Draft lesson.

Completed tensors and continuous algebra objects

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

The tensor product of two product spaces should allow infinitely many matrix entries, while every finite observation still involves only finitely many of them. Completing the ordinary tensor product does exactly this. We will prove its universal property for jointly continuous bilinear maps, and use it to turn coalgebra operations into continuous algebra operations. A continuous bidual is always available in this setting, but a tensor dual exists only in finite dimension.

Prerequisites: Pro representations and continuous duality, §§1–5; Ind representations and the locally finite coreflector, §§1–5; and the dual-pair triangles in Tensor duality and coherent inverses, §2.

1. Discrete tensor and finite observations

Fix a discrete field \(k\) and a small discrete group \(G\). Use the categories \(\mathsf{LF}(G)\) and \(\mathsf L(G)\) of the prerequisites: locally finite discrete representations, and complete Hausdorff linear representations with an invariant open finite-codimension neighbourhood basis. Their inverse contravariant functors are the full algebraic dual \(D\), with its finite topology, and the discrete continuous dual \(C\). All tensor actions below are diagonal: \(g(m\otimes n)=gm\otimes gn\).

If \(M,N\) are locally finite, then \(M\otimes_kN\) is locally finite. Each vector is a finite sum of elementary tensors. Place their first and second factors in finite-dimensional invariant subspaces \(M_0,N_0\); the finite-dimensional invariant image of \(M_0\otimes N_0\) contains the vector. Thus ordinary tensor defines a bifunctor on \(\mathsf{LF}(G)\), with all maps \(f\otimes h\).

Tensoring over \(k\) is exact. An underlying short exact vector-space sequence splits by extending a basis of its subspace; tensoring the resulting vector-space decomposition proves injectivity, the kernel equality and surjectivity. The splitting need not be equivariant: the tensor maps themselves are equivariant, and exactness in \(\mathsf{LF}(G)\) is the underlying exactness. Tensoring also preserves every small colimit. A colimit of vector spaces is the quotient of their direct sum by the span of the diagram relations. Tensor distributes over that direct sum, since every tensor has finite support, and preserves the quotient, since its kernel is the tensor of the relation subspace by exactness. These identifications send every elementary tensor and relation to the corresponding one. Colimits of locally finite representations are their ambient colimits, so the same argument applies here.

2. The completed tensor and every Pro presentation

For \(V=D(M)\), \(W=D(N)\), define

using evaluation to identify \(M=C(V)\), \(N=C(W)\). It belongs to \(\mathsf L(G)\) by §1 and continuous duality. The formula defines a bilinear equivariant map \(j\). It is jointly continuous: each finite observation in \(M\otimes N\) is a finite sum of products of continuous scalar evaluations, and multiplication in discrete \(k\) is continuous. Its linear extension \(V\otimes_kW\to V\widehat\otimes W\) is injective and has dense image.

Here is the injectivity proof. Express a nonzero tensor as \(z=\sum_{a=1}^r v_a\otimes w_a\), where the \(v_a\) are linearly independent and some \(w_a\ne0\). The evaluation map \(M\to\operatorname{span}(v_a)^*\) is onto. Otherwise its proper image would be annihilated by a nonzero vector in \(\operatorname{span}(v_a)\), by finite biduality; that vector would vanish on every \(m\in M\) and hence be zero in \(D(M)\), a contradiction. For an index \(a\) with \(w_a\ne0\), choose \(m\) with \(m(v_b)=\delta_{ab}\), then \(n\) with \(n(w_a)\ne0\). Formula (2.1) evaluates \(z\) nontrivially at \(m\otimes n\).

For density, any finite subspace of \(M\otimes N\) is contained in \(M_0\otimes N_0\) for finite-dimensional invariant \(M_0,N_0\). Restriction \(D(M)\to M_0^*\) and \(D(N)\to N_0^*\) is surjective by extension of scalar functionals. Finite biduality identifies \((M_0\otimes N_0)^*\) with \(M_0^*\otimes N_0^*\): a matrix of coefficients in dual bases is the corresponding finite sum of elementary tensors. Lift those factors to \(V,W\). The resulting ordinary tensor realizes any prescribed functional on \(M_0\otimes N_0\). It therefore realizes every possible finite observation, proving density.

