GENERALISED GELFAND–GRAEV REPRESENTATIONS IN BAD CHARACTERISTIC ?

Let G be a connected reductive algebraic group defined over a finite field with q elements. In the 1980’s, Kawanaka introduced generalised Gelfand–Graev representations of the finite group GFq\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ G\left({\mathbbm{F}}_q\right) $$\end{document}, assuming that q is a power of a good prime for G. These representations have turned out to be extremely useful in various contexts. Here we investigate to what extent Kawanaka’s construction can be carried out when we drop the assumptions on q. As a curious by-product, we obtain a new, conjectural characterisation of Lusztig’s concept of special unipotent classes of G in terms of weighted Dynkin diagrams.


Introduction
Let p be a prime and k = F p be an algebraic closure of the field with p elements.Let G be a connected reductive algebraic group over k and assume that G is defined over the finite subfield F q ⊆ k, where q is a power of p.Let F : G → G be the corresponding Frobenius map.We are interested in studying the representations (over an algebraically closed field of characteristic 0) of the finite group G F = {g ∈ G | F (g) = g}.
Assuming that p is a "good" prime for G, Kawanaka [14], [15], [16] described a procedure by which one can associate with any unipotent element u ∈ G F a representation Γ u of G F , obtained by induction of a certain one dimensional representation from a unipotent subgroup of G F .If u is the identity element, then Γ 1 is the regular representation of G F ; if u is a regular unipotent element, then Γ u is a Gelfand-Graev representation as defined, for example, in [2, §8.1] or [34, §14].For arbitrary u, the representation Γ u is called a generalised Gelfand-Graev representation (GGGR for short); it only depends on the G F -conjugacy class of u.
A fundamental step in understanding the GGGRs is achieved by Lusztig [21] where the characters of GGGRs are expressed in terms of characteristic functions of intersection cohomology complexes on G.In [21] it is assumed that p is sufficiently large; in [35] it is shown that one can reduce these assumptions so that everything works as in Kawanaka's original approach.These results have several consequences.By [11] the characters of the various Γ u span the Z-module of all unipotently supported virtual characters of G F .In addition to the original applications in [14], [15], [16], GGGRs have turned out to be very useful in various questions concerning ℓ-modular representations of G F where ℓ is a prime not equal to p; see, e.g., [12], [5].Thus, it seems desirable to explore the possibilities for a definition without any restriction on p, q.These notes arose from an attempt to give such a definition.Recall that p is "good" for G if p is good for each simple factor involved in G; the conditions for the various simple types are as follows.
A n : no condition, B n , C n , D n : p = 2, G 2 , F 4 , E 6 , E 7 : p = 2, 3, E 8 : p = 2, 3, Easy examples indicate that one can not expect a good definition of GGGRs for all unipotent elements of G F .Instead, it seems reasonable to restrict oneself to those unipotent classes which "come from characteristic 0", where the classical Dynkin-Kostant theory is available; see Section 2. This is also consistent with the picture presented by Lusztig [23], [24], [26], [27] for dealing with unipotent classes in small characteristic.Based on this framework, we formulate in Definition 3.4 some precise conditions under which it should be possible to define GGGRs for a given unipotent class.Of course, these conditions will be satisfied if p is a good prime for G, and lead to Kawanaka's original GGGRs; so then the question is how far we can go beyond this.Our answer to this question is as follows.
An essential feature of GGGRs is that they are very closely related to the "unipotent supports" of the irreducible representations of G F , in the sense of Lusztig [21].(In a somewhat different way, and without complete proofs, this concept appeared under the name of "wave front set" in Kawanaka [16].)Let C • be the set of unipotent classes of G which arise as the unipotent support of some irreducible representation of G F , or of G F n for some n 1. (Thus, since G = n 1 G F n , the set C • only depends on G but not on the particular Frobenius map F .)If p is a good prime for G, then it is known that C • is the set of all unipotent classes of G.In general, all classes in C • indeed "come from characteristic 0".Based on the methods in [25], an explicit description of the sets C • , for G simple and p bad, is given in Proposition 4.3.This result complements the general results on "unipotent support" in [13], [21] and may be of independent interest.Now, extensive experimentation (with computer programs written in GAP [9]) lead us to the expectation, formulated as Conjecture 4.4, that our conditions in Definition 3.4 will work for all the classes in C • , without any restriction on p, q.In Section 5, we work out in detail the example where G is of type F 4 .Our investigations also suggest a new characterisation of Lusztig's special unipotent classes; see Conjecture 4.10 and Corollary 5.11.
These notes merely contain examples and conjectures; nevertheless we hope that they show that the story about GGGRs is by no means complete and that there is some evidence for a further theory in bad characteristic.

