SPLITTINGS OF EXTENSIONS OF BANACH ALGEBRAS 13
1.8. THEOREM. Let A be a [commutative] Banach algebra,
(i) Let m G N. Suppose that every [commutative] singular extension of A splits
strongly. Then every [commutative] nilpotent extension of A splits strongly.
(ii) Suppose that every [commutative] singular extension of dimension at most m splits
strongly Then every [commutative] extension of dimension at most m splits strongly
PROOF: (i) By Theorem 1.7, it suffices to prove that every nilpotent [commutative] ex-
tension of dimension at most m splits strongly.
The proof of this statement is by induction on the minimum index n G N such that
In = 0. The hypothesis gives the case where n = 2; we suppose that *%2(%L\I) is an
extension of A such that / has index k 3, and assume that the result holds for each
nilpotent [commutative] extension of dimension at most m and index at most k 1. Set
J = I2 , a closed ideal of 21. Then the inductive hypothesis applies to each [commutative]
extension of A by J and to the extension 52 (21/J; I/J), and so each of these extensions
splits strongly. By Proposition 1.5(i), every [commutative] extension of A by I splits
strongly, continuing the induction.
(ii) This is the same proof, save that now we set J = T2, and note that Jk~l = 0.
Note that we have no result which proceeds from 'singular extension' to 'nilpotent
extension' in the case of algebraic splittings of extensions of a general Banach algebra; we
shall obtain a result in this direction in the case where A is commutative in §4. However,
we do have the following result, which will be used in §4.
1.9. THEOREM. Let A be a Banach algebra, and let 52(^5 -0 be an extension of A.
Suppose that I is the direct sum I = J®K of two closed ideals J and K in 21, and that
the extensions £(21/J; I/J) and £(2l/if; I/K) both split algebraically Then £(21; I)
splits algebraically.
PROOF: There exist subalgebras £j and CK , respectively, such that
21/ J = £jQK and 2l/if = ^ 0 J .
Define 03j = {a G 21 : a + J €j} and 93K = {a 21 : a + K G €K}, so that Q3j and
03K are subalgebras of 21. Then define
B = BjnB
K
,
so that 03 also a subalgebra of 21.
Take a €21. Then
a + J = 6 + J = c + I
for some b G 03j and c G 03K Since b c G / , there exist x G J and y G K with
b-c = x + y. We have 6 - x = c + y G 0 3 j n 03K = 03, and so 21 = 03 4- / .
Let x G 03 fl / . Then x + J G £j D K = {0}, and so x G J. Similarly, x G K, and so
x = 0. Thus 0 3 n / = 0.
We have shown that 21 = 03 © / , and so XX^5 I) splits algebraically.
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