For invariant open finite-codimension \(U\subset V\), \(T\subset W\), continuous duality identifies \((V/U)^*\subset M\) and \((W/T)^*\subset N\). Their finite tensor subspaces contain every finite observation after a suitable choice of \(U,T\), as above. Consequently

The map is restriction to these finite dual tensor subspaces; its inverse glues compatible scalar functionals on their union \(M\otimes N\). Thus it identifies both vectors and topology. On \(V\otimes W\), its projection is the ordinary tensor of the quotient maps, and its kernel is \(U\otimes W+V\otimes T\). To verify that kernel, choose vector-space complements to \(U,T\); the tensor decomposes into four summands, and only the complement-by-complement summand survives in the quotient. These kernels form the induced topology. Formula (2.2), together with injectivity and density, proves that the completed tensor really is the Hausdorff completion of the ordinary tensor in this topology.

More generally, if \(V=\varprojlim_iX_i\) and \(W=\varprojlim_jY_j\) are arbitrary cofiltered finite-dimensional Pro presentations, then

Indeed continuous duality gives \(C(V)=\varinjlim_iX_i^*\) and \(C(W)=\varinjlim_jY_j^*\). Section 1 identifies their tensor with \(\varinjlim_{i,j}(X_i^*\otimes Y_j^*)\). Finite tensor duality identifies \((X_i^*\otimes Y_j^*)^*\) with \(X_i\otimes Y_j\), by the same elementary evaluation formula. The Pro realization theorem identifies the dual of this colimit with the stated limit and its actual topology, even when the transitions are not epimorphisms. On elementary tensors, its projections are \(v\otimes w\mapsto p_i(v)\otimes p_j(w)\).

3. The continuous bilinear universal property

A bilinear map \(B\colon V\times W\to Z\) is equivariant when \(B(gv,gw)=gB(v,w)\). We require joint continuity in the product topology.

Lemma 3.1. A jointly continuous bilinear map into a finite-dimensional discrete space \(Q\) factors through \((V/U)\otimes(W/T)\) for suitable invariant open finite-codimension \(U,T\).

Proof. Continuity at \((0,0)\) supplies invariant basic \(U_0,T_0\) with \(B(U_0,T_0)=0\). Choose finite-dimensional vector-space complements, with bases \((v_a)\) to \(U_0\) and \((w_b)\) to \(T_0\). For fixed \(w_b\), the map \(B(-,w_b)\) is continuous linear with open kernel, because \(Q\) is discrete. Choose an invariant basic \(U\) contained in \(U_0\) and all these finitely many kernels. Then \(B(U,W)=0\), since \(W=T_0+\operatorname{span}(w_b)\). Similarly choose an invariant basic \(T\subset T_0\) in every kernel of \(B(v_a,-)\). Since \(V=U_0+\operatorname{span}(v_a)\), we have \(B(V,T)=0\). Bilinearity therefore gives a well-defined map on the two quotients, whose linear tensor extension is the required factorization. The complements need not be invariant; the refined basic subspaces are invariant by the defining topology. If \(Q\) is a representation and \(B\) equivariant, the induced tensor map is equivariant by its formula on all quotient vectors. \(\square\)

Theorem 3.2. Composition with \(j\) gives inverse natural bijections

Proof. A continuous linear map composed with the jointly continuous bilinear \(j\) is of the indicated kind. Conversely let \(B\) be such a bilinear map. For every invariant basic \(S\subset Z\), compose with \(Z\to Z/S\). Lemma 3.1 gives a linear map through some \((V/U)\otimes(W/T)\), hence a continuous equivariant map \(F_S\colon V\widehat\otimes W\to Z/S\), by (2.2). It is independent of the chosen factorization: any two choices agree on all elementary tensors, whose linear span is dense, and their equalizer is closed because \(Z/S\) is Hausdorff. The same argument proves that all \(F_S\) are compatible with quotient transitions.

