26 Nomenclature
Thus, any a ∈ g
0
, b ∈ g
0
and c ∈ M satisfy (η ([a, b])) (c) = ω ([a, b] , c) and
(a * η (b) −b * η (a)) (c)
= (a * η (b)) (c)
| {z }
=−(η(b))([a,c])
(by the deﬁnition of the dual of a g
0
-module)
− (b * η (a)) (c)
| {z }
=−(η(a))([b,c])
(by the deﬁnition of the dual of a g
0
-module)
=
−(η (b)) ([a, c])
| {z }
=ω(b,[a,c])
(by (13))
−
−(η (a)) ([b, c])
| {z }
=ω(a,[b,c])
(by (13))
= (−ω (b, [a, c])) −(−ω (a, [b, c]))
= −ω
b, [a, c]
|{z}
=−[c,a]
+ ω (a, [b, c]) = ω (b, [c, a])
| {z }
=−ω([c,a],b)
(since ω is antisymmetric)
+ ω (a, [b, c])
| {z }
=−ω([b,c],a)
(since ω is antisymmetric)
= −ω ([c, a] , b) −ω ([b, c] , a) = ω ([a, b] , c) (by (3)) ,
so that (η ([a, b])) (c) = (a * η (b) − b * η (a)) (c). Thus, any a ∈ g
0
and b ∈ g
0
satisfy
η ([a, b]) = a * η (b) − b * η (a). This shows that η is a 1-cocycle, i. e., belongs to
Z
1
(g
0
, M
∗
). Lemma 1.7.10 is proven.
Proof of Theorem 1.7.4. First notice that any a, b, c ∈ g satisfy
(14) ([a, b] , c) = ([b, c] , a) = ([c, a] , b)
4
. Moreover,
(15) there exist a, b, c ∈ g such that ([a, b] , c) = ([b, c] , a) = ([c, a] , b) 6= 0.
5
This will be used later in our proof; but as for now, forget about these a, b, c.
It is easy to see that the 2-cocycle ω on g [t, t
−1
] deﬁned by (12) is not a 2-
coboundary.
6
4
Proof. First of all, any a, b, c ∈ g satisfy
([a, b] , c) = (a, [b, c]) (since the form (·, ·) is invariant)
= ([b, c] , a) (since the form (·, ·) is symmetric) .
Applying this to b, c, a instead of a, b, c, we obtain ([b, c] , a) = ([c, a] , b). Hence, ([a, b] , c) = ([b, c] , a) =
([c, a] , b), so that (14) is proven.
5
Proof. Since g is simple, we have [g, g] = g and thus ([g, g] , g) = (g, g) 6= 0 (since the form (·, ·)
is nondegenerate). Hence, there exist a, b, c ∈ g such that ([a, b] , c) 6= 0. The rest is handled by (14).
6
Proof. Assume the contrary. Then, this 2-cocycle ω is a coboundary, i. e., there exists a linear
map ξ : g
t, t
−1
→ C such that ω = dξ.
Now, pick some a ∈ g and b ∈ g such that (a, b) 6= 0 (this is possible since the form (·, ·) is
nondegenerate). Then,
ω
|{z}
=dξ
at, bt
−1
= (dξ)
at, bt
−1
= ξ
at, bt
−1
| {z }
=[a,b]
= ξ ([a, b])
and
ω
|{z}
=dξ
(a, b) = (dξ) (a, b) = ξ ([a, b]) ,
so that ω
at, bt
−1
= ω (a, b). But by the deﬁnition of ω, we easily see that ω
at, bt
−1
= 1 (a, b)
|{z}
6=0
6= 0
and ω (a, b) = 0 (a, b) = 0, which yields a contradiction.