# Do semisimple algebraic groups always have faithful irreducible representations?

For simplicity, I will be talking only about connected groups over an algebraically closed field of characteristic zero.

The basic theorem of affine algebraic groups is that they all admit faithful, finite-dimensional representations. The fundamental theorem for semisimple groups is that these representations are all completely reducible, but unfortunately there is no reason that any irreducible summand of a faithful representation should be faithful, only that the kernels of all these representations intersect trivially.

My question is whether such a representation does, in fact, exist.

(Answered: iff the center is cyclic.)

This does not hold of general reductive groups for the following reason: if $T$ is any torus of rank $r > 1$, then its irreducible representations are all characters $\chi \colon T \cong \mathbb{G}_m^r \to \mathbb{G}_m$, which therefore have nontrivial kernels. More generally, any reductive group $G$ has connected center a torus of some rank $r$, so by Schur's lemma this center acts by a character $\chi$ in any irreducible representation of $G$ and if $r > 1$, therefore does not act faithfully.

The exceptional case $r = 1$ does have an example, namely $\operatorname{GL}_n$, whose standard representation is faithful and irreducible and whose center has rank 1. A more general version of this question might be, then:

Does any reductive group whose center has rank at most 1 have a faithful irreducible representation?

(Answered: when not semisimple, iff the center is connected.)

Another special case is that if $G$ is simple and of adjoint type, then its adjoint representation is irreducible and faithful by definition (or, depending on your definition, because the center is trivial). A constructive version of this question for any $G$ (semisimple or reductive of central rank 1) is then:

Can we give a construction of a faithful, irreducible representation of $G$ from its adjoint representation?

This is deliberately a little vague since I don't want to restrict the possible form of such a construction, only that it not start out with "Throw away the adjoint representation and take another one such that..."

Finally, suppose the answer is "no".

What is the obstruction to such a representation existing?

(Answered: for $Z$ the center, it is the existence of a generator for $X^*(Z)$.)

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In the simple semisimple case, representations of the Lie algebra "promote" to representations of the simply connected central cover. For nontrivial representations the kernel is finite, hence central (by normality). The main point is to identify the finite kernel, and that will tell you which (central) quotient of the semisimple group acts faithfully. You can play a similar game in the reductive case of central rank 1 (with simple derived group): the center of the faithful quotient must be cyclic, so the central character extends to the full center, so all real work is in the simple case. –  Boyarsky Jun 28 '10 at 18:57
Please, use the bold font sparingly: it makes the question very difficult to read and hurts the eyes. –  Victor Protsak Jun 28 '10 at 21:58
I apologize. The intent was to make it easier to read by indicating what, if anything, was the question. –  Ryan Reich Jun 28 '10 at 22:32
No worries! Thank you for changing it. Another way to unobtrusively highlight the questions is to put them in blockquotes (add > in the beginning of the line, or use <blockquote></blockquote> tags). –  Victor Protsak Jun 28 '10 at 23:17
I like bold better than italics; I wish MO supported an \em command that I could set to whichever kind of emphasis I like. I think the best way to offset a question is to put it on its own line with a greater than sign (and a space) >  in front. This should make it indented and on a gray box. –  Theo Johnson-Freyd Jun 29 '10 at 8:12

Edit: I now give the argument for general reductive $G$.

Let $G$ be a reductive algebraic group over an alg. closed field $k$ of char. 0. Fix a max torus $T$ and write $X = X^*(T)$ for its group of characters. Write $R$ for the subgroup of $X$ generated by the roots of $G$. Then the center $Z$ of $G$ is the diagonalizable subgroup of $T$ whose character group is $X/R$.

Claim: $G$ has a faithful irreducible representation if and only if the character group $X/R$ of $Z$ is cyclic.

Note for semisimple $G$, the center $Z$ is finite. Since the characteristic of $k$ is 0, in this case the group of $k$-points of $Z$ is (non-canonically) isomorphic to $X/R$. Thus $Z$ is cyclic if and only $X/R$ is cyclic.

In general, the condition that $X/R$ is cyclic means either that the group of points $Z(k)$ is finite cyclic, or that $Z$ is a 1 dimensional torus.

As to the proof, for $(\implies)$ see Boyarsky's comment following reb's answer.

For $(\Leftarrow)$ let me first treat the case where $G$ is almost simple; i.e. where the root system $\Phi$ of $G$ is irreducible. Supopse that the class of $\lambda \in X$ generates the cyclic group $X/R$. Since the Weyl group acts on $X$ leaving $R$ invariant, the class of any $W$-conjugate of $\lambda$ is also a generator of $X/R$. Thus we may as well suppose $\lambda$ to be dominant and non-0 [if $X=R$, take e.g. $\lambda$ to be a dominant root...] Now the simple $G$-module $L=L(\lambda) = H^0(\lambda)$ with "highest weight $\lambda$" will be faithful. To see this, note that since $\lambda \ne 0$, $L$ is not the trivial representation. Since $G$ is almost simple, the only proper normal subgroups are contained in $Z$. Thus it suffices to observe that the action of $Z$ on the $\lambda$ weight space of $L$ is faithful.

