whose linear spans are dense in $A_1,A_2,\Delta$. Such a set is obtained from a bounded dense sequence by multiplying its $j$-th term by $1/j$ and adjoining zero. Include adjoints when needed.

Put $\varepsilon_n=2^{-n}$. By Theorem 1.1 choose positive contractions $u_n\in A_1$ such that $\|u_nx-x\|<\varepsilon_n\quad(x\in X_1),\qquad \|[u_n,y]\|<\varepsilon_n\quad(y\in Y). \tag{2.3}$ They may be chosen as an approximate identity of $A_1$.

Using Lemmas 1.2–1.3 on $J$, choose an increasing approximate identity $v_n\in J$, with $v_0=0$, and set $b_n=(v_n-v_{n-1})^{1/2}. \tag{2.4}$ Require $\begin{aligned} \|(1-v_n)u_jx\|&<\varepsilon_n^4 && (x\in X_2,\ 1\leq j\leq n+1),\\ \|[b_n,w]\|&<\varepsilon_n && (w\in X_1\cup X_2\cup Y). \end{aligned} \tag{2.5}$ The first requirement tests compact subsets of $J$, because $u_jX_2\subset A_1A_2\subset J$. The second can be arranged prospectively: if $\eta_n$ is the square-root tolerance in Lemma 1.3 for $\varepsilon_n$, require the commutators of $v_n$ to be smaller than half both $\eta_n$ and $\eta_{n+1}$. Then the commutators of $v_n-v_{n-1}$ are smaller than $\eta_n$. The multiplier tests derive the ideal $J$, so Lemma 1.2 applies. Include an increasing dense family of $J$-tests to keep $v_n$ an approximate identity.

Define partial sums $N_r=\sum_{n=1}^r b_nu_nb_n. \tag{2.6}$ Each summand is positive and belongs to $J$. Since $u_n\leq1$, $0\leq N_s-N_r\leq v_s-v_r\leq1-v_r\qquad(s\geq r), \tag{2.7}$ and $N_r\leq v_r\leq1$. For $z\in J$, $\|(N_s-N_r)z\|^2 \leq\|z^*(N_s-N_r)z\| \leq\|z^*(1-v_r)z\|\longrightarrow0. \tag{2.8}$ Applying the same argument to $z^*$ gives convergence on the other side. Thus (2.6) converges strictly to a positive contraction $N\in M(J)$. Put $M=1-N$.

We now prove all three conclusions in (2.2), with norm estimates. For $x\in X_2$ and $n\geq2$, both $v_n$ and $v_{n-1}$ approximate $u_nx$ by the first line of (2.5); this is why that line included $j=n+1$. Thus $\begin{aligned} \|(v_n-v_{n-1})u_nx\|&<17\varepsilon_n^4,\\ \|b_nu_nx\|^2 &=\|(u_nx)^*(v_n-v_{n-1})(u_nx)\| <17\varepsilon_n^4. \end{aligned} \tag{2.9}$ Commuting the last $b_n$ past $x$, we obtain $\|b_nu_nb_nx\| \leq\sqrt{17}\,\varepsilon_n^2+\varepsilon_n. \tag{2.10}$ The series $Nx=\sum b_nu_nb_nx$ is therefore norm-convergent, apart from one harmless first term, and all its terms lie in $J$. Hence $NA_2\subset J$, by density and boundedness.

For $x\in X_1$, expand the complementary summand: $\begin{gathered} (b_n^2-b_nu_nb_n)x\\ =b_n(x-u_nx)b_n\\ \quad+b_n[b_n,x]\\ \quad-b_nu_n[b_n,x]. \end{gathered} \tag{2.11}$ Its norm is at most $3\varepsilon_n$, by (2.3) and (2.5). The norm-convergent sum of these elements belongs to $J$. Strictly, that sum is $\left(\sum b_n^2-\sum b_nu_nb_n\right)x=(1-N)x.$ Thus $MA_1\subset J$.

