TWO VECTORS AND A RECTANGULAR MATRIX 9
A short calculation yields that A + B\ = hM-i,w-i(h(^p) ® bq g a
r
) . Now that
(1.12) is established, we continue with the induction. Iff + l + wp + r and the
induction hypothesis is known to hold at w 1, then (1.12) shows that
h [hM,w(ap ®bq ®ar)} = (-l) p /iA/
)W
(a
p
g) bx Abq ® ar);
hence,
hM,w(*P ®bi® *r) = (-l)pqbq [hMMXp ® ! ® «r)] = ( - l ) p ? C ) M * P A a
r
) = 0.
Now we prove (b). Assume that w + p -f 1 9- We prove
(i.i3) x)
(-1)(P+1)(V)
(// h
A
I^MK-)])
^
=
°-
i€Z
Let A(ap) = £ ai
j ]
8) a ^ , with
o^]

A*^7*
and
aji]8
G
f\p~s
F*. Fix / , with
|J| = 2. Proposition l.l.c, together with the measuring identity, gives
(/ / \ocv A [pj(&g)](ar)J J (CJF) = / / A ap (yi(bq) A ar(ujF)j
s, [J]
= ^(-1)(?"°(P"5)+^V/[4J1(M A [//(a^1, A ar)](u/F).
It follows that the left side of (1.13) is
£(-i)«e- 53(-lr^TjVM^Ml
A [ / K ^ A O , ) ] ^ ) ,
which is zero by part (a). D
The phrase "Koszul complex" has two meanings in this paper. If X: F —* R is
a map of free R—modules, then the Koszul complex associated to X is
(i.i4) . . . - ^ A ^ - ^ A ' - 1 ^ - - . ,
where
for all bq e f\q F. Of course, if bq = b[l1 A .. . A b^\ with ftW F , then
a(fcg) = ]£(-l)
i + 1
X(bM) *M A ... A
ftl*"1!
A
6l+1l
A ... A &M;
i=l
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