Weighted Dynkin diagrams
We use Carter [2] as a general reference for results on algebraic groups and unipotent classes.Let G, k, p, . . .be as in Section 1.Also recall that G is defined over the finite field F q ⊆ k, with corresponding Frobenius map F : G → G.We fix an F -stable maximal torus T ⊆ G and an F -stable Borel subgroup B ⊆ G containing T .We have B = U ⋊ T where U is the unipotent radical of B. Let Φ be the set of roots of G with respect to T and Π ⊆ Φ be the set of simple roots determined by B. Let Φ + and Φ − be the corresponding sets of positive and negative roots, respectively.Let g = Lie(G) be the Lie algebra of G. Then G acts on g via the adjoint representation Ad : G → GL(g).

2.1.
For each α ∈ Φ, we have a corresponding homomorphism of algebraic groups for all α ∈ Φ, where G der denotes the derived subgroup of G. On the level of g, we have a direct sum decomposition where t = Lie(T ) is the Lie algebra of T and g α is the image of the differential d 0 x α : k → g; furthermore, Ad(t)(y) = α(t)y for t ∈ T and y ∈ g α .We set Then e α = 0 and g α = ke α .(For all this see, e.g., [32, §8.1].) 2.2.For α ∈ Φ, we can write uniquely α = β∈Π n β β where n β ∈ Z for all β ∈ Π.Then ht(α) := β∈Π n β is called the height of α.We fix once and for all a total ordering of Φ + which is compatible with the height, that is, if α, β ∈ Φ + are such that α β, then ht(α) ht(β).Then every u ∈ U has a unique expression u = α∈Φ + x α (u α ) where u α ∈ k (and the product is taken in the given order of Φ + ).Let α, β ∈ Φ + , α = β.Let u, v ∈ k.Then we have Chevalley's commutator relations where the constants C α,β,i,j ∈ k only depend on α, β, i, j but not on u, v (and, again, the product on the right hand side is taken in the given order of Φ + ).Furthermore, if α (Since all U α , α ∈ Φ, are contained in the semisimple algebraic group G der , this follows from [34,Lemma 15 (p. 22) and Remark (p.64)].) 2.3.It will also be convenient to fix some notation concerning the action of the Frobenius map F : G → G.By the results in [34, §10], there exist a permutation τ : Φ → Φ and signs ǫ α = ±1 (α ∈ Φ) such that Here, we can assume that τ (Π) = Π and ǫ ±β = 1 for all β ∈ Π.Now F also induces a Frobenius map on the Lie algebra g which we denote by the same symbol.We have F (uy) = u q F (y) for all u ∈ k and y ∈ g; furthermore, Finally, for g ∈ G and y ∈ g, we have Ad(F (g))(F (y)) = F (Ad(g)(y)).

2.4.
Let G 0 be a connected reductive algebraic group over C of the same type as G; let g 0 = Lie(G 0 ) be its Lie algebra.Then, by the classical Dynkin-Kostant theory (see, e.g., [2, §5.6]), the nilpotent Ad(G 0 )-orbits in g 0 are parametrized by a certain set ∆ of so-called weighted Dynkin diagrams, i.e., maps Furthermore, these nilpotent orbits in g 0 are naturally in bijection with the unipotent classes of G 0 (see [2, §1.15]).If G 0 is a simple algebraic group, then the corresponding set ∆ of weighted Dynkin diagrams is explicitly known in all cases; see [2, §13.1] and the references there.(Several examples will be given below.)For each d ∈ ∆, we denote by O d the corresponding nilpotent orbit in g 0 and set . This is a very useful invariant for distinguishing nilpotent orbits.(The number b d is also the dimension of the variety of Borel subgroups of G 0 containing an element in the unipotent class corresponding to O d ; see [2, §1.15, §5.10].).

Let us fix
Thus, as in [15, §2.1], we obtain a grading g = i∈Z g(i); note that we do have [g(i), g(j)] ⊆ g(i + j) for all i, j ∈ Z.For any i 0, we also set g( i) := j i g(j).Furthermore, we define subgroups of G as follows.
Then P is a parabolic subgroup of G with unipotent radical U 1 and Levi decomposition P = U 1 ⋊ L. The Lie algebra of P is given by p := Lie(P ) = g( 0) ⊆ g.More generally, for any integer i 1, we set Thus, we obtain a chain of subgroups P ⊇ U 1 ⊇ U 2 ⊇ U 3 ⊇ . ..; using Chevalley's commutator relations, one immediately sees that each U i is a normal subgroup of P and that U i /U i+1 is abelian.

