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It is a basic fact in representation theory of finite groups over complex numbers that the character tables of $Q_8$ and $D_8$ are identical. I believe, this implies that the corresponding categories of representations are equivalent (as tensor categories).

On the other hand, Tannakian Formalism tells us that we can reconstruct a finite group $G$ from its category of representations $\mathbf{Rep}_G$ together with the natural (foregetfull) fibre functor $F_G: \mathbf{Rep}_G\rightarrow \mathbf{Vect}$. Namely, $G$ is canonically isomorphic to the tensor automorphisms of the tensor functor $F_G$.

This implies that the fibre functors $F_{D_8}$ and $F_{Q_8}$ are different. How can one see this explicitly in terms of representations of $D_8$ and $Q_8$?

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    $\begingroup$ See mathoverflow.net/questions/155743/…. The equality of character tables tells you that $Q_8$ and $D_8$ have the same Grothendieck ring of representations, but that's it. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 29, 2017 at 4:48

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Let $V_D$ and $V_Q$ be the two dimensional simple representations of $D_4$ and $Q_8$ respectively. Let $1_D$ and $1_Q$ denote their trivial representations.

Suppose that there is a tensor equivalence between $\mathbf{Rep}(D_4)$ and $\mathbf{Rep}(Q_8)$ commuting with the fibre functor to $\mathbf{Vect}_\mathbb{C}$. This equivalence sends $1_D$ to $1_Q$ (as they're the unit object) and sends $V_D$ to $V_Q$ (as they're the unique simple of dimension 2).

In particular there is a $\mathbb{C}$-linear isomorphism $g$ from $V_D$ to $V_Q$. Consider $$g\otimes g:V_D\otimes V_D\to V_Q\otimes V_Q.$$ It must send the unique copy of $1_D$ in $V_D\otimes V_D$ to the unique copy of $1_Q$ in $V_Q\otimes V_Q$.

It is easy to see that there is no such $g$. The slickest way I can see to prove this is to note that the flip map $v\otimes w\mapsto w\otimes v$ acts by -1 on $1_Q$ and by 1 on $1_D$.

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The categories ${\rm Rep}(Q_8)$ and ${\rm Rep}(D_8)$ are not equivalent as tensor categories. They have the same Grothendieck ring, but they have non equivalent associators. As far as I am aware, it is an open problem to classify all tensor categories which have the same Grothendieck ring as ${\rm Rep}(Q_8)$ (there are at least two).

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  • $\begingroup$ This also begs another question: can we have two non-isomorphic finite groups with equivalent tensor categories of representations? $\endgroup$
    – Dr. Evil
    Commented Sep 29, 2017 at 5:11
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    $\begingroup$ @Dr. Evil: no, if by tensor you mean symmetric monoidal. This follows from stronger variants of Tannaka reconstruction which imply that the fiber functor is unique, as a symmetric monoidal functor. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 29, 2017 at 7:50
  • $\begingroup$ @Daniel - Thanks for the answer. Could you please give a reference for the associators being different? I guess this probably follows from Qiaochu's remark, but it would be good to have a more direct argument. $\endgroup$
    – Dr. Evil
    Commented Sep 30, 2017 at 9:09
  • $\begingroup$ If you want to check that the categories are not equivalent as tensor categories, you can compute the 6j symbols. I don't know a slick way to do this. $\endgroup$
    – Peter McNamara
    Commented Sep 30, 2017 at 9:26
  • $\begingroup$ @Dr.Evil: As Peter said, you need to choose a basis for each ${\rm hom}(a,b \otimes c)$ and compute the associator in terms of your chosen basis. $\endgroup$
    – Daniel Barter
    Commented Sep 30, 2017 at 22:03

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