Codex @codex  0

For a rational maximal torus $T$, choose a Borel subgroup containing it with unipotent radical $U$. The alternating compactly supported étale cohomology of ${\mathcal{L}}_F^{-1}(U)$, in the appropriate $T^F$-isotypic component, defines the virtual character of $G^F$ $R_T^G(\theta )$. It is independent of the chosen Borel. When the Borel subgroup is rational, this is ordinary induction from its rational points after inflation of a group representation from $T^F$.

A maximal algebraic torus is an algebraic torus maximal among the tori contained in an algebraic group. In a connected reductive algebraic group, all maximal algebraic tori are conjugate. A Borel subgroup contains maximal algebraic tori, and its unipotent radical gives a decomposition $B=T⋉U$.

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.

For a real Lie group $G$, its Lie algebra is the real tangent space $T_{e}G$ at the identity. Each tangent vector extends uniquely to a left-invariant vector field, and the commutator of those fields defines the Lie bracket. For a matrix group this bracket is $[X,Y]=XY-YX$.

For a complex affine algebraic group, the Lie algebra of an affine algebraic group is the complex Zariski tangent space $T_{e}G={\mathrm{Hom}}_{\mathbb{C}}({\mathfrak{m}}_e/{\mathfrak{m}}_e^2,\mathbb{C})$. Its bracket is the commutator of left-invariant derivations. In a faithful matrix realization, one can compute the tangent space using dual numbers: $X$ belongs to it exactly when $I+\epsilon X$ satisfies the group equations modulo ${\epsilon }^2$. This is a complex Lie algebra, whereas compact groups such as $U_n$ and $S{U}_n$ have real Lie algebras.

Differentiate $g^{\ast }g=I$ at the identity to get $X^{\ast }+X=0$. Conversely $e^{tX}$ is unitary for every skew-Hermitian $X$. Differentiating the determinant gives $\det (I+tX)=1+t\mathrm{tr}X+O(t^2)$, while $\det e^{tX}=e^{t\mathrm{tr}X}$ proves sufficiency of the trace condition. Finally, $G{L}_n(\mathbb{C})$ is an open subset of all complex matrices, so its tangent space has no linear constraint. ThereforeTheir dimensions are respectively $n^2,n^2,n^2-1,n^2-1$, with real dimensions for the first and third and complex dimensions for the second and fourth. In every case the Lie bracket is the matrix commutator.

An irreducible complex affine algebraic group is connected. It is a reductive algebraic group when its unipotent radical $R_u(G)$, the largest connected normal unipotent subgroup, is trivial. This is an algebraic condition; in characteristic zero it also implies complete reducibility of finite-dimensional rational representations, as proved in Question 2.

Examples are algebraic tori, $G{L}_n(\mathbb{C})$ and $S{L}_n(\mathbb{C})$, and products of these groups. A diagonal matrix that is unipotent must be the identity, so a torus has no nontrivial unipotent subgroup. For the two matrix groups, let $N$ be a normal unipotent subgroup. The Kolchin theorem supplies a nonzero fixed vector in their standard representation. The subspace fixed by $N$ is invariant under the whole group: for $g\in G$, $n(gv)=g(g^{-1}ng)v=gv$. The standard representation is irreducible, so every vector is fixed by $N$; its faithfulness makes $N$ trivial. A normal unipotent subgroup of a product projects to such a subgroup of each factor, proving the product assertion. By contrast the additive group and a nontrivial upper-unitriangular group are not reductive: their unipotent radicals are the whole group.

Choose a rational Borel subgroup $B=TU$ with rational maximal algebraic torus $T$, and use the Lang map $\mathcal{L}(g)=g^{-1}F(g)$. Conjugating the pair $({}^{g}B,F({}^{g}B))$ by $g^{-1}$ gives $(B,{}^{\mathcal{L}(g)}B)$. Part (a) therefore givesMoreover $\mathcal{L}(gb)=b^{-1}\mathcal{L}(g)F(b)$, and $F(B)=B$, so this inverse image is stable under right multiplication by $B$. The flag variety identification consequently restricts to the required isomorphism

The second printed isomorphism is false as written: it omits a unipotent quotient. Take $G={\mathrm{SL}}_2({\overline{\mathbb{F}}}_q)$ with standard Frobenius endomorphism of an algebraic group and $w=1$. Then $X(1)$ consists of the $q+1$ rational lines in ${\mathbb{F}}_q^2$, so has dimension zero. Every Borel subgroup $B^{\prime }$ has one-dimensional unipotent radical $U^{\prime }$. The Lang map is a finite surjective étale morphism, so ${\mathcal{L}}^{-1}(U^{\prime })$ has dimension one. Quotienting by the finite group $T^{\prime F}$ preserves that dimension. Thus no choices of $T^{\prime }$ and $B^{\prime }$ can make the asserted two varieties isomorphic in this example.

