# Classification of $\operatorname{Rep} D(G)$

Let $$G$$ be a finite group and $$D(G)$$ its quantum double. Its finite dimensional complex representations are classified in this Dijkgraaf et al. Quasi-Quantum Groups Related To Orbifold Models. However, in the paper, the authors claimed the examples obtained by natural inductions are complete, but without a proof. My impression is that the representation theory for a Hopf algebra is not completely known yet in general. How would one classify the representations of $$D(G)$$?

My understanding is still at the level of representation of a finite group $$G$$. In that case, the classification can be done due to

1. $$\mathbb{C}[G]$$ is semi-simple
2. $$\lvert\operatorname{Irrep}(G)\rvert =\lvert\operatorname{Conj}(G)\rvert$$.

Are there similar statements for $$D(G)$$ (better with proofs)? Pointers towards anything relevant will be appreciated. Thank you!

EDIT: While there's only one chosen answer, the others might be valuable for you. Here is a table of answers below so far.

1. Konstantinos Kanakoglou pointed out several papers that directly answered my question.

2. I wrote up a note spelling out Konstantinos Kanakoglou's wonderful answer. I am willing to discuss the detail of the proof.

3. zibadawa timmy's higher categorical view point.

• By the way, rather than entering and exiting math mode for interstitial text, it is better to use \text or \operatorname: $\text{$\operatorname{Rep} D(G)$and some text}$ $\text{$\operatorname{Rep} D(G)$and some text}$. I have edited accordingly. – LSpice Dec 10 '19 at 15:51
• Great kungfu! Thanks! – Student Dec 10 '19 at 15:58
• How high a level are you willing to go to? There are nice categorical descriptions in terms of centers of categories which describe the doubles of fairly arbitrary Hopf algebras (and more), which are pretty easy to understand and make explicit in the case of groups. Provided you don't mind the abstract nonsense angle on things. – zibadawa timmy Dec 11 '19 at 12:13
• Totally! I don't mind! Would you mind pointing them out? I started learning this for finite groups $G$ because I believe one can say so much even in this case (if we go higher), but I don't know how high we can get. – Student Dec 11 '19 at 13:05

There are some classic results on the classification of the irreducible $$D(G)$$-modules:
If the field is the complex numbers $$\mathbb{C}$$, it has been shown that a representation of the finite group $$G$$, induced from an irreducible representation of the centralizer subgroup of an element $$g$$ of $$G$$, generates an irreducible rep of $$D(G)$$ and that furthermore, all the irreducible quantum double modules are obtained in this way. Proofs for these results can be found at:
Quantum double finite group algebras and their representations, Bull. Austr. Math. Soc., 48, 1993, p.275-301, by M.D. Gould.
(See section 6, mainly theorem 6.3). In there, it is also shown that all such algebras are semisimple and their character theory is developed.

From a more general viewpoint, representations of $$D(G)$$ over algebraically closed fields of arbitrary characteristic have been studied at:
The representation ring of the quantum double of a finite group, J. of Algebra, 179, p.305-329, (1996), by S.J. Witherspoon. In there, some of the previously mentioned results have been generalized: for example an analogue of Maschke's theorem is proved; it is shown that $$D(G)$$ is semisimple if and only if the characteristic $$p$$ of the field, does not divide the order of the group $$G$$.
Furthermore, the representation ring $$R\big(D(G)\big)$$ of the quantum double is studied: it is shown to be a commutative algebra, a direct sum decomposition is described and a classification of the indecomposable $$D(G)$$-modules is also achieved (among other results as well).

• Your answer is very helpful. I am reading Gould's paper and will write up a short note. However, one point still bugs me. For a finite dimensional associative algebra $A$ over a field $k$, is the always true that the dimension of the center of $A$ is always the same as the size of the set simple modules over $A$ (mod iso)? I believe this is very important for classifying all irreps, since we can enumerate enough many of them and say we are done if we know how many there should be. – Student Dec 10 '19 at 15:43
• I figured it out! This might not hold for any associative algebra, but for Artinian semisimple ring this is correct. I will write a complete note below. Thank you so much for your answer. – Student Dec 10 '19 at 18:22
• A note has been put below. Thanks for your suggestion. Anyone who wants to discuss the details of the proof are welcome! – Student Dec 10 '19 at 20:10

There's a higher way to come at this. I'll be a little light on the rigorous details here, but everything I mention can be found in the book "Tensor Categories" by Etingof, Gelaki, Nikshych, and Ostrik. The book's a very good starting point for moving from the Hopf algebras perspective to the tensor category perspective, which is where a lot of current research is being done.

