The identity cannot be a commutator in a Banach algebra?

The Wikipedia article on Banach algebras claims, without a proof or reference, that there does not exist a (unital) Banach algebra B and elements x, y \in B such that xy – yx = 1. This is surprising to me, but maybe the proof is straightforward; anyone have a proof and/or a reference?

More generally, I would have naively thought that I could embed any ring into a Banach algebra. I guess there are actually serious restrictions to doing this; are these issues discussed anywhere?

Answer

There is a Theorem of Wielandt which asserts that if A is any normed algebra, complete or not, we can’t express I = 1_{A} in the form xy – yx. The proof is given in Rudin’s book, but it is so beautiful that I give it here. Suppose that xy -yx = I. I claim that xy^{n} – y^{n}x = ny^{n-1} for all n \in \mathbb{N}. We have the case n = 1.
Suppose that xy^k – y^kx = ky^{k-1} for some k. Then xy^{k+1} – y^{k+1}x =
(xy^{k} – y^{k}x)y +y^{k}(xy-yx) =ky^{k-1}y +y^{k}.I = (k+1)y^{k},
so the claim is established by induction. Note that y^n \neq 0 for any n, since otherwise there is a smallest value of n with y^n = 0, leading to 0 = xy^n – y^nx = ny^{n-1}, contrary to the choice of n.

But now, for any n, we have n\|y^{n-1} \| = \|xy^{n} -y^{n}x\| \leq 2\|x\|. \|y\| .
\|y^{n-1} \| .
Since y^{n-1} \neq 0, as remarked above, we have 2 \|x\| . \|y\|
\geq n
, a contradiction, as n is arbitrary.

Attribution
Source : Link , Question Author : Qiaochu Yuan , Answer Author : Geoff Robinson

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