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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    $\begingroup$ You may find of interest Halmos's exposition on commutators here. $\endgroup$ Jul 29, 2011 at 4:43
  • $\begingroup$ 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 $xy - yx = I$. $\endgroup$ Jul 29, 2011 at 11:59
  • $\begingroup$ @Geoff: ah. Let's suppose I mean "finitely generated ring" (over $\mathbb{Z}$). $\endgroup$ Jul 29, 2011 at 13:27
  • $\begingroup$ 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. $\endgroup$
    – user16299
    Jan 7, 2012 at 19:42
  • $\begingroup$ 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. $\endgroup$
    – Mike F
    Feb 8, 2016 at 6:39

3 Answers 3


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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    $\begingroup$ That's neat! The complete case also implies the general case, because every normed algebra has a Banach algebra completion. $\endgroup$ Jul 29, 2011 at 7:12
  • $\begingroup$ That (and more) is mentioned in Halmos's exposition that I already linked to in a comment. $\endgroup$ Jul 29, 2011 at 15:08
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    $\begingroup$ @ Bill: not everyone can see that link (me, for example) $\endgroup$ Jul 29, 2011 at 17:25
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    $\begingroup$ This is a very cool argument, but a little mysterious... $\endgroup$
    – Igor Rivin
    Dec 4, 2013 at 4:10
  • $\begingroup$ that's the exact answer I'm looking for. $\endgroup$
    – Tomas
    Dec 4, 2013 at 4:16

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.)

  • $\begingroup$ Oh, of course. Because $\lambda - xy$ is invertible if and only if $\lambda - yx$ is... thanks! $\endgroup$ Jul 29, 2011 at 4:42
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    $\begingroup$ 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. $\endgroup$
    – t.b.
    Jul 29, 2011 at 8:32
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    $\begingroup$ @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. $\endgroup$ Jul 29, 2011 at 8:42
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    $\begingroup$ 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. $\endgroup$
    – t.b.
    Jul 29, 2011 at 8:47
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    $\begingroup$ 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. $\endgroup$
    – t.b.
    Jul 29, 2011 at 12:16

$$ xy-yx = 1 \\ (x-\alpha 1)y-y(x-\alpha 1)=1 \\ y(\alpha 1-x)-(\alpha 1-x)y = -1 \\ (\alpha 1-x)^{-1}y-y(\alpha 1-x)^{-1}=\frac{d}{d\alpha}(\alpha 1-x)^{-1}. $$ 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 $|t| \le 2\|y\|$.


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