How big can the irreps of a finite group be (over an arbitrary field)?

Question

Let G be a nontrivial finite group. Does there exist an irreducible representation of G of dimension greater than or equal to the cardinality of G?

[Edited for clarity. -- PLC]

Answer 1

EDIT: Part 4 added. EDIT2: Second proof of Part 4 added.

1. The answer is no (as long as we are working over a field - of any characteristic, algebraically closed or not). If $k$ is a field and $G$ is a finite group, then the dimension of any irreducible representation $V$ of $G$ over $k$ is $\leq \left|G\right|$. This is actually obvious: Take any nonzero vector $v\in V$; then, $k\left[G\right]v$ is a nontrivial subrepresentation of $V$ of dimension $\leq\dim\left(k\left[G\right]\right)=\left|G\right|$. Since our representation $V$ was irreducible, this subrepresentation must be $V$, and hence $\dim V\leq\left|G\right|$.

2. Okay, we can do a little bit better: Any irreducible representation $V$ of $G$ has dimension $\leq\left|G\right|-1$, unless $G$ is the trivial group. Same proof applies, with one additional step:

If $\dim V=\left|G\right|$, then the map $k\left[G\right]\to V,\ g\mapsto gv$ must be bijective (in fact, it is surjective, since $k\left[G\right]v=V$, and it therefore must be bijective since $\dim\left(k\left[G\right]\right)=\left|G\right|=\dim V$), so it is an isomorphism of representations (since it is $G$-equivariant), and thus $V\cong k\left[G\right]$. But $k\left[G\right]$ is not an irreducible representation, unless $G$ is the trivial group (in fact, it always contains the $1$-dimensional trivial representation).

3. Note that if the base field $k$ is algebraically closed and of characteristic $0$, then we can do much better: In this case, an irreducible representation of $G$ always has dimension $<\sqrt{\left|G\right|}$ (in fact, in this case, the sum of the squares of the dimensions of all irreducible representations is $\left|G\right|$, and one of these representations is the trivial $1$-dimensional one). However, if the base field is not necessarily algebraically closed and of arbitrary characteristic, then the bound $\dim V\leq \left|G\right|-1$ can be sharp (take cyclic groups).

4. There is a way to improve 2. so that it comes a bit closer to 3.:

Theorem 1. If $V_1$, $V_2$, ..., $V_m$ are $m$ pairwise nonisomorphic irreducible representations of a finite-dimensional algebra $A$ over a field $k$ (not necessarily algebraically closed, not necessarily of characteristic $0$), then $\dim V_1+\dim V_2+...+\dim V_m\leq\dim A$.

(Of course, if $A$ is the group algebra of some finite group $G$, then $\dim A=\left|G\right|$, and we get 2. as a consequence.)

First proof of Theorem 1. At first, for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, the (left) representation $V_i^{\ast}$ of the algebra $A^{\mathrm{op}}$ (this representation is defined by $a\cdot f=\left(v\mapsto f\left(av\right)\right)$ for any $f\in V_i^{\ast}$ and $a\in A$) is irreducible (since $V_i$ is irreducible) and therefore isomorphic to a quotient of the regular (left) representation $A^{\mathrm{op}}$ (since we can choose some nonzero $u\in V_i^{\ast}$, and then the map $A^{\mathrm{op}}\to V_i^{\ast}$ given by $a\mapsto au$ must be surjective, because its image is a nonzero subrepresentation of $V_i^{\ast}$ and therefore equal to $V_i^{\ast}$ due to the irreducibility of $V_i^{\ast}$). Hence, by duality, $V_i$ is isomorphic to a subrepresentation of the (left) representation $A^{\mathrm{op}\ast}=A^{\ast}$ of $A$. Hence, from now on, let's assume that $V_i$ actually is a subrepresentation of $A^{\ast}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$.

Now, let us prove that the vector subspaces $V_1$, $V_2$, ..., $V_m$ of $A^{\ast}$ are linearly disjoint, i. e., that the sum $V_1+V_2+...+V_m$ is actually a direct sum. We will prove this by induction over $m$, so let's assume that the sum $V_1+V_2+...+V_{m-1}$ is already a direct sum. It remains to prove that $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)=0$. In fact, assume the contrary. Then, $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)=V_m$ (since $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)$ is a nonzero subrepresentation of $V_m$, and $V_m$ is irreducible). Thus, $V_m\subseteq V_1+V_2+...+V_{m-1}$. Consequently, $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ (because the sum $V_1+V_2+...+V_{m-1}$ is a direct sum, according to our induction assumption).

