Dualizable objects (and morphisms)

October 26, 2015 by Qiaochu Yuan

Let $R$ be a ring. Previously we characterized the finitely presented projective (right) $R$ -modules as the tiny objects in $\text{Mod}(R)$ : the objects $P$ such that

$\displaystyle \text{Hom}(P, -) : \text{Mod}(R) \to \text{Ab}$

preserves colimits. We also highlighted the key role that these modules play in Morita theory.

If $k$ is a commutative ring, then $\text{Mod}(k)$ has a natural symmetric monoidal structure which allows us to describe another finiteness condition called dualizability. Unlike tininess, dualizability makes no reference to colimits; instead, it is a purely equational definition involving the monoidal structure. The dualizable modules are again the finitely presented projective $k$ -modules.

Dualizability implies that we can treat finitely presented projective $k$ -modules like finite-dimensional vector spaces in many ways: for example, dualizability allows us to define the trace of an endomorphism. Moreover, since dualizability is defined using only a monoidal structure, it makes sense in very general settings, and we’ll look at some more exotic examples of dualizable objects as well.

Duals are also a special case of a 2-categorical notion of adjunction which, in the 2-category of categories, functors, and natural transformations, reproduces the usual notion of adjunction. In a suitable 2-category it will also reproduce another characterization of finitely presented projective modules, this time over noncommutative rings.

This post should, but will not, include diagrams, so pretend that I’ve inserted some string diagrams or globular diagrams where appropriate.

Conventions

All compositions will be written in “diagrammatic order”: If $f : a \to b$ is a morphism and $g : b \to c$ is another morphism, then $fg$ will denote the composite $a \xrightarrow{f} b \xrightarrow{g} c$ . This means that it’s necessary to be careful when trying to match up the definitions below to other definitions.

“2-category” means bicategory and not strict 2-category, although for simplicity we’ll be pretending that all of the 2-categories we write down are strict, and in particular we’ll ignore associators, etc.

Definition and examples

Let $C$ be a 2-category. For lack of a better notation, below $\circ$ denotes horizontal composition of 2-morphisms, while vertical composition of 2-morphisms will be denoted by concatenation.

Let $f : c \to d$ be a morphism in $C$ . $f$ is right dualizable if it has a right dual or right adjoint $g : d \to c$ , which is another morphism equipped with 2-morphisms

$\displaystyle \eta : \text{id}_c \to fg$

(the unit or coevaluation map) and

$\displaystyle \varepsilon : gf \to \text{id}_d$

(the counit or evaluation map) satisfying the zigzag identities

$\displaystyle f \xrightarrow{\eta \circ \text{id}_f} fgf \xrightarrow{\text{id}_f \circ \varepsilon} f = f \xrightarrow{\text{id}_f} f$

and

$\displaystyle g \xrightarrow{\text{id}_g \circ \eta} gfg \xrightarrow{\varepsilon \circ \text{id}_g} g = g \xrightarrow{\text{id}_g} g$ .

Switching the order of composition of morphisms, or equivalently switching the role of $f$ and $g$ , gives the definition of left dualizable and left duals / adjoints. We will write $f \dashv g$ to mean that $f$ is right dualizable with right dual / adjoint $g$ , or equivalently that $g$ is left dualizable with left dual / adjoint $f$ . If we don’t want to single out a particular interpretation, we will say that $f$ and $g$ (in that order) are a dual / adjoint pair.

Suppose in particular that $C = BM$ is the delooping of a monoidal category $(M, \otimes)$ , or equivalently that it has one object, so that the endomorphisms of that object form a monoidal category $M$ . Then right dualizability is a condition on an object $m \in M$ , and we will denote its right dual by $m^{\ast}$ . Similarly, if we need it, the left dual will be denoted by $^{\ast} m$ . If we need to distinguish this notion of dual from another one, we will use the term monoidal dual.