Finally for $y\in Y$, $[b_nu_nb_n,y] =b_nu_n[b_n,y]+b_n[u_n,y]b_n+[b_n,y]u_nb_n, \tag{2.12}$ of norm at most $3\varepsilon_n$. Its norm-convergent sum lies in $J$ and equals $[N,y]$ by strict continuity of multiplication by a fixed multiplier. Density gives $[N,\Delta]\subset J$, and $[M,\Delta]=-[N,\Delta]$. This proves the separable case. $\square$

Reduction to the stated generality. Choose strictly positive $h_0\in J$, $h_1\in A_1$, $h_2\in A_2$. Inside $A_1$, begin with $C^*(h_1)$ and repeatedly adjoin commutators with a countable dense family of $\Delta+\Delta^*$. The closed union is a separable subalgebra $B_1$ containing $h_1$ and derived by $\Delta$. Put $B_2=C^*(h_2)$.

Inside $J$, start with the separable algebra generated by $h_0$ and $B_1B_2$. Repeatedly adjoin its left and right products by $B_1,B_2,\Delta,\Delta^*$. Their closed union $J_0$ is separable and is preserved on both sides by all those multipliers. It contains the strictly positive element $h_0$ of the original $J$.

There is a canonical embedding $M(J_0)\subset M(J)$. Here are the details that make the reduction legitimate. An approximate identity of $J_0$ is also one for $J$: it approximates $h_0$, and the density of $h_0J$ and $Jh_0$ extends that approximation to every element of $J$. A bounded strict Cauchy family on $J_0$ is then strict Cauchy on $J$, by first multiplying a $J$-test by a fixed approximate-identity element of $J_0$. Consequently the bounded strict approximation $me_\lambda\in J_0$ of a multiplier $m\in M(J_0)$ extends to a multiplier of $J$. Its multiplication and adjoint are preserved by strict limits. Restriction to $J_0$ proves injectivity and identifies every original multiplier preserving $J_0$ with its extension.

Apply the proved separable case to $J_0,B_1,B_2,\Delta$. Its $M,N$ lie in $M(J)$, with $MB_1,NB_2,[M,\Delta]\subset J_0$. If $e_n=f_n(h_1)$ is an approximate identity of $A_1$, then $e_n\in B_1$, and for $a\in A_1$, $Ma=\lim_n Me_na\in J. \tag{2.13}$ Likewise $NA_2\subset J$, using $h_2$. The commutator conclusion already lies in $J_0\subset J$. This completes the theorem. $\square$

3. Grading and square roots

For graded $J$, its grading extends to $M(J)$. A graded subspace derives a graded algebra when the graded commutator is in that algebra for homogeneous inputs.

Theorem 3.1 (graded technical theorem). In Theorem 2.1, suppose $J,A_1,A_2,\Delta$ are graded and the derivating condition uses graded commutators. Then $M,N$ can be chosen even. All conclusions (2.2) hold, with graded commutators.

Proof. A graded $\sigma$-unital algebra has an even strictly positive element: replace $h$ by $h+\gamma(h)$. Here is the domination argument. If $k\ge h\ge0$, set $e_\varepsilon=k(k+\varepsilon)^{-1}$, in the unitization when necessary. Functional calculus and order give $0\le(1-e_\varepsilon)h(1-e_\varepsilon) \le(1-e_\varepsilon)k(1-e_\varepsilon) \le\frac{\varepsilon}{4}1.$ The last bound is $\sup_{t\ge0}t\varepsilon^2/(t+\varepsilon)^2=\varepsilon/4$. Hence $\|(1-e_\varepsilon)h\|\to0$, by applying the C*-identity to $h^{1/2}(1-e_\varepsilon)$ and then multiplying by $h^{1/2}$. Density of $hA$ gives $e_\varepsilon a\to a$ for every $a$; taking adjoints gives the right convergence. Since $e_\varepsilon a\in kA$, that range is dense. Apply this to $k=h+\gamma(h)$. Its functional-calculus approximate identity is even.

