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Fix a field. Given a vector space $V$ it has a dual $V^\ast$, which has its own dual $V^{\ast \ast}$, which has its own dual $V^{\ast \ast \ast}$, and so on ad infinitum.

There are lots of natural maps between these iterated duals. Most famous is the map

$$ i_V \colon V \to V^{\ast \ast} $$

given by

$$ i_V(v)(f) = f(v) \quad \forall v \in V, f \in V^\ast $$

but there are many others. For example, we have the map

$$ i_{V^{\ast }} \colon V^{\ast } \to V^{\ast \ast \ast } $$

but we also have the map

$$ i_{V}^\ast \colon V^{\ast \ast \ast } \to V^{\ast}$$

formed by taking the adjoint of

$$ i_{V} \colon V \to V^{\ast \ast}. $$

And we have various equations between these maps. For example, I believe $i_{V}^\ast$ is a left inverse of $i_{V^{\ast}}$:

$$ i_{V}^\ast \circ i_{V^{\ast}} = 1. $$

So it's natural to want to get to the bottom of this and ask what are all the natural maps between iterated duals, and all the equations involving these maps.

To formalize this it's good to treat duality as a pair of contravariant functors

$$ D \colon \mathrm{Vect} \to \mathrm{Vect}^{\rm op}$$

$$ E \colon \mathrm{Vect}^{\rm op} \to \mathrm{Vect}.$$

These are really the same functor in two disguises: namely, the contravariant functor that sends any vector space $V$ to its dual $V^\ast$, and any linear map $f\colon V \to W$ to its adjoint $f^{\ast} \colon W^\ast \to V^\ast$. Treating them as two separate functors makes it a bit easier to make sense of the fact that they are adjoint functors and thus they define a monad and also a comonad.

A bunch of natural transformations between the powers of

$$ \begin{array}{cccc} E \circ D \colon & \mathrm{Vect} &\to& \mathrm{Vect} \\ & V &\mapsto& V^{\ast \ast} \end{array} $$

and also a bunch of equations between these natural transformations, arise from the adjunction between $D$ and $E$. So, one can ask if all the natural transformations between the powers of $E \circ D$, and all the equations between these, arise from this adjunction.

To give this conjecture a chance to be true, we need to interpret 'arise from' broadly enough. For example, a linear combination of natural transformations will again be a natural transformation. So, to have a chance of getting all the natural transformations, we need to start with the unit and counit of the adjunction, and build other natural transformations from these, and the functors $D$ and $E$, by taking all composites (in the 2-categorical sense), and also linear combinations. Then we can ask whether the resulting natural transformations are all the natural transformations between powers of $E \circ D$.

We can also ask whether all equations between these natural transformations follow from the usual 'zig-zag equations' governing the unit and counit of an adjunction... where 'follow' needs to be made more precise.

The slickest way to formulate the question may use a bit of 2-category theory.

There is a 2-category $\mathbf{Adj}$ called the 'walking adjunction', such that an adjunction in $\mathbf{Cat}$ is the same as a 2-functor from $\mathbf{Adj}$ to $\mathbf{Cat}$.

The adjunction between $D$ and $E$ thus gives a particular 2-functor

$$F \colon \mathbf{Adj} \to \mathbf{Cat}$$

and we can express part of my question as a question about this 2-functor. For example, we can ask

Question 1. Is $F \colon \mathbf{Adj} \to \mathbf{Cat}$ locally faithful, meaning faithful on each hom-category?

If it's not, there are additional equations involving the unit and counit of the adjunction between $D$ and $E$, that don't hold in a general adjunction.

To ask whether we've found all the natural maps between iterated duals of vector spaces, we might ask if $F$ is locally full, meaning full on each hom-category. But we already know it's not, due the linearity issue I mentioned! So it's good to follow Peter LeFanu Lumsdaine's suggestions and work instead with something like $\mathbf{Adj}_k$, the walking additive $k$-linear adjunction. This is a locally additive $k$-linear 2-category such that a 2-functor into $\mathbf{Cat}_k$, the 2-category of additive $k$-linear categories, is the same as an adjunction in $\mathbf{Cat}_k$.

