# Equivalent condition to linear operator having trace zero

I'm trying to describe an equivalence relation on the category of finite-dimensional $$\mathbb{C}[t]$$-modules such that $$V \sim W$$ if and only if the trace of $$t$$ acting on $$V$$ is the same as the trace of $$t$$ acting on $$W$$. This is a great condition, problem is, I'm working in a categorical setting and I don't want to say the word trace $$-$$ taking the trace is the decategorification functor, and I'd like to have a more intrinsic way of comparing the modules $$V$$ and $$W$$ before decategorifying. Since trace behaves well under $$\oplus$$ and $$\otimes$$, it's enough to be able to pick out when the action of $$t$$ has trace $$0$$. Is there a way to compare these modules without mention of trace or eigenvalues?

• Well, the trace zero operators are precisely the ones that belong to the special linear Lie algebra. Aug 18, 2022 at 19:55
• Hmm, the trace of $t$ seems like an extra structure that cannot be obtained purely from the underlying rigid monoidal category. Indeed, there are many automorphisms of $\mathbf C[t]$ that do not preserve $t$, so how do you know that you're looking at the trace of $t$ rather than that of $3t-17$? Aug 18, 2022 at 20:12
• @SamHopkins, that was one of my initial thoughts. But it feels like it's hard to make it useful. Aug 18, 2022 at 21:28
• @JonAycock: vague idea, but maybe if you had some analog/generalization of the exponential map, then you can talk about the determinant being one... Aug 18, 2022 at 21:33
• @R.vanDobbendeBruyn, in fact the ring isn't $\mathbb{C}[t]$, but a group ring for a group with a canonical (topological) generator. Specifically, $V$ and $W$ should be continuous representations of $\widehat{\mathbb{Z}}$, and we're looking at the trace of $1$. (Or really, continuous representations of $D_p/I_p$, and we're looking at the trace of Frobenius.) So maybe I really need some way of using the fact that this is a group element rather than a random linear operator. Aug 19, 2022 at 21:32

Credit to Sam Hopkins above and to about ten responses in this thread: Geometric interpretation of trace for priming me to this, but I think what I'm looking for is that a linear operator $$T$$ has trace zero if and only if it is a commutator $$T = AB-BA$$. It's easy to show that a commutator has trace zero, but the equivalence can come (probably circularly) from the fact that $$\mathfrak{sl}_n$$ is semisimple and nonabelian for $$n>1$$, so its commutator subgroup is a nonzero subalgebra which is then forced to be all of $$\mathfrak{sl}_n$$. Thus every trace zero matrix is a commutator. This feels like it's categorical enough since it only references other linear maps, even if $$A$$ and $$B$$ are kind of non-canonical and auxilliary.