# Is there a notion of partial trace in a ribbon category?

I've seen some definitions of "right partial trace" and "left partial trace" in http://arxiv.org/abs/1103.1660, but these don't seem canonical in any way.

The motivation for this questions is that I'm thinking about topological quantum computing / modular tensor categories. It seems to me (from my limited physical understanding) like there should be a way to "discard" an object X, and go from a morphism $X \otimes A \to X \otimes A$ to one from $A \to A$. I can't really see what a canonical way of doing so would be, though.

First time posting a question here -- I hope this makes some sense!

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What do you mean by not canonical? That they depend on the choice of a pivotal structure? Your question references ribbon categories, which come with a canonical spherical structure via the twist, so the left and right traces are canonically defined and equal. – Evan Jenkins Aug 22 '11 at 23:33
True! I'm also interested in partial traces of (for example) functions $f: A \otimes X \otimes B \to C \otimes X \otimes D$ It seems to me (maybe I'm confuse that there would not be a canonical partial trace in this case, but that's a different question. – Julia Aug 24 '11 at 0:02

## 1 Answer

Given a rigid tensor category you have a coevaluation $\mathrm{coev}: 1 \rightarrow X \otimes X^*$ and an evaluation $\mathrm{ev}: X^* \otimes X \rightarrow 1$. If the category is pivotal (in particular, ribbon categories have a canonical pivotal structure) you have a natural isomorphism $p: X \rightarrow X^{**}$.

Ok, now suppose you have $f: X\otimes A \rightarrow X \otimes A$. Now construct

$$A = 1 \otimes A \rightarrow X^* \otimes X^{**} \otimes A \rightarrow X^* \otimes X \otimes A \rightarrow X^* \otimes X \otimes A \rightarrow 1 \otimes A = A$$

where the first map is the coevaluation for $X^*$, the second map is the inverse of the pivotal structure, the third map is $f$, and the fourth map is the evaluation.

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I think I fixed it. – Noah Snyder Aug 23 '11 at 2:58