If one considers a semisimple Hopf algebra $$H$$, and take $$\mathcal{C}=\operatorname{Rep}(H)$$ to be the category of finite dimensional left (or right) modules of $$H$$, then there is a braided tensor equivalence $$\operatorname{Rep}(D(H))\cong \mathcal{Z}(\operatorname{Rep}(H))$$, where $$\mathcal{Z}(\mathcal{C})$$ denotes the categorical center of the category $$\mathcal{C}$$. This center construction works not just for the particular choice here, but any tensor (aka monoidal) category with sufficiently similar properties. The objects of the center are the pairs $$(V,\gamma_V)$$ where $$V$$ is an object of $$\mathcal{C}$$ and $$\gamma_V$$ is a natural family of isomorphisms called a "half-braiding" (because they piece together into a braiding on the entire category).

In the case of $$H=\mathbb{C}G$$ with $$G$$ a finite group, we can go one step better. There is a Morita-equivalence between $$\operatorname{Rep}(G)$$ and $$\text{Vec}_G$$, where the latter is the space of $$G$$-graded finite-dimensional vector spaces (over $$\mathbb{C}$$). This is equivalent to saying these categories have the same centers, up to braided tensor equivalence, so we could just as well compute $$\mathcal{Z}(\text{Vec}_G)$$ instead. Once you actually write down what the half-braiding conditions are, this center becomes very easy to determine: it's $$\text{Vec}_G^G$$ (sometimes denoted $${}^G_G\mathcal{M}$$, or some variation thereof depending on the use of left/right (co)actions), the category of finite dimensional $$G$$-graded, $$G$$-equivariant vector spaces. At this point it's easy to decide the isomorphism classes of the irreducibles, and you find that they are parameterized by pairs $$(g,\chi)$$ where $$g$$ is an element in a complete set of representatives of the conjugacy classes of $$G$$, and $$\chi$$ is an element in a complete set of representatives for the irreducible representations (or characters) of $$C_G(g)$$. So the isomorphism type of the module depends only on the conjugacy class of $$g$$ and the isomorphism class of $$\chi$$.

When you understand the objects of $$\text{Vec}_G^G$$ it becomes readily apparent that the irreducible objects are just induced representations from $$C_G(g)$$ to $$G$$, but where the implicit grading of this induction via cosets of $$C_G(g)$$ is relevant to deciding the full action of $$D(G)$$.

And if you want to go even further than that, you can change the associativity morphism of $$\text{Vec}_G$$ via a normalized 3-cocycle $$\omega$$ to obtain the category $$\text{Vec}_G^\omega$$, and then we have $$\mathcal{Z}(\text{Vec}_G^\omega)\cong\operatorname{Rep}(D^\omega(G))$$, where $$D^\omega(G)$$ is the twisted Drinfeld double, and is in general a quasi-Hopf algebra and not a Hopf algebra. These objects are also quickly described in the paper you mention. The description of the irreducibles is similar, except now we're using irreducible projective representations for particular 2-cocycles of $$C_G(g)$$ obtained from $$\omega$$.

This category, as a braided tensor category, will only depend on the cohomology class of $$\omega$$, while $$D^\omega(G)$$ can have wildly different structures even for representatives of the same cohomology class. Since those structures are also quite nightmarish to deal with directly for any non-trivial 3-cocycle, most people end up gravitating towards dealing with them through their representation categories, instead.

