Finite projective modules, frames and $K_0$

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Public domain (CC0).

A Hermitian vector bundle on a compact space can be described by finitely many sections which reconstruct every section. The same reconstruction identity makes sense over a C*-algebra. It produces a matrix projection, identifies the module with the range of that projection, and explains why projective modules enter $K$-theory.

We use right Hilbert modules and inner products linear in the second variable. The preceding lessons Adjointable operators and Compact operators, multipliers and the strict topology supply the closed-range theorem and the calculus of $\mathcal L(E)$ and $\mathcal K(E)$. In particular, $\theta_{x,y}z=x\langle y,z\rangle,\qquad \theta_{x,y}^*=\theta_{y,x},\qquad \overline{\mathcal K(E)E}=E.$ The coefficient algebra is unital except where a different hypothesis is explicitly stated. The zero module is included: its frame can be empty, and its identity operator is zero.

1. Reconstruction by a finite frame

Definition 1.1. A finite Parseval frame for $E$ is a finite family $x_1,\ldots,x_n$ such that $\sum_{i=1}^n\theta_{x_i,x_i}=1_E. \tag{1.1}$ Thus every $x\in E$ satisfies $x=\sum_i x_i\langle x_i,x\rangle$. A frame need not be a basis: the coefficients in a representation of $x$ can be nonunique, and the vectors need not be orthogonal.

For an arbitrary finite family, define its synthesis and analysis operators by $\Phi:A^n\longrightarrow E,\quad \Phi(a)=\sum_i x_i a_i, \qquad \Lambda:E\longrightarrow A^n,\quad \Lambda x=(\langle x_i,x\rangle)_i.$ They are bounded. For example, each coordinate of $\Lambda x$ has norm at most $\|x_i\|\|x\|$, and the square of the finite-column norm is bounded by the sum of the squared coordinate norms. Direct calculation gives $\langle\Phi a,x\rangle=\sum_i a_i^*\langle x_i,x\rangle =\langle a,\Lambda x\rangle.$ Consequently $\Phi^*=\Lambda$. This calculation also works for a nonunital coefficient algebra.

Theorem 1.2 (The frame projection). For any C*-algebra $A$, a finite Parseval frame determines a projection $P=\Lambda\Phi=(\langle x_i,x_j\rangle)_{i,j}\in M_n(A). \tag{1.2}$ The analysis operator is a unitary $E\longrightarrow PA^n$, with inverse the restriction of $\Phi$.

Proof. The frame identity says $\Phi\Lambda=1_E$. Hence $P^2=\Lambda\Phi\Lambda\Phi=P,\qquad P^*=(\Phi^*\Phi)^*=P.$ Moreover $P\Lambda=\Lambda$, so the analysis operator takes values in $PA^n$. For $x,y\in E$, $\langle\Lambda x,\Lambda y\rangle =\langle x,\Phi\Lambda y\rangle =\langle x,y\rangle.$ If $a=Pa$, then $\Lambda\Phi a=Pa=a$. The two restricted maps are inverse and preserve the inner product. ∎

Conversely, let $p\in M_n(A)$ be a projection. Its range $pA^n$ is closed, since it is the range of a bounded projection on the complete module $A^n$. It has the inherited inner product. The columns $z_i$ of $p$ belong to $pA^n$, even when $A$ is nonunital. Writing these columns formally as $pe_i$, we have $\sum_i z_i\langle z_i,a\rangle=p^2a=a \quad(a\in pA^n). \tag{1.3}$ Here the symbols $e_i$ refer to coordinate columns in the unitization; the actual vectors $pe_i$ have entries in $A$. Thus every such projection range has a finite frame.

For $p=1_n$ over a unital algebra this is the usual coordinate frame. For $p=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}$ over $\mathbb C$, its two frame vectors are equal. Both are needed in the displayed Parseval identity with this normalization, but they are certainly not linearly independent.

2. Compact identity and algebraic finiteness

Lemma 2.1 (Normalizing generators). If a Hilbert $A$-module is algebraically generated by $x_1,\ldots,x_n$, then it has a finite Parseval frame. This implication does not require $A$ to be unital.

