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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. :(

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2 Answers 2

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

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

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  • $\begingroup$ 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 $\endgroup$
    – fosco
    Jan 15, 2011 at 13:48
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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 $G = \hat{G} = \mathbb{Z}/2\mathbb{Z}$.

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  • $\begingroup$ 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! $\endgroup$
    – fosco
    Jan 15, 2011 at 23:54
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    $\begingroup$ @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. $\endgroup$ Jan 16, 2011 at 0:09

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