# How do small central extensions drop the dimension of a faithful representation?

Apologies in advance that this is a very soft question. I might be talking complete nonsense. But I hope I am talking about something that has even been studied...

I am interested in the phenomenon in representation theory that a group whose faithful representations (over $\mathbb{C}$, say) have minimum dimension $n$ often has a central extension with faithful representations in dimensions smaller than $n$. For example, $S_4$ has no faithful representation in dimension less than 3, but it has a double cover $\tilde S_4$ with a faithful 2-dimensional representation. This phenomenon is striking to me because it takes "more information" to write down the elements of the bigger group, since an element of $\tilde A_5$ specifies an element of $A_5$, but not vice versa. So the dimension of the representation does not correspond to the amount of information it contains.

One might speculate that the "extra information" in the smaller-dimensional representation of the bigger group is housed in the (smallest) field over which the representation is defined; in my example, the faithful three-dimensional representations of $S_4$ are defined over $\mathbb{Q}$, but the faithful two-dimensional representations of $\tilde S_4$ I believe require $\mathbb{Q}(i)$.

But I'm unsatisfied by this explanation. First of all, although I do not have an example on hand, I am told (by Derek Holt) that one observes the same phenomenon with permutation representations, with degree replacing dimension, where there is no question of the field of definition. Secondly, it begs the question (to me anyway) of "how" the central extension "causes" the information to be moved from the size of the space to the size of the field.

My question is this:

Has this phenomenon attracted any research attention? Can you point me to any exploration of it in the literature?

• Infinite group examples: $SO(3)$'s smallest complex rep is $\mathbb C^3$, but its double cover $SU(2)$ acts on $\mathbb C^2$. Worse, let $\tilde H$ be the $3\times 3$ upper triangular real matrices with $1$s on the diagonal (acting on $\mathbb C^3$) and $H$ be its quotient by $\begin{pmatrix} 1&0&\mathbb Z\\ &1&0\\ &&1 \end{pmatrix}$, with no finite-dimensional representations. Feb 25, 2015 at 22:53
• @AllenKnutson If one thinks about real representations, the situation changes: $SO(3)$ acts on $\mathbb{R}^3$ because the action on $\mathbb{C}^3$ has an invariant real structure, while the $SU(2)$ action on $\mathbb{C}^2$ does not have real structure and thus $SU(2)$'s smallest real faithful rep is on $\mathbb{R}^4$ and not $\mathbb{R}^2$. I wonder if something similar happens to the finite groups in the original question? Feb 26, 2015 at 13:19
• Ben, you might be interested in this paper - arxiv.org/pdf/1408.1649.pdf - by Britnell, Saunders and Skyner. It looks at a similar phenomenon for $p$-groups - when the quotient of a $p$-group has minimal permutation representation larger than the original. Feb 26, 2015 at 16:59
• @NickGill - thank you, this is just the sort of thing I was looking for. Mar 6, 2015 at 18:33

I think the basic reason that this happens is quite simple. Matrix groups, such as ${\rm GL}(n,{\mathbb C})$ have a centre, consisting of the scalar matrices and so it natural for groups with nontrivial centres to have representations in which their centre maps to the scalars.

More to the point, the perfect group ${\rm SL}(n,{\mathbb C})$ has a cyclic centre of order $n$. So finite groups $G$ with cyclic subgroups $Z < Z(G) \cap [G,G]$ of order dividing $n$ are good candidates for mapping into ${\rm SL}(n,{\mathbb C})$.

So, for example, ${\rm SL}(2,5)$ with centre of order $2$ maps into ${\rm SL}(2,{\mathbb C})$, whereas the smallest dimension of a faithrful representation of $A_5={\rm PSL}(2,5)$ is $3$.

Or $3.A_6 < {\rm SL}(3,{\mathbb C})$ and $2.A_6 = {\rm SL}(2,9) < {\rm SL}(4,{\mathbb C})$, but the smallest degree nontrivial representation of $A_6$ is $5$.

One mechanism for this phenomenon comes from corresponding central extensions of Lie groups. E.g. the three dimensional rep. of $A_4$ comes from $A_4$ acting as rotations of the tetrahedron, and so is a homomorphism $A_4 \to SO(3).$

There is a double cover $SU(2) \to SO(3)$, and the preimage of $A_4$ under this surjection is a double cover $\widetilde{A}_4$, which now admits a two-dimensional representation.

This leaves the question as to why these double covers exist on the Lie group level.

One way to think about it is to start with $SU(2)$: it has not just its standard $2$-dim'l rep., but irreps. of all dimensions, and its odd (e.g. $3$) dim'l reps. factor through $SU(2)/\{\pm I\} = SO(3)$. So if you have a $2n+1$-dimensional rep. of a group which happens to land in the image of $SU(2)$ under its $2n+1$-dimensional rep. (e.g. which happens to land in $SO(3)$, taking $n = 1$), then after possibly passing to a double cover we reduce'' this $2n+1$-dim'l rep. to a $2$-dim'l rep. (into $SU(2)$).