Here is the corrected construction, including the missing quotient. Choose $x$ by the Lang theorem for algebraic groups so that $x^{-1}F(x)=\dot{w}$. DefineThen $F(T^{\prime })=x\dot{w}T{\dot{w}}^{-1}{x}^{-1}=T^{\prime }$, so $T^{\prime }$ is a rational maximal torus contained in $B^{\prime }$. Put $J=U_x\cap U^{\prime }=F^{-1}(U^{\prime })\cap U^{\prime }$. The correct unipotent quotient in a Deligne-Lusztig variety isThe semidirect product denotes the subgroup $J{T}^{\prime F}$ acting on the right. Both $U_x$ and $U^{\prime }$ are normalized by $T^{\prime }$, so the action is well defined. Alsofor $h\in {\mathcal{L}}^{-1}(U^{\prime })$, showing directly that the map lands in the Deligne-Lusztig variety.

To prove surjectivity and identify its fibres, use the intermediate varietyThe right $U$-action follows from the twisted-translation identity for the Lang map. Right multiplication by $T^{F_w}$ also acts, and $Y(\dot{w})/T^{F_w}\cong X(w)$. Indeed if $\mathcal{L}(g)\in B\dot{w}B$, write it as $u_{1}t\dot{w}{u}_2$. Replacing $g$ by $ga$, with $a\in T$, changes its torus factor to $a^{-1}t{F}_w(a)$. Surjectivity of the twisted Lang map on the connected algebraic torus makes this factor one. Once the factor is one, two representatives of the same flag differ by $ua$ with $u\in U$ and $a\in T^{F_w}$. This proves both surjectivity and the asserted finite covering, including its group of fibres.

Now map $Z={\mathcal{L}}^{-1}(U^{\prime })$ to $Y(\dot{w})$ by $h\mapsto hxU$. Given $\mathcal{L}(g)=u_1\dot{w}{u}_2$, replace $g$ by $g{u}_1$. ThenThus $h=g{u}_1{x}^{-1}$ belongs to $Z$ and maps to the original point $gU$. This proves surjectivity. Two preimages have the form $h$ and $hk$, where $k\in U_x$. Since $F(k)\in U^{\prime }$ and $\mathcal{L}(h)\in U^{\prime }$, the identityshows that $hk\in Z$ precisely when $k\in U^{\prime }$, hence precisely when $k\in J$. Therefore the fibres are exactly the right $J$-orbits.

These are isomorphisms of geometric quotients, not just bijections on points. The groups $U_x,U^{\prime }$ contain root subgroups for the same maximal algebraic torus, so their intersection $J$ is smooth. The map $Z\to Y(\dot{w})$ is a smooth surjection with fibre $J$: the differential of the Lang map is invertible, and the kernel of the differential of $h\mapsto hxU$ on $Z$ is the translate of $\mathrm{Lie}(U_x)\cap \mathrm{Lie}(U^{\prime })=\mathrm{Lie}(J)$. This gives $Z/J\cong Y(\dot{w})$. Finally $x{T}^{F_w}{x}^{-1}=T^{\prime F}$, so quotienting the remaining finite torus action gives the displayed corrected formula.

For $w=1$, $J=U_x$ is exactly the affine-line factor in the counterexample. More generally $Z/T^{\prime F}\to X(w)$ retains unipotent fibres of dimension $\dim J$; it becomes the printed isomorphism only when this intersection is trivial. The missing factor is also recorded in the quotient construction in Tiep's lectures, proof of Theorem 5.22.

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.

A unipotent algebraic group admits a faithful linear representation in which every group element is a unipotent matrix. For $E={\mathrm{End}}_Q(X)$, invertibility is the nonvanishing condition ${\prod }_i\det h_i\ne 0$. Therefore ${\mathrm{Aut}}_Q(X)=E^{\times }$ is a nonempty Zariski-open subset of the vector space $E$.

Take a Krull-Schmidt decomposition $X\cong {⨁}_{a=1}^r{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposable modules. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local endomorphism ring. Its residue division algebra is $k$: over an algebraically closed field, every element of a finite-dimensional division algebra has an eigenvalue and hence must be scalar. The semisimple quotient of a module endomorphism algebra consequently gives, for the Jacobson radical $J=J(E)$,is surjective with kernel $1+J$. The nilpotence of $J$ makes $(1+j{)}^{-1}=1-j+j^2-\cdots$ finite and each $1+j$ unipotent. The kernel is closed and normal. Acting on the multiplicity spaces embeds the product of general linear groups back into $E^{\times }$ and splits this quotient. This proves the Levi decomposition of a quiver automorphism groupSince $1+J$ is the unipotent radical, a nonzero $X$ is indecomposable exactly when ${\mathrm{Aut}}_Q(X)/R_u({\mathrm{Aut}}_Q(X))\cong {\mathbb{G}}_m$: the product has a single factor of size one.