The adjunction between $D$ and $E$ thus gives a particular 2-functor

$$F_k \colon \mathbf{Adj}_k \to \mathbf{Cat}_k$$

and we can attempt to formulate my whole question as follows:

Question 2. Is $F_k \colon \mathbf{Adj}_k \to \mathbf{Cat}_k$ locally full and locally faithful?

I should add that the story feels different if we restrict to finite-dimensional vector spaces, because then $i_V \colon V \to V^{\ast \ast}$ is an isomorphism. I don't want to restrict to finite-dimensional vector spaces.

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    $\begingroup$ A related (but much more naive) question: math.stackexchange.com/q/730995/660. $\endgroup$ Jun 27, 2021 at 21:31
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    $\begingroup$ Should we be looking at these not just as 2-categories, but locally $k$-additive 2-categories (where $k$ is our fixed field), including using $\mathbf{Adj}_k$, the walking $k$-additive adjunction? On the one hand, that gives $F_k : \mathbf{Adj}_k \to \mathbf{Cat}_k$ a chance at being locally full, which it certainly isn’t if we look at this just in 2-categories (since we have at least scalar multiples of the unit and counit, etc). On the other hand, if local faithfulness holds in the plain 2-cat world then it holds here, since your original $F$ factors through $F_k$. $\endgroup$ Jun 27, 2021 at 21:47
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    $\begingroup$ By coincidence I have shown at math.SE/4180665 a few days ago that $\hom(E \circ D,\mathrm{id}_{\mathbf{Vect}})=0$. This is like one piece of the puzzle for an answer. It is also not hard to show that $\hom(D,D) \cong \hom(E,E) \cong k$. $\endgroup$ Jun 27, 2021 at 23:38
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    $\begingroup$ Actually, John only asked about local faithfulness, but this is not hard to prove with the explicit description of $\mathrm{Adj}_k$ (but very tedious to write down). The hard and perhaps even more interesting question is local fullness (and this would answer Pierre-Yves' question). $\endgroup$ Jun 28, 2021 at 20:31
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    $\begingroup$ $\hom(T,T^2)$ is the set (?) of natural transformations $V^{**} \to V^{****}$, which correspond naturally to natural transformations $V^{***} \to V^{***}$. There are two candidates: the identity and $i_{V^*} \circ (i_V)^*$ (which is split idempotent). I suggest to ask for natural transformations $V^{***} \to V^{***}$ in a separate thread, since this special case is already quite interesting and challenging. A negative answer (for example, it could be infinite-dimensional, or not a set!) directly answers the general case, and a positive answer will probably show what to do in the general case. $\endgroup$ Jun 29, 2021 at 11:19

2 Answers 2

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I think I can give a concrete description of these maps, and have an approach to proving faithfulness.

Let $V^{*n}$ be the $n$'th dual of $V$.

A map $V^{*n} \to V^{*m}$ is, by definition, the same thing as a bilinear form $V^{*n} \times V^{*(m-1)} \to k$.

There are two obvious ways to produce such a bilinear form:

  1. from a map $V^{*n} \to V^{* (m-2)}$, i.e. from a bilinear form $V^{*n} \times V^{* (m-3)} \to k$,

  2. from a map $V^{* (m-1)} \to V^{*(n-1)}$, i.e. from a bilinear form $V^{*(m-1)} \times V^{* (n-2)} \to k$.

Note that each of these methods reduces one of the "exponents" by two. If we iteratively apply these methods, we can stop at the obvious bilinear form $V^{*} \times V \to k$ or $V \times V^* \to k$. Which one we stop at will depend on which of $n$ or $m-1$ is odd (exactly one must be odd and one even for a natural map to exist, even in the finite-dimensional case).

So we apply method one $\lfloor \frac{m-1}{2} \rfloor$ times and method two $\lfloor \frac{n}{2} \rfloor$ times. The total number of maps produced this way is $$\binom{ \lfloor \frac{n}{2} \rfloor + \lfloor \frac{m-1}{2} \rfloor}{ \lfloor \frac{m-1}{2} \rfloor},$$ and this set of orders is naturally in bijection with the order-preserving maps appearing in the walking adjunction in the description Martin Brandenburg linked in the comments. Thus I think these maps are the ones produced by the walking adjunction.