• In the case of $SU(2)$ vs. $SO(3)$ something funny is happening, though: the finite-dimensional irreps of $SO(3)$ are in fact precisely the finite-dimensional irreps of $SU(2)$ on which the center acts trivially, and in particular passing to the central quotient does not give us any unexpected new reps. It's unclear why this should happen on the level of finite groups. Feb 26, 2015 at 4:02
• Qiaochu, am I misreading your comment? Of course a quotient doesn't have new reps. It's $SU(2)$ that has the new reps that $SO(3)$ doesn't have. Feb 26, 2015 at 13:31

One of the most natural ways that these central extensions arise is in Clifford theory of group representations, a well-studied area. I limit my discussion to working with representations over algebraically closed fields $k$.

If we have an irreducible $kG$ module $V$ for a finite group $G$, and $G$ has a normal subgroup $N,$ there are two basic Clifford reductions. If ${\rm Res}^{G}_{N}(V)$ is not homogeneous ( ie not a sum of isomorphic irreducible $kN$-modules), then $V$ is induced from a proper subgroup $H$ of $G$ which contains $N$ (more can be said, but that is not so relevant here).

On the other hand, if ${\rm Res}^{G}_{N}(V) \cong eU$ for some irreducible $kN$-module $U$ and positive integer $e$, then while $U$ may not extend to an irreducible $kG$-module, it does extend to an irreducible $k{\hat G}$-module for some central extension ${\hat G}$ of $G$. The standard way of doing this is as follows: let $\sigma$ be the irreducible matrix representation of $N$ afforded by $U.$ Then in the case under consideration, for each $g \in G,$ the representation of $N$ given by $n \to \sigma(gng^{-1})$ is equivalent to the representation $\sigma,$ so there is an invertible matrix $T_{g}$ such that $\sigma(gng^{-1}) = T_{g}\sigma(n)T_{g}^{-1}$ for all $n \in N.$ However, the matrix $T_{g}$ is not quite unique : by Schur's Lemma, using the algebraic closure of $k,$ the matrix $T_{g}$ is unique up to a scalar multiple. Note then that for each $g,h \in G,$ we have $T_{gh} = \alpha(g,h)T_{g}T_{h}$ for some scalar $\alpha(g,h) \in k.$

Clifford's second reduction gives a factorization of $V$ as $U \otimes W$ as $k{\hat G}$-module, where $W$ is an $e$-dimensional irreducible $k\tilde{G/N}$-module, where $\tilde {G/N}$ is a finite cyclic central extension of $G/N.$ There are various normalizations required to enable us to deal with finite central extensions, and to deal with a $2$-cocycle of $G/N.$ Also, this allows us to choose each $\alpha(g,h)$ to a root of unity.

So one answer to this question is that the field necessary to obtain a non-trivial irreducible representation of a central extension of $G/N$ which may be of smaller dimension than any possessed by $G/N$ is explained by a $2$-cocycle of $G/N$ arising in Clifford's second reduction.

Later edit: Another way to look at this is that (in the second Clifford reduction case), the fact that $U$ is $G$-stable gives us an action of $G$ on ${\rm End}_{k}(U) \cong U \otimes U^{*}$, and the fact that all automorphisms of that matrix algebra are inner allows us to define an action of $G$ on $U$ "up to scalars".

In a slightly different direction, a somewhat related phenomenon is that ( for $p$ odd), the group ${\rm Sp}(2n,p)$ has "unreasonably small" irreducible characters of degree $\frac{p^{n} \pm 1}{2}$, known as the Weil characters (for reasons clear to specialists). These can be obtained as follows: An extra-special group $E$ of order $p^{2n+1}$ and exponent $p$ has a faithful irreducible character $\theta$ of degree $p^{n}$ which is unique up to Galois conjugation, and is ${\rm Sp}(2n,p)$-stable. It turns out that $\theta$ does extend to the semi-direct product $E{\rm Sp}(2n,p)$. But on restricton back to ${\rm Sp}(2n,p)$, the character decomposes into two pieces, one for each eigenspace of the central involution of ${\rm Sp}(2n,p).$ These eigenspaces have dimensions $\frac{p^{n} \pm 1}{2}.$ Notice that when $p = 5$ and $n = 1,$ we obtain $2$ and $3$-dimensional representations of ${\rm Sp}(2,5) \cong {\rm SL}(2,5)$- the $2$-dimensional representation is faithful, and the $3$-dimensional representation has central kernel of order $2$. In general, the faithful Weil character of ${\rm Sp}(2n,p)$ has degree $\frac{p^{n}+ \varepsilon}{2}$ where the sign $\varepsilon$ is chosen to make this degree even.

Even later edit: Another interesting case occurs with $A_{7}.$ We have an isomorphism $A_{8} \cong {\rm GL}(4,2).$ The $4$-dimensional representation in characteristic $2$ restricts irreducibly to $A_{7}.$ This representation of $A_{7}$ does not lift to a characteristic $0$-representation of $A_{7}$ since $A_{7}$ has no non-trivial complex irreducible character of degree less than $6$. However the representation does lift to a complex representation of the double cover ${\hat A_{7}}$.

• Could you break up this text into paragraphs? Feb 26, 2015 at 3:59
• As someone who regards himself as not entirely a non-specialist in $p$-adic representation theory, which (per Yu) involves the Weil representations quite intimately, it's not clear to me why finite symplectic groups have such a representation (aside from the fact that, well, one can prove it!). Could you elaborate for this not-specialist-enough? Nov 7, 2019 at 18:45