In this special case right dualizability takes the following form. If $1$ denote the identity object of $M$ , the unit and counit maps are

$\displaystyle \eta : 1 \to m \otimes m^{\ast}$

$\displaystyle \varepsilon : m^{\ast} \otimes m \to 1$

and the zigzag identities are

$\displaystyle m \xrightarrow{\eta \otimes \text{id}_m} m \otimes m^{\ast} \otimes m \xrightarrow{\text{id}_m \otimes \varepsilon} m = m \xrightarrow{\text{id}_m} m$

$\displaystyle m^{\ast} \xrightarrow{\text{id}_{m^{\ast}} \otimes \eta} m^{\ast} \otimes m \otimes m^{\ast} \xrightarrow{\varepsilon \otimes \text{id}_{m^{\ast}}} m^{\ast} = m^{\ast} \xrightarrow{\text{id}_{m^{\ast}}} m^{\ast}$ .

If $C = BM$ where $M$ is at least braided monoidal, then the braiding turns left duals into right duals, so the term “dualizable” is unambiguous in this case.

An important observation about this definition is that it is purely equational: it asks only that some objects, morphisms, and 2-morphisms satisfy some equations between compositions, and checking that this is the case is “local” in the sense that it doesn’t involve any other objects, morphisms, and 2-morphisms in the ambient 2-category. It also follows that adjoints are preserved by arbitrary 2-functors and that duals are preserved by arbitrary monoidal functors.

Example. Let $M = (\text{Vect}, \otimes)$ be the symmetric monoidal category of vector spaces over a field $k$ equipped with the usual tensor product. Then a vector space $V$ is dualizable iff it is finite-dimensional, in which case $V^{\ast}$ is the usual vector space dual $\text{Hom}(V, k)$ , and the unit and counit map are the familiar coevaluation and evaluation maps respectively. This is the basic source of the intuition that dualizability is a “finiteness” condition on an object.

Example. Let $C$ be the 2-category of categories, functors, and natural transformations. Then the above is precisely the definition of a unit-counit adjunction (provided that we interpret composition of functors in diagrammatic order, to get the lefts and rights to match up).

Example. Let $C$ be a category with finite limits and let $M = \text{Span}(C)$ be the symmetric monoidal category whose objects are the objects of $C$ but whose morphisms are (isomorphism classes of) spans in $C$ , with monoidal structure given by the product in $C$ . Then every object is self-dual. If $c \in C$ is an object, then the unit map can be taken to be the span

$\displaystyle 1 \leftarrow c \xrightarrow{\Delta} c \times c$

(where $\Delta$ is the diagonal), and similarly the counit map can be taken to be the span

$\displaystyle c \times c \xleftarrow{\Delta} c \rightarrow 1$ .

As defined, $\text{Span}(C)$ fails to be locally small if $C$ isn’t small. One way to fix this is to notice that $\text{Span}(C)$ is naturally a symmetric monoidal 2-category whose morphisms are spans and whose 2-morphisms are morphisms of spans, and as a 2-category it is locally small (if $C$ is) in the sense that between morphisms there is a set of 2-morphisms.

Example. Let $k$ be a commutative ring and let $\text{Mor}(k)$ be the Morita 2-category, whose

objects are $k$ -algebras $A$ ,
morphisms $A \to B$ are $(A, B)$ -bimodules $_A M_B$ over $k$ (meaning that $M$ is equipped with a left action of $A$ and a commuting right action of $B$ such that the two resulting actions of $k$ agree), with composition given by tensor product, and
2-morphisms are bimodule homomorphisms.

This example strictly generalizes the example of the monoidal category of vector spaces, which we get when $k$ is a field and we look at the endomorphism category of $k$ in $\text{Bim}(k)$ . By the Eilenberg-Watts theorem, an equivalent description of this 2-category is that its

objects are the $k$ -linear categories $\text{Mod}(A)$ ,
morphisms $\text{Mod}(A) \to \text{Mod}(B)$ are the cocontinuous $k$ -linear functors, which are precisely the functors of the form $(-) \otimes_A M$ for an $(A, B)$ -bimodule $M$ over $k$ , and
2-morphisms are natural transformations.

As we’ll see, a bimodule $_A M_B$ is right dualizable iff it is finitely presented projective as a $B$ -module, in which case its right dual is $_B N_A = \text{Hom}_B(M, B)$ .

