Let $K$ be a field. Are there non-scalar endomorphisms of the endofunctor $$ V\mapsto V^{**}/V $$ of the category of $K$-vector spaces?

I asked a related question on Mathematics Stackexchange, but got no answer.

EDIT (Apr 15'14). Here is a closely related question which involves only basic linear algebra (and no category theory):

Let $K$ be a field and $V$ an infinite dimensional $K$-vector space. Is the $K$-algebra $$ \operatorname{End}_{\operatorname{End}_K(V)}(V^{**}/V) $$ isomorphic to $K$?

EDIT (Apr 26'14). To avoid any misunderstanding, let me say explicitly that I'm unable to prove any of the following two statements:

(1) There is a pair $(K,V)$, where $K$ is a field and $V$ an infinite dimensional $K$-vector space, such that $$ \operatorname{End}_{\operatorname{End}_K(V)}(V^{**}/V)\simeq K. $$

(2) There is a pair $(K,V)$, where $K$ is a field and $V$ an infinite dimensional $K$-vector space, such that $$ \operatorname{End}_{\operatorname{End}_K(V)}(V^{**}/V)\not\simeq K. $$

EDIT (May 17'14). Here is a slight amplification of the above edit:

For any field $K$ and any infinite cardinal $\alpha$ put $$ d(K,\alpha):=\dim_K\operatorname{End}_{\operatorname{End}_K(V)}(V^{**}/V), $$ where $V$ is an $\alpha$-dimensional $K$-vector space.

Let $\kappa$ be the cardinal of $K$. The Erdős-Kaplansky Theorem implies $$ 1\le d(K,\alpha)\le\kappa\wedge(\kappa\wedge(\kappa\wedge\alpha)), $$ where $\wedge$ denotes exponentiation.

There is no pair $(K,\alpha)$ for which I can prove that the first inequality is strict, and no pair $(K,\alpha)$ for which I can prove that the second inequality is strict.

EDIT (Jun 12'14). Here is a positive result. Unfortunately, it is very weak, and I hope users will be able to improve it.

Recall that $K$ is a field. For any vector space $V$ put $V':=V^{**}/V$. Let $V$ be an infinite dimensional vector space, and form the $K$-algebras $$ A:=\operatorname{End}_KV,\quad B:=\operatorname{End}_KV',\quad C:=\operatorname{End}_AV', $$ so that $C$ is the commutant of $A$ in $B$. Recall that the dimension of $B$ is $$ \dim B=\operatorname{card}K\wedge\Big(\operatorname{card}K\wedge\big(\operatorname{card}K\wedge\dim V\big)\Big), $$ where $\wedge$ denotes exponentiation. We claim

The codimension $c$ of $C$ in $B$ satisfies $\aleph_0\le c\le\dim B$.

To prove this, it suffices to show that, for each integer $n\ge2$, there is a subspace $W$ of $B$ satisfying $\dim W/(W\cap C)\ge n$.

Let $n$ be $\ge2$, let $V_1,\dots,V_n$ be subspaces of $V$ such that $V=V_1\oplus\cdots\oplus V_n$ and $\dim V_i=\dim V$ for all $i$, and let us define isomorphisms $\varphi_{ij}:V_j\to V_i$ for $1\le i,j\le n$ as follows.

Firstly $\varphi_{ii}$ is the identity of $V_i$. Secondly $\varphi_{i+1,i}:V_i\to V_{i+1}$, for $1\le i < n$, is any isomorphism. Thirdly $\varphi_{ij}$ for $j < i$ is the appropriate composition of the $\varphi_{k+1,k}$ previously defined. Fourthly we set $\varphi_{ij}:=(\varphi_{ji})^{-1}$ for $i < j$.

Define the morphism of $K$-algebras $\psi:M_n(K)\to B$ by $$ \psi(a)(v)_i:=\sum_ja_{ij}\ \varphi'_{ij}(v_j) $$ for $v=v_1+\cdots+v_n\in V'$ with $v_i\in V'_i$. Then define $W$ as the image of $\psi$. It is easy to see that $\psi$ is injective and that $W\cap C$ is the image under $\psi$ of the scalar matrices, so that $\dim W/(W\cap C)=n^2-1\ge n$.

There is not a single case in which I'm able to improve the above inequalities $\aleph_0\le c\le\dim B$.

