Adjoints of completed tensor products

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

Completed tensor products of linearly compact representations have an adjoint on one side for every factor. On the other side, finite dimension is necessary and sufficient. The obstruction comes from a family of finite invariant subspaces and their coevaluation tensors. It applies to every small group, including groups with infinitely many generators.

The complete prerequisites are Ind representations and the locally finite coreflector, Sections 2–5, Pro representations and continuous duality, Theorem 2.1 and Sections 4–5, and Completed tensors and continuous algebra objects, Sections 1–4 and Theorem 6.1. The internal Hom construction for Ind categories, Theorem 2.2 gives the general closed-category setting. The explicit representation formula below identifies its actual objects and maps.

1. The two representation categories

Fix a discrete field \(k\), a small discrete group \(G\), and a universe containing all spaces and index sets below. Use choice for vector-space bases. Let \(\mathcal C=\mathsf{LF}(G)\) be the category of left representations whose vectors each lie in a finite-dimensional invariant subspace. Its morphisms are all equivariant linear maps. For an arbitrary representation \(A\), write \(rA\) for the sum of its finite-dimensional invariant subspaces. Each vector of that sum lies in a finite sum of those subspaces, so \(rA\) is locally finite. An equivariant map from a locally finite representation has locally finite image. This gives the coreflection \(r\) and, in particular, the intrinsic product

The product on the right is the algebraic product with coordinatewise action. A map from a locally finite source into that product factors uniquely through its locally finite part, proving the product property.

Let \(\mathcal L=\mathsf L(G)\) consist of Hausdorff complete linear topological representations with a basis of invariant open subspaces of finite codimension. Completeness means that the canonical map to the inverse limit of the finite-dimensional discrete quotients is a topological isomorphism. Morphisms are continuous equivariant linear maps. The invariant-basis hypothesis is part of this definition. A continuous action on a linearly compact space need not have such a basis.

Write

The full dual \(D(M)\) has the finite topology: annihilators of finite-dimensional subspaces of \(M\) form a zero-neighbourhood basis. The continuous dual \(C(V)\) is discrete. Both carry the contragredient action. The cited duality theorem supplies the actual natural evaluation inverses

It applies to all fields, without topological compactness of \(k\). Full duality and the completed-tensor theorem give

All identifications below use these evaluation and tensor maps. We denote morphism sets by \(\mathcal C(A,B)\) and \(\mathcal L(U,V)\).

The category \(\mathcal L\) has all small limits and coproducts. Under duality, the coproduct of the \(D(A_s)\) is \(D(\prod_s^{\mathcal C}A_s)\), with the duals of the product projections as its injections. These are the intrinsic coproducts used here.

2. The locally finite internal Hom

For \(N,M\in\mathcal C\), give the algebraic Hom representation the action

The expression before applying \(r\) contains every linear map, not just the equivariant ones.

Proposition 2.1. There are natural inverse bijections

The internal Hom is therefore \(H(N,M)\). Equivariant maps from \(N\) to \(M\) are its invariant vectors.

Proof. Tensor products are locally finite: a finite sum of elementary tensors lies in the tensor product of two finite-dimensional invariant subspaces. Given an equivariant bilinear map \(b:X\times N\to M\), define \(f(x)(n)=b(x,n)\). Linearity in the two variables gives a linear map to the algebraic Hom, and

Thus this map is equivariant. Its image is locally finite because its source is, so it factors through \(H(N,M)\). Conversely an equivariant \(f:X\to H(N,M)\) gives the bilinear map \(b(x,n)=f(x)(n)\). Its equivariance follows by evaluating the action (2.1) at \(gn\). The formulas are inverse on every pair \((x,n)\), and precomposition or postcomposition gives the same formulas; this proves naturality. A linear map \(N\to M\) is fixed by (2.1) exactly when it is equivariant. Such a fixed vector is locally finite, proving the final assertion. \(\square\)

For fixed \(N\), the unit and counit of this algebraic adjunction are

The map \(\alpha_X\) is equivariant by (2.1), so its image lies in the indicated locally finite part. Evaluation makes \(\beta_M\) equivariant as well. On \(x\otimes n\), the first triangle evaluates to \(x\otimes n\). At \(f\), the second triangle is the map \(n\mapsto\beta_M(f\otimes n)=f(n)\). These prove both triangle identities.

