## spectra of sums and products in (Banach) algebras [was: Spectrum in Banach Algebra]

Let a,b be 2 elements in a Banach Algebra.Let Spec(x) denote the spectrum of an element x. If a,b commute with each other, then by Gelfand Transformation, we have Spec(a+b) is a subset of Spec(a)+Spec(b)& Spec(ab) is a subset of Spec(a)Spec(b) (Where the definition of addition and multipication of two sets is obvious). My Questions, if we drop the the condition "a,b commute", then to what extent is the above relations about spectra still hold?

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For sums you may be interested in the following: mathoverflow.net/questions/4224/…. For products, consider the case where a is a nonzero nilpotent matrix and b is its conjugate transpose. – Jonas Meyer Feb 5 2010 at 7:41
They typically won't hold, but I don't know what you're looking for in an answer. – Jonas Meyer Feb 5 2010 at 7:48
@Jonas: sorry, I spent quite a while typing and didn't see your comment - hence the duplicate effort below. Regarding your 2nd comment: I think that in the earlier years of the subject, there was some work on showing that conditions like subadditivity and submultiplicativity of spectral radius come close to characterizing commutativity of a Banach algebra. I'd have to look this up, but my gut feeling is that in NC settings one has to give up hope of similar results – Yemon Choi Feb 5 2010 at 7:59
@Yemon: That is interesting, but to me it isn't clear exactly what the question above is. I mean, your answer covers it as far as I can tell, but a more interesting question would be the one (I think) you just raised: If these properties of the spectrum are satisfied for all a and b in a Banach algebra A, must A be commutative? – Jonas Meyer Feb 5 2010 at 8:03
@Jonas: good point. I don't have a definitive answer but will amend my answer below to say a bit more about this – Yemon Choi Feb 5 2010 at 8:22
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In general Banach algebras the spectral radius is neither subadditive nor submultiplicative; in particular, neither of the two properties you mention holds.

$2\times 2$ (real or complex) matrices should suffice to give examples, so this is not to do with any subtleties of infinite-dimensional algebras.

For example, take $a= \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)$. Note that this is nilpotent, so the only point in the spectrum is zero, and hence $\sigma(a)\sigma(b)= \{0\}$ for any other matrix $b$. On the other hand, we can find $b$ for which $ab$ is not nilpotent, so that $\sigma(ab)\not\subseteq {0}$. A simple choice which works is $b=\left(\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right)$, since then $ab$ is a non-zero projection (=idempotent) and so contains $1$ in its spectrum.

The same pair also works as a counter-example for the "additive question". For since $\sigma(a)=\sigma(b)=\{0\}$, we have $\sigma(a)+\sigma(b)=\{0\}$. On the other hand, $a+b$ is a reflection and hence its spectrum is $\{-1,1\}$

(edited 11-02-10 for a couple of typos/omissions)

Jonas Meyer points out in comments that one can pose the following converse question: let $A$ be a Banach algebra with identity, and suppose that we have

($*$) $\sigma(a+b)\subseteq\sigma(a)+\sigma(b)$ and $\sigma(a)\sigma(b)\subseteq\sigma(ab)$ for all $a,b\in A$.

Must $A$ be commutative?

As Jonas also pointed out in comments, the answer is in general `no': the algebra of strictly upper-triangular matrices (or, more precisely, the algebra of scalar+strictly upper triangular $m\times m$ matrices, for some fixed $m$) gives a counterexample, at least when $m\geq 3$.

A more careful version of this argument shows, I think, that if $A$ is a finite-dimensional algebra with identity, such that $A/{\rm Rad}(A)$ is commutative, then $A$ will satisfy condition $(*)$. I suspect that the same might be true for any unital Banach algebra that is commutative modulo its radical, i.e. that the finite-dimensional hypothesis is unnecessary; but at present I'm a bit too tired to check this properly.

We could therefore modify the question yet further, and ask if a Banach algebra that satisfies condition $(*)$ must be commutative modulo its radical. The answer turns out to be yes, after a wander down memory lane and a forage on MathSciNet:

MR0461139 (57 #1124) J. Zemánek. Spectral radius characterizations of commutativity in Banach algebras. Studia Math. 61 (1977), no. 3, 257--268.

The MR is short and informative enough to give in full, for those without access:

It is standard that the spectral radius is subadditive and submultiplicative on any commutative complex Banach algebra. The author proves that, for a complex Banach algebra $A$, the following three conditions are equivalent: (1) the spectral radius is sub-additive on $A$, (2) the spectral radius is submultiplicative on $A$, (3) $A$ is commutative modulo its radical. Some applications of this result to other problems in Banach algebras are given, along with references to a number of related papers.
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Incidentally, to me "upper triangular" just means zero below the diagonal, and "strictly upper triangular" means also zero on the diagonal. I thought that was fairly standard. Not that it affects your answer, but we were thinking of slightly different examples. Also incidentally, +1'd. – Jonas Meyer Feb 11 2010 at 20:17
(With upper triangular instead of scalar+strictly ut, m=2 works too.) – Jonas Meyer Feb 11 2010 at 21:20