Iterated duals of a vector space

Question

Let $K$ be a field and $\mathcal U$ a universe such that $K\in\mathcal U$. (Here, "universe" means "uncountable Grothendieck universe".) Let $\mathcal C$ be the category of $K$-vector spaces belonging to $\mathcal U$, and let $i\ge0$ be an integer. If $i$ is even, put $\mathcal C_i:=\mathcal C$; if $i$ is odd, put $\mathcal C_i:=\mathcal C^\text{op}$. Let $F_i:\mathcal C_i\to\mathcal C$ be the $i$-th dual functor. For integers $i,j\ge0$ of same parity, $\operatorname{Hom}(F_i,F_j)$ is a $K$-vector space. In particular $$ d(K,\mathcal U,i,j):=\dim\operatorname{Hom}(F_i,F_j) $$ is a well-defined cardinal.

Can one compute this cardinal?

Does $d(K,\mathcal U,i,j)$ depend on $K$ and $\mathcal U$?

Is $d(K,\mathcal U,i,j)$ finite?

Have these questions been asked before?

Edit. As an illustration, here is a proof of the equality $d(K,\mathcal U,2,0)=0$. This case can be handled without universes.

Assume that, for each vector space $V$ (over the field $K$ chosen once and for all), we have a linear map $\theta_V:V^{**}\to V$, and suppose that, for each linear map $f:V\to W$, we have $$ f\circ\theta_V=\theta_W\circ f^{**}. $$

We claim: $\theta_V=0$ for all $V$.

Proof. As a general notation, put $V_1:=V^*,V_2:=V^{**},f_2:=f^{**}$, and, for each vector space $V$, let $\varepsilon_V:V\to V_2$ be the natural map. In particular, the above display becomes $$ f\circ\theta_V=\theta_W\circ f_2. $$ It is easy to see that there is a scalar $\lambda\in K$ such that $\theta_V\circ\varepsilon_V=\lambda\operatorname{id}_V$ for all $V$, and that we can assume either $\lambda=0$ or $\lambda=1$.

Case $\lambda=0$. Since $\varepsilon_K$ is an isomorphism, we get $\theta_K=0$. For $v_1$ in $V_1$ we have $v_1\circ\theta_V=\theta_K\circ v_{12}=0$. Since $v_1$ is arbitrary, this implies $\theta_V=0$.

Case $\lambda=1$. We are seeking a contradiction. For $v_1$ in $V_1$ we have $$ v_{12}=\varepsilon_K\circ\theta_K\circ v_{12}=\varepsilon_K\circ v_1\circ\theta_V, $$ that is $$ \big(v_{12}(v_2)\big)(k_1)=\bigg(\varepsilon_K\Big(v_1\big(\theta_V(v_2)\big)\Big)\bigg)(k_1) $$ for $v_2$ in $V_2$ and $k_1$ in $K_1$. The last equality can be rewritten as $$ v_2(k_1\circ v_1)=k_1\Big(v_1\big(\theta_V(v_2)\big)\Big). $$ Taking the identity of $K$ as $k_1$, we get $$ v_2(v_1)=v_1\big(\theta_V(v_2)\big) $$ for all $v_1$ in $V_1$ and all $v_2$ in $V_2$. Let $V$ be infinite dimensional. By the Erdös-Kaplansky Theorem, there is a nonzero $v_2$ in $V_2$ such that $\theta_V(v_2)=0$. Since $v_2\ne0$, there is a $v_1$ in $V_1$ such that $v_2(v_1)\ne0$, contradiction.

Answer 1

This is only a very partial answer. I hope there will be more complete answers in the future.