Now, according to Theorem 2.2 and Remark 2.3 of Etingof's "Introduction to representation theory", any subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be a direct sum of the form $r_1V_1\oplus r_2V_2\oplus ...\oplus r_{m-1}V_{m-1}$ for some nonnegative integers $r_1$, $r_2$, ..., $r_{m-1}$. Hence, every irreducible subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. Since we know that $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$, we conclude that $V_m$ is isomorphic to one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. This contradicts the non-isomorphy of the representations $V_1$, $V_2$, ..., $V_m$. Thus, we have proven that the sum $V_1+V_2+...+V_m$ is actually a direct sum. Consequently, $\dim V_1+\dim V_2+...+\dim V_m=\dim\left(V_1+V_2+...+V_m\right)\leq \dim A^{\ast}=\dim A$, and Theorem 1 is proven.

Second proof of Theorem 1. I just learnt the following simpler proof of Theorem 1 from §1 Lemma 1 in Crawley-Boevey's "Lectures on representation theory and invariant theory":

Let $0=A_0\subseteq A_1\subseteq A_2\subseteq ...\subseteq A_k=A$ be a composition series of the regular representation $A$ of $A$. Then, by the definition of a composition series, for every $i\in \left\lbrace 1,2,...,k\right\rbrace$, the representation $A_i/A_{i-1}$ of $A$ is irreducible.

Let $T$ be an irreducible representation of $A$. We are going to prove that there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A_I/A_{I-1}$ (as representations of $A$).

In fact, let $I$ be the smallest element $i\in \left\lbrace 1,2,...,k\right\rbrace$ satisfying $A_iT\neq 0$ (such elements $i$ exist, because $A_kT=AT=T\neq 0$). Then, $A_IT\neq 0$, but $A_{I-1}T=0$. Now, choose some vector $t\in T$ such that $A_It\neq 0$ (such a vector $t$ exists, because $A_IT\neq 0$), and consider the map $f:A_I\to T$ defined by $f\left(a\right)=at$ for every $a\in A_I$. Then, this map $f$ is a homomorphism of representations of $A$. Since it maps the subrepresentation $A_{I-1}$ to $0$ (because $f\left(A_{I-1}\right)=A_{I-1}t\subseteq A_{I-1}T=0$), it gives rise to a map $g:A_I/A_{I-1}\to T$, which, of course, must also be a homomorphism of representations of $A$. Since $A_I/A_{I-1}$ and $T$ are irreducible representations of $A$, it follows from Schur's lemma that any homomorphism of representations from $A_I/A_{I-1}$ to $T$ is either an isomorphism or identically zero. Hence, $g$ is either an isomorphism or identically zero. But $g$ is not identically zero (since $g\left(A_I/A_{I-1}\right)=f\left(A_I\right)=A_It\neq 0$), so that $g$ must be an isomorphism, i. e., we have $T\cong A_I/A_{I-1}$.

So we have just proven that

(1) For every irreducible representation $T$ of $A$, there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A_I/A_{I-1}$ (as representations of $A$).

Denote this $I$ by $I_T$ in order to make it clear that it depends on $T$. So we have $T\cong A_{I_T}/A_{I_T-1}$ for each irreducible representation $T$ of $A$. Applying this to $T=V_i$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, we see that $V_i\cong A_{I_{V_i}}/A_{I_{V_i}-1}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$. Hence, the elements $I_{V_1}$, $I_{V_2}$, ..., $I_{V_m}$ of the set $\left\lbrace 1,2,...,k\right\rbrace$ are pairwise distinct (because $I_{V_i}=I_{V_j}$ would yield $V_i\cong A_{I_{V_i}}/A_{I_{V_i}-1}=A_{I_{V_j}}/A_{I_{V_j}-1}\cong V_j$, but the representations $V_1$, $V_2$, ..., $V_m$ are pairwise nonisomorphic), and thus

$\sum\limits_{i=1}^{m}\dim\left(A_{I_{V_i}}/A_{I_{V_i}-1}\right)=\sum\limits_{\substack{j\in\left\lbrace 1,2,...,k\right\rbrace ;\ \\ \text{there exists }\\ i\in\left\lbrace 1,2,...,m\right\rbrace \\ \text{ such that }j=I_{V_i}}}\dim\left(A_j/A_{j-1}\right)$ $\leq \sum\limits_{j\in\left\lbrace 1,2,...,k\right\rbrace}\dim\left(A_j/A_{j-1}\right)$ (since $\dim\left(A_j/A_{j-1}\right)\geq 0$ for every $j$, so that adding more summands cannot decrease the sum) $=\sum\limits_{j=1}^{k}\dim\left(A_j/A_{j-1}\right)=\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)$.