To check that these maps are linearly independent, it suffices to prove that the linear span of the canonical maps produced by method 1 does not intersect the linear span of the canonical maps produced by method 2. Then these two linear spaces would be independent, and we would break them further into independent summands, and so on, until finally breaking them into independent one-dimensional spaces.

To prove this, we can try the following lemma:

Let $W$ and $U$ be vector spaces over $k$. A bilinear form $W^* \times U^* \to k$ that arises from a map $W^* \to U$ and arises from a map $U^* \to W$ in fact arises from an element of $U \otimes W$.

Let $B$ be such a bilinear form satisfying $B(w,u)= u(a(w))=w(b(u))$ for suitable maps $a: W^\vee \to U$ and $b: U^\vee \to W$. Fix bases $\{e_i \}_{i \in I}$ of $W$ and $\{f_j\}_{j \in J}$ of $U$. Let $e_i^\vee$ and $f_j^\vee$ be the dual elements of $W^\vee$ and $U^\vee$, not necessarily forming a dual basis. Form a graph with vertices $I \cup J$ where $i$ and $j$ are connected by an edge if $B( e_i^\vee, f_j^\vee) \neq 0$. Then each vertex has finite degree since $i$ is only connected to indices of basis vectors summing to $a( e_i^\vee)$ and $j$ is only connected to indices of basis vectors summing to $b(f_j^\vee)$.

So either there are finitely many edges or there exists an infinite sequence of edges where there vertices of distinct edges in the sequence are not adjacent.

In the first case, there is an element $x\in V \otimes W$ whose value on $e_i^\vee \times e_j^\vee$ is equal to $B(e_i^\vee, e_j^\vee)$ for all $i,j$, hence by the continuity of $B$ in $W^\vee$, the value of $x$ on $w \times e_j^\vee$ is equal to $B(w, e_j^\vee)$ for all $w, j$ and thus by the continuity of $B$ in $U^\vee$, the value of $x$ on $w \times u$ is equal to $B(w,u)$ for all $w,u$, and we are done.

In the second case, let $S$ be set of vertices of the edges in that sequence that lie in $W$. Let $w$ be the linear form on $W$ that sends $e_i$ to $1$ if $i$ is in $S$ and $0$ otherwise. Then $B(w, e_j^\vee) = w ( b(e_j^\vee))$ is the sum of all the entries of $w(e_j^\vee)$ with indices in $S$. For $j$ any vertex of the edge in the sequence lying in $U$, exactly one nonzero entry of $w(e_j^\vee)$ lies in $S$ so $B(w, e_j^\vee) \neq 0$ and thus $B(w,e_j^\vee)$ is nonzero for infinitely many $j$, contradicting $B( w,e_j^\vee) =e_j^\vee ( a(w))$. QED

In addition, we need the following plausible claim:

There are no nonzero canonical (i.e. functorial) elements of $V^{*a} \otimes V^{*b}$ for any $a,b$.

Combining these, any canonical map in the intersection of the two subspaces of the space of bilinear forms on $V^{*n} \times V^{*(m-1)}$ must come from an element of $V^{ * (n-1) } \otimes V^{* (m-2)}$, which is unique, hence also canonical, and thus must be zero.

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    $\begingroup$ What do you mean by continuity of $B$ in $W^{\vee}$? $\endgroup$ Jun 29, 2021 at 23:49
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    $\begingroup$ @MartinBrandenburg I just mean that for fixed $u$, $B(u, w)$, as a function on $W^\vee$, arises by an element of $W$, so if we know its value on $e_i^\vee$ for all $i$ then we know that element and thus we know its value on all dual vectors. $\endgroup$
    – Will Sawin
    Jun 30, 2021 at 0:08
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Here is a proof of the "plausible claim" in Will Sawin's answer.

The "plausible claim" is:

"There are no nonzero canonical (i.e. functorial) elements of $V^{*a}\otimes V^{*b}$ for any $a,b$."