Completeness of \(Z\) now gives the unique map \(F\) with these coordinates. It is linear and equivariant coordinatewise, and continuous since the finite quotient kernels form the target topology. Every coordinate of \(Fj\) equals the corresponding coordinate of \(B\), so Hausdorffness gives \(Fj=B\). Uniqueness follows again from density and a closed equalizer. Starting with \(F\), the construction recovers its existing finite quotient coordinates; starting with \(B\), composition recovers \(B\). These are the two inverse maps in (3.1). Precomposing in either input, or postcomposing in the output, gives the same maps on every elementary tensor, proving naturality. \(\square\)

4. Functoriality, coherence and exact limits

For morphisms \(f\colon V\to V'\), \(h\colon W\to W'\), define

The evaluation identifications give \((f\widehat\otimes h)j(v,w)=j(fv,hw)\). Hence identity, composition and the interchange law hold on dense elementary tensors and therefore everywhere. The associator, unit maps and symmetry are the continuous maps

which send \((v\otimes w)\otimes z\) to \(v\otimes(w\otimes z)\), scalar tensors to scalar multiplication, and \(v\otimes w\) to \(w\otimes v\). They exist and are invertible by identifying the bracketed tensors with \(D(C(V)\otimes C(W)\otimes C(Z))\), and dualizing the ordinary associativity, scalar and flip isomorphisms. Those identifications use both evaluation inverses, so are natural on all maps. Elementary tensors with any finite number of factors have dense linear span: the finite-observation density proof of §2 works with that number of finite subspaces and dual bases.

This also proves all coherence equations. Both paths around the pentagon on four factors send every elementary fourfold tensor to the same ordered fourfold tensor. Both triangle paths remove the unit by the same scalar multiplication. Both hexagon paths send an elementary triple to the same permuted triple. Naturality is checked on the same tensors; symmetry squared returns each pair unchanged. The maps in each comparison are continuous and their target is Hausdorff, so density proves equality on the whole completed space. Thus \(\mathsf L(G)\) is symmetric monoidal, and continuous duality identifies it with \(\mathsf{LF}(G)^{\mathrm{op}}\) as a symmetric monoidal category, with \(C(V\widehat\otimes W)\simeq C(V)\otimes C(W)\).

For fixed \(W\), the functor \(-\widehat\otimes W\) is exact and preserves every small limit. Exactness follows by applying \(C\), tensoring the reversed short exact sequence with \(C(W)\) using §1, and dualizing back. For a limit diagram, its continuous dual is the colimit of the reversed diagram by the actual inverse duality maps. Tensor preserves that colimit by §1; dualizing back gives the proposed tensor-of-limit comparison isomorphism. Its maps to the factors are the original projections tensored with identity, as one sees on elementary tensors. This proves preservation with all its universal maps, not just an object identification.

5. Continuous algebras and locally finite coalgebras

A continuous algebra object in \(\mathsf L(G)\) has \(m\colon A\widehat\otimes A\to A\) and \(\eta\colon k\to A\), associative and unital under (4.2). By Theorem 3.2, its multiplication is equivalently a jointly continuous bilinear equivariant multiplication on \(A\). A coalgebra object in \(\mathsf{LF}(G)\) has \(\Delta\colon H\to H\otimes H\) and \(\epsilon\colon H\to k\), with

Here and below the canonical associativity and unit maps are inserted in these equations.

Theorem 5.1. Continuous duality gives inverse contravariant equivalences between continuous algebra objects in \(\mathsf L(G)\) and coalgebra objects in \(\mathsf{LF}(G)\), including all their morphisms. Commutative algebras correspond to cocommutative coalgebras.

Proof. Starting with \(H\), put \(A=D(H)\) and

Their domains are the indicated ones by (4.2) and evaluation. Starting with \(A\), put \(H=C(A)\), \(\Delta=C(m)\) and \(\epsilon=C(\eta)\), using \(C(A\widehat\otimes A)\simeq H\otimes H\). Associativity of \(m\) dualizes to coassociativity of \(\Delta\), since reversing composition turns the two triple-product paths into the two triple-coproduct paths. The two unit equations dualize respectively to the two counit equations. These implications are reversible because the duality is faithful and its associators and units agree by §4. Similarly \(m\tau=m\) is dual to \(\tau\Delta=\Delta\).