3. The left adjoint for every factor

Fix \(W\in\mathcal L\), and put \(N=C(W)\). Define

For \(a:U\to U'\), the map \(L_W(a)\) is the full dual of postcomposition by \(C(a)\) on the internal Hom. Its direction is \(L_W(U)\to L_W(U')\), since \(C(a):C(U')\to C(U)\). This defines a continuous equivariant functor, by the functoriality of \(H\) and \(D\).

Theorem 3.1. The functor \(L_W\) is left adjoint to \(T_W\) for every \(W\in\mathcal L\).

Proof. Write \(X=C(V)\) and \(M=C(U)\). The specified duality and Proposition 2.1 give

Each bijection is natural in \(U,V\), giving the adjunction on all continuous equivariant maps.

Its unit is the dual of \(\beta_M\) in (2.3), and its counit is the dual of \(\alpha_X\), transported through (1.3)–(1.4). Explicitly, the unit sends \(u\in U\) to the functional

This is a vector of \(D(H(N,M)\otimes N)\), the identified object \(T_W(L_W(U))\). The counit sends

The dual-functor construction proves continuity, equivariance and naturality of these maps. Dualizing the two evaluated triangles after (2.3), and using the evaluation triangles in the duality prerequisite, proves their two triangles. Thus the formulas specify the actual adjunction, including its inverse Hom maps. \(\square\)

In particular \(T_W\) preserves every small limit. Indeed maps into its value on a limit identify by (3.2) with maps from \(L_W(U)\) into that limit, hence with compatible families of maps from \(U\) into the \(T_W\)-images. This is exactly the limit universal property. The construction does not require \(W\) to have a tensor dual.

4. The right adjoint criterion

We first give a product test entirely in \(\mathcal C\).

Lemma 4.1. For \(N\in\mathcal C\), let \(I\) be the small set of its finite-dimensional invariant subspaces. Set

The canonical comparison

is injective. If \(N\) is infinite-dimensional, it is not surjective.

Proof. The map has coordinate \(p_E\otimes1_N\), where \(p_E\) is a product projection. Its image lies in \(Q\) since it is an equivariant image of a locally finite object. If a tensor in its kernel is written \(\sum_{a=1}^s p_a\otimes n_a\), choose the nonzero \(n_a\)'s linearly independent by collecting terms in a basis of their finite-dimensional span. Vanishing in each coordinate implies \(p_E(p_a)=0\) for every \(E,a\): apply functionals on \(N\) taking the prescribed values on that independent list. All coordinates of each \(p_a\) vanish, so \(p_a=0\). This proves injectivity.

For every \(E\), inclusion \(E\hookrightarrow N\) corresponds under the finite-dimensional tensor-Hom bijection to the coevaluation tensor

This is basis-independent and invariant: the tensor-Hom bijection intertwines the diagonal action with conjugation on maps, and inclusion is equivariant. Consequently the tuple \(u=(u_E)_{E\in I}\) is an invariant vector of the full product. Its orbit span has dimension at most one, so \(u\in Q\).

Suppose \(u=\gamma(\sum_{a=1}^s p_a\otimes n_a)\), and put \(N_0=\operatorname{span}(n_1,\ldots,n_s)\). Every coordinate \(u_E\) would then belong to \(E^*\otimes N_0\). Evaluating its first factor at \(e\in E\) gives \(e\in N_0\), by the defining inclusion represented by (4.3). Thus every \(E\) is contained in \(N_0\). Their union is \(N\), by local finiteness, so \(N=N_0\) would be finite-dimensional. This contradiction proves nonsurjectivity. The zero subspace contributes its zero tensor and causes no exception. \(\square\)

Theorem 4.2. For \(W\in\mathcal L\), the following are equivalent:

1. \(W\) is finite-dimensional.
2. \(T_W=-\widehat\otimes W\) has a right adjoint.
3. \(T_W\) preserves all small intrinsic coproducts.