• Very interesting! I will definitely check that book. Before then, I hope my question does not appear to be superficial: is it possible to have similar story with higher cohomologies involved *without going upward to 2-groups, 3-groups..."? My hope is that the infinity structure (to be made precise) of a finite group $G$ should be complicated enough. – Student Dec 12 '19 at 14:59
• @Student That is, honestly, going beyond my expertise. This paper is the only one that I can think of that seems to deal with what you're getting at (but it does use higher categories). – zibadawa timmy Dec 12 '19 at 16:38
• @Student 1) it might be worth saying that Rep G is the motivating example of so-called fusion categories (semi-simple tensor categories with finitely many simple objects), and a lot is known about those (and their center). 2) as for your other question: yes there is definitely a nice story involving higher cohomologies, which are perhaps best understood in the language of topological field theories. You might look at the intro of arxiv.org/abs/math/0503266 for a nice geometric setup, and say ncatlab.org/nlab/show/Dijkgraaf-Witten+theory or arxiv.org/abs/hep-th/9212115. – Adrien Dec 12 '19 at 20:24
• @zibadawatimmy you mentioned that the category of representations of $D(H)$ is isomorphic to the center of the category of $H$ representations. Is this isomorphism canonical? If so, is there a known formula to spell out how $D(H)$ acts on the image of $(V,\gamma_V)$? Also, there are induction and forgetful (adjoint) functors between Rep$(H)$ and the center of it. Under that isomorphism, if canonical, is there a known description of what we get as a $D(H)$ representation? – Student Dec 19 '19 at 3:56
• @Student Starting from a given category and going to its center is just a definition, and yes you can pull the action out in this case. More generally a representing object for a given category can be obtained via Tannaka-Krein reconstruction on the fiber functor (provided there is one, which there is in this case). The category Rep(D(H)) need not uniquely determine the $H$, however, even in the case where $H$ can be taken to be a group algebra. There's a paper by Naidu and Nikshych showing that non-isomorphic groups can have doubles with equivalent representation categories. – zibadawa timmy Dec 19 '19 at 13:15

This is a study note that spells out @Konstantinos's answer explicitly.

### Preface

Our goal is to classify all finite dimensional representations over the complex number field for the quantum double $$D(G)$$ for a fixed finite group $$G$$, with proofs. We will use [G] as out main reference, while auxiliary results can be found in [S] and [CR].

For other considerations, see [W], [L], and [B]. For the representation theory of $$D(G)$$ over other fields, see [W]. For the representation theory of other Hopf algebras, see [L], which deals with a class of (possibly infinite dimensional) Hopf algebra with a technical condition: co-semi-simple + involutive). For more applications, see [B].

### Abstract

In what follows, semi-simplicity allows us to focus on the simple modules. We can get lots of them by induction from the underlying group $$G$$. Character theory for $$D(G)$$ distinguishes the simple modules we get from induction, showing the abundance. The structure theorem of $$D(G)$$ predicts how many non-isomorphic simple modules there should be. Examining how many different simple modules we've got from allows us to complete the proof.

### Semi-simplicity of $$D(G)$$

Following [G] and its notations, the first main result is the semi-simplicity of $$D(G)$$. Theorem 2.3 says that any finite dimensional Hopf algebra $$A$$ is semi-simple if and only if there exists a left integral $$x \in A$$, this is a powerful criterion for semi-simplicity. A left integral of $$D(G)$$ is given in [G. (16)], where $$x = E_\iota 1^*$$, so $$D(G)$$ is semi-simple.

The proof of theorem 2.3, the powerful semi-simplicity criterion, can be found in [S. Theorem 5.18]. There, Sweedler first defined the left integrals for $$H^*$$. As $$H$$ is finite-dimensional, $$H$$ is isomorphic to $$H^{**}$$ naturally, whose left integrals can be doubly-dualed back to $$H$$. This definition coincides with that of [G]. Anyway, one can use a left integral to "average" an arbitrary linear projection and get a Hopf linear projection from any larger module to any smaller submodule, proving semi-simplicity. An explicit averaging formula is given in the proof of [S. Theorem 5.18]. The other side is easy: if $$H$$ is semi-simple, than the complement of $$ker(\epsilon)$$ is the set of left integrals. A few immediate corollaries are

1. $$D(G)$$, $$\operatorname{Fun}(G)$$, and $$\mathbb{C}[G]$$ are all semisimple.

2. $$k[G]$$ is semisimple if and only if $$\epsilon(x=\Sigma g) = |G|$$ is not zero, which in turn is equivalent to that $$|G|$$ is not divisible by \$\operatorname(char)k.

