*,*            *
             A subspace of ExtA   (Zp , Zp )

                           Zheng Qi-Bing

         Department of Mathematics, Nankai University

                       Tianjin 300071, China



                             July 10,2002



                               Abstract

        In this paper, we compute the homology and cohomology of some
     Hopf algebras and find a subspace of the cohomology of the Steenrod
     algebra that include the representative for the Greek letter families.



    In this paper, all discussions are based on a given odd prime p. Zp is the

group of integers modular p and   means tensor product over Zp.

    Let Gn be the graded Lie algebra over Zp with basis {xi,j| 06 i< j6 n}

and Lie brackets defined by that [xj,k, xi,j] = -[xi,j, xj,k] = xi,kfor i< j< k

and [xi,j, xk,l] = 0 otherwise.  kxi,jk = 2(p - 1)(pi+ . .+.pj-1).  Then, Gn

is a sub-Lie algebra of Gn+1 and we define G = [nGn.  Let U(Gn) be the

enveloping algebra of Gn. Then U(Gn) is an associative and co-commutative

Hopf algebra with xi,jprimitive for all i, j.  So is U(G).  The subalgebra

of U(Gn) generated by xpi,jis the center of U(Gn) and we define U(n) to

be the quotient algebra of U(Gn) modular the ideal generated by its center.

U = [nU(n). We always regard U(n) as a subspace of U(Gn).
______________________________
   *Supported by National Natural Science Foundation of China
                                   1



   In this paper, we give a minimal free resolution of U(Gn) (Theorem 1.6)

and compute H*,*(Gn+1) and Hn+1,*(G) for n < p (Theorem 2.8) and com-

pute H*,*(U(n)) for n < p (Theorem 2.9) and prove that Hn,*(G) survives to

infinity in May spectral sequence for n < p (Theorem 2.10) and find a sub-

space of the cohomology of the Steenrod algebra (Theorem 3.7) that includes

the representative of the Greek letter families.


1     Hopf Algebra's Free Resolution



   Definition 1.1 Let G(U(n)) be the set of all upper triangular matrix

                          0                            1
                            a0,1  a0,2 a0,3 . . .a0,n
                          BB                           C
                          B   0   a1,2 a1,3 . . .a1,n  CC
                 (ai,j) = BB  0    0   a2,3 . . .a2,n  CC
                          B@ . . .. . .. . .. . . . . .CA

                              0    0    0   . . .an-1,n


with 0 6 ai,j< p for i < j. We use such a matrix A = (ai,j) 2 G(U(n)) to

denote an element

             P0a0,1,1P1a1,2,2P0a0,2,2......Pnan-1,n-1,n.P.a.1,n1,nP0a0,n,n

in U(n), where Pis,j= _1_s!xsi,j(Pi0,j= 1).  Then, G(U(n)) is a basis of U(n).

The total degree `(A) of A is an (n+ 1)-tuple (`0(A), . .,.`n(A)) defined by
        P n         P k-1
`k(A) =   s=k+1ak,s-  s=0 as,kfor 0 6 k 6 n.  The weight of A is the n-

tuple w(A) = (i0, i0+ i1, . .,.i0+ . .+.in-1) if `(A) = (i0, i1, . .,.in). The *
 *total

degree is a multi-degree of U(n) in the sense that let Un(i0, . .,.in) be the

subspace of U(n) spanned by those matrices A with `(A) = (i0, . .,.in), then,

Un(i0, . .,.in)Un(j0, . .,.jn)   Un(i0+ j0, . .,.in+ jn).


                                   2



   Similarly, let G(U) be the set of all upper triangular matrix

                              0                    1
                                a0,1  a0,2 a0,3 . . .
                              B   0   a    a    . .C.
                     (ai,j) = BB@      1,2  1,3    CC
                                  0    0   a2,3 . .A.

                                 . . .. . .. . .. . .


with 0 6ai,j<p for i< j and there are only finite number of ai,j's with ai,j>0.

We use such a matrix A = (ai,j) 2 G(U) to denote an element

                     P0a0,1,1P1a1,2,2P0a0,2,2P2a2,3,3P1a1,3,3P0a0,3,3. . .

in U.  Then, G(U) is a basis of U.  The total degree `(A) of A is an infi-
                                                  P 1         P k-1
nite tuple (`0(A), `1(A), . .).defined by `k(A) =   s=k+1ak,s-  s=0 as,k. The

weight of A is w(A) = (i0, i0+ i1, . .).if `(A) = (i0, i1, . .)..  We also have

that as a multi-degree, `(ab) = `(a)+ `(b) and w(ab) = w(a)+ w(b) for all

a, b 2 U.

    It is easy to check that for a 2 U(n) and `(a) = (i0, . .,.in) and w(a) =

(j1, . .,.jn), i0+ . .+.in = 0 and jk > 0 for k = 1, . .,.n. For a 2 U and `(a)*
 * =

(i0, i1, . .).and w(a) = (j0, j1, . .)., there is an n such that i0+ . .+.ik = *
 *jk = 0

for k > n.


                                  P
    Definition 1.2 For n > 1, let   (n) = {(i0, i1, . .,.in) | (i0, . .,.in) is*
 * a
                                                        P
permutation of (0, . .,.n)}. For 0 < n < p_and ff, fi 2!  (n), ff = (i0, . .,.i*
 *n),

                                             i0, . .,.in
fi = (j0, . .,.jn), we define D(fi, ff)= D               2 U(n)   U(Gn) as
                                            j0, . .,.jn

follows.  D(ff, fi) 6= 0 only if there are 0 6 s < t 6 n and 0 6 M < N 6 n

such that is = jt = N and it = js = M and ik = jk otherwise.

               _                                          !
                 . . .N,  is+1, is+2, . .,. it-1, M,  . . .
             D
                 . . .M,  is+1, is+2, . .,. it-1, N,  . . .


                                   3

                            0                            1
     0                     1   as,s+1  as,s+2  . . .as,t
  X     t-1Y                BB        as+1,s+2 . . .as+1,tCC
=    @      Z(M, N; iu, xu)AB                            C , (0 omitted)
       u=s+1                @                  . . . . . .A
                                                    at-1,t

           P u-1
where xu =   v=sav,uand the sum is taken throughout all matrices (ai,j) 2

G(U(n)) such that `t(ai,j) = M- N and `s(ai,j) = N- M and `k(ai,j) = 0

otherwise. The coefficient is defined by ( 1 if t = s+ 1 )

                8 _         !
                >>> N- k- x  -1
                >>>                                  ifk < M
                ><   M - k
                                      _        ! -1
Z(M, N; k, x) = >              k- N    k- M- 1
                >>>(-1)N-M-x  ________               ifk > N, x 6 N- M
                >>>            k- M        x
                : 0                                  otherwise


                          _                           !
                            . .,.N, is+1, . .,.it-1, M, . . .
    Notice that for N> M, D                             6= 0 if and only if
                            . .,.M, is+1, . .,.it-1,_N, . . . !

                                                 . .,.N, M, . . .   N-M
for all s < k < t, either ik < M or ik > N and D                = Ps,s+1.
                                                 . .,.M, N, . . .