Proof. The associated synthesis operator $\Phi:A^n\to E$ is surjective and adjointable. Its range is therefore closed. The closed-range theorem gives an orthogonal decomposition $A^n=\ker\Phi\oplus\operatorname{ran}\Phi^*,$ and the restriction of $\Phi$ to $\operatorname{ran}\Phi^*$ is invertible onto $E$, with adjointable inverse. It follows that $S=\Phi\Phi^*$ is positive and invertible in $\mathcal L(E)$. Equivalently this is the surjective case of that theorem. Put $z_i=S^{-1/2}x_i$. Then $\sum_i\theta_{z_i,z_i} =S^{-1/2}\left(\sum_i\theta_{x_i,x_i}\right)S^{-1/2} =S^{-1/2}SS^{-1/2}=1_E.$ These are the required frame vectors. ∎

The word “algebraically” matters: every element must be an exact finite coefficient combination of the displayed generators. A dense coefficient span is a weaker assertion.

Theorem 2.2 (Four descriptions). For a Hilbert module over a unital C*-algebra, the following conditions are equivalent:

$E$ has a finite Parseval frame.
$\mathcal K(E)$ is unital.
$E$ is unitarily isomorphic to $pA^n$ for a projection $p\in M_n(A)$.
The underlying module is algebraically finitely generated and projective.

In this case $\mathcal K(E)=\mathcal L(E)$. In fact, algebraic finite generation alone implies all four conditions.

Proof. A frame writes $1_E$ as a finite sum of rank-one operators, so $1_E\in\mathcal K(E)$. Since $\mathcal K(E)$ is an ideal of $\mathcal L(E)$, it then contains every $T=T1_E$.

If $\mathcal K(E)$ has a unit $u$, then $u$ acts as the identity on $\mathcal K(E)E$. This subspace is dense in $E$, so $u=1_E$. Choose a finite-rank operator $T=\sum_{i=1}^n\theta_{a_i,b_i}, \qquad \|1_E-T\|<1.$ The Neumann series makes $T$ invertible in $\mathcal L(E)$. For $x\in E$, $x=T(T^{-1}x)=\sum_i a_i\langle b_i,T^{-1}x\rangle.$ Thus $a_1,\ldots,a_n$ generate $E$ algebraically, and Lemma 2.1 supplies a frame.

Theorem 1.2 gives the projection description from a frame. A projection range is an algebraic direct summand $A^n=pA^n\oplus(1-p)A^n$ of a finite free module, so it is finitely generated projective. Such a module is in particular finitely generated, and Lemma 2.1 closes the cycle. ∎

The first three conditions and their equivalence remain valid for nonunital $A$. The algebraic terminology for projectivity over a nonunital ring depends on the chosen category of modules; we shall use projection ranges and finite frames when discussing that case.

For a unital algebra, a finite projective Hilbert module is unusually rigid: every algebraic module map from it to another Hilbert module is bounded and adjointable. If $x_i$ is a frame and $T$ is such a map, then $Tx=\sum_i Tx_i\langle x_i,x\rangle,\qquad T^*y=\sum_i x_i\langle Tx_i,y\rangle. \tag{2.1}$ The finite sums prove boundedness and the adjoint identity directly. General Hilbert modules do not have this property.

3. Existence and uniqueness of the metric

An algebraic finite projective module is initially a summand of a finite free module. Its presenting idempotent need not be self-adjoint. We first replace it by a projection without assuming a Hilbert structure on the module.

Lemma 3.1. Every idempotent $e\in M_n(A)$, for unital $A$, is similar to a projection.

Proof. Set $s=e^*e+(1-e^*)(1-e).$ The identity $2s-1=(2e-1)^*(2e-1)\geq0$ shows that $s\geq\frac12 1$, so $s$ is positive and invertible. Using $e^2=e$, multiplication gives $se=e^*e=e^*s.$ Therefore $p=s^{1/2}es^{-1/2}$ satisfies $p^2=p$ and $p^*=s^{-1/2}e^*s^{1/2} =s^{1/2}es^{-1/2}=p.$ Multiplication by $s^{1/2}$ identifies $eA^n$ with $pA^n$. ∎

Theorem 3.2 (Hilbert structures). Every algebraic finite projective module over a unital C*-algebra admits a complete Hilbert-module inner product. Any two such inner products are related by a unitary module isomorphism. More precisely, on a fixed underlying module there is a positive invertible $R$, adjointable for the first inner product, such that $\langle x,y\rangle_2=\langle x,Ry\rangle_1. \tag{3.1}$

Proof. Write the algebraic module as $eA^n$. Lemma 3.1 identifies it with the closed Hilbert module $pA^n$. Pulling back its inner product gives existence and completeness.