Use the bounded-functional vector representations and Hilbert support projections constructed in Theorem 1.1, with an even approximate identity. For homogeneous $d$ and $a$, the identity $\pi(d)\pi(a)\eta=(-1)^{|d||a|}\pi(a)\pi(d)\eta+\pi([d,a]_{\mathrm{gr}})\eta$ shows that $\overline{\pi(A)H}$ is invariant under $\pi(d)$; the same holds for $d^*$. Thus this subspace reduces $\pi(d)$, and its projection commutes with it. Represented approximate identities converge strongly to that projection. Their commutators with $d$ therefore converge strongly to zero in every such representation; vector-coefficient testing gives Banach weak convergence. These are also the graded commutators, since the approximate identity is even. The convex-separation argument of Theorem 1.1 now gives even convex combinations, and the increasing construction and square-root functional calculus retain that parity.

In the separable reduction use grading-invariant subalgebras and homogeneous dense families, adjoining graded commutators. This keeps $B_1$ invariant and graded-derived. Choose the $u_n,v_n,b_n$ even. Every commutator involving one of them is then ordinary, so each estimate (2.3)–(2.12) applies unchanged. The strict sum $N$ and its complement $M$ are even. Their commutators with homogeneous tests are the required graded commutators. $\square$

Corollary 3.2 (the square-root partition). Under the hypotheses of Theorem 3.1, $\begin{gathered} M^{1/2}A_1\subset J,\qquad N^{1/2}A_2\subset J,\\ [M^{1/2},\Delta]\subset J,\qquad [N^{1/2},\Delta]\subset J. \end{gathered} \tag{3.1}$

Proof. For $a\in A_1$, $(M^{1/2}a)^*(M^{1/2}a)=a^*Ma\in J$. Passing to $M(J)/J$ shows that $M^{1/2}a$ has zero quotient norm, so it lies in $J$. The same argument applies to $N$ and $A_2$. In the quotient, the images of $M,N$ commute with $\Delta$. Polynomial approximation to the square root preserves that commutation. Lifting back gives the two commutator conclusions. $\square$

This is the form used to combine product operators: $M^{1/2}F_1+N^{1/2}F_2$. The partition disposes of defects; the next lesson must also verify the connection and positivity conditions for that product.

4. A multiplier lifting bridge and the simpler separation

The absence of commutator tests admits another proof. To make that proof usable, we supply the needed lifting statement here.

Theorem 4.1 (multiplier quotient lifting). If $D$ is $\sigma$-unital and $I$ is an ideal, the quotient map extends to a surjective homomorphism $\widetilde q:M(D)\longrightarrow M(D/I). \tag{4.1}$ Every positive contraction in the target has a positive contractive lift.

Proof. Write $R=D/I$. Every multiplier $m_0$ of $D$ preserves $I$. Indeed, for $i\in I$ and an approximate identity $a_\lambda$ of $I$, the elements $(m_0i)a_\lambda$ belong to $I$, since $m_0i\in D$, and they converge to $m_0i$, since $ia_\lambda\to i$. Taking adjoints gives the other side. Thus the two multiplier actions descend to $R$. This defines $\widetilde q$; on bounded families it is strictly continuous, because its products with $q(d)$ are the quotients of the products with $d$.

Take $0\leq m\leq1$ in $M(R)$. Begin with the nested approximate identity (1.3) in $D$. Its image is an approximate identity of $R$. The multiplier $m$ derives $R$. By Lemma 1.2 choose convex combinations $v_n\in D$, increasing as in (1.4), such that, for $b_n=(v_n-v_{n-1})^{1/2}$, $\|[q(b_n),m]\|<2^{-n}. \tag{4.2}$ Use Lemma 1.3 to prescribe the commutator tolerances for both consecutive $v_n$, exactly as in (2.5).

For each $n$, an approximate identity $w_k$ of $R$ satisfies $\|(w_k^{1/2}mw_k^{1/2}-m)q(b_n)\|\longrightarrow0.$ This follows by approximating $q(b_n)$ and $mq(b_n)$ by the same approximate identity; square roots are themselves an approximate identity. Choose $k=k(n)$ making the norm less than $2^{-n}$. Lift the positive contraction $w_k^{1/2}mw_k^{1/2}\in R$ to a positive contraction $a_n\in D$. Such an element lift follows by first lifting self-adjointly and then clipping by the continuous function $t\mapsto\min(1,\max(0,t))$.