Let $V$ be a vector space, let $V_i$ be its $i$-th dual, let $\varepsilon_i:V_i\to V_{i+2}$ be the natural morphism, and let $\varepsilon_{i1}:V_{i+3}\to V_{i+1}$ be the dual of $\varepsilon_i$. We claim $$ \varepsilon_{01}\circ\varepsilon_1=\operatorname{id}_{V_1}. $$ Indeed, for $v_1$ in $V_1$ and $v$ in $V$ we get $$ \Big(\varepsilon_{01}\big(\varepsilon_1(v_1)\big)\Big)(v)=\big(\varepsilon_1(v_1)\big)(\varepsilon_0(v))=\big(\varepsilon_0(v)\big)(v_1)=v_1(v). $$ This shows that, in the notation of the question, there is a subfunctor $G_3$ of $F_3$ such that $F_3\simeq F_1\oplus G_3$, and, more generally, that there are subfunctors $G_i$ of $F_i$ for $i\ge3$ such that, if we put $G_1:=F_1,G_2:=F_2$, we have $$ F_i\simeq G_1\oplus G_3\oplus\cdots\oplus G_i $$ for $i$ odd, and $$ F_i\simeq G_2\oplus G_4\oplus\cdots\oplus G_i $$ for $i$ even, $i\ge2$.

In view of this post and that post, the question comes down to computing the cardinals $\dim\operatorname{Hom}(G_i,G_j)$ for $i,j$ positive odd integers.

The most naive hope would be to have $\dim\operatorname{Hom}(G_i,G_j)=\delta_{ij}$ for $i,j$ positive odd integers.

Answer 2

As a very partial answer, here are soft categorical arguments that the dimension is $1$ in case $i = 1$ or $j = 2$, or in case $i = 0$ or $j = 1$. The first result shows we can reduce to the cases $i = 1$, $i = 0$.

Lemma: For $i, j$ of the same parity, there is an isomorphism $\hom(F_i, F_j) \cong \hom(F_{j-1}, F_{i+1})$.

Proof: The left side corresponds to dinatural or extranatural transformations of the form $F_i \otimes F_{j-1} \to K$, and similarly the right corresponds to those of the form $F_{j-1} \otimes F_i \to K$.

Proposition: $\hom(F_1, F_j)$ is $1$-dimensional for all odd $j$.

Proof: Note $F_1 \cong \hom_{Vect}(-, K)$. By Yoneda, $\hom(F_1, F_j) \cong Nat(\hom(-, K), F_j) \cong F_j(K) \cong K$.

Proposition: $\hom(F_0, F_j)$ is $1$-dimensional for all even $j$.

Proof: Note $F_0 \cong \hom_{Vect}(K, -)$. By Yoneda, $\hom(F_0, F_j) \cong Nat(\hom(K, -), F_j) \cong F_j(K) \cong K$.

Beyond this point, it gets trickier. The first non-trivial case is where $i = j = 3$, and here I'm not even sure $d(K, \mathcal{U}, i, j)$ is finite-dimensional. I can say that the dimension is greater than $1$, because we have at least two linearly independent transformations,

$$V^{\ast\ast\ast} \stackrel{\text{id}}{\to} V^{\ast\ast\ast}, \qquad, V^{\ast\ast\ast} \stackrel{d^\ast V}{\to} V^\ast \stackrel{d V^\ast}{\to} V^{\ast\ast\ast}$$

where $d V: V \to V^{\ast\ast}$ is the double dual embedding, with adjoint $d^\ast V$. (Of course these transformations differ since $V^\ast$ has lower dimension than $V^{\ast\ast\ast}$ in general.)

As an aside, my PhD thesis was on the structure of free symmetric monoidal closed categories, where the problem was how to distinguish between morphisms that can be canonically built up from the smc structure, e.g., the morphisms in the last display line. If I knew that all morphisms $F_i \to F_j$ were the ones built up from smc structure (oh, and $K$-linear structure of course), then I could probably say a lot more (although I didn't actually compute the cardinality of the analogue of $\hom(F_i, F_j)$ in the free case). But maybe there are weird non-canonical transformations in the case of the smc category $Vect$.