Chapter 5 Construction of the spin groups
5.2.3 Proof of correctness of the generators
The following lemma and its corollary are straightforward but will be used frequently. Lemma 5.2.2. Let A = Id+a P (i,j)∈X Ei,j, B = Id+b P (k,l)∈Y Ek,l, where X, Y ⊂
{1, . . . , d}2, with (i, i)∈/ X∪Y for all i. Let Z
A,B ={(i, j, l)|(i, j) ∈X,(j, l)∈Y}. Then AB=Id+a X (i,j)∈X Ei,j+b X (k,l)∈Y Ek,l+ab X (i,j,l)∈ZA,B Ei,l.
(i) If ZA,A is empty (equivalently, if for every i, there exists at most one j such
that(i, j)∈X or (j, i)∈X (but not both)), then A−1 =Id−a P
(i,j)∈X Ei,j. (ii) [A, B] =Id+ab P (i,j,l)∈ZAB Ei,l−ab P (i,j,l)∈ZBA Ei,l.
(iii) A and B commute if and only if, for every pair(i, l):
|{j : (i, j, l)∈ZAB}|=|{j: (i, j, l)∈ZBA}|.
The next lemma will be useful when proving the irreducibility of the repre- sentation.
Lemma 5.2.4. [8, Lemma 1.8.11] Let V1, . . . , Vk be nonisomorphic irreducible G-
modules. Then the only nontrivial submodules of⊕k
i=1Viare⊕j∈JVi forJ ⊂ {1, . . . , k}.
Theorem 5.2.5. Let n≥3, and q=pe be a prime power. Let B denote a basis of
Fq, viewed as a Fp-vector space. Then the group generated by{yi(n)(b) :b∈B} and
{y(in)(b)T :b∈B} as given in Section 5.2.2 is HSpin+ 2n(q).
We prove Theorem 5.2.5 using the following lemmas.
Lemma 5.2.6. The matrices as given in Section 5.2.2 satisfy the relations of The- orem 5.1.1.
Proof. Note that every generator yi(n)(b) satisfies the conditions of Lemma 5.2.2. In particular, each row and each column has at most two nonzero entries, one of which is a 1 on the diagonal, and the other (if it exists) takes the value±b. Thus it follows from Lemma 5.2.2 that yi(n)(a)yi(n)(b) = y(in)(a+b) for every choice of
i, a, b, since Z is empty. In particular, this means that y(in)(b)k =y(in)(kb), so that
yi(n)(b)p = yi(n)(pb) = I2n−1, and so the first set of relations in Theorem 5.1.1 hold for the fundamental roots. Also, sinceyi(n)(a)y(in)(b) =yi(n)(a+b) =yi(n)(b)y(in)(a), the second set of relations also hold for the fundamental roots.
In the context of Theorem 5.1.1, we can determine explicitly the roots in Φi,j.
This consists only of{±αi,±αj}unlessj =i+1 for 1≤i≤n−2 ori=n−2, j =n,
in which case Φi,j ={±αi,±αj,±(αi+αj)}. In the first case, there are four relations
to check:
[yi(n)(a), y(jn)(b)] = 1 [yi(n)(a), y(−nj)(b)] = 1
[y(−nj)(a), y−(ni)(b)] = 1 [y(jn)(a), y−(ni)(b)] = 1
In the second case, the full list of relations to check, using the structure constants given in Example 5.1.4, are:
