# iterated dual vector spaces

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 1. As an illustration, here is a proof of the equality $d(K,\mathcal U,2,0)=0$. I can't believe that this observation had not been made before. 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 $$f\circ\theta_V=\theta_W\circ f^{**}.\qquad(*)$$

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

Proof. As a general notation, put $V_1:=V^*,V_2:=V^{**},f_1:=f^*,f_2:=f^{**}$, and, for each vector space $V$, let $\varepsilon_V:V\to V_2$ be the natural map.

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 $\lambda=0$. By $(*)$ we have $v_1(\theta_V(v_2))=0$ for all $v_1$ in $V_1$ and all $v_2$ in $V_2$. This implies $\theta_V=0$.

Case $\lambda=1$. We are seeking a contradiction. Let $v_1$ be in $V_1$. We have $v_{12}=\varepsilon_K\circ v_1\circ\theta_V$ by $(*)$, that is, for $v_2$ in $V_2$ and $k_1$ in $K_1$, we have $$v_{12}(v_2)(k_1)=\varepsilon_K(v_1(\theta_V(v_2)))(k_1),$$ or $$v_2(k_1\circ v_1)=k_1(v_1(\theta_V(v_2))).$$ Taking the identity of $K$ as $k_1$, we get $$v_2(v_1)=v_1(\theta_V(v_2)).$$ Remember that this holds for all $v_1$ in $V_1$ and all $v_2$ in $V_2$. Let $V$ be infinite dimensional. By the Erdos-Kaplansky Theorem, there is a nonzero $v_2$ in $V_2$ such that $\theta_V(v_2)=0$. Since there is obviously a $v_1$ in $V_1$ such that $v_2(v_1)\neq0$, we are done. Q.E.D.

I tried unsuccessfully to prove $d(K,\mathcal U,2,2)=1$. Here is an elementary formulation. Instead of $\theta_V:V_2\to V$ we start with $\theta_V:V_2\to V_2$, we assume that the analog of $(*)$ holds, and that we have $\theta_V\circ\varepsilon_V=0$ for all $V$. Then the question is: does this imply $\theta_V=0$ for all $V$?

Edit 2. Here is a proof of the equality $d(K,\mathcal U,2,2)=1$. Let us use the same notation as in Edit 1. Assume that we have, for each vector space $V$, an endomorphism $\theta_V:V_2\to V_2$ of the second dual $V_2$ of $V$. Suppose also that we have $$\theta_W\circ f_2=f_2\circ\theta_V\qquad(**)$$ for all linear map $f:V\to W$, and that $\theta_V$ vanishes on $V$ when $V$ is viewed as a subspace of $V_2$. We claim: $\theta_V=0$ for all $V$. It is easy to see that this implies $d(K,\mathcal U,2,2)=1$.

Proof: Let $v_1$ be in $V_1$, that is, $v_1$ is a linear map $v_1:V\to K$. Using $(**)$ and the assumption that $\theta_K=0$, we get $$0=v_{12}(\theta_V(v_2))=\theta_V(v_2)\circ v_{11}$$ for all $v_2$ in $V_2$, or $$0=\theta_V(v_2)(v_{11}(k_1))=\theta_V(v_2)(k_1\circ v_1)$$ for all $v_2$ in $V_2$ and all $k_1$ in $K_1$. If $k_1$ is the identity of $K$, this gives $$\theta_V(v_2)(v_1)=0.$$ As $v_1$ is arbitrary, the proof is complete.

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I asked a related question on Mathoverflow. –  Pierre-Yves Gaillard Apr 2 '14 at 6:59

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 $$\varepsilon_{01}(\varepsilon_1(v_1))(v)=\varepsilon_1(v_1)(\varepsilon_0(v))=\varepsilon_0(v)(v_1)=v_1(v).$$

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, $i\ge2$.

Then the question comes down to computing the cardinals $\dim(G_i,G_j)$.

The most naive hope would be to have $\dim(G_i,G_j)=\delta_{ij}$.

One can also wonder if $G_i$ has nontrivial subfunctors.

By the way, wouldn't it be convenient to decree that the expressions "$1$-variant" and "$(-1)$-variant" are synonyms for "covariant" and "contravariant"? Then $F_i$ and $G_i$ would be $(-1)^i$-variant.

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