2.6.
Let us fix a weighted Dynkin diagram d ∈ ∆ as above.For any integer i 1, we have a corresponding subgroup U i ⊆ G and a corresponding subspace g(i) ⊆ g.Following Kawanaka [15, (3.1.1)],we define a map as follows.Let u ∈ U 1 .As in 2.2, we have a unique expression where the product is taken in the given order on Φ + .Then we set Proof.This is a rather straightforward application of Chevalley's commutator relations.For (b), we use the fact that [e α , e β ] = C α,β,1,1 e α+β if α+β ∈ Φ; see 2.2.For (c), we use the formulae in 2.3.We omit further details.
2.8.The general idea for defining GGGRs corresponding to a fixed d ∈ ∆ is as follows.(In the following discussion we avoid any reference to the characteristic of k.) First of all, we assume that d is invariant under the permutation τ : Φ → Φ induced by F .Consequently, all the subgroups P , U i (i 1) of G are F -stable and all the subspaces g(i) (i 0) are F -stable.
Let us fix a non-trivial character ψ : F + q → C × .Let us also consider a linear map λ : g(2) → k defined over F q , that is, we have λ(F (y)) = λ(y) q for all y ∈ g(2).Then Lemma 2.7(a) shows that , is a group homomorphism and so, by Lemma 2.7(c), we also obtain a group homomorphism We shall require that λ is in "sufficiently general position" (where this term will have to be further specified; see Definition 3.4 below).Let us assume that this is the case.If g(1) = {0}, then the GGGR corresponding to d, λ will simply be given by the induced representation Now consider the case where g(1) = {0}.Since [g(1), g(1)] ⊆ g(2), we obtain a well-defined alternating bilinear form Assume also that the radical of this bilinear form is zero.Then we choose an F -stable Lagrangian subspace in g(1) and pull back this subspace to an F -stable subgroup U 1.5 ⊆ U 1 via the map f .Using Lemma 2.7(b) we see that ker(χ λ ) is normal in U F 1.5 and U F 1.5 / ker(χ λ ) is an abelian p-group.(See also the proof of [15,Lemma 3.1.9].)So we can extend χ λ to a character χλ : U F 1.5 → C × .In this case, the GGGR corresponding to d, λ will be given by the induced representation which shows that the definition of Γ d,λ does not depend on the choice of the Lagrangian subspace or the extension χλ of χ λ (cf.[14, 1.3.6],[15, 3.1.12]).
In Kawanaka's set-up [14, §1.2], [15, §3.1], the above assumption on the radical of σ λ is always satisfied.(See also Remark 3.5 below.)Our plan for the definition of GGGRs in bad characteristic is to follow the above general procedure, but we have to find out in which situations this still makes sense at all.The following two examples show that there is a serious issue concerning the radical of σ λ when g(1) = {0}.
2.10.To simplify the notation for matrices, we define antidiag(x 1 , . . ., x n ) to be the n × n-matrix with entry x i at position (i, n + 1 − i) for 1 i n, and entry 0 otherwise.Thus, for example, where Π = {α, β}; here, α is a short simple root and β is a long simple root.By [2, p. 401], there are 5 weighted Dynkin diagrams d ∈ ∆, where: ).The determinant of this matrix is ±9x 4  1 .Hence, if p = 3, then there is no λ such that the radical of σ λ is zero.On the other hand, if p = 3, then the radical of σ λ is zero whenever x 1 = λ(e 3α+2β ) = 0.

Nilpotent and unipotent pieces
We keep the set-up of the previous section.Given a weighted Dynkin diagram d ∈ ∆, our main task is to find suitable conditions under which a linear map λ : g(2) → k may be considered to be in "sufficiently general position" (cf.2.8).For this purpose, we use Lusztig's framework [23]- [27] for dealing with unipotent elements in G and nilpotent elements in g when p (the characteristic of k) is small.
In general, H d is the union of C d and a finite number of unipotent classes of dimension strictly smaller than dim C d .We will say that the unipotent classes {C d | d ∈ ∆} "come from characteristic 0".(Alternatively, the latter notion can be defined using the Springer correspondence, see [23, 1.3, 1.4], or the results of Spaltenstein [31]; see also [13, §2].All these definitions agree as can be checked using the explicit knowledge of the unipotent classes and the Springer correspondence in all cases.)3.2.We recall some further notation and some results from [27, §2].There is a coadjoint action of G on the dual vector space g * which we denote by g.ξ for g ∈ G and ξ ∈ g * ; thus, (g.ξ)(y) = ξ(Ad(g −1 )(y)) for all y ∈ g.We denote by G ξ the stabilizer of ξ ∈ g * under this action.As in [27], an element ξ ∈ g * is called nilpotent if there exists some g ∈ G such that the Lie algebra of the Borel subgroup B ⊆ G is contained in Ann(g.ξ).Let For any Y ⊆ g, we denote Ann(Y ) := {ξ ∈ g * | ξ(y) = 0 for all y ∈ Y }.Let us fix a weighted Dynkin diagram d ∈ ∆.As in 2.5, we have a corresponding grading g = i∈Z g(i).In order to indicate the dependence on d, we shall now write g d (i) = g(i) for all i ∈ Z; similarly, we write P d = P for the corresponding parabolic subgroup of G. Now, we also have a grading g * = j∈Z g d (j) * where we set for any j ∈ Z.
We note that the subspace g d ( j) * := j ′ ∈Z: j ′ j g d (j ′ ) * is stable under the coadjoint action of P d .Let