[yi(n)(a), yj(n)(b)] =y(i+n)j(ab) [yi(n)(a), y(−nj)(b)] = 1 [yi(n)(a), y(i+n)j(b)] = 1 [yi(n)(a), y−(ni)−j(b)] =y(−nj)(−ab) [yj(n)(a), y(in)(b)] =y(i+n)j(−ab) [yj(n)(a), y(−ni)(b)] = 1 [yj(n)(a), y(i+n)j(b)] = 1 [yj(n)(a), y−(ni)−j(b)] =y(−ni)(ab) [yi(+n)j(a), y(jn)(b)] = 1 [yi(+n)j(a), y(−nj)(b)] =y(in)(ab) [yi(+n)j(a), y(in)(b)] = 1 [yi(+n)j(a), y(−ni)(b)] =y(jn)(−ab) [y−(ni)(a), y(jn)(b)] = 1 [y−(ni)(a), y−(nj)(b)] =y(−ni)−j(−ab) [y−(ni)(a), yi(+n)j(b)] =y(jn)(ab) [y−(ni)(a), y(−ni)−j(b)] = 1 [y−(nj)(a), y(in)(b)] = 1 [y−(nj)(a), y−(ni)(b)] =y(−ni)−j(ab) [y−(nj)(a), yi(+n)j(b)] =y(in)(−ab) [y−(nj)(a), y(−ni)−j(b)] = 1 [y−(ni)−j(a), yj(n)(b)] =y(−ni)(−ab) [y−(ni)−j(a), y(−nj)(b)] = 1 [y−(ni)−j(a), yi(n)(b)] =y(−nj)(ab) [y−(ni)−j(a), y(−ni)(b)] = 1
Note that some of these relations are superfluous. For instance, inverting the relation [yi(n)(a), yj(n)(b)] = y(i+n)j(ab) gives the relation [y(jn)(b), y(in)(a)] =yi(+n)j(−ab). Thus we need only check half of these relations. Further, by transposing the relation [yi(n)(a), y(jn)(b)] = yi(+n)j(ab), we obtain the relation [y−(nj)(b), y−(ni)(a)] = y−(ni)−j(ab); thus we again only need to check the former. We list the reduced number of relations below; the left hand column records the relations to be checked, and the right hand column consists of the transposes of these relations.
[yi(n)(a), yj(n)(b)] =yi(+n)j(ab) [y(in)(a), yi(+n)j(b)] = 1 [y(i+n)j(a), yj(n)(b)] = 1 [yi(n)(a), y−(nj)(b)] = 1 [y(−ni)−j(a), yi(n)(b)] =y−(nj)(ab) [y(i+n)j(a), y−(nj)(b)] =yi(n)(ab) [y−(nj)(a), y−(ni)(b)] =y−(ni)−j(ab) [y(−ni)−j(a), y(−ni)(b)] = 1 [y(−nj)(a), y−(ni)−j(b)] = 1 [yj(n)(a), y(−ni)(b)] = 1 [y−(ni)(a), yi(+n)j(b)] =yj(n)(ab) [y(jn)(a), y−(ni)−j(b)] =y−(ni)(ab)
We will take as our definition y(i+n)j(b) := [yi(n)(1), yj(n)(b)] and y−(ni)−j(b) := [y−(nj)(b), y−(ni)(1)] when the latter matrices do not commute; in other words, when
i < n−1 and j =i+ 1, or i=n−2 andj =n. Note with these definitions that yi(+n)j(b)T =y−(ni)−j(b). When i < n−1 we have yi(+(n)i+1)(b) = [yi(n)(1), y(i+1n)(b)] = [yi(−n−11)(1)⊕y(i−n−11)(1), y(in−1)(b)⊕y(in−1)(b)] = [yi(−n−11)(1), y(in−1)(b)]⊕[y(i−n−11)(1), yi(n−1)(b)] =y((ni−−1)+1) i(b)⊕y((in−−1)+1) i(b). We also have y((nn)−2)+(n−1)(b) = [yn(n−)2(1), yn(n−)1(b)] = [y(nn−−31)(1)⊕y(nn−−31)(1), yn(n−−21)(b)⊕yn(n−−11)(−b)] = [y(nn−−31)(1), yn(n−−21)(b)]⊕[yn(n−−31)(1), yn(n−−11)(−b)] =y((nn−−1)3)+(n−2)(b)⊕y((nn−−1)3)+(n−1)(−b)
and similarlyy((nn−)2)+n(b) =y((nn−−1)3)+(n−1)(b)⊕y((nn−−3)+(1) n−2)(−b). We can then check inductively forn >3, or by hand for n= 3, that in all cases y(±n()i+j)(a)y(±n()i+j)(b) =
y(±n()i+j)(a+b) and so the first and second sets of relations in Theorem 5.1.1 also hold for these generators.