    Definition 1.3 For 0 < n < p, we define M(n) to be the free right U(n)
                            P               P
module generated by the set   (n). For ff 2   (n), we still use ff to denote the

corresponding generator in M(n). The right U(n)-module homomorphism ffi
                                P     <ff,fi>
on M(n) is defined by ffi(ff) =   (-1)     fiD(fi, ff), where the sum is taken

throughout all non-zero D(fi, ff)'s and the degree is defined as follows.  If

ff = (i0, . .,.in), fi = (j0, . .,.jn) and for s < t, is= jt > it= js, then <ff*
 *, fi >=

the number of (u, v)'s such that u < v < t but iu > iv.

    Similarly, we define M(Gn) to be the free right U(Gn)-module generated
           P
by the set   (n). Since U(n) is a subspace of U(Gn), ffi also induces a right

U(Gn)-module homomorphism on M(Gn) which we still denote by ffi.

                                   4



   The following theorems are to show that ffi is a differential on M(Gn) and

M(n).



    Theorem 1.4 Suppose for i< M< N< j6 n, the following D(ff, fi)'s are

all non-zero. Then, we have in U(Gn)
                  _                  !   _                   !
                   . .,.M, i, . .,.N, . . .. .,.N , i, . .,.M, . . .
                D                      D
                   . .,.i, M, . .,.N, . . .. .,.M, i, . .,.N, . . .
                  _                   !  _                  !
                   . .,.i, N , . .,.M, . ....,.N, i, . .,.M, . . .
            =   D                      D
                   . .,.i, M, . .,.N , . ....,.i, N, . .,.M, . . .

                 _                   !   _                  !
                  . .,.N, . .,.i, M, . . .. .,.N, . .,.M, i, . . .
               D                       D
                  . .,.M, . .,.i, N, . . .. .,.N, . .,.i, M, . . .
                 _                  !   _                  !
                  . .,.M, . .,.N, i, . . .. .,.N, . .,.M, i, . . .
            =  D                      D
                  . .,.M, . .,.i, N, . . .. .,.M, . .,.N, i, . . .



                 _                   !   _                  !
                  . .,.N , j, . .,.M, . . .. .,.j, N, . .,.M, . . .
               D                       D
                  . .,.M, j, . .,.N , . . .. .,.N, j, . .,.M, . . .
                 _                  !   _                   !
                  . .,.j, M, . .,.N, . . .. .,.j, N , . .,.M, . . .
            =  D                      D
                  . .,.M, j, . .,.N, . . .. .,.j, M, . .,.N, . . .

                 _                  !   _                   !
                  . .,.M, . .,.j, N, . . .. .,.N , . .,.j, M, . . .
               D                      D
                  . .,.M, . .,.N, j, . . .. .,.M, . .,.j, N, . . .
                 _                   !   _                  !
                  . .,.N , . .,.M, j, . . .. .,.N, . .,.j, M, . . .
            =  D                       D
                  . .,.M, . .,.N , j, . . .. .,.N, . .,.M, j, . . .

 Proof.  It is easy to check that for k+ l < p and s < t < u, we have

            i   j                       P min(k,l)
Psk,tPsl,t=  k+lkPsk+l,tand Ptk,uPsl,t=   i=0   Psl-i,tPtk-i,uPsi,uin U(G). So,


             (Psas+1,s+2+1,s+2Psas+1,s+3+1,s+3.P.a.s+1,s+ts+1,s+t)PsN-i,s+1
             X               as+1,s+2-x1         as+1,s+t-xt-1x
          =     PsN-i-x,s+1Ps+1,s+2   Psx1,s+2.P.s.+1,s+t  Ps,st-1+t
                   0   _        !                                 1
                    X    N- i- x  -1             as+1,s+2-x1   x
          =  PsM-i,s+1@             PsN-M-x,s+1Ps+1,s+2   . .P.st-1,s+tA
                          M- i



                                   5



where the_sum is taken throughout!all_x1+ . .+.xt-1 = x.!This implies that

           . .,.N , i, . .,.M, . . .. .,.i, N , . .,.M, . .N.-i
PsM-i,s+1D                      = D                     Ps,s+1, which is just
           . .,.M, i, . .,.N, . . . . .,.i, M, . .,.N , . . .

the first equality .

    Similarly, we have



        PsN-i+t,s+t+1(Psas+t-1,s+t+t-1,s+tPsas+t-2,s+t+t-2,s+t.P.a.s,s+ts,s+t)
        X    as+t-1,s+t-x1  as,s+t-xt
     =     Ps+t-1,s+t  . .P.s,s+t  PsN-i-x+t,s+t+1Psx1+t-1,s+t+1.P.x.ts,s+t+1
        0    _        !                                        1
          X   N- i- x  -1  as+t-1,s+t-x1
     =  @                Ps+t-1,s+t  . .P.N-M-xs+t,s+t+1.P.x.ts,s+t+1APsM-i+t,s*
 *+t+1
                M- i


where_the sum is taken!throughout all x1+ . .+.xt_= x.  This implies!that

  . .,.N , . .,.i, M, . .M.-i       N-i       . .,.N , . .,.M, i, . . .
D                     Ps+t,s+t+1= Ps+t,s+t+1D                     , which
  . .,.M, . .,.i, N, . . .                    . .,.M, . .,.N , i, . . .

is just the second equality .

    It is easy to check that for 0 6 z 6 b- a, the following formula holds.


               Xz      _ b- k!-1_ z!        a _  b- 1 !-1
                  (-1)k              = (-1)z__
               k=0        a       k         b  b- a- z


So, we have



        Psj-M,s+1(Psas+1,s+2+1,s+2Psas+1,s+3+1,s+3.P.a.s+1,s+ts+1,s+t)
        X         as+1,s+2-x1         as+1,s+t-xt-1x       j-M-x
    =      (-1)xPs+1,s+2   Psx1,s+2.P.s.+1,s+t  Ps,st-1+tPs,s+1
        8        _         !                                      9
        < X        j- M- x  -1  a      -x        x                =
    =   :   (-1)x             Ps+s+1,s+211,s+2.P.s.t-1,s+tPsN-M-x,s+1Psj-N,s+1
                    j- N                                          ;
        8        _         !  _       !   _          !
        < X        j- M- x  -1  x1+ y1      xt-1+ yt-1
    =   :   (-1)x                       . . .          PsN-M-x-y,s+1
                    j- N          y1           yt-1
                                                                               *
 *       )

             . Psas+1,s+2-x1-y1+1,s+2Psx1+y1,s+2.P.a.s+1,s+t-xt-1-yt-1s+1,s+tPs*
 *xt-1+yt-1,s+tPsj-N,s+1

           8 0        _         !  _  ! 1
        X  <  Xz        j- M- k  -1  z
    =      : @   (-1)k                  A PsN-M-z,s+1
              k=0        j- N        k



                                   6

                                                                      )

            . Psas+1,s+2-z1+1,s+2Psz1,s+2.P.a.s+1,s+t-zt-1s+1,s+tPszt-1,s+tPsj-*
 *N,s+1

          8            _         !
       X  <       j- N   j- M- 1   -1
    =     : (-1)z______              PsN-M-z,s+1
                 j- M   N- M- z
                                                                      )

            . Psas+1,s+2-z1+1,s+2Psz1,s+2.P.a.s+1,s+t-zt-1s+1,s+tPszt-1,s+tPsj-*
 *N,s+1



where_the sum is taken!throughout all_u1+ . .+.ut-1 = u, u!= x, y, z.  So,

  . .,.N , j, . .,.M, . .j.-N   j-M   . .,.j, N , . .,.M, . . .
D                     Ps,s+1= Ps,s+1D                     , which is just
  . .,.M, j, . .,.N, . . .            . .,.j, M, . .,.N , . . .

the third equality .

