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An irreducible complex $G^F$-module is cuspidal if its Harish-Chandra restriction is zero for every proper rational parabolic subgroup. By adjunction, it occurs in no Harish-Chandra induction from a proper rational Levi subgroup. This is a finite-group notion, distinct from a cuspidal automorphic representation.

For a rational Levi subgroup $M$ of a rational parabolic subgroup and a rational maximal torus $T$, the torus case of the induction-restriction formula isIn particular it vanishes if there is no rational conjugate of $T$ in $M$. If $T\subset M$, transitivity gives $R_T^G=R_M^G{R}_T^M$. Together with adjunction these identities show that an irreducible character up to sign arising from $T$ is cuspidal exactly when $T$ is elliptic.

In characteristic zero, a finite-dimensional Lie algebra is a semidirect sum of its solvable radical and a semisimple Lie subalgebra. The group analogue splits appropriate groups as a semidirect product with a Levi subgroup.

A Levi subgroup appears in a group version of a Levi decomposition. A Levi subgroup is a reductive complement to the unipotent radical of a parabolic subgroup.

Let $B$ be a nondegenerate alternating bilinear form on $V$ of dimension $2m$ over ${\mathbb{F}}_q$. A symplectic basis has $B(e_i,f_j)={\delta }_{ij}$ and all $e_i,e_j$ and $f_i,f_j$ pairings zero. Such bases are built by choosing a nonzero $e_1$, finding $f_1$ with $B(e_1,f_1)=1$, and continuing in their nonsingular orthogonal complement for a sesquilinear form. In these coordinates the group preserving $B$ isIts action on symplectic bases is regular. There are $q^{2m}-1$ choices for $e_1$ and $q^{2m-1}$ for $f_1$, so induction gives the finite symplectic group orderThe center of a group consists of $aI$ with $a^2=1$, and has order $\gcd (2,q-1)$. To see there are no other central elements, a central element commutes with every symplectic transvection and hence fixes every symplectic transvection center line; a linear map fixing every line is scalar. Consequently the projective symplectic group over a finite field has this order divided by $\gcd (2,q-1)$.

Here is a proof of the main ingredients in simplicity. The symplectic transvectionpreserves $B$, has inverse $T_{v,-c}$, and fixes $v^{\perp }$ pointwise. These transformations generate $S{p}_{2m}(q)$. Indeed if $B(u,w)\ne 0$, then $T_{w-u,1/B(u,w)}$ sends $u$ to $w$. If the pairing is zero, choose an intermediate vector having nonzero pairing with both, so two symplectic transvections suffice. This aligns the first vector of any symplectic basis. For partners $f,f^{\prime }$ of the same $e$, their difference lies in $e^{\perp }$. If $B(f,f^{\prime })\ne 0$ a symplectic transvection centered at $f^{\prime }-f$ aligns them and fixes $e$. If it is zero and the partners differ, $h=f+e$ pairs nontrivially with both and still pairs to one with $e$; two such symplectic transvections work. After fixing the pair, induction in its orthogonal complement for a sesquilinear form aligns the rest of the basis.

The projective action on one-spaces is faithful. For $m\ge 2$, a line stabilizer has three orbits: the line itself, the other lines perpendicular to it, and the lines not perpendicular to it. Witt's lemma proves transitivity within the latter two orbits by matching the two-dimensional restricted form. This action is primitive. A block containing distinct perpendicular lines $L,M$ is invariant under both stabilizers and therefore contains a line perpendicular to $M$ but not to $L$. A block containing distinct nonperpendicular lines likewise contains a line perpendicular to $L$ but not to $M$, by using the $M$ stabilizer. Such lines exist by nonsingularity in dimension at least four. Either possibility makes the block contain both nontrivial stabilizer orbits, hence every line. For $m=1$ the action on the projective line is two-transitive and thus primitive.

Symplectic transvections with one fixed center line form an Abelian normal subgroup of its stabilizer, and their conjugates generate the whole group. The Iwasawa simplicity lemma now reduces simplicity to being perfect. Its short proof is useful: any nontrivial normal subgroup of a faithful primitive group is transitive, since its orbits are blocks. Writing $G=NH$, where $H$ is a point stabilizer, shows that all conjugates of the specified Abelian subgroup have the same image modulo $N$. Thus $G/N$ is Abelian. If $G$ is perfect, $N=G$.

The groups can also be shown to be perfect concretely. For $q>3$ choose $a$ with $a^2\ne 1$ and let $D$ scale $e$ by $a$ and $f$ by $a^{-1}$. Thenso every symplectic transvection is a group commutator. For $q=3,m\ge 2$, symplectic transvections of coefficient one are conjugate, have order three, and the four with centers $e_1,e_2,e_1+e_2,e_1-e_2$ in a two-dimensional totally isotropic subspace multiply to the identity. They commute, and the sum of their rank-one terms is zero modulo three. Their common class in the abelianization is therefore annihilated by both three and four, hence is zero. For $q=2,m\ge 3$, use the seven nonzero vectors of a three-dimensional totally isotropic subspace. The symplectic transvections commute and their product is the identity, because each diagonal coefficient occurs four times and each cross coefficient twice. Their common abelianized class is annihilated by two and seven, hence is zero. Generation now proves these groups are perfect.