We will now proceed to prove by induction that the third set of relations in Theorem 5.1.1 hold.
The base case n = 3 is a tedious but straightforward computation; thus, supposen >3. We will check all of the relations involving generators in Φi,j.
Firstly, supposei= 1. We first consider the case j= 2. By Lemma 5.2.2 we have: y1(n)(b) =I2n−1+b 2n−3 X i=1 E2n−2+i,2n−3+i y2(n)(b) =I2n−1+b 2n−4 X i=1 E2n−3+i,2n−4+i+b 2n−4 X i=1 E2n−2+2n−3+i,2n−2+2n−4+i y(1+2n)(b) =I2n−1+b 2n−4 X i=1 E2n−2+i,2n−4+i−b 2n−4 X i=1 E2n−2+2n−3+i,2n−3+2n−4+i
By considering the rows and columns where each matrix has a nonzero entry, it follows from Corollary 5.2.3 that y(1+2n)(a) commutes with both y1(n)(b) and y2(n)(b). It is easy to check the remaining relations by inspection and using Lemma 5.2.2,
and we will only provide the details for one of the checks, namely for the relation [y−(n1)−2(a), y1(n)(b)] =y−(n2)(ab). We have that y(−n1)−2(a)y(1n)(b) = (y−(n1)−2(a) +y1(n)(b)−I2n−1) +ab 2n−4 X i=1 E2n−4+i,2n−3+i ; and y1(n)(b)y−(n1)−2(a) = (y−(n1)−2(a) +y1(n)(b)−I2n−1)−ab 2n−4 X i=1 E2n−2+2n−4+i,2n−2+2n−3+i hence [y−(n1)−2(a), y1(n)(b)] =I2n−1 +ab 2n−4 X i=1 E2n−4+i,2n−3+i+ab 2n−4 X i=1 E2n−2+2n−4+i,2n−2+2n−3+i =y2(n)(ab)T =y−(n2)(ab).
Next, we will show a sufficient condition fory1(n)(b) to commute with a matrix
g satisfying the conditions of Lemma 5.2.2. Suppose we can write g = gA⊕gB⊕
gC ⊕gD. Let ukr (for k ∈ {A, B, C, D}) denote the column of the unique nonzero
entry in the r-th row of gk away from the diagonal (considering r modulo 2n−3),
or ukr = 0 if no such entry exists. Take a pair (r, s) with 1 ≤ r, s ≤ 2n−1 and
r 6= s. Suppose that we can find a t such that the (r, t)-th entry of y(1n)(a) is nonzero, as is the (t, s)-th entry ofg. Then by the definition ofy(1n)(a) we must have
r ∈ {2n−2+ 1, . . . ,2n−2+ 2n−3} and t =r−2n−3. It follows that s =uBr + 2n−3, and in particular we requireuBr 6= 0.
Conversely, suppose that there exists a t such that the (r, t)-th entry of g
is nonzero, as is the (t, s)-th entry of y1(n)(a). For t to exist we requireukr 6= 0 for somek, and then t =ukr +2rn−−13
2n−3, the second term in this equation ensuring that the coordinate (r, t) lies in one of the blocksgk of g. For the (t, s)-th entry of
y(1n)(a) to be nonzero, we requiret∈ {2n−2+ 1, . . . ,2n−2+ 2n−3}, which in particular implies thatr∈ {2n−2+ 1, . . . ,2n−2+ 2n−3} so thatk=C, and thent= 2n−2+uCr
and s = t−2n−3 = 2n−3+uCr. Thus if gB = gC, in both cases we get the same
restrictions on (r, s), and so by Corollary 5.2.3(3) the matrices commute.
Suppose 3 ≤ j ≤ n−2. Then we have yj(n)(b) = yj(n−−22)(b)⊕yj(n−−22)(b)⊕
y(jn−−22)(b)⊕yj(n−−22)(b), and it follows from the above argument thaty1(n)(a) commutes with both y(jn)(b) andyj(n)(b)T.