    We have



      (Psas+t-1,s+t+t-1,s+tPsas+t-2,s+t+t-2,s+t.P.a.s,s+ts,s+t)Psj-M+t,s+t+1
      X        j-M-x     as+t-1,s+t-x1  as,s+t-xt
  =     (-1)xPs+t,s+t+1Ps+t-1,s+t  . .P.s,s+t  Psx1+t-1,s+t+1.P.x.ts,s+t+1
               8         _         !                                           *
 *  9
               < X         j- M- x  -1                 a       -x              *
 *  =
  =   Psj-N+t,s+t+1:(-1)x             PsN-M-x+t,s+t+1Ps+s+t-1,s+t1t-1,s+t.P.x.t*
 *s,s+t+1
                            j- N                                               *
 *  ;
               8         _         !  _       !   _      !
               < X         j- M- x  -1  x1+ y1      xt+ yt
  =   Psj-N+t,s+t+1:(-1)x                       . . .
                            j- N          y1          yt
                                                                               *
 *                      )

          . Psas+t-1,s+t-x1-y1+t-1,s+t.P.a.s,s+t-xt-yts,s+tPsN-M-x-y+t,s+t+1Psx*
 *1+y1+t-1,s+t+1.P.x.t+yts,s+t+1

               8               _         !
               < X        j- N   j- M- 1   -1
  =   Psj-N+t,s+t+1:(-1)z______
                         j- M   N- M- z
                                                                               *
 *           )

          . Psas+t-1,s+t-z1+t-1,s+t.P.a.s,s+t-zts,s+tPsN-M-z+t,s+t+1Psz1+t-1,s+*
 *t+1.P.z.ts,s+t+1



where the sum_is taken throughout all!u1+_. .+.ut = u, u = x,!y, z.  So,

               . .,.N , . .,.j, M, . . .. .,.N , . .,.M, j, . .j.-M
Psj-N+t,s+t+1D                      = D                     Ps+t,s+t+1, which
               . .,.M, . .,.j, N, . . . . .,.M, . .,.N , j, . . .

is just the fourth equality .                                        Q.E.D.



    Theorem 1.5 ffi is a differential on M(Gn) and M(n).



                                   7

                               P       <fl,fi><fi,ff>
    Proof.  We must prove that   fi(-1)          D(fl, fi)D(fi, ff) = 0 for

any ff and fl. For given ff and fl, there are at most two fi1 and fi2 such that

both D(fl, fi1)D(fi1, ff) and D(fl, fi2)D(fi2, ff) are non-zero (otherwise, they

are all zero).  These are the following cases for i < j < k and s < t and in

(6) and (7), either j < s or i > t .
                    _       !_       !   _       ! _       !
                       j, i, k  k, i, j     i, k, j   k, i, j
        (1)         D        D         = D         D
                       i, j, k  j, i, k     i, j, k   i, k, j
                    _       ! _       !   _       ! _       !
                       j, k, i   k, j, i     k, i, j   k, j, i
        (2)         D         D         = D         D
                       i, k, j   j, k, i     i, k, j   k, i, j
                    _       !_       !   _       ! _       !
                       j, k, i  k, j, i     k, i, j   k, j, i,
        (3)         D        D         = D         D
                       j, i, k  j, k, i     j, i, k   k, i, j
                    _       ! _       !   _       !_       !
                       i, k, j   j, k, i     j, i, k  j, k, i
        (4)         D         D         = D        D
                       i, j, k   i, k, j     i, j, k  j, i, k
                 _        ! _        !   _        ! _        !
                    j, i, s, t j, i, t, s   i, j, t, s j, i, t, s
        (5)      D          D          = D          D
                    i, j, s, t j, i, s, t   i, j, s, t i, j, t, s
                 _        ! _        !   _        ! _        !
                    j, s, i, t j, t, i, s   i, t, j, s j, t, i, s
        (6)      D          D          = D          D
                    i, s, j, t j, s, i, t   i, s, j, t i, t, j, s
                 _        ! _        !   _        ! _        !
                    s, j, i, t t, j, i, s   t, i, j, s t, j, i, s
        (7)      D          D          = D          D
                    s, i, j, t s, j, i, t   s, i, j, t t, i, j, s


                    _                         !    _        !
                      . .j., . .,.i, . .,.t, . .,.s . .j., i, t, s
where we abbreviate D                           to D          and so on.
                      . .j., . .,.i, . .,.s, . .,.t . .j., i, s, t

    (5) always holds by definition. Now, we prove (1) and (2). In (1) and (2),

there are only three degrees i, j, k whose positions in the permutations are

moved.  We call them moving degrees and call other degrees fixed degrees.

We use induction on the number of fixed degrees that lies between the first

and second moving degrees to prove (1) and (2). If there is no fixed degree

that lies between the first and second moving degrees, then (1) and (2) are

respectively the first and third equality in Theorem 1.4.  Suppose both (1)

and (2) holds if there are less than m fixed degrees between the first and



                                   8



second moving degrees and regardless of u, the following permutations have

m- 1 fixed degrees that lies between the first and second moving degrees. If

u < i, we have
          _         !_         !_         !
             i, u, j, k i, u, k, j k, u, i, j
          D          D          D
             u, i, j, k i, u, j, k i, u, k, j
          _         !_         !_         !
             u, i, k, j i, u, k, j k, u, i, j
      =   D          D          D             (by (5))
             u, i, j, k u, i, k, j i, u, k, j
          _         !_         !_         !
             u, i, k, j u, k, i, j k, u, i, j
      =   D          D          D             (by induction on (1))
             u, i, j, k u, i, k, j u, k, i, j
          _         !_         !_         !
             u, j, i, k u, k, i, j k, u, i, j
      =   D          D          D             (by induction on (1))
             u, i, j, k u, j, i, k u, k, i, j
          _         !_         !_         !
             u, j, i, k j, u, i, k k, u, i, j
      =   D          D          D             (by induction on (1))
             u, i, j, k u, j, i, k j, u, i, k
          _         !_         !_         !
             i, u, j, k j, u, i, k k, u, i, j
      =   D          D          D             (by induction on (1))
             u, i, j, k i, u, j, k j, u, i, k


Notice that U(Gn) is a divisible ring. That is, ab = 0 implies a = 0 or b = 0.

So, by cancelling the left factor in the first and last line, we get (1) for u

between the first and second moving degrees. If u > k, then
          _         !_         !_         !
             i, u, k, j k, u, i, j u, k, i, j
          D          D          D
             i, u, j, k i, u, k, j k, u, i, j
          _         !_         !_         !
             i, u, k, j u, i, k, j u, k, i, j
      =   D          D          D             (by induction on (2))
             i, u, j, k i, u, k, j u, i, k, j
          _         !_         !_         !
             u, i, j, k u, i, k, j u, k, i, j
      =   D          D          D             (by (5))
             i, u, j, k u, i, j, k u, i, k, j
          _         !_         !_         !
             u, i, j, k u, j, i, k u, k, i, j
      =   D          D          D             (by induction on (1))
             i, u, j, k u, i, j, k u, j, i, k
          _         !_         !_         !
             j, u, i, k u, j, i, k u, k, i, j
      =   D          D          D             (by induction on (2))
             i, u, j, k j, u, i, k u, j, i, k
          _         !_         !_         !
             j, u, i, k k, u, i, j u, k, i, j
      =   D          D          D             (by induction on (2))
             i, u, j, k j, u, i, k k, u, i, j



                                   9



For the same reason as above, we may cancel the right factor in the first

and last line and get (1) for u between the first and second moving degrees.

