#
The Invariants of the Symplectic Groups
The Invariants of the Symplectic Groups
The Invariants of the Symplectic Groups
The Invariants of the Symplectic Groups
The Invariants of the Symplectic Groups
Mara D. Neusel
Mara D. Neusel
Mara D. Neusel
Mara D. Neusel
Mara D. Neusel
AG INVARIANTENTHEORIE
GERMANY
EMAIL:MDN@SUNRISE.UNIMATH.GWDG.DE
MARAMARA@STEENROD.MAST.QUEENSU.CA
MNEUSEL@CFGAUSS.UNIMATH.GWDG.DE
NEUSEL@MATH.UMN.EDU
THESE ARE LECTURE NOTES OF THE TALKS Invarianten klassischer Gruppen III AND IV, HELD AT THE
OBERSEMINAR Algebraische Topologie und Invariantentheorie, UNIVERSITY OF G ˜
OTTINGEN, GERMANY,
WINTER SEMESTER 1998/9
written high up in the mountains of Appenzell, 1998
AMS CODE: 13A50 Invariant Theory
KEYWORDS: Invariant Theory of Finite Groups, Symplectic Groups, Classical Groups
Typeset by LST E X
SUMMARY :
SUMMARY :
SUMMARY :
SUMMARY :
SUMMARY : In this notes we study the invariant rings of the symplectic groups in odd characteristic in
their tautological representation, and try to make the original paper by Carlisle and Kropholler more
readable and understandable, i.e., the only new thing is the expository, in particular that/how and
where the Steenrod algebra is used is my contribution.
§1. Notation and Introduction
Let IF be a Galois field of odd 1 characteristic p with q = p s elements. Let V be a vector space over IF
of even dimension n = 2l with basis x 1 , y 1 , . . . , x l , y l . Denote by
f : V × V ## IF
a non degenerate alternating bilinear form on V . Then the symplectic group Sp(n, IF) is defined
to be the group of isometries of f , or the maximal subgroup of the general linear group, GL(n, IF),
stabilizing 2 f
Sp(n, IF) = Isom(V , f )
= {T # GL(n, IF)#f (Tu, Tv) = f (u, v) " u, v # V}
< GL(n, IF).
The matrix A f associated to f is given by
(A f ) i,j = f (e i , e j ),
for the standard basis e 1 , . . . , e n # V . After a suitable change of bases, we can assume that the matrix
A f has the following form
A f =
2 6 6 6 6 6 6 4
0 1 · · · 0 0
1 0 · · · 0 0
0 0 · · · 0 0
: : : : :
0 0 · · · 0 1
0 0 · · · 1 0
3 7 7 7 7 7 7 5
.
So we could write the symplectic group also as
Sp(n, IF) = {T # GL(n, IF)#T t A f T = A f },
i.e., as the stabilizer subgroup of A f in GL(n, IF).
The order of the symplectic group was calculated by L. E. Dickson:
THEOREM 1.1
THEOREM 1.1
THEOREM 1.1
THEOREM 1.1
THEOREM 1.1 (L. E. Dickson): The order of the symplectic group in dimension n = 2l over a field
with q = p s elements is given by the formula
ï Sp(n, IF) ï= q l 2
l
Y i=1
(q 2i  1).
PROOF PROOF PROOF PROOF PROOF : The proof is by counting the number of symplectic bases for V , see 3 Theorem 115 in[6] .
Thanks to Dickson we also have an exlicite set of generators for our group:
1 In characteristic two there are some problems in defining a group, see [12] and [4].
2 This indeed does not depend on the choice of f since all non degenerate alternating bilinear forms are equivalent, meaning lie
in the same GL(n, IF)orbit. For a proof of this and an explicite introduction to groups associated to forms we refer to the fine
book [12], in particular pp. 234 246.
3 Careful: Dickson calls the symplectic groups special abelian groups and denotes them by SA(n, p s ).
MARA D. NEUSEL
THEOREM 1.2
THEOREM 1.2
THEOREM 1.2
THEOREM 1.2
THEOREM 1.2 (L.E. Dickson): The symplectic group Sp(n, IF) is generated by the following matri
ces:
S(k) = (s i,j (k)), " k = 1 , . . . , l,
where
s i,j (k) =
8 > > < > > :
1 if i = j #= k
1 if i = 2k  1, j = 2k
1 if i = 2k, j = 2k  1
0 otherwise,
S(k, l) = (s i,j (k, l)), " k = 1 , . . . , l and l # IF × ,
where
s i,j (k, l) = 8 < :
1 if i = j
l if i = 2k  1, j = 2k
0 otherwise,
and
S(k, l, l) = (s i,j (k, l, l)), " k, l = 1 , . . . , l and l # IF × ,
where
s i,j (k, l, l) =
8 > > < > > :
1 if i = j
l if i = 2k  1, j = 2l
l if i = 2l  1, j = 2k
0 otherwise.
PROOF PROOF PROOF PROOF PROOF : See Theorem 114 in [6] .
As Dickson remarks, we can read off this explicite description that the symplectic group is generated
by symplectic transvections, and therefore every element has determinant one. This will turn out to
be of great use later.
During this manuscript we will have to struggle with a lot of awful formulae. Therefore I have ex
plicitely calculated most of them for the case n = 4 (well, n = 2 isn't really 4 exciting). So, if you feel
overwhelmed by indices, you are welcome to consult the final Section 7.
§2. A Bunch of Invariants
We are going to find a collection of polynomial invariants of our symplectic group, which will, together
with the Dickson classes and the Euler class, lead to a complete set of algebra generators of the ring of
invariants.
PROPOSITION 2.1
PROPOSITION 2.1
PROPOSITION 2.1
PROPOSITION 2.1
PROPOSITION 2.1 : For any natural number i # IN
j n, i :=
l
X j=1
(x j y q i
j  y j x q
i
j ) # IF[x 1 , y 1 , . . . , x l , y l ] Sp(n, IF) .
4 In dimension two the symplectic group Sp(2, IF) is nothing but the special linear group SL(2, IF). To see this you could use
Dickson's theorem to find that the symplectic group is contained in the special linear, because all elements have determinant
one, and then use again Dickson to find that they have the same order. On the other hand you could replace the first step by
show by ordinary explicite matrix calculations that the special linear matrices are symplectic.
2
SYMPLECTIC INVARIANTS
PROOF PROOF PROOF PROOF PROOF : Observe that
j n := x 1 Ù y 1 + · · · + x l Ù y l # L 2 (V * ) Sp(n, IF)
# IF[V ] Ä IF L * (V * )
Sp(n, IF)
is a non degenerate alternating bilinear form and hence, by definition of the symplectic group, invariant
under the action of Sp(n, IF). Since
j n,1 :=
l
X j=1
(x j y q
j  y j x q
j )
= b P 1 b (j n ),
where b denotes the Bockstein operator
b : IF[V ] Ä IF L * (V * ) ## IF[V ] Ä IF L * (V * )
(z i , 0) ## (0, 0)
(0, z i ) ## (z i , 0)
(0, z i 1
Ù . . . Ù z i
j ) # j
P
k=1
(1) k1 (z i k
, z i 1
Ù . . . Ù
c z i k
Ù . . . Ù z i
j ),
for z i # {x 1 , y 1 , . . . , x l , y l }, and P 1 the first Steenrod reduced power operation, this is also invariant,
hence
j n,1 # IF[V ] Sp(n, IF) .
Moreover we have 5
j n,i+1 = P q
i
(j n,i ),
and therefore our statement is proved .
Here is another gorgeous observation by Carlisle and Kropholler:
LEMMA 2.2
LEMMA 2.2
LEMMA 2.2
LEMMA 2.2
LEMMA 2.2 : The symplectic group is precisely the stabilizer subgroup of j n,1 in the general linear
group
Sp(n, IF) = GL(n, IF) j n,1
.
PROOF PROOF PROOF PROOF PROOF : By the preceding lemma we know that j n,1 is an invariant and therefore
Sp(n, IF) # GL(n, IF) j n,1
.
On the other hand the map 6
b P 1 b : L 2 (V * ) # IF[V ] (q+1)
is injective. Therefore for any g # GL(n, IF) j n,1
, i.e., gj n,1 = j n,1 , also
gj n = j n .
This in turn means that g # Sp(n, IF), since the symplectic group is defined to be the stabilizer of j n .
5 If you can't do this calculation by yourself, you might look it up (Yes! There is a back of the book!) in Lemma 4.1, where all
Steenrod powers of the j's are calculated.