For uniqueness, let $E_1,E_2$ denote the same algebraic module with the two complete metrics. Each has a finite frame by Theorem 2.2. Apply (2.1) to the algebraic identity $S:E_1\to E_2$. It is bounded and adjointable, and the same argument applies to its inverse. In particular $R=S^*S$ is positive and invertible in $\mathcal L(E_1)$, since $R^{-1}=S^{-1}(S^{-1})^*.$ The adjoint identity gives (3.1). Now $U=SR^{-1/2}:E_1\longrightarrow E_2$ is invertible and satisfies $U^*U=1_{E_1}$; its inverse is $U^*$, so $UU^*=1_{E_2}$. This is the required unitary. ∎

Uniqueness concerns the isomorphism class of the metric, not equality of metrics. On $A$, for example, $\langle a,b\rangle_h=a^*hb$ is a complete Hilbert structure for any positive invertible $h\in A$. Multiplication by $h^{1/2}$ is unitary from this metric to the standard one.

Similarity of arbitrary idempotents must not be confused with unitary conjugacy. In $M_2(\mathbb C)$, the non-self-adjoint idempotent $\begin{pmatrix}1&1\\0&0\end{pmatrix}$ is similar to a projection, but cannot be unitarily conjugate to one: unitary conjugation preserves self-adjointness.

4. Projection equivalence and the module definition of $K_0$

For projections $p\in M_n(A)$ and $q\in M_m(A)$, write $p\sim q$ if there exists $v\in M_{m,n}(A)$ such that $v^*v=p,\qquad vv^*=q. \tag{4.1}$ This is Murray–von Neumann equivalence, with different matrix sizes allowed. Such a $v$ automatically satisfies $v=qvp$.

Theorem 4.1. Over a unital C*-algebra, $pA^n\cong qA^m\text{ unitarily}\quad\Longleftrightarrow\quad p\sim q.$ An algebraic module isomorphism between these modules exists if and only if a unitary one exists.

Proof. A matrix $v$ satisfying (4.1) maps $pA^n$ into $qA^m$, preserves inner products, and has inverse $v^*$.

Conversely, extend a unitary $U:pA^n\to qA^m$ to $A^n$ by $Up$. Since $A$ is unital, its values on coordinate vectors are the columns of a matrix $v\in M_{m,n}(A)$. We have $v=qvp$. On $pA^n$, $v^*v$ is the identity; it is zero on $(1-p)A^n$. Thus $v^*v=p$, and similarly $vv^*=q$.

Finally let $S:pA^n\to qA^m$ be an algebraic isomorphism. Formula (2.1) makes $S$ and $S^{-1}$ adjointable. The positive normalization $S(S^*S)^{-1/2}$ is unitary by the proof of Theorem 3.2. ∎

Direct sum corresponds to block sum: $pA^n\oplus qA^m\cong(p\oplus q)A^{n+m}.$ Consequently the commutative monoid of finite projective Hilbert-module isomorphism classes is exactly the projection monoid $V(A)$. We do not identify projections merely because they become equal in a group: their equivalence here is the actual relation (4.1).

Define the Grothendieck group of this monoid using pairs $([E],[F])$. Two pairs $([E],[F])$ and $([E'],[F'])$ represent the same element if for some finite projective $G$, $E\oplus F'\oplus G\cong E'\oplus F\oplus G. \tag{4.2}$ Addition is componentwise direct sum, and the inverse exchanges the two entries. The projection correspondence identifies (4.2) with exactly the stabilization relation in the projection definition of $K_0(A)$. We have therefore proved $K_0(A)=\operatorname{Groth}\{\text{finite projective Hilbert }A\text{-modules}\}. \tag{4.3}$ No cancellation assumption on the monoid is used.

For nonunital $A$, put $\widetilde A=A\oplus\mathbb C1$ and let $\pi:\widetilde A\to\mathbb C$ be the quotient. The definition is $K_0(A)=\ker\!\left(K_0(\widetilde A) \xrightarrow{\pi_*}K_0(\mathbb C)\cong\mathbb Z\right). \tag{4.4}$ Thus a typical class is $[p\widetilde A^n]-[q\widetilde A^m]$ whose scalar quotient projections have equal rank. The modules over $\widetilde A$ can retain geometry at infinity which projections inside $M_n(A)$ do not detect.