Similarly, we can prove (2). By using induction on the number of fixed de-

grees that lies between the second and third moving degrees, we can similarly

prove (3) and (4).

    Now, we prove (6). If j < s, we have

                  _        !_        !_        !
                     i, s, j, tj, s, i, tj, t, i, s
                  D         D         D
                     i, j, s, ti, s, j, tj, s, i, t
                  _        !_        !_        !
                     j, i, s, tj, s, i, tj, t, i, s
              =   D         D         D            (by (4))
                     i, j, s, tj, i, s, tj, s, i, t
                  _        !_        !_        !
                     j, i, s, tj, i, t, sj, t, i, s
              =   D         D         D            (by (1))
                     i, j, s, tj, i, s, tj, i, t, s
                  _        !_        !_        !
                     i, j, t, sj, i, t, sj, t, i, s
              =   D         D         D            (by (5))
                     i, j, s, ti, j, t, sj, i, t, s
                  _        !_        !_        !
                     i, j, t, si, t, j, sj, t, i, s
              =   D         D         D            (by (4))
                     i, j, s, ti, j, t, si, t, j, s
                  _        !_        !_        !
                     i, s, j, ti, t, j, sj, t, i, s
              =   D         D         D            (by (1))
                     i, j, s, ti, s, j, ti, t, j, s


By cancelling the left factor, we_get (6)!for j < s. If t < i, the case is simi*
 *lar

                                  j, i, t, s
by product with a right factor D          .
                                  j, t, i, s

    Now, we prove (7). For j < s, we have

                  _        !_        !_        !
                     s, i, j, ts, j, i, tt, j, i, s
                  D         D         D
                     i, s, j, ts, i, j, ts, j, i, t
                  _        !_        !_        !
                     j, s, i, ts, j, i, tt, j, i, s
              =   D         D         D            (by (2))
                     i, s, j, tj, s, i, ts, j, i, t
                  _        !_        !_        !
                     j, s, i, tj, t, i, st, j, i, s
              =   D         D         D            (by (1))
                     i, s, j, tj, s, i, tj, t, i, s
                  _        !_        !_        !
                     i, t, j, sj, t, i, st, j, i, s
              =   D         D         D            (by (6))
                     i, s, j, ti, t, j, sj, t, i, s



                                   10

                 _        !_        !_        !
                    i, t, j, st, i, j, st, j, i, s
              =  D         D         D            (by (2))
                    i, s, j, ti, t, j, st, i, j, s
                 _        !_        !_        !
                    s, i, j, tt, i, j, st, j, i, s
              =  D         D         D            (by (1))
                    i, s, j, ts, i, j, tt, i, j, s


By cancelling the left factor, we_get (7)!for j < s. The case t < i is similar

                                  j, t, i, s
by product with a right factor D          .
                                  t, j, i, s

    .                                                              Q.E.D.



    This theorem implies that the dimension of H*,*(Gn) must be greater than

the number of generators of M(Gn), since we may extend M(Gn) to a larger

acyclic free right module chain complex M by adding new generators and

then we have that dim H*,*(Gn) = dim M  U(Gn)Zp > dim M(Gn)  U(Gn)Zp =

number of generators of M(Gn). To compute homology, we have to compute

its dual cohomology.



    Theorem 1.6 For n < p, M(Gn) is a minimal free resolution of U(Gn),

that is, Hom  U(Gn)(M(Gn), Zp) = H*,*(Gn) and M(Gn)  U(Gn)Zp = H*,*(Gn).



    Proof. See the proof of Theorem 2.8.                           Q.E.D.
2     Symmetry in Cohomology



Let R = E(hi,j| 06 i< j), where E is the exterior algebra over Zp.  We

define R to be a bigraded DGA with cohomological degree |hi,j| = 1 and

internal degree khi,jk = 2(p-1)(pi+. .+.pj-1) and the differential defined by
       P j-1
dhi,j=   k=i+1hi,khk,j  (dhi,i+1= 0). For n > 0, let Rn be the sub-DGA of R

defined by Rn = E(hi,j| 06 i< j6 n). Then, by definition, H*,*(R) = H*,*(G)

and H*,*(Rn) = H*,*(Gn).


                                   11



   Definition 2.1 Let G(Rn) be the set of all upper triangular matrix

                          0                            1
                            a0,1  a0,2 a0,3 . . .a0,n
                          BB                           C
                          B   0   a1,2 a1,3 . . .a1,n  CC
                 (ai,j) = BB  0    0   a2,3 . . .a2,n  CC
                          B@ . . .. . .. . .. . . . . .CA

                              0    0    0   . . .an-1,n


with ai,j= 0 or 1 for i < j.  We use such a matrix A = (ai,j) 2 G(Rn) to

denote a monomial

     (h0,1)a0,1(h1,2)a1,2(h0,2)a0,2. ....(.hn-1,n)an-1,n. .(.h1,n)a1,n(h0,n)a0,n

in Rn, where (hi,j)1 = hi,j, (hi,j)0 = 1.  Then, G(Rn) is a basis of Rn.

The total degree `(A) of A is an (n+ 1)-tuple (`0(A), `1(A), . .,.`n(A)) de-
                 P k-1             P n
fined by `k(A) =   s=0(1 - as,k) +   s=k+1ak,sfor 0 6 k 6 n.  We denote

Rn(i0, i1, . .,.in) by the subvector space of Rn spanned by those monomials

A 2 G(Rn) with `(A) = (i0, i1, . .,.in).

    Similarly, let G(R) be the set of all upper triangular matrix

                              0                    1
                                a0,1  a0,2 a0,3 . . .
                              B   0   a    a    . .C.
                     (ai,j) = BB@      1,2  1,3    CC
                                  0    0   a2,3 . .A.

                                 . . .. . .. . .. . .


with ai,j= 0 or 1 for i < j and there are only finite number of ai,j's such

that ai,j= 1. We use such a matrix A = (ai,j) 2 G(R) to denote a monomial

           (h0,1)a0,1(h1,2)a1,2(h0,2)a0,2(h2,3)a2,3(h1,3)a1,3(h0,3)a0,3. . .

in R. Then, B(R) is a basis of R. The total degree `(A) of A is an infinite
                                             P k-1             P 1
tuple (`0(A), `1(A), . .).defined by `k(A) =   s=0(1 - as,k) +   s=k+1ak,sfor

k > 0. We denote R(i0, i1, . .).by the subvector space of R spanned by those

monomials A 2 G(R) with `(A) = (i0, i1, . .)..
                                   12



   Theorem 2.2 For n > 1, Rn(i0, i1, . .,.in) is a subchain complex of Rn

and there is a chain complex isomorphism



                        Rn =  Rn(i0, i1, . .,.in),



where (i0, i1, . .,.in) is taken throughout all possible (n+ 1)-tuples such that

Rn(i0, i1, . .,.in) is non-empty.  So, there is a chain complex isomorphism

R =  R(i0, i1, . .).with (i0, i1, . .).taken throughout all possible tuples such

that R(i0, i1, . .).is non-empty.