3.3.
In order to apply the above results to the situation in Section 2, we need a mechanism by which we can pass back and forth between the vector spaces g d (i) and g d (i) * .If there exists a G-equivariant vector space isomorphism g ∼ → g * , then there is a canonical way of doing this, as explained in [27, 2.3].However, such an isomorphism will not always exist.To remedy this situation, we follow Kawanaka [14, §1.2], [15, §3.1] and fix an F q -opposition automorphism g → g, y → y † .This is a linear isomorphism, defined over F q , such that t † = t and e † α = ±e −α for all α ∈ Φ. (See also [35,Lemma 5.2].)If ξ ∈ g * , then we define ξ † ∈ g * by ξ † (y) := ξ(y † ) for y ∈ g.Definition 3.4.Let d ∈ ∆ be a weighted Dynkin diagram and consider the corresponding grading g = i∈Z g d (i).Let λ : g d (2) → k be a linear map.We regard λ as an element of g * by setting λ equal to zero on g d (i) for all i = 2.We say that λ is in "sufficiently general position" if the following conditions hold.
(K1) We require that , then we also require that the radical of the corresponding alternating form σ λ : g d (1) × g d (1) → k in 2.8 is zero.Note that (K1), (K2) only refer to the algebraic group G, but not to the Frobenius map F .If (K1), (K2) hold and if λ is defined over F q , then we can follow the procedure in 2.8 and define the corresponding GGGR Γ d,λ of the finite group G F .Remark 3.5.Kawanaka's work [14], [15] fits into this setting as follows.Assume that p is a good prime for G and that there exists a non-degenerate, symmetric and G-invariant bilinear form κ : g×g → k.We also need to make a certain technical assumption on the isogeny type of the derived subgroup of G. (For details see [35, 3.22], [29].)Let d ∈ ∆.Then there is a dense open orbit under the adjoint action of P d on g d (2).Let e be an element of this orbit and define a linear map λ e : g d (2) → k as follows.).The entries of G λ are given as follows.We set x l := λ(e γ l ) for 1 l m.For 1 i, j n, we define an element ν ij ∈ k as follows.If β i + β j ∈ Φ, then ν ij := 0. Otherwise, there is a unique l(i, j) ∈ {1, . . ., m} and some ν ij ∈ k such that [e β i , e β j ] = ν ij e γ l(i,j) .Then we have In order to work this out explicity, we may assume without loss of generality that G is semisimple (since U α ⊆ G der for all α ∈ Φ; see 2.1).But then, by [34,Remark (p. 64)], the structure constants of the Lie algebra g are obtained from those of a Chevalley basis of g 0 by reduction modulo p.Thus, we can explicitly determine the elements ν ij , via a computation inside g 0 .Using one of the two canonical Chevalley bases in [10, §5] (the two bases only differ by a global sign), one can even avoid the issue of choosing certain signs.Hence, there is a purely combinatorial algorithm for computing G λ , and this can be easily implemented in GAP [9].In particular, we have: ( * ) Up to a global sign, the Gram matrix G λ only depends on the root system Φ and the values λ(e α ) (α ∈ Φ 2 ).
The radical of σ λ is zero if and only if det(G λ ) = 0. Now we notice that this determinant is given by evaluating a certain m-variable polynomial at x 1 , . . ., x m .In particular, we see that condition (K2) is an "open" condition: either there is no λ at all for which (K2) holds, or (K2) holds for a non-empty open set of linear maps λ : g d (2) → k.Combining this with the discussion concerning (K1) in Remark 3.6, we immediately obtain the following conclusion.With these preparations, we now obtain our first example where bad primes exist but (K2) holds without any restriction on the field k.
In the above examples, it was easy to compute the determinant of the Gram matrix G λ .However, we will encounter examples below where the computation of det(G λ ) becomes a very serious issue.

Unipotent support
We are now looking for a unifying principle behind the various examples that we have seen so far.In Conjecture 4.4 below, we propose such a unification.First we need some preparations.