For example, when $X$ is connected and noncompact, every projection in $M_n(C_0(X))$ is zero. Indeed its pointwise norm is either zero or one; its nonzero set is both open and closed, while vanishing at infinity makes that set compact. Connectedness leaves only the empty set. This explains why the Grothendieck group of projection ranges over the nonunital algebra is insufficient as a general definition of $K_0$.

5. Bundle and smooth examples

To construct a frame from local bundle coordinates, we need scalar weights whose supports stay inside the coordinate domains. We give the topological construction explicitly.

Lemma (Cutoffs and finite partitions). A finite open cover of a compact Hausdorff space admits a continuous subordinate partition of unity. For a compact set $K$ in an open subset $U$ of a locally compact Hausdorff space, there is $\chi\in C_c(U)$, $0\leq\chi\leq1$, with $\chi=1$ on $K$. A finite open cover of such a compact set admits subordinate compactly supported weights whose sum is one near that set.

Proof. First a compact Hausdorff space is normal. To separate a point from a closed set, use disjoint neighborhoods of the point and each point of the closed set, then take a finite intersection on the first side and a finite union on the second. Apply this construction to each point of a second disjoint closed set and use its compactness. It follows that whenever $F\subset U$, with $F$ closed and $U$ open, one can insert an open $V$ with $F\subset V\subset\overline V\subset U$.

For disjoint closed sets $F_0,F_1$, repeatedly use this insertion to choose open sets $V_r$, indexed by dyadic rationals in $[0,1]$, with $F_0\subset V_0$, $V_1=X\setminus F_1$, and $\overline V_r\subset V_s$ for $r<s$. The function $f(x)=\inf\{r:x\in V_r\}$, with empty infimum one, is continuous, equals zero on $F_0$, and equals one on $F_1$. Indeed $\{f<a\}=\bigcup_{r<a}V_r$ and $\{f>a\}=\bigcup_{r>a}(X\setminus\overline V_r)$. This is the required separating function.

Shrink a finite open cover $(U_i)$ to an open cover $(V_i)$ with $\overline V_i\subset U_i$. One can shrink in order: the closed set not covered by the other current members lies in the member being replaced, and insertion retains the cover. Insert once more between $\overline V_i$ and $U_i$, and use the separating function to obtain $f_i=1$ on $\overline V_i$ with support inside $U_i$. Then $\sum_i f_i>0$, and $\rho_i=f_i/\sum_jf_j$ is the desired partition.

For a locally compact Hausdorff space use its one-point compactification, or the space itself if compact. Cover $K$ by finitely many relatively compact neighborhoods whose closures lie in $U$; these neighborhoods exist by local compactness and regularity, the latter obtained from the same compact Hausdorff argument. Apply the separating-function construction twice between $K$, those closures and $U$. The resulting cutoff has compact support in $U$. For a finite cover of $K$, choose finitely many smaller neighborhoods with compact closure inside the respective members, construct their cutoffs $f_i$, and put $F=\sum_i f_i$, positive near $K$. Choose a cutoff $\eta=1$ near $K$ with support in $\{F>0\}$. The functions $\eta f_i/F$, extended by zero, are continuous, have the prescribed compact supports, and sum to $\eta$, hence to one near $K$. ∎

Local orthonormal frames require no additional bundle theorem: the Gram matrix of a local continuous frame is positive and invertible, and multiplying the frame by its continuous inverse square root orthonormalizes it. These scalar and matrix constructions also apply in the later locally compact bundle and field examples.

Example 5.1 (A Hermitian bundle). Let $V\to X$ be a finite-rank Hermitian vector bundle on a compact Hausdorff space. Its continuous sections form a Hilbert $C(X)$-module with pointwise inner product. Take a finite trivializing cover, continuous local orthonormal frames $e_{i,\alpha}$, and a partition of unity $\rho_i$ with support contained in the corresponding trivializing open set. Define $x_{i,\alpha}(t)=\sqrt{\rho_i(t)}\,e_{i,\alpha}(t)$ there and extend by zero. These are continuous global sections. At each point, $\sum_{i,\alpha}x_{i,\alpha}(t) \langle x_{i,\alpha}(t),v\rangle =\sum_i\rho_i(t)v=v.$ They are a finite Parseval frame. The Gram projection presents the section module as $pC(X)^N$.