Since $\dim\left(A_{I_{V_i}}/A_{I_{V_i}-1}\right)=\dim V_i$ for each $i$ (due to $A_{I_{V_i}}/A_{I_{V_i}-1}\cong V_i$) and $\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)=\dim A$ (in fact, the sum $\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)$ is a telescopic sum and simplifies to $\dim A_k-\dim A_0=\dim A-\dim 0=\dim A-0=\dim A$), this inequality becomes $\sum\limits_{i=1}^{m}\dim V_i\leq\dim A$. This proves Theorem 1.

Answer 2

I just want to add that for the $\sqrt{|G|}$ bound, we don't need the field to be algebraically closed, we only require that it be a splitting field for the group (i.e., all irreducible representations over the algebraic closure are realized in the group). For a finite group of exponent d, any field over which $x^d - 1$ splits completely into distinct linear factors is a splitting field for the group. (Such fields are called "sufficiently large fields" in some places). The converse doesn't hold -- there could be splitting fields that are not sufficiently large.

If the field is not a splitting field for the group, but has a degree r extension that is a splitting field for the group, then the degrees of irreducible representations are bounded by $r\sqrt{|G|}$ because each irreducible representation over the field can splitinto at most $r$ irreducibles when we pass to the splitting field.

So, over the real numbers, the degree of any irreducible representation is at most $2\sqrt{|G|}$, which is smaller than $|G| - 1$ for groups of size greater than five.

On the other hand, over the rational numbers, the cyclic group of prime order has an irreducible representation of degree equal to the order minus one, so the bound is tight for the rationals.

Answer 3

In the modular case (i.e., when the characteristic of the field divides the order of the group), it is more interesting to look for indecomposable representations (because there are very few irreps or, as they are known in this context, simple modules: the 'worst' case is that of a $p$-group, which has one simple module!), and those can be as big as you want, in general, as soon as the group is not of finite representation type. This follows from the first Brauer-Thrall conjecture, proved by Maurice Auslander et al.; see, for example, [Ringel, Claus Michael. On algorithms for solving vector space problems. I. Report on the Brauer-Thrall conjectures: Rojter's theorem and the theorem of Nazarova and Rojter. Representation theory, I (Proc. Workshop, Carleton Univ., Ottawa, Ont., 1979), pp. 104--136, Lecture Notes in Math., 831, Springer, Berlin, 1980. MR0607142] and the references therein.

The smallest non-trivial example is the Klein Klein four group $\mathbb Z_2\oplus\mathbb Z_2$ in characteristic two, which is of infinite tame representation type, so has indecomposable modules of arbitrary high dimension. They have been known for ages, in various forms; they are described nicely, e.g., in [Benson, D. J. Representations and cohomology. I. Basic representation theory of finite groups and associative algebras. Second edition. Cambridge Studies in Advanced Mathematics, 30. Cambridge University Press, Cambridge, 1998. xii+246 pp. MR1644252].

Answer 4

The $\sqrt{|G|}$ bound is true [for algebraically closed fields -- PLC] in any characteristic. In fact if $R$ is the Jacobson radical of $k[G]$, then $k[G]/R\cong\prod_iM_{n_i}(k)$ where $i$ runs over the irreducible representations and $n_i$ is the degree of the corresponding representation. This gives $\sum_in_i^2=\dim k[G]/R\leq|G|$. In the modular case we always have that $R\neq0$ so that in particular we have strict inequality. Also for every irreducible representation in charateristic $p$ there is an irreducible representation in characteristic $0$ whose degree is $\ge$ than the degree of the characteristic $p$ representation: Choose a number field $K$ which is a splitting field for $G$ and let $R$ be its ring of integers and let further $P$ be a maximal ideal of $R$ dividing $p$. We may filter $K[G]$ by a Jordan-Hölder filtration $W'_i$ so that $W'_i/W' _{i-1}$ is irreducible. Put $W_i:=W'_i\cap R[G]$ so that in particular $W_i/W_{i-1}$ is $R$-torsion free. Hence reducing modulo $P$ we get a filtration $\overline{W}_i$ of $k[G]$. This filtration can be refined to a Jordan-Hölder filtration showing that every irreducible $k[G]$-module which appears in any Jordan-Hölder filtration of $k[G]$ must appear in any Jordan-Hölder filtration of some $\overline{W}_i/\overline{W} _{i-1}$ and thus its degree is $\leq \dim(\overline{W}_i/\overline{W} _{i-1})=\mathrm{rank}(W_i/W _{i-1})=\dim(W'_i/W' _{i-1})$. Hence the degree of any $k[G]$-representation is $\le$ the degree of some $K[G]$-representation. It is rare (but does happen) that $\overline{W}_i/\overline{W} _{i-1}$ is irreducible in the modular case.