(b) E PK®fK(*p) = (-i)k{p-k) E M«P)®PK,
[K\=k \K\=p-k
(c) E
^ A W Q P ) =
(d) ap(bqAbr) = Z Zt-^-'hii^Aifda^br),
(e) E VK® IK I\bqAbr= E bq(ipK) ® fie Kbr,
\K\=k \K\=k+q
(0 E I M ^ ) ] ( / / c A M - ( ' * ! : % A fcr,and
(g) E
E (-i) f t ( 1 + / ) /ic(^)®^M .
Proof. To prove (a), fix i and project onto f\l F* ® /\p~l F*. Both sides become
bi(ap), upon application of (&* g) 1) * . Apply (6^ ® 1) * to both sides of (b).
The left side becomes E M^-tf) fx(^p) = &fc(ap). The right side becomes
J2 (bk */*)(*)-PK
£ /jc[6fc(«p)]-^ = Wap).
|K|=p-fc |JC|=p-fc
Part (c) follows from (a), together with the well-known fact that the composition
F* A
F* ®
F* A
is equal to multiplication by (£). Part (d) is an immediate consequence the mea-
suring identity, [8, Proposition A.2], together with (a). Apply (bk 0 1)* to both
sides of (e). The left side becomes
y i M ^ K ) 7 K A ^ A ^ = bk A bq A br.
The right side becomes
E^ tk\bq(pK)y fK Abr= E^ (bkAbq}(tpK)-fKAbr = bkAbqAbr.
|K|=fc+7 \K\ = k+q
Part (e) shows that the left side of (f) is equal to E VKUK A bq A br). One
may finish the proof of (f) by establishing the assertion when bq A br is a basis
vector from
F. Apply (bk 8 1) * to each side of (g); then use part (b) of
Proposition 1.1.
Remark 1.10. With the exception of section 6, binomial coefficients play only a
minor role in this paper. Nonetheless, it should be mentioned, at the beginning, that
(™) is defined for all integers m and i. This binomial coefficient is zero whenever
i 0 or 0 m i See [14,15] for more details.
Each complex 1 ^ of section 2 is obtained by splicing together two smaller com-
plexes. The next result is the multilinear algebra which is used in the proof of
Theorem 2.11 at this splice. It is not apparent, at first glance, but identities (a)
and (b) are actually dual to one another. The proof we have given of (b) emphasizes
this realtionship. On the other hand, one can give a proof of (b) which mimics the
proof of (a).
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