Conversely a continuous projection-valued matrix $p(t)$ gives a vector bundle with fibre $\operatorname{ran}p(t)$. To see local triviality, fix $t_0$, choose a basis of $\operatorname{ran}p(t_0)$, and apply $p(t)$ to these vectors. They remain independent near $t_0$. The rank remains constant there: projections at distance less than one have isomorphic ranges by the corner-inverse argument used below. These sections are therefore a local basis. Its section module is exactly $pC(X)^N$. This proves the existence part of the compact-space Serre–Swan correspondence needed here.

Example 5.2 (The Bott line). On $S^2=\{(x,y,z):x^2+y^2+z^2=1\}$, set $p(x,y,z)=\frac12 \begin{pmatrix} 1+z&x-iy\\ x+iy&1-z \end{pmatrix}. \tag{5.1}$ The off-diagonal entries are conjugate. Multiplication, using $x^2+y^2=1-z^2$, gives $p^2=p$. Its trace is one, so its range is a line at every point. The two columns of $p$ form a finite Parseval frame for $pC(S^2)^2$.

Away from the south pole and north pole respectively, normalized local spanning vectors are $u_N=\frac{(1+z,x+iy)^{\mathsf T}}{\sqrt{2(1+z)}}, \qquad u_S=\frac{(x-iy,1-z)^{\mathsf T}}{\sqrt{2(1-z)}}.$ On the equator $x+iy=e^{it}$, their relation is $u_S=e^{-it}u_N$. This computes the transition function with the stated convention. It also proves nontriviality: a global unit section would write as $u_Na_N=u_Sa_S$ on the two hemispheres, with circle-valued $a_N,a_S$ extending over disks. Their boundary winding numbers are zero, whereas the transition requires $a_N=e^{-it}a_S$, whose winding number differs by $-1$.

Here are the elementary winding facts used in that argument. A continuous circle-valued path has a continuous real argument: subdivide its parameter interval into finitely many pieces whose images lie in proper arcs, choose an argument on each arc, and adjust by multiples of $2\pi$ at successive endpoints. For a loop the argument's endpoint difference is $2\pi$ times an integer, its winding number. Arguments add under multiplication, so winding numbers add. Two uniformly close loops have the same winding number: their quotient avoids $-1$, hence has a periodic principal argument and winding zero. Uniform continuity of a homotopy then proves invariance by subdividing its homotopy parameter into such close pairs. A loop extending over a disk contracts by radial restriction to a constant loop, so has winding zero. These facts justify the contradiction above.

Theorem 5.3 (Dense smooth algebras). Let $\mathcal A\subseteq A$ be a dense unital *-subalgebra stable under holomorphic functional calculus at all matrix levels. Extension of scalars gives a bijection between isomorphism classes of algebraic finite projective $\mathcal A$-modules and finite projective $A$-modules, preserving direct sums.

Proof. An idempotent over $\mathcal A$ can be replaced by the projection of Lemma 3.1 inside $M_n(\mathcal A)$: the positive square root and inverse square root of its $s$ are holomorphic functions on a neighborhood of its positive spectrum. Thus we can work with projections.

Let $p\in M_n(A)$ be a projection. Approximate it by self-adjoint $b\in M_n(\mathcal A)$ with $\delta=\|b-p\|<1/8$. The spectrum of $b$ lies in the $\delta$-neighborhood of $\{0,1\}$: outside it the resolvent of $p$ and a Neumann series give the resolvent of $b$. The holomorphic function which is zero near the first spectral component and one near the second yields a projection $q\in M_n(\mathcal A)$. On the contour $|z-1|=1/2$, $\|(z-p)^{-1}\|\leq2,\qquad \|(z-b)^{-1}\|\leq\frac{2}{1-2\delta}.$ The resolvent identity and the contour length give $\|q-p\|\leq\frac{2\delta}{1-2\delta}<1.$ Both $pqp$ and $qpq$ are invertible in their corners, since they differ from the corner units by norm less than one. Thus $qp:pA^n\to qA^n$ has a left inverse $(pqp)^{-1}pq$ and a right inverse $pq(qpq)^{-1}$. They coincide, proving that $pA^n\cong qA^n$. Moreover the natural scalar-extension map $q\mathcal A^n\otimes_{\mathcal A}A\to qA^n$ is an algebraic isomorphism: it is the restriction of the free-module isomorphism $\mathcal A^n\otimes_{\mathcal A}A\cong A^n$ to a direct summand.