$\text{Bim}(k)$ is itself symmetric monoidal with monoidal structure the tensor product $A \otimes_k B$ , and so we can also ask when an object $A$ is dualizable. (In a monoidal 2-category this means we only ask that the zigzag identities hold up to 2-isomorphism.)

It turns out that every object $A$ is dualizable, with dual $A^{op}$ . The unit morphism is $A$ regarded as a $(k, A^{op} \otimes_k A)$ -bimodule and the counit morphism is $A$ regarded as a $(A \otimes_k A^{op}, k)$ -bimodule.

Traces and dimensions

Suppose $m$ is a dualizable object in a symmetric monoidal category $M$ and that $f : m \to m$ is an endomorphism.

Definition: The trace $\text{tr}(f) \in \text{End}(1)$ of $f$ is the composite

$\displaystyle 1 \xrightarrow{\eta} m \otimes m^{\ast} \xrightarrow{\beta_{m, m^{\ast}}} m^{\ast} \otimes m \xrightarrow{\text{id}_{m^{\ast}} \otimes f} m^{\ast} \otimes m \xrightarrow{\varepsilon} 1$

where $\beta$ denotes the braiding in $M$ . The dimension $\dim(m)$ of $m$ is the trace $\text{tr}(\text{id}_m)$ .

We won’t use traces in the rest of the post, but it’s good to know that they’re there. The fact that they can be defined and manipulated using string diagrams is an aspect of $1$ -dimensional topological field theory.

Example. Both of these names come from the fact that in the special case that $M$ is the symmetric monoidal category of vector spaces over a field $k$ , we get back the usual notion of the trace of an endomorphism of a finite-dimensional vector space $V$ and the dimension of $V$ respectively (as an element of $k = \text{End}(1)$ , which loses some information if $k$ has positive characteristic).

Example. Let $M = \text{Span}(C)$ as above, and let $f : c \to c$ be an endomorphism of an object $c \in C$ in the usual sense, regarded as the span $c \xleftarrow{\text{id}_c} c \xrightarrow{f} c$ . The trace $\text{tr}(f)$ turns out to be the pullback of the diagram $c \xrightarrow{\Delta} c \times c \xleftarrow{(\text{id}_c, f)} c$ ; in other words, it’s the intersection of the diagonal and the graph of $f$ inside $c \times c$ , which can in turn be identified with the object $\text{fix}(f)$ of fixed points of $f$ (e.g. if $C = \text{Set}$ we get back the set of fixed points in the usual sense), regarded as a span $1 \leftarrow \text{fix}(f) \rightarrow 1$ .

Example. Let $M = \text{Mor}(k)$ as above. If $A$ is a $k$ -algebra, then an endomorphism of $A$ in this 2-category is an $(A, A)$ -bimodule $_A N_A$ , and its trace turns out to be

$\displaystyle \text{tr}(N) = A \otimes_{A^{op} \otimes_k A} N \cong N / \{ an - na : a \in A, n \in N \}$

also known as the zeroth Hochschild homology $HH_0(A, N)$ . It can be thought of as the universal quotient of $N$ on which the left and right $A$ -actions agree. In particular, the dimension of $A$ turns out to be the zeroth Hochschild homology

$\displaystyle \dim(A) = A \otimes_{A^{op} \otimes_k A} A \cong A/[A, A]$

where $[A, A]$ denotes the subspace (not ideal!) of commutators.

Dualizable bimodules

Let $k$ be a commutative ring as above. We will now classify the left and right dualizable 1-morphisms in $\text{Mor}(k)$ ; that is, we will classify left and right dual pairs of bimodules $_B L_A, _A R_B$ . Throughout this section it may be useful to keep in mind the special case that $A = B = k$ and that $k$ is a field, in which case we are just saying some familiar facts about duality for finite-dimensional vector spaces.

