Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Is there a monoidal category $\mathcal C$ whose unit object is $I$ (i.e. $I\otimes A\cong A\cong A\otimes I$ for all $A\in \text{Ob}_\mathcal C$), with an object "$-1$" such that $$ (-1)\otimes(-1)\cong I ? $$

(Edit: no one misunderstood, but I'm also asking that $-1\neq I$)

I'm struggling with that since I read this math.SE post... Martin, if you see me, your server rejected every mail I tried to send you. :(

share|cite|improve this question
up vote 7 down vote accepted

what about $\mathbb{Z}_2$ vector spaces over a field $k$?

$(k,0)$ is the unit in this category and $(0,k)\otimes (0,k) = (k,0)$.

share|cite|improve this answer
Great, thanks! ...No hope to find such an object in $k-\mathbf{Vect}$, huh? In fact, it is plainly false in $\mathbf{Sets}$ with the monoidal structure induced by the product and the identity given by $\{*\}$... So I'm looking for some (counter)examples in more stuctured categories – Fosco Loregian Jan 15 '11 at 13:48

For any locally compact abelian group $G$, the monoidal category of continuous unitary representations of $G$ is generated by the characters $G \to \mathbb{C}$, e.g. the Pontrjagin dual group $\hat{G}$, with tensor product corresponding to the group operation and duals corresponding to the inverse (and the trivial representation corresponding to the identity). Prometheus's example is the case $G = \hat{G} = \mathbb{Z}/2\mathbb{Z}$.

share|cite|improve this answer
Unfortunately I don't know almost (=some wiki stuff) anything about group representations... In particular what about my previous question (finding such an object in $k-\mathbf{Vect}$)? :( Nonetheless I can catch the whole idea, thanks! – Fosco Loregian Jan 15 '11 at 23:54
@tetrapharmakon: you know what every object in k-Vect looks like, right? Any category where dimension makes sense has the property that only a 1-dimensional object can have tensor square the identity, and all 1-dimensional objects in k-Vect are isomorphic. – Qiaochu Yuan Jan 16 '11 at 0:09

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.