For injectivity, suppose smooth projections $q,r$ give isomorphic $A$-modules. By Theorem 4.1 take a rectangular partial isometry $v$ with $v^*v=q$, $vv^*=r$. Approximate $v$ by $w\in rM_{m,n}(\mathcal A)q$. For sufficiently close approximation, $w^*w$ and $ww^*$ are invertible in the respective corners. Their inverses belong to the smooth corners: add the complementary projection, invert in the full matrix algebra, and compress. Therefore $(w^*w)^{-1}w^*$ is a left inverse over $\mathcal A$, and $w^*(ww^*)^{-1}$ is a right inverse. The two coincide. Hence the smooth modules were already isomorphic. Direct sum commutes with scalar extension. ∎

The theorem asserts a bijection of isomorphism classes, not an equivalence of all module categories with all morphisms. For instance, the endomorphisms of the rank-one free modules are $\mathcal A$ and $A$, which generally differ. Matrix-level holomorphic stability is the precise condition used in the proof. For smooth elements of a norm-continuous Lie group action it holds: the orbit map of an inverse is the inverse of the orbit map, and differentiation under the contour integral proves smoothness of holomorphic functions. The same argument applies entrywise to matrices.

Example 5.4 (Schwartz modules for a rotation algebra). Let $A_\theta$ be the universal unital C*-algebra generated by unitaries with $UV=e^{2\pi i\theta}VU$. Its smooth algebra $A_\theta^\infty$ consists of the series $\sum_{n,m}a_{n,m}U^nV^m$ whose coefficients are rapidly decreasing: for every $N\geq0$, $\sup_{n,m}(1+|n|+|m|)^N|a_{n,m}|<\infty$. These are the smooth vectors for the torus action multiplying $U,V$ by scalar phases; the preceding inverse-orbit argument applies. The construction and its connection are also developed in Connections and curvature from symmetries of an algebra, Sections 6–7. Let $q\geq1$, $p\in\mathbb Z$, and $\varepsilon=p/q-\theta\ne0$. On $\mathcal S(\mathbb R\times\mathbb Z/q)$, set $(\xi U)(s,h)=\xi(s+\varepsilon,h+p),\qquad (\xi V)(s,h)=e^{2\pi i(s-h/q)}\xi(s,h). \tag{5.2}$ The right-action operators satisfy $R_UR_V=e^{-2\pi i\theta}R_VR_U$; reversed operator composition is exactly what a right action requires. With the $L^2$ inner product linear in the second variable, define $\langle\xi,\eta\rangle_A =\sum_{n,m\in\mathbb Z} \langle R_{U^nV^m}\xi,\eta\rangle_{L^2}\,U^nV^m. \tag{5.3}$ Translation decay and integration by parts in the modulation variable give rapidly decreasing coefficients. Unitarity and reversed composition give right linearity and Hermitian symmetry, by comparing coefficients of unitary monomials.

Here is a concrete finite frame. Choose finitely many compactly supported smooth functions $g_j$, each supported in an interval of length less than one, such that $\sum_{j,n}|g_j(s+n\varepsilon)|^2=1. \tag{5.4}$ They exist by taking a finite bump cover of the circle of circumference $|\varepsilon|$ and dividing the bumps by the square root of their periodized squared sum. Put $g_{j,r}(s,h)=g_j(s)\mathbf1_{h=r}$ for $r\in\mathbb Z/q$. Then $\eta=\sum_{j,r}g_{j,r}\langle g_{j,r},\eta\rangle_A. \tag{5.5}$ Indeed expand (5.3) in this sum. Fourier summation in $m$ restricts the integration variable to $s+(k-h)/q-\ell$, $\ell\in\mathbb Z$. The finite-coordinate supports of the two window factors force $k=h$. Their real arguments then differ by $\ell$; support length less than one forces $\ell=0$. The remaining expression is $\eta(s,h)\sum_{j,n}|g_j(s+n\varepsilon)|^2$, proving (5.5). Fourier summation here is the Poisson identity on Schwartz functions; equivalently one can use its Fejér-regularized form.

The Gram matrix of these vectors is a self-adjoint idempotent over $A_\theta^\infty$, so the same analysis and synthesis calculation as Theorem 1.2 gives $\mathcal S(\mathbb R\times\mathbb Z/q)\cong P(A_\theta^\infty)^N.$ It also proves positivity of (5.3), since (5.5) gives $\langle\eta,\eta\rangle_A=\sum_i\langle g_i,\eta\rangle_A^*\langle g_i,\eta\rangle_A$. If this is zero all coefficients vanish, and (5.5) gives $\eta=0$. Its Hilbert completion is $PA_\theta^N$. This is the frame construction in Connections and curvature from symmetries of an algebra, Section 7; the general frame theorem explains its projection formula.