Forj=n−1 or j=n we get that
y(nn−)1(b) =y(nn−−21)(b)⊕y(nn−−11)(−b) =yn(n−−32)(b)⊕yn(n−−22)(−b)⊕y(nn−−22)(−b)⊕yn(n−−32)(b). yn(n)(b) =yn(n−−11)(b)⊕y(nn−−21)(−b) =yn(n−−22)(b)⊕yn(n−−32)(−b)⊕yn(n−−32)(−b)⊕y(nn−−22)(b).
In both cases the middle two summands are the same, so again these generators commute with y±(n1)(a). Hence we have shown that all the relations involving the generatorsy(±n1)(a) hold.
When 1 < i, j < n −1, checking all of the relations hold follows im- mediately by induction, since yk(n)(b) = y(kn−−11)(b) ⊕ yk(n−−11)(b) for k = i, j, and
yi(+n)j(b) = y((in−−1)+(1) j−1)(b)⊕y((in−−1)+(1) j−1)(b). If 1 < i < n−2 and j = n−1 or
j=n, or wheni=n−1 andj=n, it again follows immediately by induction that all the generators involved commute.
Thus, the only remaining set of relations to check is when i = n−2 and
j=n−1 or j=n. These again generally follow by straightforward induction, and we will only prove one of the relations here.
[y−(n()n−2)−(n−1)(a), yn(n−)2(b)]
= [y−(n(−n1)−3)−(n−2)(a)⊕y(−n(−n−1)3)−(n−1)(−a), yn(n−−31)(b)⊕yn(n−−31)(b)] = [y−(n(−n1)−3)−(n−2)(a), y(nn−−31)(b)]⊕[y(−n(−n−1)3)−(n−1)(−a), yn(n−−31)(b)] =y(−n(−n−1)2)(ab)⊕y−(n(−n1)−1)(ab)
=y(−n()n−1)(ab).
Hence all the relations of the CST presentation hold.
Lemma 5.2.7. The centre of the group generated by the matrices in Section 5.2.2 is the same as the centre of the half-spin representation as given in the table in Section 5.1.1.
Proof. It is a standard result (see for instance [9, Lemma 6.4.4]) that the diagonal elements of the group are generated by elements of the form
hi(n)(t) =y(in)(t)y−(ni)(−t−1)yi(n)(t)yi(n)(−1)y−(ni)(1)y(in)(−1).
that the following formulae hold:
h(1n)(t) =I2n−3 ⊕t−1I2n−3 ⊕tI2n−3 ⊕I2n−3
hi(n)(t) =hi(n−−11)(t)⊕h(i−n−11)(t) for 2≤r ≤n−2
hn(n−)1(t) =h(nn−−21)(t)⊕hn(n−−11)(−t)
h(nn)(t) =h(nn−−11)(t)⊕h(nn−−21)(−t)
Combining these with a direct computation which shows that
h(3)1 (t) = diag(1, t−1, t,1)
h(3)2 (t) = diag(t−1, t,1,1)
h(3)3 (t) = diag(1,1, t−1, t) we can explicitly construct the elementsh(in)(t) in all cases.
A direct computation, or induction, gives that
h(nn−)1(t)h(nn)(t) = diag(t−1, t, t−1, t, . . . , t−1, t).
Whenq is odd, we can taket=t−1 =−1 and obtain a central element of order 2. We can obtain an additional central element precisely when n is odd and
q ≡1 mod 4. Since q ≡1 mod 4, there exists an element λ=νq−41 ∈F∗q of order 4. Define
s(n)(t) =h1(n)(−1)h3(n)(−1). . . h(nn−)4(−1)hn(n−)2(−1)hn(n−)1(t−1)h(nn)(t).