6 For a graded algebra A we denote the homogeneous part of degree d by A (d) . By the way, Carlisle und Kropholler call this map
symmetrizing map.
3
MARA D. NEUSEL
REMARK REMARK REMARK REMARK REMARK : In the same way we could observe that also the maps
b P q i
P q i1
· · · P 1 b : L 2 (V * ) # IF[V ] (q i+1 +1)
are injective, sending j n to j n,i+1 . So that the symplectic group is also the stabelizer of all the other
j n,i+1 's
Sp(n, IF) = GL(n, IF) j n,i+1
" i,
what in turn could replace the preceding Proposition 2.1.
Next we show that the maximal possible number of the invariants just constructed, namely n of them,
are algebraically independent.
LEMMA 2.3
LEMMA 2.3
LEMMA 2.3
LEMMA 2.3
LEMMA 2.3 : The polynomials
j n,1 , . . . , j n,n # IF[V ] Sp(n, IF)
are algebraically independent.
PROOF PROOF PROOF PROOF PROOF : We show this by calculating the determinant of the Jacobian matrix, Jac (j n,1 , . . . , j n,n ),
associated to these polynomials. To this end note that for all j = 1 , . . . , l and i # IN
#j n,i
# x j
= y q i
j
and
#j n,i
# y j
= x q i
j .
Therefore we get
det Jac (j n,1 , . . . , j n,n ) = det
2 6 6 6 6 4
y q
1 x q
1 · · · y q
l x q
l
: : · · · : :
: : · · · : :
: : · · · : :
y q n
1 x q n
1 · · · y q n
l x q n
l
3 7 7 7 7 5
.
However, the latter is, up to sign, the qth power of the Euler class E n (of the orbit V * \ 0 of SL(n, IF)
on V * ) and in particular non zero. That's all we need .
We note the obvious corollary:
COROLLARY 2.4
COROLLARY 2.4
COROLLARY 2.4
COROLLARY 2.4
COROLLARY 2.4 :
IF[j n,1 , . . . , j n,n ] # IF[V ] Sp(n, IF)
is a polynomial subalgebra.
Lets remark at this point also that a symplectic matrix has determinant one, because the symplectic
group is generated by symplectic transvections, see Theorem 114 in [6] or, for a more modern treatment,
e.g., [8] Lemma 1 on Section 6.9, and therefore
IF[E n , d n, 1 , , . . . , d n, n1 ] = IF[V ] SL(n, IF)
# IF[V ] Sp(n, IF) ,
where E n denotes the Euler class of the orbit V * \ 0 of SL(n, IF) on V * .
In a later section it will turn out that these invariants, namely
j n,1 , . . . , j n,n , E n , d n, 1 , . . . , d n, n1
4
SYMPLECTIC INVARIANTS
form a complete set of algebra generators 7 of the ring of invariants IF[V ] Sp(n, IF) , in other words, we
will show that
A := IF < j n,1 , . . . , j n,n , E n , d n, 1 , . . . , d n, n1 >,
where IF <  > denotes the IFalgebra generated by the stuff in the <  >brackets (which is possibly not
a polynomial algebra), is precisely the ring of invariants.
We do this by determine also the relations among these generators, which leads to a dozen pages of
calculations. Then we still have to prove that that's it: that takes another 6 pages, i.e., it's probably
not really pleasant to read all this (and it is certainly not pleasant to give a talk about it).
§3. A Bunch of Relations
In this section we calculate some relations between the invariants we have so far.
We need some preliminaries.
Define an alternating bilinear form on the nfold direct product × n
IF[V ] by (n=2l remember)
< F , H >:=
l
X j=1
(f 2j1 h 2j  f 2j h 2j1 ),
for ntuples F = (f 1 , . . . , f 2l ) and H = (h 1 , . . . , h 2l ). Denote by
t := (12)(34)(· · ·)(2l  1 2l) # S 2l
and take its centralizer in the symmetric group
C(t) < S 2l .
Next define
ïïF (1) , . . . , F (2l) ïï := X s
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A ,
where the sum runs over a set of coset representatives of C(t) in S 2l .
LEMMA 3.1
LEMMA 3.1
LEMMA 3.1
LEMMA 3.1
LEMMA 3.1 : With the preceding notation we have
(1) ïï · · · ïï is independent of the choice of the coset representatives.
(2) ïï · · · ïï is multilinear and alternating.
(3) ïïe (1) , . . . , e (2l) ïï = 1 where e (i) denotes the ith standard basis vector (0 , . . . , 0, 1, 0 , . . . , 0).
PROOF PROOF PROOF PROOF PROOF :
AD AD AD AD AD (1) : The centralizer of the element t = (12)(34) · · · (2l  1 2l) # S 2l can be expressed in the
following way: Take an embedding
S l # S 2l , s # •
s,
where we define the image for any k = 1 , . . . , l by
•
s(2k  1) := 2s(k)  1
•
s(2k) := 2s(k).
7 Note that these generators arise in a very natural way: the Dickson classes, an Euler class, the form j n that defines the group
and its Steenrod powers.
5
MARA D. NEUSEL
Then
C(t) = á(12), S l ñ # S 2l
= Z/2 × · · · × Z/2
####### l ###### # S l
= Z/2 # S l ,
so the S l permutes the l transpositions (12) , . . . , (2l  1 2l) of t and the Z/2 permutes the entries
2k  1, 2k, " k of the transpositions. That this is indeed the full centralizer is an easy calculation. An
equally straightforward calculation shows that our sum is the same if we replace s by sp where p is
one of the generators of C(t), because we just permute the order of the product or the order of the pairs
in the factors.
AD AD AD AD AD (2) : Since < ,  > is bilinear, we have that ïï · · · ïï is multilinear. To prove that the form is
alternating assume that F (j 0 ) = F (k 0 ) = F for some k 0 , j 0 = 1 , . . . , n. If
s 1 (2i 0  1) = j 0 and s 1 (2i 0 ) = k 0
(or vice versa) for some i 0 = 1 , . . . , l and some s 1 , then
< F s 1 (2i 0 1) , F s 1 (2i 0 ) > = < F , F > = 0,
i.e., the product Q j
< F s 1 (2j1) , F s 1 (2j) > = < F , F > Q
j#=i 0
< F s 1 (2j1) , F s 1 (2j) > vanishes. If
s 2 (2i 0  1) = j 0
and
s 2 (2i #
0  1) = k 0 or s 2 (2i #
0 ) = k 0
(or vice versa) for some i 0 , i # 0 = 1 , . . . , l and some s 2 , then
Y j
< F s 2 (2j1) , F s 2 (2j) > = < F , F s 2 (2i 0 ) > < F , F s 2 (2i
#
0 ) > Y
j#=i 0 , i # 0
< F s 2 (2j1) , F s 2 (2j) > .
Consider the element s 2 (2i 0 2i #
0 ) # S 2l / C(t). It gives the same summand as s 2 , namely
Y j
< F ( s 2 (2i 0 2i
#
0 ) ) (2j1) , F ( s 2 (2i 0 2i # 0 ) ) (2j) >
= < F , F s 2 (2i # 0 ) > < F , F s 2 (2i 0 ) > Y
j#=i 0 , i # 0
< F ( s 2 (2i 0 2i #
0 ) ) (2j1) , F ( s 2 (2i 0 2i #
0 ) ) (2j) >,
but with the opposite sign, because sign(s 2 ) = sign(s 2 (2i 0 2i # 0 )). So the two summands cancel. Alto
gether we have
ïïF (1) , . . . , F (2l) ïï = X s
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A
= X s 1
0 @ sign(s 1 ) < F , F > Y j#=i 0
< F s 1 (2j1) , F s 1 (2j) > 1 A
+ X s 2
sign(s 2 ) < F , F s 2 (2i 0 ) >< F , F s 2 (2i # 0 ) >
6
SYMPLECTIC INVARIANTS
0 B @ Y
j#=i 0 , i #
0
< F ( s 2 (2i 0 2i # 0 ) ) (2j1) , F ( s 2 (2i 0 2i #
0 ) ) (2j) >
 Y
j#=i 0 , i # 0
< F s 2 (2j1) , F s 2 (2j) >
1 C A
= 0.
AD AD AD AD AD (3) : We have
< e (i) , e (j) >= 8 < :
+1 if i is odd and j = i + 1
1 if i is even and j = i  1
0 otherwise.
Hence all summands vanish except the one where s = id
ïïe (1) , . . . , e (2l) ïï = X s
0 @ sign(s)
l
Y j=1
< e s(2j1) , e s(2j) > 1 A
=
l
Y j=1
< e (2j1) , e (2j) >
= 1.