4.1.
Let Irr(G F ) be the set of complex irreducible representations of G F (up to isomorphism).Then there is a canonical map defined in terms of the notion of "unipotent support".To explain this, we need to introduce some notation.Let C be an Since F (u i ) = u i , the Frobenius map F induces an automorphism of A(u i ) which we denote by the same symbol.Let A(u i ) F be the group of fixed points under F .Then we set AV(ρ, C) := for any ρ ∈ Irr(G F ).Note that this does not depend on the choice of the representatives u i .Now the desired map is obtained as follows.Let ρ ∈ Irr(G F ) and set a ρ := max{dim C | AV(ρ, C) = 0} (where the maximum is taken over all F -stable unipotent classes C of G).By the main results of [13], [21], there is a unique C such that dim C = a ρ and AV(ρ, C) = 0.This C will be denoted by C ρ and called the unipotent support of ρ.
By [13, Remark 3.9], it is known that C ρ comes from characteristic 0 (see 3.1) and, hence, equals C dρ for a well-defined weighted Dynkin diagram d ρ ∈ ∆.We set k,F consists precisely of those weighted Dynkin diagrams for which the corresponding unipotent class of G occurs as the unipotent support of some irreducible representation of G F .In order to obtain a subset of ∆ which only depends on G and not on the choice of the particular Frobenius map F , we set (Note that, if F 1 : G → G is another Frobenius map, then there always exist integers n, n 1 1 such that is precisely the set of unipotent classes mentioned at the end of Section 1.
Remark 4.2.Recall from Lusztig [19], [22] the notion of special unipotent classes.The precise definition of these classes is not elementary; it involves the Springer correspondence and the notion of special representations of the Weyl group of G.By [13, Prop.4.2], an F -stable unipotent class is special if and only if it is the unipotent support of some unipotent representation of G F .Thus, we conclude that special unipotent classes come from characteristic 0 and we have: Proof.This follows by analogous methods as in [25].The special feature of the case where G is of classical type and p = 2 is the fact that then the centraliser of a semisimple element is a Levi subgroup of some parabolic subgroup (see, e.g., [1, §4]).If G is of exeptional type, then one uses explicit computations completely analogous to those in [25, §7]; here, one also needs Lübeck's tables [18] concerning possible centralisers of semisimple elements in these cases.We omit further details.k be invariant under the permutation τ : Φ → Φ induced by F .Then there exist linear maps λ : g d (2) → k which are defined over F q and are in "sufficiently general position" (see Definition 3.4).Hence, following the general procedure described in 2.8, we can define the corresponding GGGRs Γ d,λ of G F .By Remarks 3.5, 3.7( * ) and 4.2(b), the conjecture holds for all d ∈ ∆ • k = ∆ if p is a good prime for G.In particular, it holds when G is of type A n .We will now discuss a number of examples supporting the conjecture in cases where p is a bad prime.As already explained in the previous section, the main issue is the validity of condition (K2) in Definition 3.4.
Example 4.5.(a) Let G = Sp 4 (k) and assume that k has characteristic 2. By Proposition 4.3, or by inspection of the known character table of G F (see [6]), we note that the unipotent class corresponding to the weighted Dynkin diagram d 0 in Example 2.9 is not the unipotent support of any irreducible character of G F .Thus, d 0 ∈ ∆ • k and so this critical case does not enter in the range of validity of Conjecture 4.4.In fact, ∆  1.We have d 0 (α 4 ) = d 0 (α 7 ) = 1 and d 0 (α i ) = 0 for i = 4, 7; see [2, p. 406].By Table 1, we have ∆ → k be a linear map and consider the Gram matrix G λ of the alternating form σ λ .We claim: (a) If p = 5, then det(G λ ) = 0 for some λ : First, we find that dim g d 0 (1) = 24 and dim g d 0 (2) = 21.As explained in Remark 3.7, we then explicitly work out G λ .We have where f ij are certain polynomials with integer coefficients in 21 indeterminates.In order to verify (a), we argue as follows.If p > 5, then (a) holds by Remark 3.5.If p = 2, 3, then we simply run through all vectors of values (x 1 , . . ., x 21 ) ∈ {0, 1} 21 (starting with the vector 1, 1, . . ., 1 and then increasing step by step the number of zeroes) until we find one such that det(G λ ) = 0.It turns out that this search is successful just after a few steps.The verification of (b) is much harder.Let p = 5 and denote by fij the reduction of f ij modulo p. Then we need to check that det( fij ) = 0.It seems to be practically impossible to compute such a determinant directly, or even just the rank.(Using special values of the x i as above, one quickly sees that the rank of ( fij ) is at least 22.) Now, since G λ is anti-symmetric, we can use the fact that the desired determinant is given by Pf( fij ) 2 , where Pf( fij ) denotes the Pfaffian of the matrix ( fij ); see, for example, [4], [17].(I am indebted to Ulrich Thiel for pointing this out to me.)A simple recursive algorithm (via row expansion, as in [4, 1.5]) is sufficient to compute Pf( fij ) = 0 in this case and yields (b).(Over Z, the Pfaffian of (f ij ) is a non-zero polynomial which is a linear combination of 1386 monomials in 21 indeterminates, where all coefficients are divisible by 5.) The  1.We have 2) → k be a linear map and consider the Gram matrix G λ of the alternating form σ λ .Here, we find that dim g d 1 (1) = 22 and dim g d 1 (2) = 18.Using computations as in the previous example, we obtain: (a) If p = 2, 3, then det(G λ ) = 0 for some λ : Finally, we state the following conjecture concerning special unipotent classes.As in 2.4, let G 0 be a connected reductive algebraic group over C of the same type as G; let g 0 be its Lie algebra.For d ∈ ∆ and i = 1, 2, we set As in 2.8, given a homomorphism λ : g Z,d (2) → Z, we obtain an alternating form σ λ : g Z,d (1) × g Z,d (1) → Z and we may consider its Gram matrix with respect to the Z-basis {e α | α ∈ Φ 1 } of g Z,d (1).If this Gram matrix has determinant ±1, then we say that σ λ is non-degenerate over Z.Note that, if g d (1) = {0}, then we certainly have d ∈ ∆ spec .(This easily follows from [28, Prop.1.9(b)].)Hence, in order to verify the above conjectures for a given example, it is sufficient to consider the cases where g d (1) = {0}.This will be further discussed in the following section.