When these hold, let \(W^\vee=C(W)\) with its finite-dimensional discrete topology. Both the left and the right adjoint of \(T_W\) are isomorphic to \(-\widehat\otimes W^\vee\).

Proof of necessity. A functor with a right adjoint preserves coproducts: its Hom adjunction turns maps from the image of a coproduct into a family of maps from the images of its summands. Thus condition 2 implies condition 3.

Put \(N=C(W)\). Under the specified duality, the canonical coproduct comparison for the family \(D(A_E)\) of Lemma 4.1 is

The two displayed arrows are the same map under the evaluation and tensor identifications: the structural injections are the duals of the coordinate projections, which are precisely the components defining \(\gamma\).

If \(N\) is infinite-dimensional, \(\gamma\) is injective and not surjective. Full duality over a field makes \(D(\gamma)\) surjective, by extension of functionals, and gives it a nonzero kernel \(D(Q/\operatorname{im}\gamma)\). A nonzero quotient has a nonzero functional by a basis. Hence (4.4) is a proper quotient, not an isomorphism, contradicting condition 3. Thus \(N\) is finite-dimensional, and \(W\simeq D(N)\) is finite-dimensional. This proves condition 3 implies condition 1.

Proof of sufficiency and the maps. A finite-dimensional Hausdorff linear topology is discrete: successive finite intersections of open subspaces can be chosen to lower the dimension until zero, using Hausdorffness. Completed tensoring with such a factor has the usual finite-biproduct topology, as proved in Section 6 of the completed-tensor prerequisite.

For a basis \((w_i)_{i=1}^d\) and dual basis \((w^i)\), put \(R_W(U)=U\widehat\otimes W^\vee\). The unit and counit for \(T_W\dashv R_W\) are

They are continuous finite-coordinate maps. Evaluation is equivariant, and the coevaluation tensor is invariant because it represents the identity of \(W\); thus both maps are equivariant and basis-independent. Applying any continuous equivariant map to the first tensor coordinate commutes with both formulas, proving naturality. Their triangles are the finite identities

Applied to each finite tensor coordinate, these prove the triangles on \(T_W(V)\) and \(R_W(U)\). The induced Hom bijection is therefore a natural adjunction. This proves condition 1 implies condition 2.

For the left adjoint, the usual finite tensor-Hom identification gives \(H(N,M)\simeq M\otimes N^*\). Every map from finite-dimensional \(N\) into \(M\) has image in a finite-dimensional invariant subspace of \(M\), so the algebraic Hom is already locally finite. Substitution in (3.1), followed by finite biduality, gives \(L_W(U)\simeq U\widehat\otimes W^\vee\). This proves the last assertion. If \(W=0\), both tensor and both adjoints are the constant zero functor; the triangles are identities on zero. \(\square\)

The invariant coevaluation tuple is important when \(G\) is infinite. A family of unrelated vectors in \(N\) need not be a locally finite vector of a product, whereas the tuple (4.3) always is. The criterion therefore imposes no finite-generation, semisimplicity or finiteness assumption on the group.

5. Two examples of the left adjoint

Example 5.1 (a finite trivial factor). Take \(W=k^d\) with trivial action. Then \(T_W(V)=V^d\), with the finite-product topology. Its left and right adjoints are both \(U\mapsto U^d\). A map \(U^d\to V\) is the sum of its restrictions to the \(d\) summands; those restrictions also form its transpose \(U\to V^d\). Projections and injections give the inverse, continuously and equivariantly. For \(d=0\), both Hom sets consist of their unique zero maps.