    Proof. Direct checking.                                        Q.E.D.



    It is easy to check by induction that if Rn(i0, i1, . .,.in) is non-empty, *
 *then

i0+ i1+ . .+.in = 1_2n(n + 1) and that if R(i0, i1, . .).is non-empty, then the*
 *re

is an integer n such that i0 + i1 + . .+.in = 1_2n(n + 1) and ij = j for j > n.



    Theorem 2.3 We have the following four isomorphisms.



   1. There is a vector space isomorphism from Rn(i0, . .,.in) to Rn(j0, . .,.j*
 *n)

      if (j0, . .,.jn) is a permutation of (i0, . .,.in).



   2. There is a DGA isomorphism ~n from Rn to itself (we call it reflection

      isomorphism) that induces graded vector space isomorphism

           H*,*(Rn(i0, i1, . .,.in))= H*,*(Rn(n- in, n- in-1, . .,.n- i0)).



   3. There is a chain complex isomorphism oen from Rn to itself (we call it

      rotation isomorphism) that induces ungraded vector space isomorphism

           H*,*(Rn(i0, i1, . .,.in-1, in))= H*,*(Rn(i1, i2, . .,.in, i0)).



   4. There is a chain complex isomorphism ~n from Rn to its dual complex

      R*n(we call it dual isomorphism) that induces ungraded vector space



                                   13



     isomorphism

          H*,*(Rn(i0, i1, . .,.in))= H*,*(R*n(n- i0, n- i1, . .,.n- in)).

    Proof. For 0 6 s < n, we define vector space isomorphism  s from Rn

to itself as follows. For A = (ai,j) 2 G(Rn),  s(A) = (bi,j) 2 G(Rn) satisfies

that ai,j= bi,jexcept the following cases. For i = 0, . .,.s- 1, bi,s=bi,s+1=ai*
 *,s

if ai,s=ai,s+1and bi,s=1- bi,s+1=1- ai,sif ai,s6=ai,s+1.  For j  = s+ 2, . .,.n,

bs,j=bs+1,j=as,jif as,j=as+1,jand bs,j=1- bs+1,j=1- as,jif as,j6=as+1,j. bs,s+1=

1 - as,s+1. It is obvious that  s interchanges the s-th and the s+ 1-th degree

of the total degree of Rn.

    We define algebra isomorphism ~n by ~n(hi,j) = hn-j,n-i.  It is obvious

that ~n is a DGA isomorphism and reverses the order of the total degree of

Rn.

    Since Rn-1 is a sub-DGA of Rn, we regard Rn as a left DGA-module over

this sub-DGA and denote this module by M1.  By definition, M1 is a free

left module over Rn-1 generated by the following elements h"10,nh"21,n.h.".nn-1*
 *,n,

"i = 0 or1, i = 1, 2, . .,.n.

    Define DGA monomorphism j: Rn-1 ! Rn by j(hs,t) = hs+1,t+1for all

hs,t2 Rn-1. It is obvious that imj is another sub-DGA of Rn. We regard Rn

as a left DGA-module over the sub-DGA imj and denote this module by M2.

By definition, M2 is a free left module over Rn-1 generated by the following

elements h"10,1h"20,2.h.".n0,n, "i = 0 or1, i = 1, 2, . .,.n.

    Now define the left DGA-isomorphism oen: M1 ! M2 by

        oen(xh"10,nh"21,n.h.".nn-1,n) = (-1)fij(x)h1-"10,1h1-"20,2.h.1.-"n0,n,
                       P
where x 2 Rn-1 and ø =   "i=1i. It is a direct checking that oen is a DGA-

module isomorphism and rotates the total degree of Rn.

    We still use upper triangular matrices to denote the dual basis of R*n. In



                                   14



this way, the total degree of R*nis defined just in the same way as that of Rn.

Define ~n: Rn ! R*nby that for A = (ai,j) 2 G(Rn), ~n(A) = (bi,j) 2 G(R*n)

satisfies that bi,j= 1 - ai,jfor i < j. It is a direct checking that ~nd = ffi~n

(ffi is the dual map of d) and ~n changes the total degree as shown in the

theorem.                                                          Q.E.D.



    Definition 2.4 For m, n > 0, if (k0, . .,.km+n+1 ) is a permutation of

(i0, . .,.im , j0+ m+ 1, . .,.jn+ m+ 1) such that the order of (i0, . .,.im ) a*
 *nd

(j0 + m + 1, . .,.jn + m + 1) is unchanged in (k0, . .,.km+n+1 ), then we call

Rm+n+1 (k0, . .,.kn+m+1 ) reducible with lower factor Rm (i0, . .,.im ) and upp*
 *er

factor Rn(j0, . .,.jn) and we denote it by

      Rm+n+1 (k0, . .,.km+n+1 ) = Rm (i0, . .,.im ) o Rn(j0, . .,.jn),

where we define Zp = R0(0) and thus have



   R0(0) o Rn(i0, . .,.in-1)=  Rn(i0+ 1, . .,.ik-1+ 1, 0, ik+ 1, . .,.in-1+ 1)


   Rn(i0, . .,.in-1) o R0(0)=  Rn(i0, . .,.ik-1, n+ 1, ik, . .,.in-1).



    Notice that R1 o R2 in fact represents a class of chain complexes but we

always use this symbol to denote a given chain complex.



    Theorem 2.5 There is always an ungraded vector space isomorphism

                     H*,*(R1 o R2) = H*,*(R1)   H*,*(R2)



    Proof. If R1= Zp or R2= Zp, the theorem is a direct checking.

    Suppose m, n  >  0 and R  =  Rm+n+1 (k0, . .,.km+n+1 )  =  R1 o R2  =

Rm (i0, . .,.im ) o Rn(j0, . .,.jn) and for 0  6  s0 <  . .<. sm  6  m+ n+ 1,

ksr = ir; for 0 6 t0 < . . .< tn 6 m+ n+ 1, ktr = jr+ m+ 1.  Then for any

0 6 u 6 m+ n+ 1, either u = sr or u = tr0.

    Define algebra monomorphism ~1: R1 ! R and ~2: R2 ! R respectively



                                   15

                                                   Q
by ~1(hu,v) = hsu,svand ~2(hu,v) = htu,tv. Let c =   u,vhtv,su2 R, then it is

obvious that dc = 0. For a 2 R1 and b 2 R2, we define a o b = ~1(a)~2(b)c.

Then this o-product is a monomorphism from R1   R2 to R. It is a direct

checking that d(~1(a)c) = ~1(da)c and d(~2(b)c) = ~2(db)c and therefore,

d(a o b) = (da) o b+ (-1)|a|a o (db). This means that o-product sends R1  R2

to a subchain complex of R and to prove the theorem, we need only prove

that o-product is an epimorphism.  For an upper triangular matrix C  =

(ci,j) 2 G(Rm+n+1 ) such that `(C) = (k0, . .,.km+n+1 ), define A = (ai,j) 2

G(Rm ) and B = (bi,j) 2 G(Rn) by au,v= csu,svand bu,v= ctu,tv.  Suppose

`(A) = (i00, . .,.i0m) and `(B) = (j00, . .,.j0n). Compare the total degree of C

with A o B and we have that ksr= `sr(C) > `sr(A o B) = `r(A) = i0r, but we

have that
                Xm      1            Xm      1
                   ksr= __m(m+ 1),      i0r= __m(m+ 1),
                r=0     2            r=0     2

so ksr = i0rfor r = 0, . .,.m.  Analogously, ktr = j0r+m+ 1 for r = 0, . .,.n.