The positive sum $L=\sum_{n=1}^\infty b_na_nb_n \tag{4.3}$ converges strictly to a positive contraction in $M(D)$: its partial sums and tails are dominated as in (2.7), and (2.8) proves strict convergence. By strict continuity, $\widetilde q(L)=\sum q(b_n)q(a_n)q(b_n)$. The differences $\begin{aligned} q(b_n)q(a_n)q(b_n)-q(b_n)mq(b_n),\\ q(b_n)mq(b_n)-q(b_n)^2m \end{aligned} \tag{4.4}$ belong to $R$ and each have norm less than $2^{-n}$, by the selection of $a_n$ and (4.2). Their sums consequently converge in norm in $R$. Meanwhile $\sum q(b_n)^2m=m$ strictly. Hence $\widetilde q(L)-m\in R$.

Choose a self-adjoint $d\in D$ with $q(d)=m-\widetilde q(L)$. Then $L+d$ is a self-adjoint multiplier lifting $m$. Clipping it to $[0,1]$ gives a positive contractive multiplier lift, since functional calculus commutes with the quotient and fixes $m$. Finally every multiplier is a linear combination of positive multipliers, by splitting its real and imaginary parts into positive and negative parts. This proves surjectivity. $\square$

Corollary 4.2 (separation without commutator tests). Theorem 2.1 with $\Delta=0$ follows from Theorem 4.1.

Proof. Let $D=C^*(J,A_1,A_2)\subset M(J)$. This algebra is $\sigma$-unital. To verify that point, add strictly positive elements of $J,A_1,A_2$. The sum is positive and its functional-calculus approximate identity approximates each generator algebra; for example domination of the strictly positive element $h_i$ gives $\|(1-e_n)h_i^{1/2}\|\to0$, then density of $h_iA_i$ gives approximation on $A_i$. Approximation extends to their generated algebra.

The quotient is $q(D)=q(A_1)\oplus q(A_2)$, since these two subalgebras are orthogonal. In its multiplier algebra choose the central projection $r=(0,1)$. Theorem 4.1 lifts $r$ to a positive contraction $M\in M(D)$. Restricting multipliers to the essential ideal $J$ embeds $M(D)$ in $M(J)$: multiplication on $J$ determines a multiplier, and essentiality detects its kernel. Put $N=1-M$. Their quotient products with $q(A_1)$ and $q(A_2)$ vanish in the required order, so $MA_1,NA_2\subset J$. $\square$

This proof has supplied the multiplier lifting, rather than presupposing it. It does not produce a multiplier commuting modulo $J$ with a specified $\Delta$; that additional constraint is the content of the series selection in Section 2.

5. Extending an operator homotopy from an ideal

Let $A$ be separable and graded, $I\subset A$ a graded ideal, and $E$ a countably generated graded Hilbert $B$-module. Let $\phi:A\to\mathcal L(E)$ be graded. Write $K=\mathcal K(E)$ and define the two pseudolocal algebras $\begin{aligned} C&=\{x:[x,\phi(a)]_{\mathrm{gr}}\in K\ (a\in A)\},\\ D&=\{x:[x,\phi(i)]_{\mathrm{gr}}\in K\ (i\in I)\}. \end{aligned} \tag{5.1}$ Use homogeneous components in this notation. Their locally compact ideals are $\begin{aligned} J_A&=\{x\in C:x\phi(a),\phi(a)x\in K\ (a\in A)\},\\ J_I&=\{x\in D:x\phi(i),\phi(i)x\in K\ (i\in I)\}. \end{aligned} \tag{5.2}$ These are closed graded C*-ideals in the respective algebras, by the commutator Leibniz rule and adjoints, as in the quotient formulation of the preceding cycle lesson.

Lemma 5.1 (pseudolocal decomposition). One has $D=C+J_I$. Consequently inclusion induces a surjective map $C/J_A\longrightarrow D/J_I. \tag{5.3}$

Proof. Take homogeneous $x\in D$. For homogeneous $a\in A,i\in I$, the graded Leibniz rule gives $[x,\phi(a)]_{\mathrm{gr}}\phi(i) =[x,\phi(ai)]_{\mathrm{gr}} -(-1)^{|x||a|}\phi(a)[x,\phi(i)]_{\mathrm{gr}}\in K. \tag{5.4}$ Adjoints give compactness in the other order. Thus the algebra $A_1=C^*(K,\phi(I))$ has compact products with $A_2=C^*(K,\{[x,\phi(a)]_{\mathrm{gr}}:a\in A\}). \tag{5.5}$ Indeed the localized commutators, their adjoints, and their finite products annihilate $\phi(I)$ modulo $K$, by (5.4), and products involving $K$ are compact.

