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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?

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You may find of interest Halmos's exposition on commutators here. –  Bill Dubuque Jul 29 '11 at 4:43
    
By "ring" here, did you intend "algebra over $\mathbb{R}$"? I suppose there is a subtlety in the question. I think the axiom of choice tells us that every linear space over $\mathbb{R}$ admits a (subadditive) norm. But we see below that not every algebra over $\mathbb{R}$ admits a submultiplicative norm, since there are real algebras which contains solutions of $xy - yx = I$. –  Geoff Robinson Jul 29 '11 at 11:59
    
@Geoff: ah. Let's suppose I mean "finitely generated ring" (over $\mathbb{Z}$). –  Qiaochu Yuan Jul 29 '11 at 13:27
    
A small comment on your last question: if you take the algebra ${\mathbb C}^{\mathbb N}$ with pointwise product, then I think it cannot be given a submultiplicative norm. On the other hand, it is an old result of Allan that there is an embedding (no continuity) of ${\mathbb C}[[X]]$ into the unitization of some radical Banach algebra. –  user16299 Jan 7 '12 at 19:42
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Here's a sketch of a proof. Let $\sigma(x)$ denote the spectrum of $x$. Then $\sigma(xy)\cup\{0\} = \sigma(yx)\cup\{0\}$. On the other hand, $\sigma(1+yx)=1+\sigma(yx)$. If $xy=1+yx$, then the previous two sentences, along with the fact that the spectrum of each element of a Banach algebra is nonempty, imply that $\sigma(xy)$ is unbounded. But every element of a Banach algebra has bounded spectrum.

(I don't remember where I first learned this proof, nor do I have a reference for it off-hand, but I did not come up with it myself.)

The proof that $\sigma(xy)\cup\{0\}=\sigma(yx)\cup\{0\}$ reduces to showing that $1-xy$ is invertible if and only if $1-yx$ is invertible, a problem that was the subject of a MathOverflow question.


There's a proof using derivations in section 2.2 of Sakai's book, Operator algebras in dynamical systems: the theory of unbounded derivations in $C^*$-algebras. A bounded derivation on a Banach algebra $A$ is a bounded linear map $\delta$ on $A$ such that $\delta(ab)=\delta(a)b+a\delta(b)$ for all $a$ and $b$ in $A$. Theorem 2.2.1 on page 18 shows that if $\delta$ is a bounded derivation on $A$, and if $a$ is an element of $A$ such that $\delta^2(a)=0$, then $\lim\limits_{n\to\infty}\|\delta(a)^n\|^{1/n}=0$. The proof uses induction with a neat computation to show that $\delta^2(a)=0$ implies that $n!\delta(a)^n=\delta^n(a^n)$, and then the result follows from boundedness of $\delta$ and the fact that $\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0$.

Corollary 2.2.2 concludes that the identity is not a commutator. If $ab-ba=1$, then the bounded derivation $\delta_a:A\to A$ defined by $\delta_a(x)=ax-xa$ satisfies $\delta_a^2(b)=\delta_a(1)=0$. By the preceding theorem, this implies that $1=\lim\limits_{n\to\infty}\|1^n\|^{1/n}=\lim\limits_{n\to\infty}\|\delta_a(b)^n\|^{1/n}=0$.

(Completeness is not used in this approach. An element $x$ of $A$ satisfies $\lim\limits_{n\to\infty}\|x^n\|^{1/n}=0$ if and only if $\sigma(x)=\{0\}$, and such an $x$ is called a generalized nilpotent. Incidentally, this also gives an approach to answering the example problem in the MathOverflow question Linear Algebra Problems? The remainder of Section 2.2 has a number of interesting results on bounded derivations and commutators of bounded operators.)

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Oh, of course. Because $\lambda - xy$ is invertible if and only if $\lambda - yx$ is... thanks! –  Qiaochu Yuan Jul 29 '11 at 4:42
    
@Geoff: In Arveson's A short course on spectral theory, the spectral radius formula is attributed to Gelfand and Mazur (with the remark that special cases were found by Beurling). I know that it appears in Gelfand's earliest work on commutative Banach algebras, but I don't know the rest of the story. I speculate that it was discovered independently by Mazur, but Arveson doesn't give a reference and I haven't read any of Mazur's work. –  Jonas Meyer Jul 29 '11 at 7:37
    
@Geoff: Żelazko's book is usually great for finding definitive references on such matters, and I'd turn there to find out if I had it on me. –  Jonas Meyer Jul 29 '11 at 7:45
    
@Geoff: Usually "Gelfand-Mazur theorem" means something else. I had this in mind when speculating above; Mazur's version was earlier and more general, but its proof was not published when Gelfand independently proved his case (which initiated Gelfand's theory of commutative Banach algebras). Part of the reason I strongly suspect that Żelazko will be a good source here is that Mazur's proof of the Gelfand-Mazur theorem appeared there for the first time in published form (and is there called the Mazur-Gelfand theorem). –  Jonas Meyer Jul 29 '11 at 7:56
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Sorry to pursue this further: Is there such a thing as a detailed and reliable early history of Banach algebras/operator algebras? What I have in mind is something comparable to Dieudonnés History of Functional Analysis but with more focus on the algebra aspect? Of course one could just say: read the Gel'fand school (and what I've seen from that is well worth it). Dixmier has comprehensive references in both his big books, but the history parts are rather terse. –  t.b. Jul 29 '11 at 12:16
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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.

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That's neat! The complete case also implies the general case, because every normed algebra has a Banach algebra completion. –  Jonas Meyer Jul 29 '11 at 7:12
    
That (and more) is mentioned in Halmos's exposition that I already linked to in a comment. –  Bill Dubuque Jul 29 '11 at 15:08
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@ Bill: not everyone can see that link (me, for example) –  Geoff Robinson Jul 29 '11 at 17:25
    
This is a very cool argument, but a little mysterious... –  Igor Rivin Dec 4 '13 at 4:10
    
that's the exact answer I'm looking for. –  sun Dec 4 '13 at 4:16
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