That's all we claimed .
So, the first statement of the lemma tells us that our definition of ïï · · · ïï is welldefined, while the
second and third imply that
ïïF (1) , . . . , F (2l) ïï = det 2 4
F (1)
:
F (2l)
3 5 .
We are now prepared to prove the following lemma.
PROPOSITION 3.2
PROPOSITION 3.2
PROPOSITION 3.2
PROPOSITION 3.2
PROPOSITION 3.2 : With the notation above we have:
(1) The Euler class E n # IF[j n,1 , . . . , j n,n1 ] is irreducible in IF[V ] Sp(n, IF) .
(2) E n d n, i # IF[j n,1 , . . . , j n,n ] for all i = 1 , . . . , n  1.
(3) If for some polynomial f #IF[V ] we have that E n f #IF[j n,1 , . . . , j n,n1 ], then f #IF[j n,1 , . . . , j n,n1 ].
PROOF PROOF PROOF PROOF PROOF : We take the things in order.
AD AD AD AD AD 1 : Define 2ltuples F (j) := (x q j1
1 , y q j1
1 , . . . , x q j1
l , y q j1
l ) # ×
2l
IF[V ] of polynomials for j # IN. Then
together with the preceding observation the Euler class E n is, according to Dickson [6], given by
E n = det
2 6 6 4
x 1 y 1 · · · x l y l
: : · · · : :
: : · · · : :
x q 2l1
1 y q 2l1
1 · · · x q 2l1
l y q 2l1
l
3 7 7 5
= det 2 4
F (1)
:
F (2l)
3 5
= ïïF (1) , . . . , F (2l) ïï
7
MARA D. NEUSEL
= X s
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A .
Since
< F (i) , F (j) >= j q min(i, j)1
n,ïjiï
as one easily calculates, and for our choice ïj  iï £ 2l  1 we get the desired inclusion
E n # IF[j n,1 , . . . , j n,n1 ],
i.e., there exists a polynomial E = E(X 1 , . . . , x n1 ) # IF[X 1 , . . . , X n1 ] in n  1 indeterminants such that
E n = E(j n,1 , . . . , j n,n1 ).
The Euler class is by construction the product of linear forms where we take for each one dimensional
vector subspace of V * exactly one form. Moreover the symplectic group Sp(n, IF) acts transitively 8 on
the set of hyper planes of V . Hence E n is irreducible in IF[V ] Sp(n, IF) and a fortiori in IF[j n,1 , . . . , j n,n1 ].
AD AD AD AD AD 2 : Since (again thanks to Dickson)
E n d n, i = det
2 6 4
x q i 1
1 y q i 1
1 · · · x q i 1
l y q i 1
l
: : · · · : :
x q i n
1 y q
i n
1 · · · x q i n
l y q i n
l
3 7 5
for 0 £ i 1 < i 2 < · · · < i n1 < i n £ n where i j #= i we can proceed as in (1) and get
E n d n, i = X s
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A # IF[j n,1 , . . . , j n,n ].
In other words there exist polynomials in n indeterminants
D n,i = D n,i (X 1 , . . . , X n ) # IF[X 1 , . . . , X n ]
such that
D n,i (j n,1 , . . . , j n,n ) = E n d n, i ,
for all i = 0 , . . . , n  1.
AD AD AD AD AD 3 : Consider the composition of maps
v : IF[j n,1 , . . . , j n,n1 ] # IF[V ] # IF[x 1 , y 1 , . . . , x l1 , y l1 ],
where the first is just the canonical inclusion of algebras while the second is induced by the inclusion
of vector spaces
Span IF (e 1 , . . . , e 2l2 ) # V ,
where e 1 , . . . , e 2l2 denotes the dual basis to x 1 , y 1 , . . . , x l1 , y l1 . By Lemma 2.3 the polynomials
v(j n,1 ) = j n2,1 , . . . , v(j n,n2 ) = j n2,n2 are algebraically independent. Therefore the image of v gener
ates a subalgebra of Krull dimension at least n  2. Since n 2 = dim(IF[x 1 , . . . , y l1 ] the kernel of v is a
prime ideal of height 1. Obviously the Euler class E n is in the kernel. By (1) the class E n is irreducible,
and therefore prime, because IF[j n,1 , . . . , j n,n1 ] is a unique factorization domain, i.e.,
(E n ) = ker(v) # IF[j n,1 , . . . , j n,n1 ].
8 This is just an application of Witt's Lemma, see, e.g., Section 20 in [1], or Lemma 3 in Section 6.9 of [8].
8
SYMPLECTIC INVARIANTS
Denote by < E n > # IF[V ] the principle ideal generated by the Euler class in the full polynomial ring
IF[V ]. Certainly we have
(E n ) # < E n > Ç IF[j n,1 , . . . , j n,n1 ].
Since for every element in the big ideal f E n # < E n > Ç IF[j n,1 , . . . , j n,n1 ] (i.e., for all f # IF[V ]) we have
that v(f E n ) = 0, the two ideals in question must be equal, i.e., the polynomial f # IF[j n,1 , . . . , j n,n1 ],
as we wanted to show .
We need to be more precise about the polynomials D n,i occuring in part (2).
LEMMA 3.3
LEMMA 3.3
LEMMA 3.3
LEMMA 3.3
LEMMA 3.3 : For i = 0 , . . . , n  1 the polynomials D n,i (X 1 , . . . , X n ) are linear in X n with leading
coefficient
D n2,i1 (X 1 , . . . , X n ) q .
PROOF PROOF PROOF PROOF PROOF : Recall that
E n d n, i = det
2 6 6 4
x q i 1
1 y q
i 1
1 · · · x q i 1
l y q i 1
l
: : · · · : :
: : · · · : :
x q i n
1 y q i n
1 · · · x q i n
l y q i n
l
3 7 7 5
= det 2 4
F (1)
:
F (2l)
3 5
= ïïF (1) , . . . , F (2l) ïï
= X s
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A
= X
{s,ïs(2j1)s(2j)1ï#=n,"j}
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A
+ X
{s,ïs(2j 0 1)s(2j 0 )1ï=n}
0 @ sign(s)
l
Y
j=1,#=j 0
< F s(2j1) , F s(2j) > j n,n
1 A
= X
{s,ïs(2j1)s(2j)1ï#=n,"j}
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A
+
0 B @ X
{s,ïs(2j 0 1)s(2j 0 )1ï=n}
0 @ sign(s)
l
Y
j=1,#=j 0
< F s(2j1) , F s(2j) > 1 A
1 C A j n,n .
Therefore E n d n, i is linear in j n, n . To find the leading coefficient we have to work a bit harder and
calculate the sum
X
{s,ïs(2j 0 1)s(2j 0 )1ï=n}
0 @ sign(s)
l
Y
j=1,#=j 0
< F s(2j1) , F s(2j) > 1 A .
Note first of all that we can assume without loss of generality that j 0 = l (if not we replace s by sc
where c # C(t) interchanges j 0 and l). So we have to calculate
X
{s,ïs(2l1)s(2l)1ï=n}
0 @ sign(s)
l1
Y j=1
< F s(2j1) , F s(2j) > 1 A ,
9
MARA D. NEUSEL
where the exponents s(2j  1), s(2j) # {2 , . . . , n  1}. That means that the involved vectors F all are
qth powers (because F (1) does not occur anymore) of, say, ”
F , i.e., we have
0 B @ X
{s,ïs(2l1)s(2l)1ï=n}
0 @ sign(s)
l1
Y j=1
< ”
F s(2j1) , ”
F s(2j) > 1 A
1 C A
q
.
Note that we are still summing over coset representatives s of the centralizer C(t) in S 2l , with the
only restriction that our s's look 9 like (n  1 1)(n) ”
s, where ”
s # S 2l2 . We have to convince ourselves
that the elements ”
s run over a complete set of coset representatives (exactly once) of C(t 2l2 ) in S 2l2 ,
where we set t 2l2 := (12) · · · (2l  3 2l  2) # S 2l2 . Define a map
S 2l2 / C(t 2l2 ) # (n  1 1)(n) ”
sC(t 2l ) , ”
sC(t 2l2 ) # (n  1 1)(n) ”
sC(t 2l ).
This map is obviously injective. We define a splitting via
(n  1 1)(n) ”
sC(t 2l )
# S 2l2 / C(t 2l2 ), (n  1 1)(n) ”
sC(t 2l ) # ”
sC(t 2l2 ).