Therefore, C =  (A o B).                                         Q.E.D.



    Notice that by the above theorem, if (i0, i1, . .,.in) is a permutation of

(0, 1, . .,.n), then Rn(i0, . .,.in) is one dimensional and so H*,*(Rn(i0, . .,*
 *.in))

= Rn(i0, . .,.in). Thus, we have the following algebra which is always a sub-

space of the cohomology of G.



    Definition 2.6 Sn is a bigraded algebra over Zp defined as follows. As a

vector space, Sn =  Rn(i0, . .,.in), where (i0, . .,.in) is taken throughout all

permutation of (0, 1, . .,.n). That is, for e(i0, . .,.in) 2 Sn with total degr*
 *ee

(i0, . .,.in), its bidegree is defined by



     |e(i0, . .,.in)|= the number of pairs(s, t) such thats< t but is> it


    ke(i0, . .,.in)k=  2(p- 1)(a0+ a1p+ . .+.an-1pn-1)



                                   16



where ak = i0+ . .+.ik- 1- . .-.k.  The product of Sn is defined by that

e(i0, . .,.in)e(j0, . .,.jn) is their product in Rn if (i0 + j0, i1 + i1 - 1, .*
 * .,.in +

jn - n) is still a permutation of (0, 1, . .,.n) and 0 otherwise.

    We regard Sn as a subalgebra of Sn+1 by identifying e(i0, . .,.in) with

e(i0, . .,.in, n+ 1). Thus, we have S = [nSn which has a basis of e(i0, i1, . .*
 *).

with only finite number of is's such that is6=s.



    Lemma 2.7 If H*,*(Rn) = Sn and for all e = e(i0, . .,.in) 2 Sn with

it = is+ 1 (s < t), the Massey product < e, hs,s+1, . .,.ht-1,t>is non-trivial,

then H*,*(Rn+1) = Sn+1.

    If Hk,*(Rn) = Sk,*nfor k < p and for all e = e(i0, . .,.in) 2 Sk,*nwith

it = is+ 1 (s < t), the Massey product < e, hs,s+1, . .,.ht-1,t>is non-trivial,

then Hk,*(Rn+1) = Sk,*n+1for k < p.



    Proof. We prove that if H*,*(Rn+1(i0, . .,.in+1)) 6= 0, then (i0, . .,.in+1)

is a permutation of (0, . .,.n+ 1).  If the smallest of i0, . .,.in+1 is 0, then

Rn+1(i0, . .,.in+1)  =  R0(0) o Rn(j0, . .,.jn).  by Theorem 2.5 and induc-

tion hypothesis, (j0, . .,.jn) can only be a permutation of (0, . .,.n).  So,

(i0, . .,.in+1) is a permutation of (0, . .,.n+ 1). Analogously, if the biggest*
 * of

i0, . .,.in+1 is n+ 1, then (i0, . .,.in+1) is also a permutation of (0, . .,.n*
 *+ 1).

    Now suppose Rn+1(i0, . .,.in+1) is non-empty and 1 6 is 6 n for s =

0, . .,.n+ 1. To prove that its cohomology is 0, we may suppose that in+1 is

the biggest of i0, . .,.in+1, since there is the rotation isomorphism in Theo-

rem 2.3.  Define factor degree | . |f on Rn+1 as follows.  |hi,n+1|f = n- i+ 1,

i = 0, . .,.n, |hi,j|f = 0 if j 6 n, and |ab|f = |a|f+|b|f for all a, b 2 Rn+1.*
 * De-

fine a filtration E0   E1   . .o.f Rn+1 by Er = {a 2 Rn+1 | |a|f 6 r}. Then,

we get a spectral sequence Es,trconverging to H*,*(Rn+1) with a chain com-


                                   17



plex isomorphism E*,*1= Rn   E(h0,n+1, . .,.hn,n+1), where E is the exterior

algebra and d1(hi,n+1) = 0 for i = 0, . .,.n. So, by the induction hypothesis,

E*,*2= Sn   E(h0,n+1, . .,.hn,n+1). Notice that for e(j0, . .,.jn) 2 Sn,the tot*
 *al

degree of e(j0, . .,.jn)h"00,n+1.h.".nn,n+1is (j0+ "0, . .,.jn+ "n, n+ 1- "0- .*
 * .-."n).

So, if e(j0, . .,.jn)h"00,n+1.h.".nn,n+12 Rn+1(i0, . .,.in+1) with the biggest *
 *of

i0, . .,.in+1 to be in+1, then there is only one "s = 1 and so (i0, . .,.in+1) *
 *is a

permutation of (1, 1, 2, . .,.n- 1, n, n). Suppose for 0 < s < t 6 n, is = it =*
 * 1.

Then E*,*2restricted on Rn+1(i0, . .,.in+1) is two dimensional spanned by e1 =

e(i0, . .,.is-1, 0, is+1, . .,.in)hs,n+1and e2 = e(i0, . .,.it-1, 0, it+1, . .,*
 *.in)ht,n+1.

By definition, dt-s+1(e1hs,n+1) = < e1, hs,s+1, . .,.ht-1,t>= ce2, where c 2 Zp

and c 6= 0.  Therefore, E*,*1restricted on Rn+1(i0, . .,.in+1) is 0 and so,

H*,*(Rn+1(i0, . .,.in+1)) = 0.  The proof of the second conclusion is just an

analogue.                                                         Q.E.D.



    Before we prove the following theorem, we must point out the structure

of U(n)*, the dual of U(n).  It a commutative coassociative Hopf algebra.

As an algebra, U(n)* = P (,i,j| 06 i< j6 n)=(,pi,j), the truncated polynomial

algebra generated by ,i,j. The dual basis G(U(n)*) of G(U(n)) is to substitute

Psk,tin U(n) with ,ks,tin U(n)*.  The diagonal map of U(n)* is defined by
                           P t-1
 (,s,t) = 1  ,s,t+,s,t 1 +   k=s+1,s,k ,k,t.  We have similar definitions for

U*.



    Theorem 2.8 For n < p, H*,*(Gn+1) = Sn+1 and Hn+1,*(G) = Sn+1,*.



    Proof. We use induction on n to prove that H*,*(Gn) = Sn for n < p. If

n = 1, it is a direct checking that H*,*(G1) = S1. Suppose H*,*(Gk) = Sk for

k 6 n. Then, H*,*(Gn) = S*n(* denotes dual space). Thus, the free resolution

M(Gn) in Definition 1.3 is a minimal free resolution of U(Gn). By definition,
                                   18



we have that_for (i0, . .,.in) a permutation!of (0, 1, . .,.n) with it = is+ 1,

               i0, . .,.it, . .,.is, . .,.in1
(s < t), if D                         = cPs,t+ . .,.then
               i0, . .,.is, . .,.it, . .,.in


                 < e(i0, . .,.is, . .,.it, . .,.in), hs,s+1, . .,.ht-1,t>


             =   < e(i0, . .,.is, . .,.it, . .,.in), ,s,t>


             =   c e(i0, . .,.it, . .,.is, . .,.in)


                              Q   ik - it
For n < p, we have that c =    ik________ 6= 0, where the product is taken
                                  ik - is

throughout all s < k < t such that ik > it (c =  1 if t = s + 1). Thus, by

Lemma 2.7, H*,*(Gn+1) = Sn+1.  The proof of the second conclusion is just

an analogue.                                                      Q.E.D.