A worked example: type F 4
In this section, we work out in detail the example where G is of type F 4 .We believe that the results of our computations are strong evidence for the truth of Conjectures 4.4 and 4.10; the discussion of the various cases will also provide a good illustration of the computational issues involved.Let Π = {α 1 , α 2 , α 3 , α 4 } be a set of simple roots such that the Dynkin diagram of G looks as follows: There are eight weighted Dynkin diagrams which satisfy g d (1) = {0}.We now consider these eight cases in detail, where we just focus on the validity of condition (K2) in Definition 3.4.(In particular, the Frobenius map F : G → G will not play a role in this section.) . We have det(G λ ) = 64x 14  1 .Hence, if p = 2, then the radical of σ λ is not zero.If p = 2, then the radical is zero whenever x 1 = 0.
In principle, we could work out det(G λ ) and then try to find out for which values of x 1 , . . ., x 8 it is non-zero.However, this determinant is already quite complicated; it is a linear combination of 34 monomials in x 1 , . . ., x 8 .But we can just notice that, if we set x 4 := 1 and x i := 0 for all i = 4, then det(G λ ) = 1.So the radical of σ λ will be zero for this choice of λ, and this works for any field k.Let λ : g d (2) → k be any linear map, and denote by x 1 , . . ., x 6 the values of λ on the 6 basis vectors of g d (2) (ordered as above).Then we obtain Here we notice that, if we set x 3 := 1, x 4 := 1 and x i := 0 for i = 3, 4, then det(G λ ) = 1.Hence, the radical of σ λ is zero for this choice of λ, and this works for any field k.Let λ : g d (2) → k be any linear map.Let x 1 , . . ., x 9 be the values of λ on the 9 basis vectors of g d (2) (ordered as above).Then we obtain We have det(G λ ) = 16(x 1 x 5 x 9 − x 1 x 7 x 8 − x 2 x 3 x 9 + x 2 x 6 x 7 + x 3 x 4 x 8 − x 4 x 5 x 6 ) 2 .So, if p = 2, then det(G λ ) = 0 (for any λ).If p = 2, then we notice that det(G λ ) = 16 for x 4 := 1, x 5 := 1, x 6 := 1 and x i := 0 for i = 4, 5, 6.Hence, the radical of σ λ is zero for this choice of λ.Let λ : g d (2) → k be any linear map.Let x 1 , . . ., x 6 be the values of λ on the 6 basis vectors of g d (2) (ordered as above).Then we obtain G λ = ±2x 6 • antidiag(−1, 1, −1, 1).We have det(G λ ) = 16x 4  6 .Hence, if p = 2, then the radical of σ λ is not zero.If p = 2, then the radical is zero whenever x 6 = 0. Let λ : g d (2) → k be any linear map.Let x 1 , . . ., x 5 be the values of λ on the 5 basis vectors of g d (2) (ordered as above).Then we obtain We have det(G λ ) = 9(x 1 x 2 4 x 5 − x 2 x 3 x 2 4 ) 2 .Hence, if p = 3, then the radical of σ λ is not zero.Now assume that p = 3.Then we notice that det(G λ ) = 9 for x 1 := 1, x 4 := 1, x 5 := 1 and x i := 0 for i = 2, 3.So the radical is zero for this choice of λ.Let λ : g d (2) → k be any linear map.Let x 1 , . . ., x 5 be the values of λ on the 5 basis vectors of g d (2) (ordered as above).Then we obtain 3 ) 2 .Hence, if p = 2, then the radical of σ λ is not zero.Now assume that p = 2. Then we notice that det(G λ ) = 4 for x 1 := 1, x 5 := 1 and x i := 0 for i = 2, 3, 4.So the radical of σ λ is zero for this choice of λ.Let λ : g d (2) → k be any linear map.Let x 1 , x 2 , x 3 be the values of λ on the 3 basis vectors of g d (2) (ordered as above).Then we obtain Hence, we see that the radical of σ λ is zero for any linear map λ : g d (2) → k such that λ(e 1110 ) = x 2 = 0, and this works for any field k.
Similar computations can, of course, be performed for other types of groups.The results are summarized as follows.For types G 2 , F 4 , we can see all this immediately from the results of the computations in Example 2.11 and in 5.1-5.8, by comparing with the entries in the last column of Table 2.However, there are critical cases in type E 8 (for example, A 2 +3A 1 , 2A 2 +A 1 , 2A 2 +2A 1 , A 3 +2A 1 , A 3 +A 2 +A 1 ) where the computation of the Pfaffian appears to be practically impossible.In these cases, some more sophisticated computational methods are required.(1) = {0}, then we must show that there exists a homomorphism λ : g Z,d (2) → Z such that σ λ : g Z,d (1) × g Z,d (1) → Z is non-degenerate over Z.The verification is similar to that in 5.9(a), but now we work over Z. Again, we systematically run through all possibilities where λ(e α ) ∈ {0, 1} for α ∈ Φ 2 , until we find one such that the Gram matrix of σ λ has determinant equal to 1.Even in type E 8 , this just works in a few seconds.Conversely, assume that either g Z,d (1) = {0}, or there exists a homomorphism λ : g Z,d (2) → Z such that σ λ : g Z,d (1) × g Z,d (1) → Z is non-degenerate over Z.If g Z,d (1) = {0}, then d ∈ ∆ spec (see again the remark at the end of Section 4).Now assume that g Z,d (1) = {0}.By reduction modulo p, we obtain a linear map λ k : g d (2) → k and a corresponding alternating form σ λ k : g d (1) × g d (1) → k.Since σ λ is non-degenerate over Z, the radical of σ λ k will be zero and so d ∈ ∆ • k , since we already know that Conjecture 4.9 is true for G.Note that this holds for all choices of k.So Table 1 shows that d ∈ ∆ spec .Remark 5.12.Assume that G is simple of type A n .Then there are no bad primes for G, and all unipotent classes of G are special.Now Conjecture 4.9 is known to hold in this case; see Remarks 3.5 and 3.7( * ).As far as Conjecture 4.10 is concerned, it remains to show that if d ∈ ∆ is such that g Z,d (1) = {0}, then there exists a homomorphism λ : g Z,d (2) → Z such that σ λ : g Z,d (1) × g Z,d (1) → Z is non-degenerate over Z.At the moment, we do not see a general argument but, by similar methods as above, we have at least checked this holds for 2 n 15.
In order to verify Conjecture 4.4 in full, one would also need a description of the sets g d (2) * ! ; this will be discussed elsewhere.(For G of type A n , B n , C n , D n , such a description is available from [27], [36].)Furthermore, one would need to check whether or not there exists some λ which is defined over F q and is in sufficiently general position.-It would be highly desirable to find a more conceptual explanation for all this.