Recall that dualizability means that we have a unit 2-morphism

$\displaystyle \eta : B \to L \otimes_A R$

(here a morphism of $(B, B)$ -bimodules) and a counit 2-morphism

$\displaystyle \varepsilon: R \otimes_B L \to A$

(here a morphism of $(A, A)$ -bimodules) satisfying the zigzag identities. Explicitly, $\eta$ is completely determined by a particular element

$\displaystyle \eta(1) = \sum \eta_i \otimes \eta_j \in L \otimes_A R$

of $L \otimes_A R$ satisfying $\sum b \eta_i \otimes \eta_j = \sum \eta_i \otimes \eta_j b$ , and $\varepsilon$ is a bilinear map

$\displaystyle R \times L \ni (r, \ell) \mapsto \varepsilon(r, \ell) \in A$

satisfying $\varepsilon(a r, \ell) = a \varepsilon(r, \ell)$ and $\varepsilon(r, \ell a) = \varepsilon(r, \ell) a$ , as well as $\varepsilon(r b, \ell) = \varepsilon(r, b \ell)$ . The first zigzag identity is

$\displaystyle L \ni \ell \mapsto \sum \eta_i \otimes \eta_j \otimes \ell \mapsto \sum \eta_i \varepsilon(\eta_j, \ell) = \ell \in L$

and the second is

$\displaystyle R \ni r \mapsto \sum r \otimes \eta_i \otimes \eta_j \mapsto \sum \varepsilon(r, \eta_i) \eta_j = r \in R$ .

At this point we need the following lemma.

Dual basis lemma: A right $A$ -module $P$ is finitely presented projective iff there exist $e_1, \dots e_n \in P$ and $e_1^{\ast}, \dots e_n^{\ast} \in P^{\ast} \cong \text{Hom}_A(P, A)$ (which is a left $A$ -module) such that, for any $p \in P$ , we have

$\displaystyle p = \sum_{i=1}^n e_i \left( e_i^{\ast}(p) \right)$ .

Under these hypotheses, we furthermore have $(P^{\ast})^{\ast} \cong P$ as right $A$ -modules, and for any $p^{\ast} \in P^{\ast}$ , we have

$\displaystyle p^{\ast} = \sum_{i=1}^n p^{\ast}(e_i) e_i^{\ast}$ .

In particular, $P^{\ast}$ is also finitely presented projective.

Proof. The first statement just a restatement of the condition that $P$ is a retract of a finite rank free module $A^n$ . The “basis” $e_1, \dots e_n$ define a morphism $f : A^n \to P$ , the “dual basis” $e_1^{\ast}, \dots e_n^{\ast}$ define a morphism $g : P \to A^n$ , and the above condition is precisely the condition that $gf = \text{id}_P$ (keeping mind that we’re still writing compositions in diagrammatic order).

Applying $\text{Hom}_A(-, A)$ to the previous paragraph, we obtain dual morphisms $f^{\ast} : P^{\ast} \to A^n, g^{\ast} : A^n \to P^{\ast}$ satisfying $f^{\ast} g^{\ast} = \text{id}_{P^{\ast}}$ , hence $P^{\ast}$ is finitely presented projective and the $e_i^{\ast} \in P^{\ast}$ , together with the images $e_i \in (P^{\ast})^{\ast}$ under the canonical map $P \to (P^{\ast})^{\ast}$ , form a dual basis of $P^{\ast}$ .

It remains to show that the canonical map $P \to (P^{\ast})^{\ast}$ is an isomorphism. Here is a cheap trick for doing so: it is clearly true if $P$ is finite free, and because splitting idempotents is an absolute colimit, it is preserved by all functors, including $((-)^{\ast})^{\ast}$ . Hence it’s true for retracts of finite free modules. $\Box$

Theorem: A bimodule $_B L_A$ has a right dual $_A R_B$ iff it is finitely presented projective as a right $A$ -module, in which case $R \cong \text{Hom}_A(L, A)$ . Dually, a bimodule $_A R_B$ has a left dual $_B L_A$ iff it is finitely presented projective as a left $A$ -module, in which case $L \cong \text{Hom}_A(R, A)$ .

