MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've seen some definitions of "right partial trace" and "left partial trace" in, 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!

share|cite|improve this question
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

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.

share|cite|improve this answer
I think I fixed it. – Noah Snyder Aug 23 '11 at 2:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.