Both $A_1,A_2$ are $\sigma$-unital: $K$ is $\sigma$-unital, and each adjoins a separable set of generators. Explicitly, if $h_K$ is strictly positive in $K$ and $s_j$ is a bounded countable generating set for the adjoined part, the positive sum $h_K+\sum_j2^{-j}(s_j^*s_j+s_js_j^*) \tag{5.6}$ has functional-calculus approximate identity converging on $K$ and each $s_j,s_j^*$, by domination. It therefore approximates the generated algebra. Take homogeneous generators to make the sum even.

The separable graded subspace spanned by $x,x^*,\phi(A)$ derives $A_1$. Commutators with $K$ remain in $K$; commutators of $\phi(A)$ with $\phi(I)$ remain in $\phi(I)$; and the commutators of $x,x^*$ with $\phi(I)$ are compact. Apply Theorem 3.1 in $M(K)=\mathcal L(E)$. It gives even $M,N$ such that $M\phi(I)\subset K,\quad N[x,\phi(A)]_{\mathrm{gr}}\subset K,\quad [M,x],[M,\phi(A)]\subset K. \tag{5.7}$ The graded product rule shows $Nx\in C$. The other part $Mx$ belongs to $J_I$: commuting $x$ across $\phi(i)$ leaves a compact error and a factor $M\phi(i)$, and the same argument with adjoints gives the other side. Its pseudolocality for $I$ follows from the same product rule. Thus $x=Nx+Mx\in C+J_I$.

Decompose an arbitrary $x$ into its homogeneous parts. This proves the equality and the surjectivity in (5.3); the inclusion maps $J_A$ into $J_I$. $\square$

The localized compactness conditions in (5.2) must be stated inside the pseudolocal algebras. A one-sided compactness set of arbitrary adjointable operators need not itself be a C*-algebra.

Lemma 5.2 (lifting an involution path). For a surjective graded homomorphism between unital C*-algebras, a norm-continuous path of odd self-adjoint unitaries lifts to such a path, starting at any prescribed odd self-adjoint unitary lift.

Proof. First recall an elementary Banach quotient argument. Approximate a continuous quotient path by finite polygonal paths, retaining its endpoints and with geometrically decreasing uniform errors. Lift the differences of consecutive polygonal paths by lifting their finitely many vertices to elements of at most twice their quotient norm and interpolating. The differences vanish at the endpoints, so their lifts can vanish there too. The uniform sum of those correction paths converges, giving a lift with prescribed endpoint values.

If instead only the initial value zero is prescribed, and the quotient path has norm at most $\delta$, begin with a polygonal approximation of error at most $\delta/4$. Its vertex lifts have norms at most $2\delta$. Subsequent errors at most $\delta/2^{n+2}$ yield uniformly summable correction lifts of total norm at most $2\delta$. A bound $8\delta$ therefore holds for the full lift. Contractive averaging with the adjoint and grading makes every lift self-adjoint and odd when the quotient path has those properties.

Subdivide the involution path until each piece stays within a sufficiently small $\delta$ of its initial value. Lift its difference from that initial value by this argument, starting at zero.

Choose the subdivision so these lifted differences have norm less than $1/2$. Adding one to the current self-adjoint unitary lift gives an odd self-adjoint invertible path $h_t$. Replace it by $h_t|h_t|^{-1}. \tag{5.8}$ It is an odd self-adjoint unitary path. Its quotient is the original involution path, because (5.8) fixes a self-adjoint unitary. At the start it equals the prescribed lift. Continue from the new endpoint on the next subinterval; the resulting finitely many paths glue. $\square$

For logical independence, the elementary polygonal lifting used here needs no result about KK. Its proof is the summable finite-vertex correction argument, valid in every Banach quotient; the self-adjoint and odd parts are obtained by contractive averaging.