3. 1 .
Let U be the variety of unipotent elements of G.In [23, §1], Lusztig introduced a natural partition U = d∈∆ H d where each H d is an irreducible locally closed subset stable under conjugation by G.A general, case-free proof for the existence of this partition was given by Clarke-Premet [3, Theorem 1.4].The sets {H d | d ∈ ∆} are called the unipotent pieces of G.In each such piece H d , there is a unique unipotent class

Corollary 3 . 8 .
Let d ∈ ∆ and assume that g d (1) = {0}.Then either there is a non-empty open set of linear maps λ : g d (2) → k in "sufficiently general position", or there is no such linear map at all.

Proposition 4 . 3 .
(a) ∆ spec ⊆ ∆ • k , where ∆ spec denotes the set of all d ∈ ∆ such that the corresponding unipotent class C d of G is special.Explicit descriptions of the sets ∆ spec are contained in the tables in [2, §13.4].Using the above concepts, one can give an alternative description of the map ρ → C ρ ; see [20, 13.4], [21, 10.9], [25, §1], [13, §3.C].This alternative description allows one to compute ∆ • k explicitly, without knowing any character values of G F .In particular, this yields the following statement:(b) If p is a good prime for G, then ∆ • k = ∆.This was first stated (in terms of the alternative description of ρ → C ρ and for p large) as a conjecture in[19,  §9]; see also[20, 13.4].A full proof eventually appeared in[25, Theorem 1.5].Assume that G is simple and p is a bad prime

Conjecture 4 . 4 .
Let d ∈ ∆ • → k.(In (b), there are λ such that G λ has rank 20.)The examples suggest the following characterisation of the set ∆ • k .Conjecture 4.9.Let d ∈ ∆.Then d ∈ ∆ • k if and only if either g d (1) = {0}, or there exists a linear map λ : g d (2) → k such that the radical of σ λ is zero.