The general case is more-or-less the same, but with a bit more book-keeping. Write the root system $\Phi$ of $G$ as a disjoint union $\Phi = \cup \Phi_i$ of its irreducible components. There is an isogeny $$\pi:\prod_i G_{i,sc} \times T \to G$$ where $T$ is a torus and $G_{i,sc}$ is the simply connected almost simple group with root system $\Phi_i$. Write $G_i$ for the image $\pi(G_{i,sc}) \subset G$.

The key fact is this: a representation $\rho:G \to \operatorname{GL}(V)$ has $\ker \rho \subset Z$ if and only if the restriction $\rho_{\mid G_i}$ is non-trivial for each $i$.

Now, as before pick $\lambda \in X$ for which the coset of $\lambda$ generates the assumed-to-be cyclic group $X/R$. After replacing $\lambda$ by a Weyl group conjugate, we may suppose $\lambda$ to be dominant. After possibly repeatedly replacing $\lambda$ by $\lambda + \alpha$ for dominant roots $\alpha$, we may suppose that $\lambda$ has the following property: $$(*) \quad \text{for each i, there is \beta_i \in \Phi_i with \langle \lambda,\beta_i^\vee \rangle \ne 0}$$

Now let $L = L(\lambda)$ be the simple module with highest weight $\lambda$. Condition $(*)$ implies that $G_i$ acts nontrivially on $L$ for each $i$, so by the "key fact", the kernel of the representation of $G$ on $L$ lies in $Z$. but since $\lambda$ generates the group of characters of $Z$, the center $Z$ acts faithfully on the $\lambda$ weight space of $L$.

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@Ryan: I don't see any reason to limit your question to "simple" types, though the proof for a general semisimple group reduces to this crucial case. (See my added paragraph.) –  Jim Humphreys Jun 29 '10 at 11:42
BCnrd edited it, possibly to reflect the way the discussion was going. I have put it back since George's answer covers my original more general question. –  Ryan Reich Jun 29 '10 at 14:10

This is an elaboration of George's answer (and my own deleted semi-answer), along with a comment on the further questions raised. George has already addressed the basic question about existence of faithful irreducible representations for semisimple groups. So this is just a supplement, not a separate answer.

For a connected semisimple algebraic group $G$ (over any algebraically closed field), the Chevalley classification determines $G$ up to isomorphism in terms of the way the weight lattice of a maximal torus of $G$ lies between the full weight lattice (of a simply connected group) and the root lattice. The related classification of irreducible representations of $G$ by highest weights is also given in terms of dominant integral weights lying in this weight lattice. In particular, if such a weight can be chosen to lie in no smaller intermediate lattice, it will produce a faithful irreducible representation of $G$: otherwise it gives a faithful irreducible representation of a proper quotient group and thus has highest weight in a proper sublattice.

Such a weight exists unless the "co-fundamental group" (quotient of weight lattice by root lattice) is noncyclic. For an irreducible root system this happens just for even type D, where you get a Klein 4-group (for Spin groups); in other cases the co-fundamental group (= center of $G$ in characteristic 0) is cyclic. In general $G$ is an almost-direct product of groups with irreducible root systems; so one has to be careful about factors of even type D.

ADDED: For any connected semisimple $G$ (and any characteristic), the argument just outlined leads routinely to the conclusion that a faithful irreducible representation exists precisely when no simple factor of $G$ is simply connected of even type D. That exception was pointed out in earlier comments/answer and uses only Schur's Lemma. To treat all simple types, the (Killing-Cartan/Chevalley) classification is used to avoid cases when the co-fundamental group is noncyclic. Then the proof for a general $G$ reduces to this one: $G$ is an almost-direct product of simple groups, so you just have to find suitable highest weights for the individual factors to get a suitable highest weight for $G$. (In effect, you find an irreducible representation of the simply connected covering group, a direct product of simple groups, which induces a faithful irreducible representation of $G$ just in case $G$ has no Spin factors.)

As Ryan points out there is a problem for reductive groups if the center is a torus of dimension $>1$. Otherwise the situation for general linear groups can probably be imitated: start with a faithful irreducible representation of the derived group (if available), then throw in the scalars (which may of course already overlap the image of the derived group in a finite subgroup).

The further question about "construction" of $G$ (in a suitable representation) from knowledge of its adjoint representation depends on what "construction" means in this context. For semisimple groups, there are indirect ways to construct each irreducible representation of a given highest weight (in characteristic 0) by working with tensor products of fundamental representations. This does not give direct information of the sort contained in Weyl's character and dimension formulas, however. Anyway, if there exists a suitable highest weight (as discussed above) giving a faithful representation of $G$, this rather abstract procedure does "construct" the corresponding faithful irreducible representation. But I think you are asking for a more "natural" construction based on the adjoint representation, which I can't visualize.

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I wish I could vote up this answer more than once; it deserves some recognition. –  Ryan Reich Jun 29 '10 at 14:11

Spin groups in even dimensions have center a non-cyclic group (of order 4) and so have no faithful irreducible representations.

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So in other words, in view of the existence of a central character, a necessary condition in general is that the character group of the center should be a cyclic $\mathbb{Z}$-module (so the center can be embedded into $\mathbb{G}_ m$). That rules out those spin groups and cases with $r > 1$ as in the question. –  Boyarsky Jun 28 '10 at 18:28