Theorem 5.3 (ideal homotopy extension). Let $(E,\phi,F)$ be a cycle over $(A,B)$, with $A$ separable. Suppose $F_t$ is a norm-continuous operator homotopy for its restriction to $I$, starting with $F_0=F$. There is an operator homotopy $G_t$ for the original representation of $A$, with $G_0=F$, and $(G_t-F_t)\phi(i),\quad\phi(i)(G_t-F_t)\in\mathcal K(E) \qquad(t\in[0,1],\ i\in I). \tag{5.9}$

Proof. The quotient formulation of cycle conditions identifies the images of $F_t$ in $D/J_I$ as odd self-adjoint unitaries. The image of $F$ in $C/J_A$ is one such lift of their initial value. By Lemmas 5.1–5.2, lift the whole path to odd self-adjoint unitaries in $C/J_A$. The Banach-space path lifting argument then lifts this to a norm-continuous odd path $G_t\in C$, starting at the original $F$. Self-adjointness is required only modulo $J_A$, so the prescribed $F$ need not be exactly self-adjoint.

In $C/J_A$, $G_t^2=1$ and $G_t=G_t^*$, so its square and adjoint defects are locally compact for all $a\in A$. Membership in $C$ gives its compact graded commutators. It is therefore the required operator homotopy. Its image in $D/J_I$ equals that of $F_t$, proving (5.9). $\square$

6. Examples and exercises

If $A_2=0$, take $M=0,N=1$. This meets every condition, including all commutators. If $A_1\subset J$, take $M=1,N=0$: then $MA_1\subset J$, $NA_2=0$, and all commutators vanish. The interesting case is when the two algebras are only orthogonal after passing to the quotient.

For a concrete compact-operator example, take $J=\mathcal K(L^2(S^1))$, the Hardy projection $P$, and multiplication representation $\pi$ of $C(S^1)$. Put $A_1=\mathcal K(L^2(S^1))+\mathbb C(1-P),\quad A_2=\mathbb CP,\quad \Delta=\pi(C(S^1)). \tag{6.1}$ These are separable and $\sigma$-unital. Their products are compact. The compact Hardy commutators show that $\Delta$ derives $A_1$. Here the explicit partition $M=P,N=1-P$ already satisfies the theorem. The general construction replaces this visible projection when no exact geometric projection is available.

6.1. Finite-rank quasicentrality. For a fixed $T\in\mathcal B(H)$, where $H$ is separable, construct positive finite-rank contractions $u_n$, each a convex combination of finite-rank projections from an increasing sequence converging strongly to $1$, such that $\|[u_n,T]\|\to0$.

6.2. Why projections are too restrictive. Let $S$ be the unilateral shift. Prove that finite-rank projections $P_n\to1$ strongly cannot satisfy $\|[P_n,S]\|\to0$. Increasingness is not needed for this obstruction.

6.3. Separation through multiplier lifting. For $\Delta=0$, form $D=C^*(J,A_1,A_2)$ and use the central projection in $M(D/J)$ to deduce the technical theorem. Verify the $\sigma$-unitality and essential-ideal steps.

6.4. Even partitions. Prove the graded theorem, explaining why merely averaging an arbitrary ungraded partition is insufficient when the given derivating hypothesis is only graded.

7. Solutions

Solution to 6.1. Let $p_n$ project onto the first $n$ vectors of an orthonormal basis. They form an approximate identity of $\mathcal K(H)$. The subspace spanned by $T,T^*$ derives that ideal, since multiplication by a bounded operator preserves compactness. Apply Theorem 1.1 to successive tails, making the two commutators smaller than $1/n$. The selected convex combinations are positive finite-rank contractions. Choosing each tail beyond $n$ retains strong convergence to $1$, since all projections in the combination fix the first $n$ basis vectors. In particular the commutator with $T$ tends to zero in norm.