This map is equally injective, because if we take two different elements
(n  1 1)(n) ”
s 1 C(t 2l ) #= (n  1 1)(n) ”
s 2 C(t 2l ),
then also ”
s 1 C(t 2l2 ) #= ”
s 2 C(t 2l2 ).
Coming back to our coefficient we summarize
E n d n, i = X
{s,ïs(2j1)s(2j)1ï#=n,"j}
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A
+
0 B @ X
{s,ïs(2j 0 1)s(2j 0 )1ï=n}
0 @ sign(s)
l
Y
j=1,#=j 0
< F s(2j1) , F s(2j) > 1 A
1 C A j n,n
= X
{s,ïs(2j1)s(2j)1ï#=n,"j}
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A
+
0 B @ X
{s,ïs(2l1)s(2l)1ï=n}
sign(s)
l1
Y j=1
< ”
F s(2j1) , ”
F s(2j) > q 1 C A j n, n
= X
{s,ïs(2j1)s(2j)1ï#=n,"j}
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A
+ 0 @ X
”
s#C(t 2l2 )
0 @ sign( ”
s)
l1
Y j=1
< ”
F ”
s(2j1) , ”
F ”
s(2j) > 1 A 1 A
q
j n, n
= X
{s,ïs(2j1)s(2j)1ï#=n,"j}
0 @ sign(s)
l
Y j=1
< F s(2j1) , F s(2j) > 1 A
+ D n2, i1 (j n, 1 , . . . , j n, n2 ) q
j n, n ,
9 The (n) emphasizes that the s's fix n.
10
SYMPLECTIC INVARIANTS
where in the last step we used that
”
F j = (x q i j 1
1 , . . . , y q
i
j 1
l ) " j
by construction. Since the invariants j n,1 , . . . , j n,n are algebraically independent by Lemma 2.3 this
proves, for all i = 0 , . . . , n  1,
D n,i (X 1 , . . . , X n ) = D n1,i1 (X 1 , . . . , X n2 ) q X n + junk,
where junk does not depend on X n .
A similar construction leads to another relation. For that we need the following lemma.
LEMMA 3.4
LEMMA 3.4
LEMMA 3.4
LEMMA 3.4
LEMMA 3.4 : Let F (1) , . . . , F (n+2)
# ×
n
IF[V ] be n + 2 ntuples of polynomials. Then
##F (1) , . . . , F (n+2)
## :=
n+2
X k=2
(1) k < F (1) , F (k) > ïïF (2) , . . . ,
d
F (k) , . . . , F (n+2) ïï
defines an alternating multilinear form.
PROOF PROOF PROOF PROOF PROOF : Turn the F's into (n + 2)tuples of polynomials by adding two zero entries at the end and
note that this does not change the value of < F j , F k > for any j, k. We have seen in Lemma 3.1 that
ïïF (1) , . . . , F (n+2) ïï := X s
0 @ sign(s)
l+1
Y j=1
< F s(2j1) , F s(2j) > 1 A
is a multilinear alternating form. In every product the factor < F (1) , F (k) > occurs for some k, i.e.,
s k (2j 0  1) = 1 and s k (2j 0 ) = k
(or vice versa) for some j 0 = 1 , . . . , l. Without loss of generality we can assume that s k (1) = 1 and
s k (2) = k for otherwise we replace s k by s k (2j 0  1 1)(2j 0 2), which represents the same coset. Hence
we have
##F (1) , . . . , F (n+2)
## = X s
sign(s) < F (1) , F (k) > 0 @
l+1
Y j=2
< F s(2j1) , F s(2j) > 1 A .
Like in the preceding lemma we replace s k by ”
s where ”
s # S 2l permutes the set {3 , . . . , 2l + 2} and we
observe that the permutations ”
s run through a complete set of coset representatives of C(t 2l ) in S 2l .
Moreover observe that sign(s) = (1) k sign( ”
s), because the number of descents of s is precisely k plus
the number of descents of ”
s. Therefore we have
##F (1) , . . . , F (n+2)
##
= X s
sign(s) < F (1) , F (k) >
l+1
Y j=2
< F s(2j1) , F s(2j) >
=
n+2
X k=2
(1) k < F (1) , F (k) > 0 @ X ”
s
sign( ”
s)
l+1
Y j=2
< F ”
s(2j1) , F ”
s(2j) > 1 A
=
n+2
X k=2
(1) k < F (1) , F (k) > ïïF (2) , . . . ,
d
F (k) , . . . , F (n+2) ïï
as claimed .
11
MARA D. NEUSEL
CONVENTION CONVENTION CONVENTION CONVENTION CONVENTION : Denote d n,n = 1 and d n,j = 0, whenever j ## {0 , . . . , n}. Then setting D n, n = E n makes
the whole story consistent.
PROPOSITION 3.5
PROPOSITION 3.5
PROPOSITION 3.5
PROPOSITION 3.5
PROPOSITION 3.5 : We have
P 0 :=
n
X j=1
(1) j j n,j d n,j = 0
in IF[V ] Sp(n, IF) .
PROOF PROOF PROOF PROOF PROOF : We set
F = F (1) = F (2) = (x 1 , y 1 , . . . , x l , y l )
and for j = 3 , . . . , n + 2
F (j) = (x q j2
1 , y q j2
1 , . . . , x q j2
l , y q j2
l ).
Since ## · · · ## is alternating we get
0 = ##F , F , F (3) , . . . , F (n+2)
##
=
n+2
X j=2
(1) j < F , F (j) > ïïF , F (3) , . . . ,
d F (j) , . . . , F (n+2) ïï
=
n+2
X j=3
(1) j < F , F (j) > E n d n, j2
=
n+2
X j=3
(1) j j n,j2 E n d n, j2
= 0 @
n+2
X j=3
(1) j
j n,j2 d n, j2
1 A E n .
Since E n #= 0 # IF[V ] Sp(n, IF) we have
P 0 =
n
X j=1
(1) j j n,j d n,j = 0,
as claimed .
§4. Steenrod plays his game
We are going to calculate the Steenrod powers of our invariants, exploit Steenrod to find further rela
tions, and to show that the IFalgebra generated by
j n,1 , . . . , j n,n , E n , d n, 1 , . . . , d n, n1
is closed under the action of the Steenrod algebra induced from IF[V ], i.e., we will show that
A := IF < j n,1 , . . . , j n,n , E n , d n, 1 , . . . , d n, n1 >
is an unstable algebra over the Steenrod algebra.
Since the Steenrod powers of the Dickson and Euler classes are known and as well polynomials in the
Dickson and Euler classes, see [11] Appendix A.2 and the references there, we are left to deal with the
new polynomials j n,i . For simplicity of notation we make the following conventions:
12
SYMPLECTIC INVARIANTS
CONVENTION CONVENTION CONVENTION CONVENTION CONVENTION : Let P i º 0 whenever i ## IN 0 . Moreover, let j n,i = 0 for i ## IN.
LEMMA 4.1
LEMMA 4.1
LEMMA 4.1
LEMMA 4.1
LEMMA 4.1 : The Steenrod powers of the new classes j n,i , i ³ 1, are given by the following formulae
P j (j n,i ) =
8 > > > < > > > :
j q
n,i if j = q i + 1
j n,i+1 if j = q i
j q
n,i1 if j = 1
0 otherwise
.
PROOF PROOF PROOF PROOF PROOF : By straightforward calculation:
P j (j n,i ) =
l
X k=1
P j (x k y q i
k )  P j (x q i
k y k )
=
l
X k=1
x k P j (y q
i
k )  P j (x q i
k )y k + x q
k P j1 (y q i
k )  P j1 (x q i
k )y q
k
=
l
X k=1
x k (P j/q i
(y k ) q i
 P j/q
i
(x k ) q i
y k + x q
k P (j1)/q i
(y k ) q i
 P (j1)/q i
(x k ) q i
y q
k
=
8 > > > > > > > > > > < > > > > > > > > > > :
l
P k=1
x k y q i+1
k  x q i+1
k y k if j = q i
l
P k=1
x q
k y q i+1
k  x q i+1
k y q
k if j = q i + 1
l
P k=1
x q
k y q i
k  x q i
k y q
k if j = 1
0 otherwise,
which was to be shown .
We evaluate the Steenrod derivations on our j's in the next lemma.
LEMMA 4.2
LEMMA 4.2
LEMMA 4.2
LEMMA 4.2
LEMMA 4.2 : The Steenrod derivations act on the j's by
P D
j (j n,i ) =
8 > > < > > :
(1) j j q i
n,ji if j > i
0 if j = i
(1) j+1 j q
j
n,ij if j < i
.
