Let $Vect_k$ denote the category of (not necessarily finite-dimensional) $k$-vector spaces. Clearly, this category is closed symmetric monoidal with internal hom $[X,Y]=Hom_k(X,Y)$.

Is it true that the coend $\int^{X\in Vect_k}\, [X,X]$ doesn't exist? I'm pretty sure that it doesn't, since in general there is no trace $tr_X:[X,X] \to k$, but I want to be sure. What makes me uncertain is that I read that the category of $k$-vector spaces is cocomplete..

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    $\begingroup$ Cocompleteness is usually with respect to small diagrams, i.e. diagrams indexed by a small category. $\mathsf{Vect}_k$ is not a small category. $\endgroup$ Jan 23, 2019 at 22:54
  • $\begingroup$ that's what I thought, thanks! so the coend doesn't exist? $\endgroup$ Jan 23, 2019 at 23:02
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    $\begingroup$ Actually proving that it doesn't exist would probably be pretty tricky (and necessarily involves some analysis of size issues--the coend "morally" exists as a certain class vector space, and the question is essentially whether that class actually is a set). $\endgroup$ Jan 23, 2019 at 23:11
  • $\begingroup$ @BipolarMinds As demonstrated by Eric Wofsey, my comment only implies that we can't say it exists just because $\mathsf{Vect}_k$ is cocomplete. It does not imply that it does not exist. That is, cocompleteness states that colimits of small diagrams exist, but says nothing one way or the other about large diagrams. It's just important to be careful about this or about (co)limits over large diagrams in general. $\endgroup$ Jan 24, 2019 at 0:06

1 Answer 1


This coend actually does exist: it is just $0$. Morally, the idea is an Eilenberg swindle. Suppose we had a trace operation $tr$ defined for all vector spaces, and let $A$ be any endomorphism of a vector space. Then if you take an infinite direct sum $B$ of copies of $A$, $tr(B)$ should be an infinite sum of copies of $tr(A)$. But if you add one more term $tr(A)$ to that infinite sum, it won't change the sum. But that means $tr(B)=tr(B)+tr(A)$ and so $tr(A)$ must be $0$.

To make this idea precise, let us recall that the coend (if it exists) is the initial vector space $V$ together with a linear map $tr_X:[X,X]\to V$ for each vector space $X$ such that for any pair of linear maps $A:X\to Y$ and $B:Y\to X$, $tr_X(BA)=tr_Y(AB)$. To prove that $V=0$ has this universal property, it suffices to show that any such collection of linear maps $tr_X:[X,X]\to V$ for any $V$ satisfies $tr_X=0$ for all $X$.

So, suppose we have a vector space $V$ and such linear maps $tr_X:[X,X]\to V$. Fix a vector space $X$ and $A\in [X,X]$; we will show that $tr_X(A)=0$. To do this, let $Y=X^{\oplus\mathbb{N}}$ and define $R:Y\to Y$ by $$R(x_0,x_1,x_2,\dots)=(0,Ax_0,Ax_1,Ax_2,\dots)$$ and $L:Y\to Y$ by $$L(x_0,x_1,x_2,\dots)=(x_1,x_2,x_3,\dots).$$ We then have $tr_Y(RL)=tr_Y(LR)$, so $tr_Y(LR-RL)=0$. However, notice that $LR-RL$ is given by the formula $$(LR-RL)(x_0,x_1,x_2,\dots)=(Ax_0,0,0,\dots).$$ Now let $i:X\to Y$ be the inclusion of the first summand and $p:Y\to X$ be the projection onto the first summand. We have $$tr_X(piA)=tr_Y(iAp).$$ But $piA=A$ and $iAp=LR-RL$ by the formula above, so this means $$tr_X(A)=tr_Y(LR-RL)=0,$$ as desired.

  • $\begingroup$ cool, I didn't expect this.. Eilenberg swindle - I have to remember that $\endgroup$ Jan 23, 2019 at 23:53

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