    Theorem 2.9 For n < p, H*,*(U(n)) = Sn  P (bi,j| 06 i< j6 n), where P

is the polynomial algebra and bi,jis represented in cobar complex by


                   X   (p - 1)!
                      _________[,r1 . .,.rk|,r1 . .,.rk],
                      r1! . .r.k!i,s1   i,sk s1,j   sk,j


where the sum is taken throughout all i6 s1< . .<.sk6 j and r1+ . .+.rk = p,

k = 1, . .,.t- s, ,i,i= ,j,j= 1.

    Similarly, for n < p, Hn,*(U) = S  P (bi,j| 06 i< j).



    Proof. We only prove the first conclusion. By the May spectral sequence

(see [4]), there is a spectral sequence En,s,trconverging to H*,*(U(n)) with

E*,*,*1= Rn   P (bi,j| 06 i< j6 n). The differential d1 restricted on Rn is just

the differential of Rn and d1(bi,j) = 0.  But bi,jalready have representative

as given in the theorem, so dr(bi,j) = 0 for all r > 1. So, we need only prove

that Sn = H*,*(Rn) has representative in the cobar complex and thus, the

spectral sequence collapse from r = 2.

                                   19



   The dual space S*nof Sn is naturally a subspace of the bar complex

B(U(Gn)) of U(Gn) defined as follows. For e*(i0, . .,.in) 2 S*n, we define

                       X
     e*(i0, . .,.in) =    (-1)<ff0,ff1>+...+<ffs,ff>[D(ff0, ff1)| . .|.D(ffs, f*
 *f)],



where the sum is taken throughout all non-zero D(ffi, ffi+1) such that ff0 =

(0, . .,.n) and ff = (i0, . .,.in). Let I = span{[a1| . .|.as] | ai 2 ~U(Gn), t*
 *here

is at least one 0 6 i 6 s such that ai is in the ideal generated by the center

of U(Gn)} (i.e., ai has a factor xps,t). Then, it is obvious that I is a subcha*
 *in

complex of B(U(Gn)) and I \ S*n= 0.  Therefore, I can be extended to an

acyclic subchain complex J of B(U(Gn)) such that B(U(Gn)) = J   S*nas

chain complexes.  Dually, the cobar complex C(U(Gn)) of U(Gn) also has

a direct sum decomposition C(U(Gn)) = J*   Sn, where J* is the comple-

mentary space of S*nand Sn is the complementary space of J. Since I   J,

we have that Sn   Ic, the complementary space of I.  By definition, Ic =

span{[a1| . .|.ak] | every ai 2 G(U(n)*)}, where G(U(n)*) is the dual basis of

G(U(n)). Ic is a subchain complex of C(U(n)), the cobar complex of U(n)

which is also a subchain complex of the cobar complex of U(Gn). So, Sn have

representatives in the cobar complex of U(n).

    Similarly, we can prove that Sn,*has representatives in the cobar complex

of U.  In fact, it has representatives in the cobar complex of the Steenrod

algebra. This is proved in the following Theorem 2.10.             Q.E.D.



    Theorem 2.10 For n < p, Sn,*in the May spectral sequence in [4] sur-

vives to infinity.



    Proof. Notice that D(ff, fi) in Definition 1.3 can be naturally generalized

for n > p if |ff| < p with |ff| as defined in Definition 2.6.  Thus, the dual

space S* of S is naturally a subspace of the bar complex B(U) of U defined



                                   20



as follows. For e*(i0, i1, . .).2 S*, we define


                       X
     e*(i0, i1, . .).=    (-1)<ff0,ff1>+...+<ffs,ff>[D(ff0, ff1)| . .|.D(ffs, f*
 *f)],



where the sum is taken throughout all non-zero D(ffi, ffi+1) such that ff0 =

(0, 1, . .).and ff = (i0, i1, . .).. Let G = {a 2 U | w(a) = (i0, i1, . .)., th*
 *ere is

at least one k > 0 such that ik > p} and let I be the subchain complex of the

bar complex B(U) of U defined by I = span{[a1| . .|.as] | ai 2 ~U, there is at

least one 0 6 i 6 s such that either ai in the ideal generated by the center of

U (i.e., ai has a factor xps,t), or ai 2 G}. Notice that for all non-zero D(ff,*
 * fi),

the weight w(D(ff, fi)) = (i0, i1, . .).satisfies that ik < p for all k > 0. Th*
 *us,

it is obvious that I is a subchain complex of B(U) and I \ S* = 0 (always

at cohomology degree < p from now on!).  Therefore, I can be extended

to an acyclic subchain complex J of B(U) such that B(U)) = J   S* as

chain complexes.  Dually, the cobar complex C(U) of U also has a direct

sum decomposition C(U) = J*   S, where J* is the complementary space

of S* and S is the complementary space of J.  Since I   J, we have that

S   Ic, the complementary space of I. By definition, Ic = span{[a1| . .|.ak] |

every ai satisfies that w(ai) = (i0, i1 . .).such that ii < p for all k > 0},

where the weight of element in G(U*) is just the weight of its dual element

in G(U). Notice that the cobar complex C(U) of U is just the cobar complex

C(A) of the Steenrod algebra A as vector spaces but the differentials of the

two cobar complexes are different. However, it is easy to check that the two

differentials coincides on Ic. So, Sn,*has representative in the cobar complex

Cn,*(A) of the Steenrod algebra for n < p. So, the May spectral sequence in

[4] satisfies that En,*2= Hn,*(U) = S  P (bi,j| 06 i< j) (Excluding the direct

sum component with factor øn) and Sn,*survives to infinity in the spectral

sequence.                                                         Q.E.D.



                                   21



    We can only prove that part of Sn,* represents non-trivial cohomology

classes. This is done in the next section.
3     Subspace



   Definition 3.1 We call such numbers pi+ . .+.pj-1 (0 6 i < j) neat

numbers. An equality N = N1+ . .+.Ns is call a neat decomposition for N if

Ni is neat for i = 1, . .,.s and s is called the length of this decomposition. *
 *For

a positive integer N, the rank r(N) of N is defined to be the smallest length

of all neat decompositions of N.  Such a decomposition is called a shortest

decomposition of N. A positive integer N is called connected with component
                                 P j-1   k
(i, j) for some 0 6 i < j if N =   k=iakp  with every 0 < ak < p.  Two

connected integers M and N are called disjoint if their components are (i, j)

and (s, t) such that j < s.



    Theorem 3.2  r(a+ b) 6 r(a)+ r(b) for all positive integers. Every posi-

tive integer N has a unique decomposition N = N1+ . .+.Ns with every Nk

connected and N1, . .,.Ns are mutually disjoint.  For such a decomposition,
               P s
we have r(N) =   k=1r(Nk).