Example 5.2 (a dense finite-evaluation subspace). Let \(G=1\) and \(W=k^{\mathbb N}\) with product topology. Then \(N=k^{(\mathbb N)}\) and

The subscript means that the vector space of all sequences is given the discrete topology before taking its full dual. Define

It is injective by evaluation on coordinate vectors. Its image is dense in the finite topology. Indeed, for a finite-dimensional subspace \(E\subset k^{\mathbb N}\), the coordinate functionals restricted to \(E\) span \(E^*\). If they spanned a proper subspace, finite biduality would give a nonzero vector of \(E\) killed by every coordinate functional, which is impossible. Thus a finite linear combination of coordinate evaluations agrees with any prescribed functional on \(E\). Such agreement is exactly a basic finite-topology neighbourhood condition.

The image is proper. The class of the constant sequence \(1\) in \(k^{\mathbb N}/k^{(\mathbb N)}\) is nonzero. Extend it to a basis of that quotient and define a functional taking value one on that class and zero on the other basis vectors. Its pullback is nonzero and vanishes on every finitely supported sequence. A finite linear combination of coordinate evaluations vanishing on every coordinate vector has every coefficient zero. Therefore this pullback is outside the image of \(\iota\).

6. Four graded exercises with full solutions

Exercise 1 (standard: finite factors with a nontrivial action). For a finite-dimensional representation \(W\in\mathcal L\), identify \(H(C(W),C(U))\) explicitly, with its \(G\)-action, and compute both adjoints of \(T_W\). Include \(W=0\).

Solution. Put \(N=C(W)\), \(M=C(U)\). The map \(M\otimes N^*\to\operatorname{Hom}_k(N,M)\) takes \(m\otimes\phi\) to \(n\mapsto\phi(n)m\). A basis of \(N\) and its dual basis give its inverse, so it is bijective. The diagonal action corresponds to conjugation, since

For any linear map, finitely many basis images lie in one finite-dimensional invariant subspace \(M_0\subset M\). The map then lies in the finite-dimensional invariant space \(\operatorname{Hom}_k(N,M_0)\). Thus the entire Hom is locally finite, and \(H(N,M)=M\otimes N^*\). Taking full duals gives \(L_W(U)=D(M\otimes N^*)\simeq U\widehat\otimes D(N^*)\). Finite evaluation identifies \(D(N^*)\) with \(N=W^\vee\), including its action and discrete topology. The right adjoint is the same tensor functor by (4.5)–(4.6). For zero \(W\), all these internal Hom and tensor objects are zero and both adjunctions hold as stated.

Exercise 2 (intermediate: an explicit failed coproduct comparison). Let \(G=1\), \(W=k^{\mathbb N}\), \(N=k^{(\mathbb N)}\), and \(E_n=\operatorname{span}(e_0,\ldots,e_{n-1})\) for \(n\geq1\). Use the family \(A_n=E_n^*\) to show that the canonical coproduct comparison for \(T_W\) is a surjection with nonzero kernel. Exhibit a tuple outside its dual comparison.

Solution. Since the action is trivial, every vector of every representation is locally finite. Thus \(P=\prod_n E_n^*\) and \(Q=\prod_n(E_n^*\otimes N)\). The dual comparison is \(\gamma:P\otimes N\to Q\), with the coordinate formula of Lemma 4.1, and the same independent-coefficient argument makes it injective. The tuple

is in \(Q\). If it came from a finite tensor sum, one fixed finite-dimensional subspace \(N_0\subset N\) would contain every \(E_n\), by evaluating the first tensor factor. This is impossible because the coordinate vectors are linearly independent. Thus \(u=(u_n)\) is outside the image.

Extension of linear functionals makes \(D(\gamma)\) surjective. The class of \(u\) is nonzero in \(Q/\operatorname{im}\gamma\); choose a functional on this quotient taking value one on that class, and pull it back to \(Q\). It is a nonzero element of \(\ker D(\gamma)\). Under (4.4), \(D(\gamma)\) is exactly the required coproduct comparison. Its quotient topology and closed kernel follow from the complete continuous-duality theorem. Hence it is a proper continuous quotient.