For completeness, the Fourier identity just used follows directly from periodization. For a Schwartz function $f$, $F(t)=\sum_{\ell\in\mathbb Z}f(t+\ell)$ is smooth and periodic, with uniform convergence of every differentiated sum. Its $m$-th Fourier coefficient is $\int_{\mathbb R}f(t)e^{-2\pi imt}\,dt$. Integration by parts makes these coefficients rapidly decreasing, so their Fourier series converges uniformly. It equals $F$: their difference has all Fourier coefficients zero, hence all its Fejér means vanish. Those means converge uniformly to every continuous periodic function. Indeed their convolution kernels $K_N(t)=\frac1N\left|\sum_{j=0}^{N-1}e^{2\pi ijt}\right|^2$ are nonnegative with integral one, and satisfy $K_N(t)\leq[N\sin^2(\pi\delta)]^{-1}$ when the distance of $t$ from an integer is at least $\delta$. Splitting the convolution into that region and its complement proves uniform convergence by uniform continuity. This proves the periodization identity, including the translated test-function version in (5.5).

6. Exercises with solutions

Exercise 6.1 (Basic: the Gram projection). For a Parseval frame, verify $\Phi\Lambda=1_E$, $P^2=P=P^*$, and identify the range of $\Lambda$.

Solution. For $x\in E$, $\Phi\Lambda x=\sum_i x_i\langle x_i,x\rangle=x$. Hence $(\Lambda\Phi)^2=\Lambda(\Phi\Lambda)\Phi=\Lambda\Phi$. Since $\Lambda=\Phi^*$, the operator $\Lambda\Phi$ is self-adjoint. Its matrix entries are $\langle x_i,x_j\rangle$, so it is $P$. The inclusion $\operatorname{ran}\Lambda\subseteq PA^n$ follows from $P\Lambda=\Lambda$. If $a=Pa$, then $a=\Lambda\Phi a$, giving the reverse inclusion. Finally $\langle\Lambda x,\Lambda y\rangle=\langle x,y\rangle$, so this identification is unitary. ∎

Exercise 6.2 (Intermediate: a compact unit). Derive a finite frame from the assumption that $\mathcal K(E)$ has a unit.

Solution. Its unit is $1_E$, because it acts as the identity on the dense subspace $\mathcal K(E)E$. Approximate it within distance less than one by $T=\sum_{i=1}^n\theta_{a_i,b_i}$. Invertibility of $T$ gives the exact reconstruction $x=\sum_i a_i\langle b_i,T^{-1}x\rangle$. Thus the synthesis operator $X:A^n\to E$ for the $a_i$ is surjective. The closed-range theorem makes $XX^*$ invertible. Set $z_i=(XX^*)^{-1/2}a_i$. Multiplying the rank-one sum on both sides by $(XX^*)^{-1/2}$ gives $\sum_i\theta_{z_i,z_i}=1_E$. Neither this proof nor its conclusion needs unitality of the coefficient algebra. ∎

Exercise 6.3 (Intermediate: changing the metric). On an algebraic finite projective module, prove that two complete Hilbert metrics are unitarily equivalent.

Solution. The finite-frame formula (2.1) makes the algebraic identity $S$ and its inverse bounded and adjointable for the two metrics. Thus $R=S^*S$ is positive invertible, and $\langle x,y\rangle_2=\langle x,Ry\rangle_1$. The map $U=SR^{-1/2}$ satisfies $U^*U=1$. Because it is invertible, this identity also implies $UU^*=1$, by substituting $U^{-1}$. Hence $U$ is unitary. Existence of at least one complete metric follows separately from the idempotent-to-projection construction; it is not assumed merely from algebraic projectivity. ∎

Exercise 6.4 (Intermediate: the standard module). Show that $H_A$ is not algebraically finitely generated if $A\ne0$.