PROOF PROOF PROOF PROOF PROOF : By induction on j. For j = 1 we have
P D 1 (j n,i ) = P 1 (j n,i )
= 0 if i = 1
j q
n,i1 if i > 1 ,
where we made use of the preceding lemma. Next take an j > 1. Then we get by using again the
preceding lemma and the induction hypothesis
P D j (j n,i ) = P D j1 P q j1
(j n,i )  P q j1
P D j1 (j n,i )
=
8 > > > < > > > :
0  (1) j1 P q j1
j q
i
n,j1i if j  1 > i
P D j1 (j n,i+1 )  0 if j  1 = i
0  (1) j P q j1
j q j1
n,ij+1 if j  1 < i
13
MARA D. NEUSEL
=
8 > > > < > > > :
(1) j P q ji1
j n,j1i q
i
if j  1 > i
(1) j j q j1
n,i+1j+1 if j  1 = i
(1) j+1 P 1 j n,ij+1 q j1
if j  1 < i
=
8 > > > > > < > > > > > :
(1) j j q i
n,ji if j > i + 1
(1) j j q j1
n,1 if j = i + 1
0 if j = i
(1) j+1 j q
j
n,ij if j < i
=
8 > > < > > :
(1) j j q i
n,ji if j > i
0 if j = i
(1) j+1 j q j
n,ij if j < i
,
as we wanted .
REMARK REMARK REMARK REMARK REMARK : Since
P D i := P 1 if i = 1
[P q i
, P D i1 ] if i > 1
the the action of P D
i and P q i
P q i1
· · · P q P 1 agree on the classes in V * . This can be used to give another
proof of Lemma 4.2.
Now we can prove
PROPOSITION 4.3
PROPOSITION 4.3
PROPOSITION 4.3
PROPOSITION 4.3
PROPOSITION 4.3 : The IFalgebra A generated by j n,1 , . . . , j n,n , E n , d n,1 , . . . , d n,n1 is an unstable
algebra over the Steenrod algebra.
PROOF PROOF PROOF PROOF PROOF : The only thing we need to show is that our algebra A is closed under the action of the
Steenrod algebra. The rest is inherited from IF[V ].
First note that part (1) of Proposition 3.2 tells us that
IF < j n,1 , . . . , j n,n1 , d n,1 , . . . , d n,n1 >
contains the Euler class E n , while Proposition 3.5 gives that it also contains j n,n , i.e.,
A = IF < j n,1 , . . . , j n,n1 , d n,1 , . . . , d n,n1 > .
From [11] Appendix A.2 we have for j ³ 0 and i = 0 , . . . , n  1 that
P j (d n,i ) # IF[d n,0 , . . . , d n,n1 ] = IF[E q1
n , d n,1 , . . . , d n,n1 ] # A.
Next we consider j ³ 0 and i = 1 , . . . , n  1 and get from Lemma 4.1
P j (j n,i ) # IF[j n,1 , . . . , j n,n ] # A
by Proposition 2.1 .
This allows us to construct a second family of relations from P 0 given in Proposition 3.5 in a very natural
way.
COROLLARY 4.4
COROLLARY 4.4
COROLLARY 4.4
COROLLARY 4.4
COROLLARY 4.4 : In IF[V ] Sp(n, IF) we have
P i :=
n
X
j=i+1
(1) j j q i
n,ji d n,j 
i1
X j=0
(1) j j q j
n,ij d n,j = 0
for i = 0 , . . . , n  1.
14
SYMPLECTIC INVARIANTS
PROOF PROOF PROOF PROOF PROOF : By Proposition 3.5 we know that
P 0 = 0.
Therefore all Steenrod powers of this polynomial are zero, and, by Proposition 4.3, are again polynomi
als in the algebra generators of A. Observe that
P i+1 = P q
i
(P i ),
which is proved by a straighforward calculation, to wit:
P q i
(P i ) =
n
X
j=i+1
(1) j
P q i
j q i
n,ji d n,j 
i1
X j=0
(1) j
P q i
j q j
n,ij d n,j
=
n
X
j=i+1
(1) j
0 B @ X
a+b =q i
P a (j q
i
n,ji )P b (d n,j )
1 C A 
i1
X j=0
(1) j
0 B @ X
a+b =q
i
P a (j q j
n,ij )P b (d n,j )
1 C A
=
n
X
j=i+1
(1) j
0 B @ X
a+b =q
i
P a
q i (j n,ji ) q i
P b (d n,j )
1 C A 
i1
X j=0
(1) j
0 B @ X
a+b =q
i
P a
q j (j n,ij ) q j
P b (d n,j )
1 C A
=
n
X
j=i+1
(1) j
P 1 (j n,ji ) q i
d n,j + j q i
n,ji P q
i
(d n,j )

i1
X j=0
(1) j
j q j
n,ij P q i
(d n,j ) + P 1 (j n,ij ) q j
P q i q
j
(d n,j ) + P q ij
(j n,ij ) q j
d n,j
=
n
X
j=i+1
(1) j
j q i+1
n,ji1 d n,j + (1) i+1 j q i
n,1 d n,i 
i1
X j=0
(1) j
j q j
n,ij+1 d n,j
=
n
X
j=i+2
(1) j
j q i+1
n,ji1 d n,j 
i
X j=0
(1) j
j q j
n,ij+1 d n,j
= P i+1 ,
where we made heavily use of the Cartan formulae, Lemma 4.1 and Appendix A.2 in [11] .
Note that Proposition 4.3 tells us a priori that all Steenrod powers of P 0 are again polynomials in the
algebra generators of A. However, for their explicite description we had to calculate them anyway. In
Section 6 it will turn out that we need only to consider P 0 , . . . , P l1 . Moreover, note that
P i º j q i
n,ni MOD MOD MOD MOD MOD (j n,1 , . . . , j n,ni1 )
whenever i £ l  1.
§5. British T
Recall from Proposition 3.2 (1) that the Euler class E n is a polynomial in j n,1 , . . . , j n,n1 . So, there
exists a polynomial E n = E n (X 1 , . . . , X n1 ) such that
E n = E n (j n,1 , . . . , j n,n1 ).
The same proposition, part (2), shows that
E n d n, i # IF[j n,1 , . . . , j n,n ],
15
MARA D. NEUSEL
i.e., there exist polynomials D n,i = D n,i (X 1 , . . . , X n ) such that
E n d n, i = D n,i (j n,1 , . . . , j n,n ),
for all i = 0 , . . . , n  1. Moreover, we have seen that these polynomials D n,i are linear in their last
indeterminant, which has coefficient
D n2,i1 (X 1 , . . . , X n2 q ,
compare Lemma 3.3.
PROPOSITION 5.1
PROPOSITION 5.1
PROPOSITION 5.1
PROPOSITION 5.1
PROPOSITION 5.1 : There exist polynomials
T i,j = T i,j (X 1 , . . . , X n ) # IF[X 1 , . . . , X n ]
such that for i, j = 1 , . . . , l we have
D n,i =
li
X k=0
T q k
i,lki+1 D n,nk
and
T i,j = T i,j (X 1 , . . . , X 2(i+j)3 ) # IF[X 1 , . . . , X 2(i+j)3 ],
i.e., T i,j depents only on the first 2(i + j)  3 variables.
PROOF PROOF PROOF PROOF PROOF : We construct the T i,j by induction on j. Let j = 1. Then we define
T i,1 (X 1 , . . . , X n ) := E 2i (X 1 , . . . , X n ) q1 ,
which is in IF[X 1 , . . . , X 2i1 ], because the polynomial E 2i lives there. For i = l this polynomial satisfies
the desired relation
D n,0 = E q
n = E q1
n E n = T l,1 E n .
For i < l there is nothing more to prove.
Next take an j > 1 and assume T i,j is defined for all i = 1 , . . . , l and all j = 1 , . . . , l  i such that the
required relations hold. We then have by the induction hypothesis that
T i,li+1 E n = D n,li 
li
X k=1
T q k
i,lki+1 D n,nk # IF[X 1 , . . . , X n ].
We want to show that
T i,li+1 # IF[X 1 , . . . , X n1 ].
From Proposition 3.2 (2) it follows that
T i,li+1 E n = D n,li 
li
X k=1
T q k
i,lki+1 D n,nk
= D n,li (j n,1 , . . . , j n,n ) 
li
X k=1
T q k
i,lki+1 D n,nk (j n,1 , . . . , j n,n )
is an element in IF[X 1 , . . . , X n ]. Since k = 1 , . . . , l  i we have by induction
T i,lki+1 # IF[X 1 , . . . , X 2l2k1 ].
