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?
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ Well, the trace zero operators are precisely the ones that belong to the special linear Lie algebra. $\endgroup$– Sam HopkinsCommented Aug 18, 2022 at 19:55
-
3$\begingroup$ 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 $3t-17$? $\endgroup$– R. van Dobben de BruynCommented Aug 18, 2022 at 20:12
-
2$\begingroup$ @SamHopkins, that was one of my initial thoughts. But it feels like it's hard to make it useful. $\endgroup$– Jon AycockCommented Aug 18, 2022 at 21:28
-
2$\begingroup$ @JonAycock: vague idea, but maybe if you had some analog/generalization of the exponential map, then you can talk about the determinant being one... $\endgroup$– Sam HopkinsCommented Aug 18, 2022 at 21:33
-
1$\begingroup$ @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. $\endgroup$– Jon AycockCommented Aug 19, 2022 at 21:32
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
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.
-
2$\begingroup$ See also links on this Math.SE page: math.stackexchange.com/questions/95537/… $\endgroup$ Commented Aug 23, 2022 at 11:08