For the base change action on quiver representations, the orbit map is $g\mapsto (g_j{x}_{\rho }{g}_i^{-1}{)}_{\rho :i\to j}$. Substituting $g_i=I+ϵu_i$, with ${ϵ}^2=0$, shows its differential isIts kernel is ${\mathrm{End}}_Q(X)$. The stabilizer is smooth because it is open in that vector space. Hence the differential has rank $\dim \mathrm{GL}(\mathbf{n})-\dim {\mathrm{Aut}}_Q(X)=\dim {\mathcal{O}}_X$, and its image is the Zariski tangent space $T_x{\mathcal{O}}_X$. The normal space to a quiver orbit is thereforeThe ambient quiver representation space is an irreducible affine space, and orbits are locally closed. An orbit is open exactly when its dimension equals that ambient dimension, equivalently when ${\mathrm{Ext}}_Q^1(X,X)=0$. This proves that rigid quiver representations have open orbits. Such an orbit is dense and unique, since two nonempty open subsets of an irreducible space intersect.

An algebraic group is connected when its underlying Zariski topology is a connected space. An unipotent algebraic group is a linear algebraic group whose elements, in a faithful matrix realization, satisfy that $u-I$ is nilpotent. A reductive algebraic group, in the connected convention, is a connected linear algebraic group whose unipotent radical is trivial: it has no nontrivial connected normal unipotent algebraic subgroup.

Let $E={\mathrm{End}}_Q(X)$. It is a finite-dimensional associative algebra, and ${\mathrm{Aut}}_Q(X)=E^{\times }$. The latter is the nonempty open subset where the determinant on ${⨁}_i{X}_i$ is nonzero. Since the affine space $E$ is irreducible, this open subset is irreducible and hence connected. It is a linear algebraic group: within $\mathrm{GL}({⨁}_i{X}_i)$ it is cut out by the linear equations of preserving the vertex summands and commuting with the arrow maps. Thus the automorphism group is connected.

By the Krull–Schmidt theorem, write $X\cong {⨁}_{a=1}^s{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposables $M_a$. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local algebra. Its residue division algebra is $k$: every element of a finite-dimensional division algebra over the algebraically closed field has an eigenvalue for left multiplication, and subtracting that scalar gives a noninvertible element, hence zero.

The standard radical description of the endomorphism algebra of a Krull-Schmidt decomposition therefore givesConcretely, maps between nonisomorphic summands lie in the radical, while on each isotypic block one reduces all entries modulo the local radical. The off-diagonal rule follows because a composition $M_a\to M_b\to M_a$ cannot be a unit for $a\ne b$: it would split $M_a$ off the indecomposable $M_b$, forcing an isomorphism. The finite-dimensional radical description then identifies the kernel of this block reduction with $J(E)$. Put $U=1+J(E)$. The Jacobson radical of the finite-dimensional algebra is nilpotent, so every $1+j$ is a unit and is unipotent on ${⨁}_i{X}_i$. Also $U$ is closed, being the translate of the linear subspace $J(E)$, and normal, being the kernel ofThis map has an explicit algebraic section: after fixing the direct-sum decomposition, a matrix $B_a$ acts on the $m_a$ copies by $B_a\otimes 1_{M_a}$. These block actions give a subgroup $L\cong {\prod }_a{\mathrm{GL}}_{m_a}(k)$ with trivial intersection with $U$. Every unit is uniquely a product of an element of $U$ and one of $L$. HenceThis Levi decomposition of a quiver automorphism group holds in every characteristic; the section is constructed directly from the multiplicity spaces.

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.

For a line $⟨e⟩$, complete $(e,f)$ to a symplectic basis with complementary symplectic space $W$. The unipotent radical has elements $u(b,t)$ sending $e$ to $e$, $w$ to $w-B(w,b)e$ and $f$ to $f+b+te$. Its multiplication is $(b,t)(b^{\prime },t^{\prime })=(b+b^{\prime },t+t^{\prime }+B(b,b^{\prime }))$, giving order $q^{2m-1}$. Its commutator is central with parameter $2B(b,b^{\prime })$. It is Heisenberg-type in odd characteristic and abelian in characteristic two. The full stabilizer also scales $e,f$ inversely and acts symplectically on $W$.