Here we can assume that $a$ is a nonnegative even integer and $b$ a positive odd integer. Let $f_V$ be such an element, depending functorially on $V$, and let us derive a contradiction. The element $f_V$ can be viewed as a finite rank linear map $f_V:V^{*(b-1)}\to V^{*a}$ which is functorial in $V$. For all $V$ the image of $f_V$ is a finite dimensional subspace $F_V$ of $V^{*a}$, and there is at least one $V$ such that $F_V$ is nonzero. We claim

$(1)$ There is an infinite dimensional $V$ such that $F_V$ is nonzero.

Let us show that $(1)$ yields a contradiction. Note that $F_V$ is a nonzero finite dimensional module over the ring $A:=\operatorname{End}(V)$. Let $I\subset A$ be the two-sided ideal formed by the endomorphisms whose rank is less than $\dim V$. By Exercise 3.16 in the book

Lam, Tsit-Yuen. Exercises in Classical Ring Theory, second ed., Problem Books in Mathematics, New York, Springer-Verlag, 2003,

$A$ is the unique two-sided ideal of $A$ not contained in $I$. It is easily seen that $I$ has infinite codimension in $A$. This implies that $A$ has no nonzero finite dimensional module, contradiction. (Note that a solution to the exercise is given in the book.)

It only remains to prove $(1)$.

If $V$ is finite dimensional, we can identify $V^{*(b-1)}$ and $V^{*a}$ to $V$. Then it is easy to see that there is a scalar $\lambda\in k$ such that $f_V=\lambda\operatorname{id}_V$ for all finite dimensional $V$.

Case $\lambda=0$. By assumption there is an infinite dimensional $V$ such that $F_V\ne0$, so there is nothing to prove in this case.

Case $\lambda\ne0$. We will see that this case is impossible. We can assume $\lambda=1$. Write $D_n$ for the functor $V\mapsto V^{*n}$. Let $V$ be any infinite dimensional vector space and $v$ any nonzero vector of $V$. Regard $v$ as a linear map $k\to V$. Since, in the commutative diagram $\require{AMScd}$ $$ \begin{CD} k@>f_k>>k\\ @VD_{b-1}(v)VV@VVD_a(v)V\\ V^{*(b-1)}@>>f_V>V^{*a}, \end{CD} $$ $f_k$ is the identity of $k$, we see that $F_V$ contains the canonical copy of $V$ sitting inside $V^{*a}$, contradicting the finite dimensionality of $F_V$.

EDIT. Here are a few additional details. (We use the above notation. In particular $V$ is infinite dimensional and $F_V\subset V$ is finite dimensional and nonzero. Also recall that $D_n$ is the functor $W\mapsto W^{*n}$, and that $a$ and $b-1$ are nonnegative even integers.)

$\bullet$ $F_V$ is an $A$-module. Consider a vector in $F_V$, that is a vector of the form $f_V(v_{b-1})$ with $v_{b-1}$ in $V^{*(b-1)}$, and let $x$ be in $A$. Then $(D_a(x))(f_V(v_{b-1}))$, being equal to $f_V(D_{b-1}(v_{b-1}))$, is again in $F_V$, and the formula $xf_V(v_{b-1}):=(D_a(x))(f_V(v_{b-1}))$ defines a structure of $A$-module on $F_V$.

$\bullet$ $I$ has infinite codimension in $A$. Let $J$ be a set of cardinality $\dim V$. There is a family $(V_j)_{j\in J}$ of subspaces $V_j\subset V$ such that $\dim V_j=\dim V$ for all $j$ and $V=\bigoplus V_j$. If $e_j$ is the projection onto $V_j$ viewed as an idempotent of $A$, then any nontrivial linear combination of the $e_j$ is of rank $\dim V$. This implies $\dim A/I\ge\dim V$.

$\bullet$ Derivation of the contradiction. Using the first bullet, define $\psi:A\to\operatorname{End}(F_V)$ by $(\psi(x))(w):=xw$ for $x$ in $A$ and $w$ in $F_V$. This is a $k$-algebra morphism, and its kernel is a proper two-sided ideal of $A$ of whose codimension coincides with the rank of $\psi$, which is finite. By Exercise 3.16 mentioned above and the previous bullet, there is no such ideal.

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