Proof. With notation as above, the identity $\ell = \sum \eta_i \varepsilon(\eta_j, \ell)$ from above implies that $\eta_i \in L$ and $\varepsilon(\eta_j, \ell) \in \text{Hom}_A(L, A)$ form a dual basis, so by the dual basis lemma, if $L$ has a right dual then it is finitely presented projective as a right $A$ -module. Similarly, the identity $r = \sum \varepsilon(r, \eta_i) \eta_j$ implies that $\eta_j \in R$ and $\varepsilon(r, \eta_i) \in \text{Hom}_A(R, A)$ form a dual basis, so by the dual basis lemma, if $R$ has a left dual then it is finitely presented projective as a left $A$ -module. Now, the counit defines morphisms

$\displaystyle L \ni \ell \mapsto \varepsilon(-, \ell) \in \text{Hom}_A(R, A)$

and

$\displaystyle R \ni r \mapsto \varepsilon(r, -) \in \text{Hom}_A(L, A)$

of bimodules which we want to show are isomorphisms. We’ll do this by using the unit to exhibit morphisms of bimodules

$\displaystyle \text{Hom}_A(R, A) \ni r^{\ast} \mapsto \sum \eta_i r^{\ast}(\eta_j) \in L$

and

$\displaystyle \text{Hom}_A(L, A) \ni \ell^{\ast} \mapsto \sum \ell^{\ast}(\eta_i) \eta_j \in R$

which we want to show are their inverses. This comes down to checking four identities, two of which are the zigzag identities, and two of which are their $A$ -linear duals.

Conversely, suppose $_B L_A$ is finitely presented projective as a right $A$ -module, and set $_A R_B = \text{Hom}_A(L, A)$ . We want to show that $R$ is the right dual of $_B L_A$ . The counit of the adjunction will just be the dual pairing

$\displaystyle \varepsilon : R \otimes_B L \ni \ell^{\ast} \otimes \ell \mapsto \ell^{\ast}(\ell) \in A$

so the interesting question is how to find the unit. Explicitly, we do this by finding a basis $e_1, \dots e_n \in L$ and a dual basis $e_1^{\ast}, \dots e_n^{\ast} \in R$ , which we assemble into a proposed unit

$\displaystyle \eta : B \ni b \mapsto b \sum e_i \otimes e_i^{\ast} \in L \otimes_A R$ .

It may not be immediately clear why this is equal to $\sum e_i \otimes e_i^{\ast} b$ . To see this, apply both sides to an element $\ell \in L$ ; we get $b \ell$ either way. Then we need to use the fact that $L \otimes_A R \cong L \otimes_A \text{Hom}_A(L, A) \cong \text{End}_A(L)$ , which follows from the stronger statement that

$\text{Hom}_A(L, (-)) \cong \text{Hom}_A(L, A) \otimes_A (-)$ .

This statement follows in turn from the fact that $L$ is finitely presented projective, hence $\text{Hom}_A(L, -)$ is cocontinuous, and then from the Eilenberg-Watts theorem. This lets us interpret the unit more abstractly as the map

$\displaystyle \eta : B \to \text{End}_A(L)$

describing the left action of $B$ on $L$ .

It remains to verify the zigzag equations. The first one reads

$\displaystyle L \ni \ell \mapsto \sum e_i \otimes e_i^{\ast} \otimes \ell \mapsto \sum e_i \left( e_i^{\ast}(\ell) \right) = \ell \in L$

which is the first half of the dual basis lemma. The second one reads

$R \ni \ell^{\ast} \mapsto \sum \ell^{\ast} \otimes e_i \otimes e_i^{\ast} \mapsto \sum \ell^{\ast}(e_i) e_i^{\ast} = \ell^{\ast} \in R$

which is the second half of the dual basis lemma. $\Box$

Corollary: Let $k$ be a commutative ring. A $k$ -module $P$ is dualizable iff it is finitely presented projective as a $k$ -module, in which case its dual (both left and right) is its linear dual $M^{\ast} \cong \text{Hom}_k(P, k)$ .

