$\DeclareMathOperator\RoT{RoT}$I'm interested in the following ring. Fix a (Noetherian?) base ring $R$, and consider the category of finitely generated projective $R$-modules equipped with endomorphisms up to isomorphism. Since $\otimes$ distributes over $\oplus$, you can view this as a commutative semiring with those two operations. Then you can quotient by the equivalence that $(V,f) \sim (W,g)$ if there exist $(V^\prime,f^\prime)$ and $(W^\prime,g^\prime)$ such that $(V,f) \oplus (V^\prime,f^\prime) \cong (W,g) \oplus (W^\prime,g^\prime)$ and $f^\prime$ and $g^\prime$ are both commutators $f^\prime = AB - BA$ and $g^\prime = XY - YX$. We get a ring, which I've been calling the "ring of traces" over $R$ or $\RoT(R)$.

When all projective modules over $R$ are free (e.g. $R$ is a field or a PID), then $\RoT(R) \cong R$, and the isomorphism is realized by the trace map (which is why I'm calling it the ring of traces). This is true because all endomorphisms are represented by matrices with entries in $R$, and the commutator subalgebra of $\mathfrak{gl}_n(R)$ is $\mathfrak{sl}_n(R)$, the algebra of traceless matrices — any commutator has trace zero and any trace zero matrix is a commutator. You can also show that if $R$ is a Dedekind domain, the ring of traces over $R$ is isomorphic to $R$, though it's a little more involved, and still involves writing endomorphisms in terms of matrices.

THE QUESTION: is there a ring $R$ for which this "ring of traces" $\RoT(R)$ is not isomorphic to $R$? Such an $R$ which is Noetherian, and for which $\operatorname{Spec}R$ is smooth (or at least normal) would be ideal, but I can't find one even without those restrictions. If no, can we patch together a scheme which would give an affirmative answer to the analogous question for schemes/vector bundles/endomorphisms?

EDIT: In response to the comment below, I should add the following: Note that $(V,f)\oplus(V,g) \sim (V,f+g)$, due to the following calculation involving matrices with entries in $\operatorname{End}(V)$. $$ \begin{pmatrix} 0 & f & 0 \\ 0 & 0 & g \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 0 \\ I_V & 0 & 0 \\ 0 & -I_V & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 & 0 \\ I_V & 0 & 0 \\ 0 & -I_V & 0 \end{pmatrix}\begin{pmatrix} 0 & f & 0 \\ 0 & 0 & g \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} f & 0 & 0 \\ 0 & -f-g & 0 \\ 0 & 0 & g \end{pmatrix} $$ So $(V,f) \oplus (V,-f-g) \oplus (V,g)$ is a commutator. $$ \begin{pmatrix} 0 & f+g \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 \\ I_V & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ I_V & 0 \end{pmatrix}\begin{pmatrix} 0 & f+g \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} f+g & 0 \\ 0 & -f-g \end{pmatrix} $$ So $(V,f+g) \oplus (V,-f-g)$ is a commutator too.

Adding the second to $(V,f) \oplus (V,g)$, we get $$ (V,f) \oplus (V,g) \sim (V,f) \oplus (V,g) \oplus (V,f+g) \oplus (V,-f-g). $$ Adding the first to $(V,f+g)$, we get $$ (V,f+g) \sim (V,f+g) \oplus (V,f) \oplus (V,-f-g) \oplus (V,g). $$ The right hand sides of the two equivalences are isomorphic to each other (permute the factors), so the left hand sides are equivalent, and $(V,f) \oplus (V,g) \sim (V,f+g)$. Which is a very natural thing to want from something that is supposed to be a trace!

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    $\begingroup$ I'm not sure I understand. Does $(V,f+g)\sim (V,f)+(V,g)$ follow from your definition? $\endgroup$ Sep 18 at 0:30
  • $\begingroup$ It should, similar to how $(V,f) \oplus (V,g) \sim (V,fg)$ follows in the definition of $K_1(R)$. I need to look at my notes to find the argument though. $\endgroup$
    – Jon Aycock
    Sep 18 at 1:12
  • $\begingroup$ The OP is updated with an argument. $\endgroup$
    – Jon Aycock
    Sep 18 at 1:49
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    $\begingroup$ There is a very similar construction given by a generalization of Hochschild homology, which takes the direct sum of all endomorphism rings and quotients by the relation $fg \sim gf$ (where $f : a \to b, g : b \to a$ are morphisms). If you input the category of finitely generated projective $R$-modules into this construction, for $R$ not necessarily commutative, you get the zeroth Hochschild homology $H_0(R, R) = R/[R, R]$ (which is only an abelian group in general). If that construction is the same as this construction then you should always get $R$ if $R$ is commutative. $\endgroup$ Sep 18 at 2:34
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    $\begingroup$ I do not. The proof can be done by arguing that the construction is invariant under Morita equivalence, then applying the Morita equivalence between f.g. projective $R$-modules and the one-object category with endomorphisms $R$. $\endgroup$ Sep 18 at 3:02

1 Answer 1


$\newcommand\HH{\mathit{HH}}$I write $[V,f]$ for the class of $(V,f)$ in your ring.

Let me prove the following property of $[V,f]$ : if $f: V\to W, g: W\to V$, then $[V,gf] = [W,fg]$.

Indeed, $[V,gf] = [V\oplus W, (gf, 0)]$, because $(W,0)$ is a commutator.

Further, by your trick about sums, $[P,-h] = -[P,h]$ so that $[V,gf] - [W,fg] = [V\oplus W, (gf,0)]+[V\oplus W, (0,-fg)] $ which, by using your trick again, is $[V\oplus W, (gf, -fg)]$. Now I claim that this is a commutator of the obvious things, namely $(v,w)\mapsto (0,f(v))$ and $(v,w)\mapsto (g(w),0)$.

This is a simple computation: applying the first composite to $(v,w)$ gives $(gf(v),0)$ and the second $(0,fg(w))$.

This proves the cyclic invariance claim. Now, by definition more or less, $\HH_0(R)$ is the universal recipient of a cyclically invariant map $(V,f)\mapsto “\operatorname{tr}(V,f)”$. The trick here is to use cyclic invariance or summation with $(Q,0)$ to reduce from arbitrary projective to free, and then from free to “free on 1 generator” by using the standard basis of $\operatorname{End}(R^n)$.

Conversely, sending $(V,f)$ to its Hattori–Stallings trace in $\HH_0(R)$ satisfies your relation that commutators are sent to $0$.

So they are the same. For a commutative ring, $\HH_0(R) = R$, as pointed out in the comments.

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    $\begingroup$ I guess that you need the linearity to say that $\operatorname{HH}$ is the universal recipient of cyclically invariant map. A common nonlinear example is the characteristic polynomial. $\endgroup$
    – Z. M
    Sep 18 at 15:01
  • $\begingroup$ Yes, absolutely $\endgroup$ Sep 18 at 15:54

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