For a continuous algebra map \(f\colon A\to B\), dualizing \(fm_A=m_B(f\widehat\otimes f)\) gives \(\Delta_A C(f)=(C(f)\otimes C(f))\Delta_B\); dualizing \(f\eta_A=\eta_B\) gives \(\epsilon_A C(f)=\epsilon_B\). These are precisely the coalgebra-map equations. Conversely every coalgebra map dualizes to an algebra map by the same reversible equations. The evaluation isomorphisms commute with all structure maps by their naturality and the tensor comparison, so their inverses recover the original objects and every original morphism. They also respect identities and reversed composition. \(\square\)

The same construction reverses continuous coalgebras and locally finite algebras, when both kinds of objects are defined in these monoidal categories. It also gives bialgebra and Hopf correspondences. To be explicit, a bialgebra has both structures, and in addition to their own laws satisfies

Use completed tensors on the continuous side. Duality interchanges \(m\) with \(\Delta\) and \(\eta\) with \(\epsilon\), reverses every composition, and preserves the middle flip. Therefore the first equation becomes the same first equation for the dual structures, the second and third exchange, and the fourth is unchanged. This proves both implications of every compatibility law. An antipode \(S\) satisfies

Reversing these composites gives exactly the two antipode equations for \(C(S)\) or \(D(S)\). Morphisms preserving both structures and the antipode correspond by dualizing each of their equations as in Theorem 5.1. These give the actual inverse bialgebra and Hopf equivalences. A vector-space quotient of an algebra inherits its multiplication and unit exactly when its kernel is a two-sided ideal: well-definedness of the two input quotients is equivalent to closure under multiplying that kernel on either side.

6. Finite tensor factors and tensor duals

If \(W\) is finite-dimensional, its Hausdorff linear topology is discrete: finitely many successive intersections of open subspaces reduce the dimension to zero. Then \(V\widehat\otimes W\) is the ordinary \(V\otimes_kW\), with its finite-biproduct topology. In a basis of \(W\), both are \(V^{\dim W}\): the dual description is \(D(C(V)\otimes W^*)\), the dual of a finite direct sum of copies of \(C(V)\). The identification sends each \(v\otimes w\) to its actual coordinates. A basis change is an invertible finite scalar matrix, continuous in both directions, so the topology is intrinsic.

Theorem 6.1. An object of \(\mathsf L(G)\) has a tensor dual if and only if it is finite-dimensional.

Proof. Suppose \(V\) has a dual \(W\), with coevaluation \(b\colon k\to V\widehat\otimes W\) and evaluation \(e\colon W\widehat\otimes V\to k\). The first triangle says \((1_V\widehat\otimes e)(b\widehat\otimes1_V)=1_V\), with the unit and associativity maps. The jointly continuous bilinear evaluation factors through \((W/T)\otimes(V/U)\) for finite quotients, by Lemma 3.1. Project \(b(1)\) to \(V\widehat\otimes(W/T)\), which by the finite-factor observation is an ordinary tensor. Write it as \(\sum_{a=1}^r v_a\otimes\overline w_a\). The triangle then gives, for every \(v\),

Thus \(V\) is spanned by finitely many \(v_a\). The second triangle similarly makes \(W\) finite-dimensional. This argument permits \(b(1)\) to be a genuinely completed tensor; only its finite quotient projection is written as a finite sum.

Conversely, for finite-dimensional \(V\), take its ordinary contragredient dual \(V^*\). All topologies are discrete. Evaluation is \(\lambda\otimes v\mapsto\lambda(v)\), and in a basis \((v_i)\) coevaluation sends \(1\) to \(\sum_i v_i\otimes v^i\). Evaluation is equivariant since \((g\lambda)(gv)=\lambda(v)\). The coevaluation corresponds to the identity under \(V\otimes V^*\simeq\operatorname{End}_k(V)\); the action is conjugation, so it is invariant and basis-independent. On \(v\), the first triangle is \(\sum_i v_i v^i(v)=v\); on \(\lambda\), the second is \(\sum_i\lambda(v_i)v^i=\lambda\). These prove both triangles and the converse. \(\square\)

The full left-adjoint construction and the precise right-adjoint criterion for completed tensoring are proved in Adjoints of completed tensor products. The proof applies to every small group and identifies the actual coproduct comparison that fails for an infinite-dimensional tensor factor.