16
SYMPLECTIC INVARIANTS
Therefore, together with Lemma 3.3 our polynomial is linear in X n with leading coefficient
D n2,li1 (X 1 , . . . , X n2 ) q 
li
X k=1
T q k
i,lki+1 D n2,nk1 (X 1 , . . . , X n2 ) q =
0 @ D n2,li1 (X 1 , . . . , X n2 ) 
li1
X k=0
T q k
i,lki D n2,n2k (X 1 , . . . , X n2 ) 1 A
q
= 0,
where the last equation follows from the induction hypothesis. Therefore
T i,li+1 E n # IF[X 1 , . . . , X n1 ]
which in turn implies that
T i,li+1 # IF[X 1 , . . . , X n1 ],
where the desired relation holds by construction .
§6. Some Algebra
In this section we do some algebra and prove that we have found all generators and relations of our
ring of invariants. By what we have done so far we know that
A = IF < j n,1 , . . . , j n,n , d n,0 , . . . , d n,n1 >= IF < j n,1 , . . . , j n,n , d n,1 , . . . , d n,n1 >,
compare Proposition 3.2 (1). 10
Next we show that we can omit the Dickson classes of high degree.
LEMMA 6.1
LEMMA 6.1
LEMMA 6.1
LEMMA 6.1
LEMMA 6.1 : With the preceding notation we have
A = IF < j n,1 , . . . , j n,n , d n,l , . . . , d n,n1 > .
PROOF PROOF PROOF PROOF PROOF : Certainly A contains this algebra. So we have to show that
d n,1 , . . . , d n,l1 # IF < j n,1 , . . . , j n,n , d n,l , . . . , d n,n1 > .
Proposition 5.1 hands us equations
d n,li =
li
X k=0
T q k
i,lik+1 (j n,1 , . . . , j n,2(lk)1 )d n,nk ,
for i = 1 , . . . , l  1. Since 2l  2k  1 £ 2l  1 for k ³ 0 we get,
d n,li =
li
X k=0
T q k
i,lik+1 (j n,1 , . . . , j n,2(lk)1 )d n,nk , # IF < j n,1 , . . . , j n,n , d n,l , . . . , d n.n1 >,
where we use that i = 1 , . . . , l  1 .
Consider the remembering map #
# : B := IF[j n,1 , . . . , j n,n , d n,l , . . . , d n,n1 ] # IF[V ] Sp(n, IF) .
10 We forget for a moment that P 0 , given in Proposition 3.5, cancels j n,n .
17
MARA D. NEUSEL
Recall the remarks after Corollary 4.4
P i º j q i
n,ni MOD MOD MOD MOD MOD(j n,1 , . . . , j n,ni1 ),
whenever i £ l  1. Since the sequence
j n,1 , . . . , j n,n , d n,l , . . . , d n,n1 # B
forms a regular sequence, so does the sequence
j n,1 , . . . , j n,n1 , P 0 , d n,l , . . . , d n,n1 # B.
Therefore also j n,1 , . . . , j q
n,n1 , P 0 , d n,l , . . . , d n,n1 # B forms a regular sequence and hence so does
j n,1 , . . . , j n,n2 , P 1 , P 0 , d n,l , . . . , d n,n1 # B. Successively we get that
j n,1 , . . . , j n,l , P l1 , . . . , P 0 , d n,l , . . . , d n,n1 # B
is a regular sequence, and in particular
P 0 , . . . , P l1 # B
is a regular sequence. Hence we have shown
LEMMA 6.2
LEMMA 6.2
LEMMA 6.2
LEMMA 6.2
LEMMA 6.2 : The IFalgebra
B/ P 0 , . . . , P l1 = IF[j n,1 , . . . , j n,n , d n,l , . . . , d n,n1 ]/ P 0 , . . . , P l1
is a complete intersection of Krull dimension n. In particular it is a CohenMacaulay algebra.
PROOF PROOF PROOF PROOF PROOF : The CohenMacaulayness follows from the same calculation:
j n,1 , . . . , j n,l , P l1 , . . . , P 0 , d n,l , . . . , d n,n1 # B
is a regular sequence, and hence in the quotient algebra
j n,1 , . . . , j n,l , d n,l , . . . , d n,n1 # B/ P 0 , . . . , P l1 ,
is a regular sequence of length n .
What we are about to do is to show that this algebra B/ P 0 , . . . , P l1 is precisely the ring of invariants
we are looking for, and, moreover the same as A.
The proof of the following lemma uses Nagata's theorem, see [9], i.e., one of the few existing standard
methods to prove that a given ring is the desired ring of invariants, compare [10].
LEMMA 6.3
LEMMA 6.3
LEMMA 6.3
LEMMA 6.3
LEMMA 6.3 : The algebra B/ P 0 , . . . , P l1 is a unique factorization domain.
PROOF PROOF PROOF PROOF PROOF : We rewrite our system of relations P 0 , . . . , P l1 as a system of linear equations for the
Dickson classes, i.e., the system
P 0 = 0
. . . . . .
P i = 0
. . . . . .
P l1 = 0
18
SYMPLECTIC INVARIANTS
is by Corollary 4.4 equivalent to
n
P j=1
(1) j j n,j d n,j = 0
. . . . . .
n
P
j=i+1
(1) j j q i
n,ji d n,j 
i1
P j=0
(1) j j q
j
n,ij d n,j = 0
. . . . . .
n
P j=l
(1) j j q l1
n,jl+1 d n,j 
l2
P j=0
(1) j j q j
n,l1j d n,j = 0,
what in turn can be written as
n1
P j=l
(1) j j n,j d n,j +
l1
P j=1
(1) j j n,j d n,j + (1) n j n,n = 0
. . . . . .
n1
P j=l
(1) j j q
i
n,ji d n,j +
l1
P
j=i+1
(1) j j q i
n,ji d n,j 
i1
P j=0
(1) j j q j
n,ij d n,j + (1) n j q i
n,ni = 0
. . . . . .
n1
P j=l
(1) j j q l1
n,jl+1 d n,j 
l2
P j=0
(1) j j q
j
n,l1j d n,j + (1) n j q l1
n,nl+1 = 0.
We set
M =
2 6 6 6 6 4
(1) l j n,l · · · (1) n1 j n,n1
· · · · · · · · ·
(1) l j q i
n,li · · · (1) n1 j q i
n,n1i
· · · · · · · · ·
(1) l j q l1
n,1 · · · (1) n1 j q l1
n,l
3 7 7 7 7 5
and
N =
2 6 6 6 6 4
0 j n,1 · · · · · · (1) i1 j n,i1 · · · · · · (1) l j n,l1
· · · · · · · · · · · · · · · · · · · · · · · ·
j n,i j q
n,i1 · · · (1) i1 j q i1
n,1 0 (1) i j q i
n,1 · · · (1) l j q i
n,l1i
· · · · · · · · · · · · · · · · · · · · · · · ·
j q
n,l1 j q
n l2 · · · (1) i j q i1
n,li · · · · · · (1) l2 j q l2
n,1 0
3 7 7 7 7 5
and get a system of linear equations as follows
M
2 6 4
d n,l
. . .
d n,n1
3 7 5  N
2 6 4
d n,0
. . .
d n,l1
3 7 5 + (1) n
2 6 6 4
j n,n
. . .
j q l1
n,nl+1
3 7 7 5 =
2 6 4
0
. . .
0
3 7 5 .
Recall from Proposition 5.1 that the Dickson classes of high degree are given by
d n,j =
j
X k=0
T q k
lj,jk+1 (j n,1 , . . . , j n,2(lk)1 )d n,nk
= T lj,j+1 (j n,1 , . . . , j n,2l1 ) +
j
X k=1
T q k
lj,jk+1 (j n,1 , . . . , j n,2(lk)1 )d n,nk .
19
MARA D. NEUSEL
We put that in our system of linear equations, set T i,j (j) := T i,j (j n,1 , . . . , j n,2(i+j)1 ) for short and get
2 6 4
0
. . .
0
3 7 5 = M
2 6 4
d n,l
. . .
d n,n1
3 7 5  N
2 6 4
d n,0
. . .
d n,l1
3 7 5 + (1) n
2 6 6 4
j n,n
. . .
j q l1
n,nl+1
3 7 7 5
= M
2 6 4
d n,l
. . .
d n,n1
3 7 5  N
2 6 6 6 4
d n,0
T l1,2 (j)
. . .
T 1,l (j)
3 7 7 7 5
 N
2 6 6 6 6 4
0
T q
l1,1 (j)
. . .