    Proof. Trivial direct checking.                                 Q.E.D.


                                            P    k
    Theorem 3.3 For a connected integer N =   akp  with 0 < m 6 ak < p

for k = i, i+ 1, . .,.j- 1 (06 i< j) and ak = 0 otherwise, r(N) = m+ N0, where

N0 = N- m(pi+ . .+.pj-1).



    Proof. It is obvious that we need only prove the case m = 1. Suppose

m = 1 and N = N1+ . .+.Ns is a shortest decomposition for N. Since ai > 0,

there must be an Nk1 such that Nk1 = pi1+ . .+.pj1-1with j1 > i1 and i1 = i.
                                   22



Similarly, since aj1> 0, there must be an Nk2 such that Nk2 = pi2+ . .+.pj2-1

with j2 > i2 and i2 6 j1.  Going on like this, we finally have an Nitsuch

that Nit = pit+ . .+.pjt-1 with jt > it and jt = j and it 6 jt-1.  Take

N0k1= pi+ . .+.pj-1, N0ik= pik+ . .+.pjk-1-1, k = 2, . .,.t and N0k= Nk

otherwise. Then, N = N01+. .N.0sis still a shortest decomposition for N such

that N0k1= pi+ . .+.pj-1. We may suppose k1 = 1, then, N0 = N02+. .+.N0s

is a neat decomposition for N0 and so s- 1 > r(N0). That is, r(N) > r(N0)+ 1.

By Theorem 3.3, r(N) 6 r(N0)+ 1. So, r(N) = r(N0)+ 1.            Q.E.D.


                           P 1     k
    Definition 3.4 For N =   k=0akp  with every 0 6 ak < p and only finite

number of non-zero ak's, a pair (s, t) is call a peak if as-1 < as, at-1 > at, *
 *and

as = as+1 = . .=.at-1 (we define a-1 = 0). The critical value for this peak

is defined to be cs,t= as. For two succeeding peaks (s, t) and (u, v) (i.e., th*
 *ere

is no peak (i, j) such that t 6 i and j 6 u), the pair (t, u) is called a valle*
 *y.

The critical value for this valley is defined to be ct,u= min (at, . .,.au-1).


                                                   P      P
    Theorem 3.5 For any positive integer N, r(N) =   cs,t-   cu,v, where

the sum is taken throughout all critical values for peaks (s, t) and valleys

(u, v).



    Proof. We use induction on N. When N = 1, it has one peak (0, 1) and

c0,1= 1. The theorem holds. Suppose the theorem holds for all 1 < k < N.

Firstly, we prove the case N is connected.  Suppose the connected integer
    P    k
N =   akp  with 0 < ak < p for k = i, i+ 1, . .,.j- 1 and ak = 0 otherwise.

Take m = min (ai, . .,.aj-1), and N0 = N- m(pi+ . .+.pj-1).  By Theorem

3.3, r(N) = r(N0)+ m. IF N0 = 0, then N has one peak (i, j) and ci,j= m.

The theorem holds. If N0 6= 0, by definition, N0 and N have the same set of

peaks and valleys and all critical values of N0 are less than that of N by m.
                                   23



Notice that the number of peaks is always more than the number of valleys

by 1. So, the theorem holds by induction hypothesis. If N is not connected

and N = N1+ . .+.Ns is the unique connected component decomposition in

Theorem 3.2, then the set of peaks of N is the union of the sets of peaks of

Nk's, k = 1, . .,.s. So, the set of valleys of N is the union of the sets of va*
 *lleys

of Nk's added with those new valleys between these connected components

with critical value 0. So the theorem holds by induction hypothesis. Q.E.D.


                        P
    Definition  3.6 Let   (n) be as defined in Definition 1.2.  For ff  =
                P                                              P        s    s+1
(i0, . .,.in) 2   (n), we define the weight of ff to be kffk =   (s,t)(p  + p  *
 *  +

. .+.pt-1), where the sum is taken throughout all pairs (s, t) such that s < t

and is > it.  The length of ff is |ff| = the number of pairs (s, t) such that

s < t and is > it. The rank of ff is r(ff) = r(kffk).



    Theorem 3.7 For eff2 S such that |ff| = r(ff) < p, effrepresents a

non-trivial cohomology class in the cohomology of the Steenrod algebra.



    Proof.  It is easy to check that the representative of eff= e(i0, i1, . .).
        Q
in R is   hs,twith the product taken throughout all pairs (s, t) such that
                                                      Q        *,*,*
s < t and is > it.  We will prove that there is no efi  bi,j2 E2    such
           Q                               Q
that dr(efi  bi,j) = eff.  If so, then kefi  bi,jk = keffk and so, kffk = kfik +
P                                                               P
   kbi,jk=q (q = 2(p- 1)) is a neat decomposition for kffk. So,   |bi,j|+ |fi| >

1_P                                           Q
2   |bi,j|+ |fi| > r(ff) = |ff|. That is, |efi  hi,j| > |eff|. This is impossib*
 *le.

                                                                  Q.E.D.



    Even if we can not prove that all Sn,*represents non-trivial cohomology

classes in the cohomology of the Steenrod algebra, this theorem is also very

important.  In [10], we prove that if the n-th Greek letter family element

                                   24



ff(n)sexists, then its representative in the May spectral sequence is ~ff(n)s=

__s!_                         s-n
(s-n)!hn-1,nhn-2,n. .h.0,n(tn)    (hi,j2 Rn). Specifically, the total degree for

~ff(n)nis ~ = (1, 2, . .,.n, 0). By Theorem 3.5, r(~) = |~| = n. So, for n < p,

~ff(n)nrepresents a non-trivial cohomology class even if the geometric ff(n)ndo*
 *es

not exist.
References



 [1] Adams,J.F.,On the structure and application of the Steenrod algebra,

     Math. Helv. 32.(1958),180-247



 [2] Bousfield,A.K.,Curtis,E.B.,Kan,D.N. et al, The mod p lower central se-

     ries and the Adams spectral sequence, Topology 5 (1966), No.4 331-342



 [3] May,J.P.,A general algebraic approach to Steenrod algebra,  Lecture

     Notes in Mathematics. 168,153-231



 [4] May,J.P.,The cohomology of restricted Lie algebras and Hopf algebras,

     Journal of Algebra 3 (1966),123-146



 [5] Milnor,John.,The Steenrod algebra and its dual, Annuals of Mathemat-

     ics. 67. No. 1.(1958),150-171



 [6] Steenrod,N.E.,Cohomology operations, Lectures printed in Princeton

     University Press,1962



 [7] Zheng,Qibing., A New Massey Product on Ext Groups, Journal of Al-

     gebra 183.(1996),378-395



 [8] Zheng,Qibing., S-module and the New Massey-Product, Journal of Al-

     gebra 190.(1997),487-497



                                   25



 [9]Zheng,Qibing., Twisted Product and Cohomology, Acta Mathematica

    Sinica, New Series 13.(1997),76-80



[10]Wang, Xiangjun. and Zheng, Qibing., The convergence of ~ff(n)sh0hk, Sci-

    ence in China (Serires A) 41 No. 6 (1998),622-628

    Address of the author

Zheng Qibing

Department of Mathematics

Nankai University

Tianjin, 300071,P.R.China



                                   26