7. Four worked exercises

Exercise 7.1 (intermediate: finite rank versus completion). For \(G=1\) and \(V=W=k^{\mathbb N}\), identify the completed tensor with all infinite matrices. Identify the image of the ordinary tensor and prove that the infinite identity matrix lies in the completion but outside that image.

Solution. The continuous duals are \(k^{(\mathbb N)}\), so their tensor has basis indexed by \(\mathbb N\times\mathbb N\). Its full dual is \(k^{\mathbb N\times\mathbb N}\), with product topology. Formula (2.1) sends \(v\otimes w\) to the matrix \((v_iw_j)\). A finite sum of such matrices has its columns in the finite-dimensional span of the first factors.

Conversely, if all columns of a matrix span a finite-dimensional space with basis \(v_1,\ldots,v_r\), express column \(j\) as \(\sum_a w_{a,j}v_a\). Each \((w_{a,j})_j\) is a vector of \(k^{\mathbb N}\), and these expressions reconstruct the matrix as \(\sum_a v_a\otimes w_a\). Thus the ordinary tensor image consists exactly of matrices of finite column rank. The identity matrix has the infinitely many independent coordinate columns, so has infinite rank and is excluded. Its upper-left finite square truncations belong to the ordinary tensor and converge coordinatewise to it. More generally finitely supported matrices realize every finite list of entries, proving density directly. Rectangle projections give exactly (2.2), with every matrix recovered by its compatible finite rectangles.

Exercise 7.2 (advanced: formal-series multiplication and its coalgebra). With trivial group action, calculate \(k[[u]]\widehat\otimes k[[u]]\), its multiplication map to \(k[[u]]\), and the full dual coalgebra. Check all associativity and counit equations.

Solution. The coefficient functionals \(\epsilon_n\), \(n\geq0\), form a basis of the continuous dual. Their pairwise tensors give the basis whose full dual is \(k[[x,y]]\). The corresponding completion topology is restriction to finite rectangles of coefficients, or equivalently the \((x,y)\)-adic topology: the ideal \((x^r,y^s)\) contains \((x,y)^{r+s-1}\), and \((x^n,y^n)\subset(x,y)^n\), so the two families of neighbourhoods refine one another. On elementary tensors the identification is \(a(u)\otimes b(u)\mapsto a(x)b(y)\).

Multiplication is the continuous linear map

Each output coefficient uses a finite antidiagonal; finitely many output coefficients therefore impose only finitely many input conditions, proving continuity. On elementary tensors this is exactly the Cauchy product, and the unit is the constant series \(1\). Its dual coalgebra is the direct sum of the \(\epsilon_n\), with

Both iterated coproducts are the finite sum \(\sum_{i+j+\ell=n}\epsilon_i\otimes\epsilon_j\otimes\epsilon_\ell\). Applying either counit removes the terms with the corresponding index nonzero and returns \(\epsilon_n\). This verifies coassociativity and both counits on a basis, hence on every vector. Dually the two iterated products have coefficient \(n\) equal to the same finite sum over \(i+j+\ell=n\), and multiplying by \(1\) preserves every coefficient. The coproduct is cocommutative and the algebra commutative. The subspace spanned by \(\epsilon_0,\ldots,\epsilon_{r-1}\) is a finite subcoalgebra; its dual algebra is the actual quotient \(k[[u]]/(u^r)\), with reduction dual to the inclusion.

Exercise 7.3 (advanced: every continuous coordinate algebra map). Let nonempty sets \(S,T\) carry \(G\)-actions with finite orbits. Give \(k^S,k^T\) the product topologies, coordinatewise multiplication, and actions \((ga)_s=a_{g^{-1}s}\). Classify all continuous unital equivariant \(k\)-algebra maps \(k^S\to k^T\), and their dual coalgebra maps.