Conjecture 4 . 10 .
With the above notation, let d ∈ ∆.Then we have d ∈ ∆ spec if and only if either g Z,d (1) = {0}, or there exists a homomorphism λ : g Z,d (2) → Z such that σ λ is non-degenerate over Z.

5. 9 .
Assume that the characteristic of k is a bad prime for G. Let d ∈ ∆ be such that g d (1) = {0}.Consider the following two statements.(a)If d ∈ ∆ • k , then there exist linear maps λ :g d (2) → k such that det(G λ ) = 0 and λ(e α ) ∈ {0, 1} for all α ∈ Φ 2 .(b) If d ∈ ∆ • k , then det(G λ ) = 0 for all linear maps λ : g d (2) → k.Let G be simple of exceptional type G 2 , F 4 , E 6 , E 7 or E 8 .Then (a) can be verified by exactly the same kind of computations as in Example 4.7(a), by systematically running through all possibilities where λ(e α ) ∈ {0, 1} for α ∈ Φ 2 , until we find one such that det(G λ ) = 0.Even in type E 8 , this just works in a few seconds.The verification of (b) is much harder.As in the verification of Example 4.7(b), we need to show that the determinant of a certain matrix with entries in a polynomial ring over F p is 0. Except for some cases in type E 8 , it is sufficient to use the Pfaffian of that matrix, as in Example 4.7(b).

Proposition 5 . 10 (
).Let G be of type E 8 .Then the statement in 5.9(b) holds for all d ∈ ∆ \ ∆ • k .(The proof relies on Groebner basis techniques.)Corollary 5.11.If G is simple of exceptional type G 2 , F 4 , E 6 , E 7 or E 8 , then Conjectures 4.9 and 4.10 hold for G. Proof.First consider Conjecture 4.9.Let d ∈ ∆ • k .If g d (1) = {0}, then we must show that there exists some λ : g d (2) → k such that det(G λ ) = 0.If p (the characteristic of k) is a good prime for G, then this holds by Remarks 3.5 and 3.7( * ).If p is a bad prime, then this holds since 5.9(a) is known to hold.Conversely, assume that either g d (1) = {0}, or there exists a linear map λ :g d (2) → k such that det(G λ ) = 0.If g d (1) = {0}, then d ∈ ∆ spec ⊆ ∆ • k ,as already remarked at the end of Section 4. If g d (1) = {0}, then we have d ∈ ∆ • k by 5.9(b) and Proposition 5.10.Now consider Conjecture 4.10.Let d ∈ ∆ spec .If g Z,d Then σ * d is stable under the coadjoint action of P d on g d ( 2) * .Finally, let σ * d ⊆ g * be the union of the orbits of the elements in σ d * under the coadjoint action of G. Then ξ → ξ is a map Ψ g By [27, Theorem 2.2], the map Ψ g * is a bijection if the adjoint group of G is a direct product of simple groups of types A, C and D. By the main result of [36], this also holds if there is a direct factor of type B. In the remarks just following [27, Theorem 2.2], Lusztig expresses the expectation that Ψ g * is a bijection without any restriction on G.By Clarke-Premet [3, Theorem 7.3] it is known that Ψ g * is always surjective.

Table 1 .
in Table 1.(In Table 1, we list all d ∈ ∆ \ ∆ spec ; for each such d, the last column gives the condition on p such that d ∈ ∆ • k .)The sets ∆ • k \ ∆ spec for G of exceptional type The situation is similar for G of type G 2 .Consider the two weighted Dynkin diagrams with g d (1) = {0} in Example 2.11.By Table 1, or by inspection of the known character table of G F (see [8] for p = 2 and [7] for p = 3), we see that ∆ • k = ∆ \ {d} where d is as in Example 2.11(a) if p = 2, and d is as in Example 2.11(b) if p = 3. Example 4.6.Let again G be of type D 4 and return to the discussion in Example 3.9.The special unipotent classes are explicitly described in [2, p. 439]; there is only one class which is not special (it corresponds to elements with Jordan blocks of sizes 3, 2, 2, 1), and this is precisely the one considered in Example 3.9(b).But, by Proposition 4.3, the corresponding weighted Dynkin diagram does not belong to ∆ • k if p = 2. Example 4.7.Let G be of type E 8 with diagram α 1 Let d 0 ∈ ∆ correspond to the class denoted A 4 +A 3 in Table class A 4 +A 3 in type E 8 also plays a special role in [30, §4.2].(I thank Alexander Premet for pointing this out to me.) Example 4.8.Let again G be of type E 8 , with diagram as above.Let d 1 ∈ ∆ correspond to the class denoted A 5 +A 1 in Table

Table 2 .
By[2, p. 401], there are 16 weighted Dynkin diagrams in ∆; together with some additional information, these are printed in Table 2. (The entries in the last column are determined by Remark 4.2(a) and Table 1.) Unipotent classes in type F 4