Exercise 3 (advanced: the locally finite part of an infinite Hom). Take \(k=\mathbb Q\), \(G=\mathbb Z\) with generator \(g\), and distinct nonzero scalars \(\lambda_n=n+2\). Let

where \(g\) acts on \(k_\lambda\) by multiplication by \(\lambda\). Compute \(H(N,k)\) and \(L_W(k)\). Show that \(L_W\) is not completed tensoring by the object \(L_W(k)\).

Solution. The full Hom is the product of the one-dimensional characters \(k_{\lambda_n^{-1}}\). If a vector \(a=(a_n)\) of this product has finite-dimensional orbit span, the vectors \(a,ga,g^2a,\ldots\) are linearly dependent. A nonzero polynomial \(p\in k[t]\) therefore satisfies \(p(g)a=0\), giving \(p(\lambda_n^{-1})a_n=0\) in each coordinate. A nonzero polynomial has only finitely many roots: division by \(t-b\) lowers the degree whenever \(b\) is a root, proving this by induction on degree. The distinct \(\lambda_n^{-1}\)'s consequently force all but finitely many \(a_n\) to vanish. Conversely any finite-support tuple lies in a finite direct sum of characters and is locally finite. Hence

with the product topology in the second line. This object is infinite-dimensional. If \(L_W\) were isomorphic to completed tensoring by it, that tensor functor would have right adjoint \(T_W\), by Theorem 3.1. Theorem 4.2 excludes a right adjoint for an infinite-dimensional tensor factor. Thus no such functor isomorphism exists. The value of the left adjoint at the unit does not by itself give a tensor-dual formula.

Exercise 4 (expert: internal Hom and equivariant maps have different exactness). Let \(G=C_p\), \(\operatorname{char}k=p\), and \(A=k[C_p]\). Show that \(H(k,-)\) carries the surjective augmentation \(A\to k\) to a surjection, whereas the induced map \(\mathcal C(k,A)\to\mathcal C(k,k)\) is zero. Explain its compatibility with the tensor adjoints.

Solution. Evaluation at \(1\) identifies the conjugation Hom from the trivial representation \(k\) with the original representation: \(\operatorname{Hom}_k(k,M)\simeq M\). For finite \(G\), every representation is locally finite, since the orbit of each vector spans at most \(|G|\) vectors. Therefore \(H(k,M)=M\), and the internal-Hom map in question is the original surjective augmentation.

An element \(\sum_{i=0}^{p-1}a_i g^i\) of \(A\) is invariant exactly when cyclic permutation of its coefficients leaves it unchanged, or exactly when all \(a_i\) are equal. Thus \(A^G\) is spanned by the nonzero norm vector \(q=\sum_{i=0}^{p-1}g^i\). Its augmentation is \(p=0\) in \(k\). Since \(k^G=k\), the map \(A^G\to k^G\) is zero and not surjective. Proposition 2.1 identifies these invariant spaces with the two equivariant Hom sets. Taking invariant vectors after internal Hom has introduced a nonexact operation. Tensoring with the unit \(k\) and its left and right adjoints are all the identity; those adjunctions assert no exactness of taking invariant vectors.

References

• Ind representations and the locally finite coreflector, for the complete coreflection and intrinsic-product proofs.
• Pro representations and continuous duality, for both actual evaluation inverses, exact duality, intrinsic coproducts, and the topologies of kernels and quotients.
• Completed tensors and continuous algebra objects, for completed tensor products, their universal maps and monoidal structure, and finite-dimensional tensor duals.
• Internal Hom in Ind categories and flat module obstructions, for the general filtered-completion internal-Hom construction.
• Pietro Freni, On Vector Spaces with Formal Infinite Sums, Applied Categorical Structures 34, article 15 (2026), Lemma 5.7, for further reading on the finite-codimensional topology and dense bidual evaluation.