Solution. The continuous dual of \(k^S\) is \(k^{(S)}\), with basis \(e_s\); its action permutes this basis. Finite orbits make it locally finite, and kernels of restrictions to finite unions of orbits give the invariant finite-codimension basis of \(k^S\). The same holds for \(T\). Coordinate multiplication is jointly continuous, because each finite output list depends on the same finite list in both inputs, and the unit is the constant function \(1\). The dual coalgebra has \(\Delta e_s=e_s\otimes e_s\), \(\epsilon(e_s)=1\), verified by evaluating the product on every pair of functions.

If \(F\colon k^S\to k^T\) is continuous, each output coordinate is a continuous linear functional, so factors through finitely many source coordinates \(E\subset S\). If that functional is unital and multiplicative, write it as \(\sum_{s\in E}a_s\operatorname{ev}_s\). The coordinate idempotents imply \(a_s^2=a_s\), so each \(a_s\) is 0 or 1 in the field; their pairwise products are zero, so at most one is 1. Evaluating the constant function gives \(\sum a_s=1\), so exactly one is 1. Thus each target coordinate chooses a unique \(\phi(t)\in S\), and

Conversely every function \(\phi\colon T\to S\) gives a unital algebra map by (7.3). It is continuous because finitely many target coordinates use finitely many source coordinates. Comparing \(F(ga)_t\) with \((gF(a))_t\) for all functions \(a\) gives \(g^{-1}\phi(t)=\phi(g^{-1}t)\); this is exactly equivariance of \(\phi\). Hence the classification includes all maps in both directions.

Dualizing sends \(e_t\mapsto e_{\phi(t)}\). Every counital coalgebra map must have this form: if a finite vector \(x=\sum a_se_s\) satisfies \(\Delta x=x\otimes x\) and \(\epsilon(x)=1\), comparing off-diagonal coefficients gives \(a_sa_r=0\) for \(s\ne r\), comparing diagonal coefficients gives \(a_s=a_s^2\), and the counit gives exactly one coefficient 1. Thus the images of all \(e_t\) select exactly the same function \(\phi\). Equivariance and composition are the stated function equations, so both the algebra and coalgebra classifications agree on every morphism.

Exercise 7.4 (advanced: an infinite coevaluation cannot supply a tensor dual). For \(G=1\), \(V=k^{\mathbb N}\), the infinite identity matrix gives a continuous candidate coevaluation \(b\colon k\to V\widehat\otimes V\). Show that it cannot be paired with a continuous evaluation satisfying both duality triangles. Explain the failure of the finite-coordinate evaluations to have an inverse-limit evaluation.

Solution. The identity matrix is an element of the completed tensor by Exercise 7.1, so \(a\mapsto aI\) is a continuous map from discrete \(k\). Any continuous evaluation \(e\colon V\widehat\otimes V\to k\), composed with \(j\), factors through finite-dimensional coordinate quotients in both inputs by Lemma 3.1. After projecting the second factor of \(b(1)\) to that finite quotient, its image is a finite sum \(\sum_a v_a\otimes\overline w_a\). The first triangle would express every \(v\) in the finite span of the \(v_a\), exactly as (6.1). But the coordinate vectors in \(V\) are infinitely linearly independent, a contradiction. In particular the formal expression \(\sum_i w_iv_i\) does not define this continuous bilinear evaluation on all tuples.

On \(k^n\), evaluation \(e_n(w,v)=\sum_{i<n}w_iv_i\) and the finite identity coevaluation do satisfy both triangles. The coevaluations have compatible rectangle projections and yield the infinite identity matrix. The evaluations are not compatible with those projections: on \(k^{n+1}\times k^{n+1}\), the evaluation \(e_{n+1}\) differs from \(e_n\) after both coordinate projections by \(w_nv_n\), which is nonzero on the \(n\)-th coordinate vectors. Thus there is no compatible family of evaluation maps into constant \(k\) to which the inverse-limit universal property could apply. Continuous biduality \(V\simeq D(C(V))\) still holds, but it is a different pair of inverse maps; the infinite-dimensional discrete space \(C(V)\) is not an object of \(\mathsf L(1)\).

Further reading

Eric Marberg, Linear compactness and combinatorial bialgebras, §3, describes completed tensors and dual algebraic structures. The finite quotient and bilinear arguments above provide their explicit continuous universal maps in the representation setting.