3. $$k[X]/$$ is not semi-simple, since $$\epsilon(x^{p-1})$$ is zero.

4. $$k[X]/$$ is semi-simple, since $$\epsilon(x^{p-1})$$ is -1.

### Unitarity of representations and orthogonality of matrix elements

Every finite dimensional $$D(G)$$-module is equivalent to a unitary one [G. Lemma 4.1], so in particular $$D(G)$$ is proven again to be semi-simple. Routine arguments show the orthogonality of matrix elements [G. Theorem 4.1]. Applying this to characters, we get the orthogonal relations among them [G. Theorem 5.1]. Note that this can be generalized to a larger class of Hopf algebras (possibly infinite dimensional), which are co-semi-simple and involutive [L]. The rest of chapter 5 in [G] exhibits the character theory for $$D(G)$$ and finds an explicit basis for the center of $$D(G)$$ [G. (25) -- Thm 5.2]. This basis is in 1-1 correspondence to the number of $$G$$-equivalence classes of $$Q$$, and is also in 1-1 correspondence to the set of non-isomorphic irreducible $$D(G)$$-modules by the structure theorem for Artinian semisimple rings [G. Theorem 5.2]. We will justify the last statement later.

### Enumeration of representations of $$D(G)$$

The representations of $$D(G)$$ can be obtained by induction from the centralizer subgroups of $$G$$. This is done in chapter 6. The character theory developed in chapter 5 distinguishes one from another, showing the abundance of the results. Since we have known how large $$\operatorname{Irrep}(D(G))$$ is, we will be done by showing the structure theorem for $$D(G)$$.

### Structure theorem for $$D(G)$$

In this section, our reference is [CR. section 23 to 26]. From now on, we will assume $$R$$ to be a unital Artinian ring (associative, but not necessarily commutative). We will show the structure theorem for $$R$$ if it is semisimple. Since $$D(G)$$ obviously satisfies all the conditions, we will then be done.

Since $$R$$ is Artinian, any left ideal $$I$$ is nilpotent if and only if it has no idempotent elements. It is then easy to show that the set of nilpotent left ideals is closed under finite sum. More interestingly, the sum of all nilpotent left ideals is a nilpotent two-sided ideal, called the radical $$\sqrt(R)$$ of $$R$$. If the radical is zero, we call $$R$$ semisimple. It is easy to show that $$R/\sqrt(R)$$ is semisimple.

If $$R$$ is semisimple, then any minimal left ideal $$L$$ is not nilpotent and thus have an idempotent element $$e$$. Minimality guarantees that $$L$$ is generated by that idempotent element. Note that the generator is not unique in general. In this case, $$R = Re \oplus R(1-e) = L \oplus L'$$. One can furthur decompose $$R$$ into $$R = Re_1 \oplus \cdots Re_n$$, where the $$e_i$$'s are orthonormal idempotents. It is easy to show the uniqueness of decomposition, and also that any $$R$$ with this decomposition is in fact semisimple. The decomposition breaks the unit $$1$$ into the sum of the $$e_i$$'s, this is the key. Using this key, it is not hard to show that every left $$R$$ ideals are completely reducible [CR. 25.8], and also that any irreducible $$R$$-module is isomorphic to some minimal left ideal in $$R$$.

Therefore, the complete set of non-isomorphic simple modules can be found in the decomposition of $$_RR$$ as a left $$R$$-module! The Wedderburn structure theorem shows that the number of them is the same as the size of the center of $$R$$ (TODO: needs clarification). This completes the argument.

### References

[G]. Quantum double finite group algebras and their representations, Bull. Austr. Math. Soc., 48, 1993, p.275-301, by M.D. Gould.

[S]. Hopf algebras (Benjamin, New York, 1969), by M.E. Sweedler.

[CR]. Representation theory of finite groups and associative algebras, by C.W. Curtis and I. Reiner.

[W] The representation ring of the quantum double of a finite group, J. of Algebra, 179, p.305-329, (1996), by S.J. Witherspoon.

[L] Characters of Hopf algebras, J. Algebra 17 (1971), 352-368, by R.G. Larson.

[B] Exactly solved models in statistical mechanics (Academic press, 1982), by R.J. Baxter.