We prove that s(n)(λ) = λI2n−1. As usual the case s(3)(λ) is easy to check. For
n≥5 and i≥3 it follows immediately from the above calculations that
h(in)(t) =hi(n−−22)(t)⊕h(i−n−22)(t)⊕h(i−n−22)(t)⊕h(in−−22)(t)
h(nn−)1(t) =h(nn−−32)(t)⊕h(nn−−22)(−t)⊕h(nn−−22)(−t)⊕h(nn−−32)(t)
and hence s(n)(λ) =h(1n)(−1) h3(n)(−1). . . hn(n−)2(−1)h(nn−)1(−λ)h(nn)(λ) = (I2n−3 ⊕ −I2n−3⊕ −I2n−3⊕I2n−3) ·s(n−2)(λ)⊕s(n−2)(−λ)⊕s(n−2)(−λ)⊕s(n−2)(λ)
= (I2n−3 ⊕ −I2n−3⊕ −I2n−3⊕I2n−3) (λI2n−3⊕ −λI2n−3 ⊕ −λI2n−3 ⊕λI2n−3) =λI2n−1.
Hence whennis odd the centre of the representation equals the full Schur multiplier of O+2n(q), which is isomorphic to 2 when q ≡3 mod 4 and 4 when q ≡1 mod 4. Whennis even, it follows from Lemma 5.2.8 below that the representation is abso- lutely irreducible. Since the Schur multiplier whennis even is 2×2, it is impossible for the centre to be larger than 2, as the representation is centralised only by scalars and so the centre will be cyclic. Thus we have a representation of HSpin+2n(q) in all cases.
Lemma 5.2.8. The module generated by the matrices in Section 5.2.2 is absolutely irreducible.
Proof. We proceed by induction. Whenn= 3, we have that O+6(q)∼= L4(q), and we have the isomorphism HSpin+6(q)∼= SL4(q). In particular, the module we construct is a faithful 4-dimensional representation, and hence by [41] it must be the natural module of SL4(q), which is absolutely irreducible. Forn >3, consider the subgroup generated by y(±ni)(b) for i = 2, . . . , n. The corresponding group is a direct sum of the two half-spin representations of HSpin+2n−2(q), which by induction are absolutely irreducible and by [10, Theorem II.4.2] are non-isomorphic. Hence by Lemma 5.2.4 the only non-trivial submodules of this direct sum are the two half-spin represen- tations, and neither of these are preserved by the action of the generator y1(n)(b). Hence the half-spin representation of HSpin+2n(q) is absolutely irreducible.
Proof of Theorem 5.2.5. It follows from Lemmas 5.2.6, 5.2.7 and 5.2.8 that the ma- trices generate an absolutely irreducible module of a group isomorphic to HSpin+2n(q). It follows from [33, Proposition 5.4.11] that the only absolutely irreducible represen- tations of an extension of O+2n(q) in dimension 2n−1are the half-spin representations, and so we are done.
Note that instead of using the results of Lemmas 5.2.7 and 5.2.8, we could show that the module is the absolutely irreducible half-spin representation by the
direct computation of the elements h(in)(t) in the proof of Lemma 5.2.7. By con- sidering the first entry of the diagonal, we see that this representation has highest weight (0,0, . . . ,0,1,0), and the representation obtained by interchanging the roles of y(nn)(b) and yn(n−)1(b) has highest weight (0,0, . . . ,0,0,1). By comparing the di- mensions of these representations with the dimensions of the irreducible half-spin representations with these weights (as given for instance in [33, p.196]) and using Lemma 4.1.21 we obtain that both representations are absolutely irreducible.
We next consider which classical groups the half-spin representations embed into, and the forms that they preserve (if any).
Lemma 5.2.9. [33, Proposition 5.4.9] Let G = HSpin+2n(q). Then the half-spin representation embedsGinto a classical group Ωas described below, and no smaller classical group in dimension 2n−1.
(i) If n is odd, then Ω = SL2n−1(q).
(ii) If n is even and q is even, then Ω = Ω+2n−1(q).
(iii) If n≡0 mod 4and q is odd, then Ω = Ω+2n−1(q).
(iv) If n≡2 mod 4and q is odd, then Ω = Sp2n−1(q).
Lemma 5.2.10. Let G = HSpin+2n(q) denote the half-spin representation with n
even, generated by the generators as given above. Then G preserves the form f(n)
defined as follows: f(2) = antidiag(−1,1). f(n+2) =−f(n) ⊕ −b f (n) b ⊕ f(n) ⊕b f (n).
Proof. It is a direct computation to check that each generator preserves the appro- priate form above.