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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?

Qiaochu Yuan
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    You may find of interest Halmos's exposition on commutators here. – Bill Dubuque Jul 29 '11 at 04:43
  • 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. –  Jan 07 '12 at 19:42
  • Since nobody explicitly mentioned this, maybe I'll point it out in the comments here: In QM, the canonical commutation relation is $xy-yx= i \hbar$ so, up to rescaling, the theorem under discussion is telling us, in particular, that the canonical commutation relation cannot be satisfied by two bounded operators. Hence, this particular part of QM necessarily involves unbounded operators. – Mike F Feb 08 '16 at 06:39
  • In intros to QM, it is common to note the CCR cannot hold for finite matrices $A$ and $B$. Taking the trace of $AB-BA=I$ leads to a contradiction, so the CCR can only arise in infinite dimension. Bearing this in mind, it's not too surprising the proof given by Jonas Meyer below uses the spectrum $\sigma$. It doesn't always make sense to take the trace in a Banach algebra (which one could think of as an average of the spectrum), but $\sigma$ itself makes sense, and has a similar algebraic property to the trace. Compare $tr(xy)=tr(yx)$ to $\sigma(xy) \cup {0}=\sigma(yx)\cup{0}$. – Mike F Feb 08 '16 at 06:47

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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.

Geoff Robinson
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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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Jonas Meyer
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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 04:42
  • @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 07:37
  • @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 07:45
  • @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 07:56
  • @Jonas: Thanks again. – Geoff Robinson Jul 29 '11 at 08:01
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    For what it's worth: I know that I learned about this argument it from Appendix A, A.1, p. 409 in Pedersen's $C^{\ast}$-algebras and their automorphism groups (so this would be a reference). I would be surprised if it weren't older. – t.b. Jul 29 '11 at 08:32
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    @Theo: Thanks for the reference. I meant the proof that the identity is not a commutator. I learned about $\sigma(xy)\cup{0}=\sigma(yx)\cup{0}$ from Arveson's A short course on spectral theory, Exercises 3 and 4 on page 7. I like the exercise because it gives the seemingly nonsense series method to discover the proof, which is the subject of Bill Dubuque's question linked above. – Jonas Meyer Jul 29 '11 at 08:42
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    I see. Yes, that's really nice. The funny thing is that this nonsense series is the only way for me to remember the relevant identity. If I really needed a reference for the fact that the identity isn't a commutator, I'd probably start looking in the context of "no-go theorems" in quantum mechanics. This must go way back (von Neumann?). Added Ah, I see only now that Halmos refers to Wintnter [160] and Wielandt [158], but I can't see the detailed references. – t.b. Jul 29 '11 at 08:47
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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
  • @Theo: I don't know of anything comparable to Dieudonné's book for those subjects. For von Neumann algebras, the beginning is von Neumann's double commutant theorem in 1929 and then Murray and von Neumann's series of papers beginning in 1936. For C*-algebras, it begins with Gelfand-Neumark 1943. If one is willing to put in a lot of effort, then as you hinted much of the history can be found using extensive bibliographies such as those in Dixmier's books. – Jonas Meyer Jul 30 '11 at 06:12
  • Similarly, for Banach algebras a book like Rickart's (in addition to Żelazko) is great for its references, especially since the subject wasn't very old in 1960 when it was written. (I know I'm only saying the obvious.) I believe that Kadison has written a (not comprehensive) survey article or two on the history of C* and von Neumann algebras. If I'm not wrong I saw one of them in the proceedings to the 1980 conference in Kingston ON (I can't check now). – Jonas Meyer Jul 30 '11 at 06:20
  • Thank you very much for all these pointers. I read some of the Gelfand-Neumark papers and books and I also glanced at Murray-von Neumann but I found the latter extremely difficult to follow. I didn't know either Rickart's or Żelazko's books. I would have loved to read more on the background how these results were discovered, maybe in the spirit of Gindikin's essay on the Gelfand seminar. I recently found that there is http://www.israelmgelfand.com/, with lots of intersting stuff on it. – t.b. Jul 30 '11 at 06:26
  • @Theo: Thanks for those links! I didn't know about either of those. I agree with your sentiments, and I wish such a source existed. For example, it would be fortunate for us if Richard Kadison, who played such an important role in the development of operator algebras since the early 1950s, who shows interest in the history of the subject, and who is an excellent writer (and story teller), would choose to write a comprehensive history. (By the way, there is a whole MathOverflow thread for such fantasies.) – Jonas Meyer Jul 30 '11 at 07:09
  • The Wintner reference is http://journals.aps.org/pr/abstract/10.1103/PhysRev.71.738.2 – Philip Brooker Sep 10 '15 at 13:09
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$$ xy-yx = 1 \\ (x-\alpha 1)y-y(x-\alpha 1)=1 \\ y(\alpha 1-x)-(\alpha 1-x)y = -1 \\ (\alpha 1-x)^{-1}y-y(\alpha 1-x)^{-1}=\frac{d}{d\alpha}(\alpha 1-x)^{-1}. $$ Using the functional calculus for a function $F$ that is holomorphic on an open neighborhood of $\sigma(x)$ leads to a bounded differentiation: $$ F(x)y-yF(x)=-F'(x) \\ \|F'(x)\| \le 2\|y\|\|F(x)\|. $$ If you set $F(\alpha)=e^{t\alpha}$ for large enough real $t$, then the above yields the contradiction $|t| \le 2\|y\|$.

Disintegrating By Parts
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