  • 13
    $\begingroup$ Interesting question. I know there is a unique endomorphism of the functor $V \mapsto V^{**}$ that restricts to the identity on finite-dimensional vector spaces. (More generally, $(\ )^{**}$ is terminal among all endofunctors of Vect that restrict to the identity on FDVect.) This is similar in spirit to your question, and I was hoping to use it to answer your question, but I haven't managed. $\endgroup$ Apr 2, 2014 at 12:45
  • 2
    $\begingroup$ @TomLeinster - Dear Tom: Thanks a lot for your kind comment! Failing to prove your parenthetical remark, I did some googling, and found your paper Codensity and the ultrafilter monad, in particular Remark 7.6. I also found the links related to this paper on your webpage. Are these the good references for this fact? $\endgroup$ Apr 2, 2014 at 14:59
  • 4
    $\begingroup$ Yes, that's the paper I had in mind. I was really thinking of a simpler version of Proposition 5.4 (concerning just endofunctors, not monads). Double dualization is a codensity monad, i.e. a right Kan extension of a certain kind, and so has a certain universal property. This implies that it's terminal in the sense I mentioned. But there must surely be a more direct proof of this, and maybe it would be possible to adapt it to answer your actual question. $\endgroup$ Apr 3, 2014 at 3:24
  • 2
    $\begingroup$ @GeraldEdgar - Yes, that's what I'm saying. Don't hesitate to tell me if it's easy, even if it's humiliating for me... Thanks for your comment. $\endgroup$ May 17, 2014 at 15:02
  • 4
    $\begingroup$ @GeraldEdgar - If $A$ is a $K$-algebra and $B$ is an $A$-module, then $\mathrm{End}_A(B)$ designates the $K$-vector space of all $A$-linear endomorphisms of $B$. $\endgroup$ May 17, 2014 at 22:01

1 Answer 1


I've thought about this problem on and off for the last couple years. While I haven't solved it, I did have one comment which is too big to fit in the comments section.

Let $K$ be a field and let $V_K$ be an infinite-dimensional $K$-vector space. Let $A={\rm End}(V_K)$ as above. We may identify $V_K$ with $K^{(I)}$ for some index set $I$; this is the set of columns vectors, with entries indexed by $I$, such that only finitely many entries are non-zero. [If we think of $V=K[x]$, then this identifies $x^n$ with the $\omega$-indexed columns whose only non-zero entry is $1$ in the $n$th coordinate.]

The ring $A$ may then naturally be identified as the ring of $I\times I$ column-finite matrices. This is the set of $I\times I$ matrices, such that each column has only finitely many non-zero entries. The two-sided ideals in this ring form a well-ordered ascending chain, and the minimal non-zero ideal is the set of matrices of finite rank. Call this ideal $J$.

The ideal $J$ is idempotent. In fact, each element in $J$ is fixed, by left-multiplication, by a finite-rank idempotent (namely, the idempotent with $1$'s down the diagonal sufficiently far). It is not difficult to prove that $JV^{\ast\ast}=V$. Thus the left $A$-module structure on $V^{\ast\ast}/V$ has $J$ acting as zero. Therefore, it really is an $A/J$-module structure.

When $V$ is countable dimensional, $J$ is also the unique maximal two-sided ideal (but there are many maximal one-sided ideals). In particular, $A/J$ is simple in that case. I don't know if this helps solve the problem in any way.

  • 4
    $\begingroup$ Thank you very for this wonderful answer! I haven't completely digested it, but I'll keep on thinking about it. I noticed that Exercices 3.16 p. 46 and 10.6 p. 168 of Lam's book A first course in noncommutative rings (which is freely and legally available here) are closely related to your answer. - I'm not mentioning this for you, but for the other potential readers: I'm sure you know this book and master its contents! (By the way, in your answer, $R$ is $A$, right?) $\endgroup$ Jan 7, 2017 at 16:08
  • 1
    $\begingroup$ @Pierre-YvesGaillard Yes, exercise 3.16 covers the ideal structure of $A$. If you have access to his "Exercises in Classical Ring Theory", he works out the solution to that exercise. (Also, I fixed the typos you pointed out.) $\endgroup$ Jan 7, 2017 at 22:26
  • 4
    $\begingroup$ @Pierre-YvesGaillard: The book is no longer freely available. $\endgroup$
    – user57432
    Feb 13, 2019 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.