It follows thatThe exceptions are $S_3$, $A_4$ and $S_6$ respectively. For odd $q$ the full symplectic group over a finite field has nontrivial center of a group, so simplicity refers to its quotient group; for even $q$ the center of a group is trivial.

A symplectic parabolic subgroup stabilizes a flag of totally isotropic subspaces. The maximal such parabolic subgroups $P_r$ stabilize a single $r$-space, $1\le r\le m$. In an adapted basis their Levi subgroup is $G{L}_r(q)\times S{p}_{2m-2r}(q)$, with an arbitrary block of $r(2m-2r)$ parameters and a remaining symmetric block of $r(r+1)/2$ parameters. ThusThese are the single-member flag stabilizers, equivalently the standard maximal parabolic subgroups obtained by omitting one node of the symplectic Dynkin diagram. Two particular cases have transparent descriptions.

For the symplectic point stabilizer, write $V=⟨e,f⟩\perp W$. Its unipotent radical consists ofAll pairings are preserved by direct substitution. The product law is $(b,t)(b^{\prime },t^{\prime })=(b+b^{\prime },t+t^{\prime }+B(b,b^{\prime }))$, so this radical has order $q^{2m-1}$. With group commutator convention $uv{u}^{-1}{v}^{-1}$, its commutator has central parameter $2B(b,b^{\prime })$: it is Heisenberg-type in odd characteristic, but Abelian in characteristic two. A Levi subgroup scales $e,f$ inversely and acts as $Sp(W)$, givingThis index counts the one-spaces, all of which are isotropic for an alternating bilinear form.

For a Lagrangian subspace $E=⟨e_1,\dots ,e_m⟩$, the symplectic Lagrangian stabilizer has block formSubstituting into $g^{\mathsf{T}}Jg=J$ proves both the inverse-transpose lower block and symmetry of $S$. The unipotent radical is the additive group of symmetric matrices, and $A$ acts by $S\mapsto AS{A}^{\mathsf{T}}$. ThereforeThe last identity follows by canceling $q$-powers and $(q^j-1)$ factors in the two order formulas; it counts maximal totally isotropic subspaces.

Let $P=L⋉U$ and $Q=M⋉V$ be rational parabolic subgroups with rational Levi subgroups. Work with complex representations of their finite groups of Lie type. Here Harish-Chandra induction and Harish-Chandra restriction meanThe latter is an $M^F$-module, and it is adjoint to the former by Frobenius reciprocity.

DefineFor $x\in \mathcal{S}(M,L{)}^F$, put $H_x=M\cap {}^{x}L$ and let $\mathrm{ad}x$ transport an $L^F$-module to a $({}^{x}L{)}^F$-module. The Mackey formula for Harish-Chandra induction is the natural functor isomorphismOn characters the direct sum becomes a sum. The intersection parabolics displayed in the subscripts are important: they specify exactly which Harish-Chandra induction and Harish-Chandra restriction occur.

We use the following parabolic-intersection lemmas, stating explicitly the geometric input to the proof. First, representatives of $M^F\backslash \mathcal{S}(M,L{)}^F/L^F$ represent exactly the double cosets $Q^F\backslash G^F/P^F$. Second, for such $x$, the groups $M\cap {}^{x}P$ and $Q\cap {}^{x}L$ are parabolic subgroups of $M$ and ${}^{x}L$, respectively, with common Levi subgroup $H_x$ and respective unipotent radicals $M\cap {}^{x}U$ and $V\cap {}^{x}L$. If $K=Q\cap {}^{x}P$, projection $Q\to M$ maps $K$ onto $M\cap {}^{x}P$ with kernel $K\cap V$. Projection ${}^{x}P\to {}^{x}L$ maps $K\cap V$ onto $V\cap {}^{x}L$ with kernel $V\cap {}^{x}U$. These assertions hold on rational points as well. The surjectivity on rational points uses connectedness of the unipotent kernels and the Lang theorem for algebraic groups. These are the compatible-Levi and rational double-coset lemmas; they follow by applying the root-subgroup decomposition after choosing the common maximal algebraic torus.

For completeness, the representation-theoretic reduction is as follows. The group algebra realization of induced representation splits according to the double cosets:as a $(Q^F,P^F)$-bimodule. Tensoring on the right with an inflated $L^F$-module $E$ identifies each summand with induction from $K^F=(Q\cap {}^{x}P{)}^F$. This proves the finite-group Mackey restriction formula in the formHere ${}^x\tilde{E}$ is trivial on $({}^{x}U{)}^F$.

