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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 $\aleph_0\le c\le\dim B$.

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    $\begingroup$ 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. $\endgroup$ – Tom Leinster Apr 2 '14 at 12:45
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    $\begingroup$ @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? $\endgroup$ – Pierre-Yves Gaillard Apr 2 '14 at 14:59
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    $\begingroup$ 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. $\endgroup$ – Tom Leinster Apr 3 '14 at 3:24
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    $\begingroup$ @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. $\endgroup$ – Pierre-Yves Gaillard May 17 '14 at 15:02
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    $\begingroup$ @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 $B$. $\endgroup$ – Pierre-Yves Gaillard May 17 '14 at 22:01
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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.

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    $\begingroup$ 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?) $\endgroup$ – Pierre-Yves Gaillard Jan 7 '17 at 16:08
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    $\begingroup$ @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.) $\endgroup$ – Pace Nielsen Jan 7 '17 at 22:26
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    $\begingroup$ @Pierre-YvesGaillard: The book is no longer freely available. $\endgroup$ – user 170039 Feb 13 at 17:39

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