Solution. If it were, Lemma 2.1 would give a finite frame, hence $1_{H_A}\in\mathcal K(H_A)$. We show that this is impossible. Fix $a\in A$ with $\|a\|=1$, and let $z^{(k)}\in H_A$ have coordinate $a$ at $k$ and zero elsewhere. For a rank-one operator, $\|\theta_{x,y}z^{(k)}\| =\|x\,y_k^*a\|\leq\|x\|\|y_k\|.$ The coordinates of $y\in H_A$ satisfy $\|y_k\|\to0$, since its squared-coordinate series converges in norm. Therefore $\|Fz^{(k)}\|\to0$ for every finite-rank $F$. Since $\|z^{(k)}\|=1$, it follows that $\|1-F\|\geq1$. No finite-rank sequence or net can approximate the identity in norm, contradicting compactness. ∎

Exercise 6.5 (Advanced: noncompact coefficient spaces). Let $X$ be noncompact and locally compact Hausdorff. Show that a Hilbert $C_0(X)$-module with a finite frame cannot be full. Deduce the same for any algebraically finitely generated Hilbert module, and explain the unitization in $K_0$.

Solution. By Theorem 1.2 the module is $pC_0(X)^n$, with $p\in M_n(C_0(X))$ a projection. Its support $Y=\{t:p(t)\ne0\}=\{t:\|p(t)\|=1\}$ is open by continuity and closed as the inverse image of $\{1\}$. It is compact by vanishing at infinity. Since $X$ is noncompact, $Y\ne X$. Every inner product of vectors in the module vanishes outside $Y$, so their closed span lies in the proper ideal $C_0(Y)\subsetneq C_0(X)$. Properness follows, for example, by a nonzero compactly supported function at a point outside $Y$. Thus the module is not full. In fact its coefficient ideal is $C_0(Y)$: the entries of the Gram matrix include the entries of $p$, whose scalar sum of absolute squares is $\operatorname{tr}p$, positive throughout $Y$; these generate that ideal.

Lemma 2.1 applies to an algebraically finitely generated Hilbert module, giving the same conclusion. On a connected noncompact space $Y$ must be empty, so there are no nonzero finite-frame modules at all. Projection classes in the nonunital algebra can therefore lose the relevant information at infinity. Formula (4.4) instead uses finite projective modules over $C(X^+)$ and subtracts equal fibre ranks at the added point.

For $X=\mathbb R^2$, the Bott projection of Example 5.2 is viewed on $X^+=S^2$; subtract its constant rank-one limit at infinity. The quotient ranks agree, so this defines a relative class in $K_0(C_0(\mathbb R^2))$. The projection difference has entries vanishing at infinity, even though the individual projection generally does not. This illustrates precisely what the relative definition retains. The calculation of the whole group or a proof that this relative class generates it would require Bott periodicity and is not asserted here. ∎

The noncompact conclusion is specific to these commutative coefficient algebras. It is not a statement that every full finite-frame module over every nonunital C*-algebra is impossible.

What this lesson does not prove

We use the C*-algebra matrix norm, positivity, continuous and holomorphic functional calculus, and spectral mapping as prerequisites; precise standard references are Blackadar, Operator Algebras, II.2.3, II.3.1 and II.6.6. We use the preceding course proofs of the closed-range theorem and the essential action of compact operators.

The opening lemma of Section 5 proves compact Hausdorff partitions, locally compact cutoffs and continuous local orthonormalization. The winding and Fourier reconstruction facts needed for the two explicit examples are proved in Section 5.

We do not prove Bott periodicity, a classification of rotation-algebra projections, or Morita equivalence of rotation algebras. Those assertions are unnecessary for the frame and projectivity statements above. Nor does the dense-algebra theorem assert that every algebraic endomorphism over the completion descends to the dense subalgebra.

References

[Blackadar 1998] Bruce Blackadar, K-Theory for Operator Algebras, second edition, Cambridge University Press, 1998, Sections 5.1–5.5: the projection monoid, Grothendieck completion and the relative definition for nonunital algebras. Author's corrected second edition.

[Blackadar 2006] Bruce Blackadar, Operator Algebras: Theory of C*-Algebras and von Neumann Algebras, Springer, 2006, II.7.1–II.7.2 for module and operator foundations. Author's revised edition.

[Connes 1994] Alain Connes, Noncommutative Geometry, Academic Press, 1994, Chapter II, Appendix A, Proposition 4: compact identity, finite projectivity and change of metric. Author's electronic edition.

[Emerson 2024] Heath Emerson, An Introduction to C*-Algebras and Noncommutative Geometry, Birkhäuser, 2024, Section 5.2 for compact-space bundles and projective modules, and Section 6.2 for their relation to Morita correspondences.

Finite projective modules, frames and \(K_0\)