We next record the elementary normal-subgroup identity underlying Harish-Chandra restriction. If $N◃D$ are finite groups, with $N$ a normal subgroup, and $C\subset D$, then for a complex $C$-module $A$,To see this, replace invariants by coinvariants using the averaging idempotent $|N{|}^{-1}{\sum }_{n\in N}n$. Taking coinvariants in $\mathbb{C}[D]{\otimes }_{\mathbb{C}[C]}A$ first replaces $\mathbb{C}[D]$ by $\mathbb{C}[D/N]$, and forces $C\cap N$ to act trivially on $A$. Thus it yields $\mathbb{C}[D/N]{\otimes }_{\mathbb{C}[CN/N]}{A}_{C\cap N}$; averaging over $C\cap N$ identifies this last module with the one displayed. This also shows why characteristic zero makes the argument exact.

Apply this identity to $D=Q^F$, $N=V^F$, $C=K^F$. By the intersection lemmas, $CN/N=(M\cap {}^{x}P{)}^F$. Since ${}^x\tilde{E}$ is trivial on $({}^{x}U{)}^F$, its $(K\cap V{)}^F$-invariants are exactlyMoreover $(M\cap {}^{x}U{)}^F$ acts trivially on this module, so induction from $(M\cap {}^{x}P{)}^F$ to $M^F$ is precisely Harish-Chandra induction from $H_x$. Substituting into each double-coset summand gives the boxed formula. This proves the formula for rational parabolics; it does not assume a general Mackey formula for nonrational Deligne-Lusztig induction.

Write $\chi =\epsilon R_T^G(\theta )$, where $\epsilon \in \{1,-1\}$ and $\chi$ is the character of an irreducible representation. An irreducible representation is a cuspidal representation of a finite reductive group exactly when its Harish-Chandra restriction to every proper rational Levi subgroup of a rational parabolic subgroup is zero.

The needed identity is the Harish-Chandra restriction of a Deligne-Lusztig character. For a rational parabolic subgroup $P=M⋉V$, the torus case of the induction-restriction formula givesHere ${}^x\theta (xt{x}^{-1})=\theta (t)$. The geometric meaning of the indexing condition is that only a rational conjugate of the inducing maximal algebraic torus lying in the Levi can survive taking $V^F$-invariants. This is the established torus case of Deligne-Lusztig induction, valid without a restriction on $q$; it is not an appeal to an unrestricted induction-restriction conjecture. See Dudas and Michel's lectures, Theorem 14.4.

Suppose first that $T$ is an elliptic maximal torus, meaning it lies in no proper rational parabolic subgroup. For any proper rational $P=M⋉V$, the indexing set in the displayed sum is empty. Indeed ${}^{x}T\subset M$ with $x\in G^F$ would imply $T\subset {}^{x^{-1}}P$, itself a proper rational parabolic subgroup. Hence ${}^{\ast }{R}_{M\subset P}^G{R}_T^G(\theta )=0$, and multiplication by $\epsilon$ gives ${}^{\ast }{R}_{M\subset P}^G\chi =0$. This proves cuspidality.

Conversely suppose $T\subset P$ for a proper rational parabolic subgroup. There is a unique Levi subgroup of $P$ containing this particular maximal algebraic torus: relative to $T$, it is generated by $T$ and those root subgroups for which both the root and its negative occur in $P$. Call it $M$. Uniqueness and $F(T)=T$, $F(P)=P$ imply $F(M)=M$, so this is a proper rational Levi. Transitivity of Deligne-Lusztig induction through this rational parabolic saysSet $\psi =R_T^M(\theta )$, allowing it to be a virtual character. Adjunction of Harish-Chandra induction and Harish-Chandra restriction, extended linearly to virtual characters, givesThus this Harish-Chandra restriction cannot vanish, so $\chi$ is not cuspidal. No positivity or irreducibility of $\psi$ is needed, and the sign in $\chi$ causes no cancellation problem. We have provedThe term elliptic maximal torus is relative to the ambient group: split central factors are allowed. The actual condition is absence of a proper rational parabolic subgroup containing $T$.

Put $i_0=0$ and $d_j=i_j-i_{j-1}$. The stabilizeris the standard parabolic subgroup containing $B$ obtained by allowing arbitrary changes of basis inside the successive blocks. Its Levi subgroup isThe fiber of ${\pi }_I$ over the displayed partial flag is $P_I/B$. Choosing a complete refinement amounts to choosing complete flags in every quotient $F_{i_j}/F_{i_{j-1}}$ and a complete isotropic flag in $F_{i_r}^{\perp }/F_{i_r}$. Hencethe flag variety of an algebraic group $L_I$; explicitly it is the product of the complete flag varieties of the listed general linear and symplectic factors.

With a Lagrangian subspace first in a paired symplectic basis, its stabilizer consists of $\left(\begin{array}{cc}A& AS\\ 0& A^{-\mathsf{T}}\end{array}\right)$ with $S$ symmetric. The additive symmetric-matrix unipotent radical has order $q^{m(m+1)/2}$; the Levi subgroup is $G{L}_m(q)$, acting by $S\mapsto AS{A}^{\mathsf{T}}$. Its index in $S{p}_{2m}(q)$ is ${\prod }_{i=1}^m(q^i+1)$, the number of Lagrangian subspaces.