Solution to 6.2. Let $p_0$ project onto the first basis vector $e_0$, so $SS^*=1-p_0$. For one finite-rank projection $P$, put $C=PSP$ on the finite-dimensional space $PH$, and $\delta=\|[P,S]\|$. Its two off-diagonal blocks have norms at most $\delta$. Since $S^*S=1$, $0\leq P-C^*C=PS^*(1-P)SP,\qquad \|P-C^*C\|\leq\delta^2. \tag{7.1}$ If $\delta<1$, $C$ is invertible on $PH$. Its polar decomposition gives $\|P-CC^*\|=\|P-C^*C\|\leq\delta^2$. But $P-CC^* =Pp_0P+PS(1-P)S^*P\geq Pp_0P. \tag{7.2}$ Hence $\|Pe_0\|^2=\|Pp_0P\|\leq\delta^2$. Strong convergence $P_ne_0\to e_0$ contradicts $\delta_n\to0$. This also explains why convex positive contractions, rather than projections, appear in the technical theorem.

Solution to 6.3. The positive sum of strictly positive elements of $J,A_1,A_2$ is strictly positive in $D$, by the domination and generator argument in Corollary 4.2. The quotient $D/J$ is the orthogonal sum $q(A_1)\oplus q(A_2)$. Lift its central multiplier projection $(0,1)$ by Theorem 4.1 to a positive contraction $M\in M(D)$. The ideal $J$ is essential in $D$, since $D\subset M(J)$ and a multiplier annihilating $J$ is zero. Thus restriction embeds $M(D)$ in $M(J)$. Quotient multiplication gives $MA_1\subset J$ and $(1-M)A_2\subset J$. Taking $N=1-M$ proves every conclusion; there are no commutator tests.

Solution to 6.4. Graded derivation does not imply ordinary derivation on two odd inputs, so one cannot simply invoke the ungraded theorem and average its answer. Instead begin with even strictly positive elements and even functional-calculus approximate identities. Their commutators with homogeneous tests are ordinary commutators and graded commutators simultaneously. In each vector-coefficient representation used in Theorem 1.1, the represented approximate identity tends strongly to the projection onto the closure of the represented algebra acting on the Hilbert space. For a homogeneous test and a homogeneous algebra element, the graded derivation identity proves invariance of this subspace under the test and its adjoint, just as in that proof; the sign changes only the first term. Thus the subspace reduces the test and its support projection commutes with it. Since the approximate identity is even, its ordinary and graded commutators coincide. Testing vector coefficients proves the required Banach weak convergence and hence the convexity step. Retain evenness through the convex combinations, the separable invariant reduction and the square roots $b_n$. All factors used in the estimates are even, so no extra signs occur. The resulting positive strict sum and its complement are even and satisfy the graded conditions. This verifies each step of Theorem 3.1.

What this lesson does not prove

Continuous functional calculus, the GNS construction and the compact-module multiplier identification are the exact earlier prerequisites stated at the beginning. The bounded-functional vector representation and the support-projection argument were proved within Theorem 1.1. They provide the framework; all convex quasicentrality, separation, quotient multiplier lifting and ideal-homotopy extension arguments were proved here.

The construction and uniqueness of the Kasparov product, its associativity, boundary products and the equivariant technical theorem require further proofs in the following lessons. The partition theorem alone does not assert them.

References

B. Blackadar, K-Theory for Operator Algebras, author-posted corrected second edition. The author identifies this freely downloadable version as a slightly corrected second edition. The source retains its author copyright and the usage terms stated on that page. Sections 12.4 and 14.6 supply the free source comparison for quasicentrality and the strict-series partition. The convex separation, increasing choices, graded construction and multiplier lifting used here are proved in Sections 1–4 above; the proof uses the preceding results identified above.
Lesson 05, Lemma 2.0, for $\sigma$-unitality of $\mathcal K(E)$, $\mathcal L(E)=M(\mathcal K(E))$, and bounded strict limits.
Foundations of von Neumann algebras, Continuous functional calculus, Theorems 5.1 and 6.1.
B. Blackadar, Operator Algebras, freely readable corrected author edition dated February 8, 2017, Sections II.6.3–II.6.4, for positive state extension and GNS comparison. Theorem 1.1 gives the complete bounded-functional construction and weak-support proof used here. The proof does not require Jordan decomposition or a universal-bidual theorem. The author retains copyright and the terms stated on the linked author publications page.