T q
1,l1 (j)
3 7 7 7 7 5
d n,n1  · · ·
· · ·  N
2 6 6 6 4
0
. . .
0
T q l1
1,1 (j)
3 7 7 7 5
d n,l+1 + (1) n
2 6 6 4
j n,n
. . .
j q l1
n,nl+1
3 7 7 5 .
Denote by
L :=
2 6 6 6 6 4
0
. . .
. . .
0
0 B B B @
N
2 6 6 6 4
0
. . .
0
T q l1
1,1 (j)
3 7 7 7 5
1 C C C A
· · ·
0 B B B B @
N
2 6 6 6 6 4
0
T q
l1,1 (j)
. . .
T q
1,l1 (j)
3 7 7 7 7 5
1 C C C C A
3 7 7 7 7 5
the l × l matrix with columns
0 B B B @
0
. . .
. . .
0
1 C C C A
0 B B B @
N
2 6 6 6 4
0
. . .
0
T q l1
1,1 (j)
3 7 7 7 5
1 C C C A
· · ·
0 B B B B @
N
2 6 6 6 6 4
0
T q
l1,1 (j)
. . .
T q
1,l1 (j)
3 7 7 7 7 5
1 C C C C A
.
Then
2 6 4
0
. . .
0
3 7 5 = M
2 6 4
d n,l
. . .
d n,n1
3 7 5  N
2 6 4
d n,0
. . .
d n,l1
3 7 5 + (1) n
2 6 6 4
j n,n
. . .
j q l1
n,nl+1
3 7 7 5
= M
2 6 4
d n,l
. . .
d n,n1
3 7 5  L
2 6 4
d n,l
. . .
d n,n1
3 7 5
 N
2 6 6 6 4
d n,0
T l1,2 (j)
. . .
T 1,l (j)
3 7 7 7 5
+ (1) n
2 6 6 4
j n,n
. . .
j q l1
n,nl+1
3 7 7 5
= T
2 6 4
d n,l
. . .
d n,n1
3 7 5  N
2 6 6 6 4
E n (j n,1 , . . . , j n,n1 ) q1
T l1,2 (j)
. . .
T 1,l (j)
3 7 7 7 5
+ (1) n
2 6 6 4
j n,n
. . .
j q l1
n,nl+1
3 7 7 5 ,
where we set
T := M  L
20
SYMPLECTIC INVARIANTS
and use that the top Dickson class d n,0 is nothing but the (q  1)st power of the Euler class E n what in
turn is a polynomial in the first n  1 j's
d n,0 = (E n ) q1 = E n (j n,1 , . . . , j n,n1 ) q1
by Proposition 3.2. The matrix T is modulo j n,1 , . . . , j n,l1 upper triangular with determinant
det(T) º (1) (3l1)l
2 j n,l j q
n,l · · · j q l1
n,l =: D MOD MOD MOD MOD MOD j n,1 , . . . , j n,l1 .
By Lemma 6.2 we know that
j n,l # B/ P 0 , . . . , P l1
is not a zero divisor, hence so is D . Therefore we get by localizing at D an inclusion
B/ P 0 , . . . , P l1 # B[D 1 ]/ P 0 , . . . , P l1 .
In the bigger algebra the system of equations given by the relations P 0 , . . . , P l1 can be solved for
d n,l , . . . , d n,n1 (well, we just inverted the determinant of the matrix of our system of linear
equations) and hence
B[D 1 ]/ P 0 , . . . , P l1 = IF[j n,1 , . . . , j n,n , D 1 ]
is a polynomial algebra, and in particular an integral domain. Hence
B/ P 0 , . . . , P l1
is also an integral domain.
Next, we want to show that D is a prime element in B/(P 0 , . . . , P l1 ). Observe that the entries of the
matrix T are polynomials in j n,1 , . . . , j n,n1 . However,j n,n1 occurs only in the top right corner, i.e., we
can rewrite T as
T :=
2 6 6 6 6 4
(1) l j n,l · · · (1) n2 j n,n2 (1) n1 j n,n1
(1) n1 j q
n,n2 + N 1
T cof . . .
(1) n1 j q l1
n,l + N l1
3 7 7 7 7 5
,
where we set
N
2 6 6 6 6 4
0
T q
l1,1 (j)
. . .
T q
1,l1
3 7 7 7 7 5
(j) =
2 6 6 6 4
0
N 1
. . .
N l1
3 7 7 7 5
.
The cofactor matrix T cof has determinant
D cof
# IF[j n,1 , . . . , j n,n2 ].
Therefore
D cof
# B/(P 0 , . . . , P l1 , D )
is not a zero divisor (with a little help from Lemma 6.2). So, if we invert this determinant we get an
inclusion
B/(P 0 , . . . , P l1 , D ) # B[(D cof ) 1 ]/(P 0 , . . . , P l1 , D ).
21
MARA D. NEUSEL
In the localization the relations P 1 , . . . , P l1 can be solved for d n,l , . . . , d n,n2 and P 0 gives an equation
for j n,n . The determinant D is linear in j n,n1 and since we inverted the leading coefficient, namely
D cof , D can be solved for j n,n1 . Hence
B[(D cof ) 1 ]/(P 0 , . . . , P l1 , D ). = IF[j n,1 , . . . , j n,n2 , d n,n1 , (D cof ) 1 ]
is a polynomial ring and in particular an integral domain. Therefore the little algebra
B/(P 0 , . . . , P l1 , D )
is an integral domain, what in turn implies that
D # B/(P 0 , . . . , P l1 )
is a prime element. Therefore
B/(P 0 , . . . , P l1 )
is a unique factorization domain, because its localization at D
B[D 1 ]/ P 0 , . . . , P l1 = IF[j n,1 , . . . , j n,n , D 1 ]
is, where we use Nagata's wonderful theorem, [9] or [2] Lemma 2.2.2. .
Finally, we are going to be rewarded with the explicit description of the ring of invariants of the sym
plectic group.
THEOREM 6.4
THEOREM 6.4
THEOREM 6.4
THEOREM 6.4
THEOREM 6.4 : With the preceding notation, the ring of invariants IF[V ] Sp(n, IF) is given by
B/ P 0 , . . . , P l1 = IF[j n, 1 , . . . , j n, n1 , d n, l , . . . , d n, n1 ]/ P 1 , . . . , P l1 .
PROOF PROOF PROOF PROOF PROOF : We have the remembering map
# : B = IF[j n, 1 , . . . , j n, n1 , d n, l , . . . , d n, n ] # IF[V ] Sp(n, IF) .
By Corollary 4.4 the kernel contains the polynomials P 0 , . . . , P l1 . Hence # factorizes through
v : B/ P 0 , . . . , P l1 # IF[V ] Sp(n, IF) .
In Lemma 6.3 we have seen that the quotient
B/ P 0 , . . . , P l1
is an integral domain, so the map v is injective. By construction we have that
D * (n) # Im(v) # IF[V ] Sp(n, IF) .
Since the over all ring extension is finite, so is
Im(v) # IF[V ] Sp(n, IF) ,
and in particular integral. So, let's have a look at what we have now:
D * (n) # B/ P 0 , . . . , P l1 @ Im(v) #
integral, finite
IF[V ] Sp(n, IF)
#
integral, finite
IF[V ]
and at the level of fields of fractions:
FF(D * (n)) # FF(B/ P 0 , . . . , P l1 ) @ FF(Im(v)) #
finite
IF(V ) Sp(n, IF)
#
Galois
IF[V ].
22
SYMPLECTIC INVARIANTS
So, we have Galois groups as follows
Gal IF[V ]/F F(D * (n)) = GL(n, IF) # Gal IF[V ]/F F(Im(v) # Gal IF[V ]/IF(V ) Sp(n, IF)
= Sp(n, IF).
Since the image of v contains j n,1 , so does its field of fractions, i.e.,
Gal IF[V ]/F F(Im(v) # GL(n, IF) j n,1
= Sp(n, IF),
where we made use of Lemma 2.2. That means that our field of fractions is correct:
Im(v) #
integral
F[V ] Sp(n, IF)
# #
FF(Im(v)) = IF(V ) Sp(n, IF) .
Moreover, IF[V ] Sp(n, IF) is integrally closed, because its a ring of invariants (well, that goes back to
Emmy) and Im(v) is integrally closed, because its a unique factorization domain, as we had seen in
Lemma 6.3, [14] Example 1 of Section V.3. Since the ring extension is integral we have that both rings
must be equal .
