# trace(xy)=trace(yx) in full generality

It is well known that, for square matrix $x$ and $y$, we have $\operatorname{tr}(xy)=\operatorname{tr}(yx)$. Here of course the trace of a matrix is just the sum of the elements of the diagonal.

The notion of trace has a lot of generalization. As I know, the most general definition is the following: let $(\mathcal C, \otimes, 1, ^\vee)$ be a rigid symmetric monoidal category, $X$ an object of $\mathcal C$ and $f$ an endomorphism of $X$. Then $\operatorname{tr}(f) \in \operatorname{End}(1)$ is defined by the following composition $$1 \longrightarrow X^\vee \otimes X \stackrel{\operatorname{id}_{X^\vee} \otimes f}{\longrightarrow} X^\vee \otimes X \longrightarrow X \otimes X^\vee \longrightarrow 1$$ So my questions is: it is true, in this generality, that $\operatorname{tr}(f\circ g)=\operatorname{tr}(g \circ f)$, for $f$ and $g$ in $\operatorname{End}(X)$?

Ricky

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Have you tried drawing the appropriate string diagrams? – Qiaochu Yuan Jul 6 '11 at 21:33
Why specify "square" matrices in particular? The occasion for use of this identity with which I am most familiar depends crucially on the matrices not being square. – Michael Hardy Jul 6 '11 at 21:56
To Qiaochu: I'm completely new to string diagram, so it is possible that my question becomes trivial in that language. I just don't know. To Micheal: what is the trace of a non square matrix? I've always seen the trace of an endomorphism! – Ricky Jul 6 '11 at 22:12
If $X$ is $m \times n$ and $Y$ is $n \times m$, then $\mathrm{Tr}(XY) = \mathrm{Tr}(YX)$. – David Speyer Jul 6 '11 at 22:46
It is also true for Ribbon categories, which are in general not symmetric, but just "braided", see eg. Kassel - Quantum Groups. – Marcel Bischoff Jul 7 '11 at 3:30

## 1 Answer

Yes. The string diagram chase can be found on page 8 of this paper by Kate Ponto and Mike Shulman.

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This is a great paper, but I think the chase given obscures the underlying topological idea a little: you take two wires, feed them into each other, then rotate everything by $180^{\circ}$. I'm not completely sure what structure is necessary to make that particular argument valid, though. – Qiaochu Yuan Jul 7 '11 at 4:38
I think they could have done it that way, by first introducing movie moves which express extranaturality, e.g., one which gives an equation where $g \otimes 1_{x^\ast}$ followed by the counit $x \otimes x^\ast \to I$ equals $1_y \otimes g^\ast$ followed by the counit $y \otimes y^\ast \to I$; here I'm using $g: y \to x$ and $g^\ast: x^\ast \to y^\ast$ denotes the mate of $g$. This interprets the move where you slide the bead labeled $g$ through the bottom critical point, after which $g^\ast$ pops up on the other side. There's probably a lemma or two that needs to be proven first. – Todd Trimble Jul 7 '11 at 10:13