Proof. Apply the above theorem to the case that $A = B = k$ . $\Box$

Example. If $f : A \to B$ is a morphism of $k$ -algebras, we can consider the corresponding bimodule $_A B_B$ , where the left $A$ -module structure is provided by $f$ . This defines a functor from the ordinary category of $k$ -algebras and $k$ -algebra homomorphisms to the Morita 2-category $\text{Mor}(k)$ . This bimodule is always finitely presented projective as a right $B$ -module, so it always has a right dual, namely $_B B_A$ , where now the right $A$ -module structure is provided by $f$ . This reflects the fact that the extension of scalars functor

$\displaystyle \text{Mod}(A) \ni M \mapsto M \otimes_A {}_A B_B \in \text{Mod}(B)$

always has a cocontinuous right adjoint, namely the restriction of scalars functor

$\displaystyle \text{Mod}(B) \ni N \mapsto N \otimes_B {}_B B_A \in \text{Mod}(A)$ .

However, the condition that this bimodule has a left dual is quite strong: this is the condition that $B$ is finitely presented projective as a left $A$ -module. For example, if $A = k$ and $k$ is a PID, this means that $B$ is finite free as a $k$ -module.

12 Responses

on December 24, 2017 at 9:42 am | Reply Runyu Jing

Very helpful. Thank you!
on April 28, 2017 at 6:11 am | Reply Courant-Fischer Theorem

[…] statement about finite-dimensional vector spaces. This reflects the fact that is always dualizable with respect to the above tensor product, with dual […]
on October 13, 2016 at 9:44 am | Reply Noah Snyder

I’m a little confused, I can’t figure out where you used finitely presented rather than just finitely generated in the dual basis lemma.
- on October 13, 2016 at 8:17 pm | Reply Qiaochu Yuan
  
  I never directly use finitely presented, but only “finitely presented projective” in the form “retract of a finite free module,” which is of course equivalent to “finitely generated projective.”
  - on October 15, 2016 at 12:53 pm | Reply Noah Snyder
    
    Ah, right, in the presence of projective they’re fp and fg are equivalent. Thanks.
on May 31, 2016 at 11:03 am | Reply Higher linear algebra | Annoying Precision

[…] statement about finite-dimensional vector spaces. This reflects the fact that is always dualizable with respect to the above tensor product, with dual […]
on December 15, 2015 at 5:11 pm | Reply Monads are idempotents | Annoying Precision

[…] As previously, in this post compositions will be done in diagrammatic order, so if and are two morphisms, their composite will be denoted , or sometimes (which is independent of, but strongly suggests, the diagrammatic order). […]
on November 13, 2015 at 6:10 pm | Reply Projective representations give categorical representations | Annoying Precision

[…] construction is in fact a 2-functor from the category of -algebras to the Morita 2-category of -algebras, -bimodules, and bimodule homomorphisms, or equivalently of module categories , […]
on October 26, 2015 at 5:05 pm | Reply Theo Johnson-Freyd

Note that in general compact projectivity and dualizability can diverge badly. See for example: L. Gaunce Lewis, Jr. When projective does not imply flat, and other homological anomalies. Theory Appl. Categ., 5:No. 9, 202–250 (electronic), 1999.
- on October 26, 2015 at 6:10 pm | Reply Qiaochu Yuan
  
  Right, it seems like the key property of $\text{Mod}(k)$ here is that the monoidal unit is itself compact projective.
on October 26, 2015 at 9:25 am | Reply Oscar_Cunningham

This definition of dimension can be funny if you actually calculate it out. For example the dimension of $\mathbb{Z}/25\mathbb{Z}$ as a $ latex \mathbb{Z}/100\mathbb{Z}$ module turns out to be 76. What kinda stupid number is that?
- on October 26, 2015 at 2:26 pm | Reply Qiaochu Yuan
  
  Well, as an element of $\mathbb{Z}/100\mathbb{Z}$ , and under the CRT isomorphism $\mathbb{Z}/100\mathbb{Z} \cong \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/25\mathbb{Z}$ , this is just $(0, 1)$ . In other words, the dimension is a locally constant function on $\text{Spec } \mathbb{Z}/100\mathbb{Z}$ , which has two points, and at one point it takes the value $0$ and at the other point it takes the value $1$ . Totally reasonable!

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"A good stock of examples, as large as possible, is indispensable for a thorough understanding of any concept, and when I want to learn something new, I make it my first job to build one." – Paul Halmos