Finally let's just summarize what we know about the symplectic invariants. First of all they are ex
plicitely given by
IF[V ] Sp(n, IF) = IF[j n,1 , . . . , j n,n1 , d n,l , . . . , d n,n1 ]/ P 1 , . . . , P l1 ,
where the generators are given by Proposition 2.1
j n, i :=
l
X j=1
(x j y q i
j  y j x q
i
j ) # IF[x 1 , y 1 , . . . , x l , y l ] Sp(n, IF) ,
while the d n,i 's are the Dickson classes of degree q n  q i . The relations are explicity given by Proposition
3.5 and Corollary 4.4
P i :=
n
X
j=i+1
(1) j j q i
n,ji d n,j 
i1
X j=0
(1) j j q j
n,ij d n,j
for i = 1 , . . . , l  1. Moreover the ring has Krull dimension n and is an integral domain, which is no
surprise, since any ring of invariants has these properties. What is not always the case but holds for
the ring of invariants of the symplectic group is: it is a unique factorization domain by Lemma 6.3, and
a complete intersection by Lemma 6.2 (and a fortiori CohenMacaulay), i.e., its still a relatively nice
ring.
§7. Examples and Rational Invariants
We want to calculate in this final section also the rational invariants of the symplectic group, what is
pretty easy now after all this hard work. Moreover, we have a look at some examples.
THEOREM 7.1
THEOREM 7.1
THEOREM 7.1
THEOREM 7.1
THEOREM 7.1 : The rational invariants of the symplectic group are given by
IF(V ) Sp(n, IF) = IF(j n,1 , . . . , j n,n ).
23
MARA D. NEUSEL
PROOF PROOF PROOF PROOF PROOF : By Theorem 6.4
IF(V ) Sp(n, IF) = FF IF[j n, 1 , . . . , j n, n , d n, l , . . . , d n, n1 ]/ P 0 , . . . , P l1 ,
(Note carefully that I put back in the last j n, n which is compensated by adding the respective relation
P 0 .) However, in the field of fractions, the relations P 1 , . . . , P l1 can be solved for d n, l , . . . , d n, n1 ,
while P 0 becomes the trivial relation, i.e., with the help of P 1 , . . . , P l1 the Dickson polynomials can be
expressed as rational functions in j n,1 , . . . , j n,n . So we have 11
IF(V ) Sp(n, IF) = IF(j n, 1 , . . . , j n, n )
and this is indeed neat .
Next we look at examples. If n = 2 then the symplectic group is nothing else than the special linear
group
Sp(2, IF) = SL(2, IF),
(do Exercise 8.13 of [12] if you have doubts). So we knew the invariants from Dickson's work, namely
IF[V ] Sp(2, IF) = IF[E 2 , d 2,1 ],
which you, of course, can look up in the bible: Theorem 8.1.8 in [13].
So, a bit more interesting is the next case: n = 4, and let's take the field with 3 elements. Then our
symplectic group has order
51840 = 2 7 3 4 5,
by Dickson's calculations. The group is generated by the matrices
2 6 6 4
0 1 0 0
1 0 0 0
0 0 1 0
0 0 0 1
3 7 7 5 ,
2 6 6 4
1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0
3 7 7 5 ,
2 6 6 4
1 1 0 0
0 1 0 0
0 0 1 0
0 0 0 1
3 7 7 5 ,
2 6 6 4
1 0 0 0
0 1 0 0
0 0 1 1
0 0 0 1
3 7 7 5 ,
2 6 6 4
1 0 0 1
0 1 0 0
0 1 1 0
0 0 0 1
3 7 7 5 ,
(recall Dickson's Theorem 1.2 from the first Section) and the proof of Satz 9.25 in [7] tells you that the
bad guy, i.e., the 3Sylow sub group is non Abelian and consists of matrices of the form
2 6 6 4
1 d a b
0 1 0 0
0 b  ac 1 c
0 a 0 1
3 7 7 5 .
So, our ring of invariants is generated by
j 4,1 , . . . , j 4,4 , E 4 , d 4,1 , d 4,2 , d 4,3 .
The relations given by Proposition 3.2 are 12
E 4 = j 3
4,1 j 4,3  j 4
4,2 + j 10
4,1
E 4 d 4,1 = j 4,2 j 27
4,1 + j 9
4,1 j 4,4  j 9
4,2 j 4,3
E 4 d 4,2 = j 28
4,1 + j 3
4,2 j 4,4  j 4
4,3
E 4 d 4,3 = j 4,1 j 9
4,2  j 3
4,1 j 4,4  j 3
4,3 j 4,2 .
11 Note, that this is purely trancendental over IF.
12 The proof of this proposition gives you an explicite algorithm to find these expressions: note first that for n = 4 we have
that t = (12)(34), the centralizer C(t) = {e, (12), (34), (12)(34), (13)(24), (1324), (1423), (14)(23)} and a complete set of coset
representatives in S 4 is given by e, (13) and (14). So, if you use the formula given in Proposition 3.2 you get the above expression.
24
SYMPLECTIC INVARIANTS
Note that E 4 is a polynomial in j 4,1 , j 4,2 , j 4,3 as Proposition 3.2 (1) predicts. The relation P 0 given in
Proposition 3.5 reads as follows
P 0 = j 4,1 d 4,1 + j 4,2 d 4,2 + j 4,3 d 4,3 + j 4,4 = 0.
Corollary 4.4 hands us the remaining P 1 , which is
P 1 = j q
4,1 d 4,2  j q
4,2 d 4,3 + j q
4,3  j 4,1 d 4,0 = 0.
The british T's of Proposition 5.1, evaluated at j 4,1 , . . . , j 4,4 are given by
T 1,1 = j 2
4,1
T 1,2 = d 4,1  j 6
4,1 d 4,3 = j 27
4,1 j 4,2  j 9
4,2 j 4,3  j 7
4,1 j 9
4,2 + j 6
4,1 j 4,2 j 3
4,3
T 2,1 = d 4,0 = E 2
4 = j 3
4,1 j 4,3  j 4
4,2 + j 10
4,1 2
The model algebra B of Section 6 is
B = IF[j 4,1 , j 4,2 , j 4,3 , j 4,4 , d 4,2 , d 4,3 ]/(P 0 , P 1 )
and the system of linear equations used in Lemma 6.3 looks like
j 4,2 j 4,3
j 3
4,1 j 3
4,2  j 7
4,1
d 4,2
d 4,3
+ j 4,1 E 2
4
j 4,1 T 1,2
+ j 4,4
j 3
4,3
= 0.
So, we get
D = j 4,2 j 3
4,2 + j 7
4,1  j 3
4,1 j 4,3
and the cofactor determinant is just j 3
4,1 .
References
[1] Michael Aschbacher: Finite Groups Theory, Cambridge Studies in Advanced Mathematics 10, Cambridge
University Press, Cambridge 1993.
[2] Stanislaw Balcerzyk and Tadeusz J’ ozefiak: Commutative Rings. Dimension, Multiplicity and Homological
Methods, Ellis Horwood Limited, Chichester 1989.
[3] David J. Benson: Polynomial Invariants of Finite Groups, London Math. Soc. Lecture Note Series 190,
Cambridge University Press, Cambridge 1993.
[4] Armand Borel: Linear Algebraic Groups, 2nd edition, Graduate Texts in Mathematics 126, SpringerVerlag,
New York 1991.
[5] David Carlisle and Peter H. Kropholler: Modular Invariants of Finite Symplectic Groups, preprint, Manch
ester 1992.
[6] Leonard E. Dickson: Linear Groups, Dover Publications Inc., New York 1958.
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Verlag, Berlin 1967.
[8] Nathan Jacobson: Basic Algebra I, W. H. Freeman and Company, San Francisco 1974.
[9] Masayoshi Nagata: A Remark on the Unique Factorization Theorem, J. of the Math. Soc. of Japan 9 (1957),
143145.
[10] Mara D Neusel: Invariants of some Abelian pGroups in Characteristic p, Proceedings of the AMS 125
(1997), 19211931.
25
MARA D. NEUSEL
[11] Mara D Neusel: The Inverse Invariant Theory Problem and Steenrod Operations, preprint, Kingston &
Minneapolis 1997/8.
[12] Joseph J. Rotman: An Introduction to the Theory of Groups, 4th edition, Graduate Texts in Mathematics
148, SpringerVerlag, New York 1995.
[13] Larry Smith: Polynomial Invariants of Finite Groups, 2nd corrected printing, A.K. Peters Ltd., Wellesley
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[14] Oskar Zariski and Pierre Samuel: Commutative Algebra. Volume I, D. van Nostrand Company, Inc., Prince
ton NJ 1958.
26