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Let $A$ be a Banach algebra with the following property:

For every two nets $ x_{\alpha}$ and $y_{\alpha}$ in $A$, $x_{\alpha}y_{\alpha}$ converges if and only if $y_{\alpha}x_{\alpha}$ converges.

Is $A$ necessarily commutative?

1)Note that we do not assume that the above two nets converge to the same value.

2)Note that we do not assume that $A$ has an approximate identity.

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  • $\begingroup$ Is this true for finite-dimensional algebras? $\endgroup$
    – Will Sawin
    Dec 31, 2013 at 23:29
  • $\begingroup$ @WillSawin thanks for your comment. Does a f.d algebra have approx. id.? I ask this question because my main question is true for Banach algebras with approx. Id. The reason: $\endgroup$
    – Ali Taghavi
    Jan 1, 2014 at 21:04
  • $\begingroup$ Let $e_{n}$ be a sequence approximate identity(The net case can be modified in a similar manner). assume that $x,y\in A$. Define $s_{n}=\cases{x & n=2k\\xy & n=2k+1}$ and $t_{n}=\cases{y&n=2k\\e_{n}& n=2k+1}$. Now $S_{n}t_{n}$converges, then $t_{n}s_{n}$ must converges, So xy=yx $\endgroup$
    – Ali Taghavi
    Jan 1, 2014 at 21:10

1 Answer 1

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No. Some algebras satisfy the identity $xy=-yx$ but are not commutative. Such an algebra clearly satisfies your condition. Obviously, it never has an approximate identity.

To construct one, let $V,W$ be two Banach spaces and let $f: V \times V \to W$ be a continuous symplectic bilinear form. Then define a multiplication on $V + W$ where anything in $W$ times anything is zero, and the multiplication on $V$ is given by $f$. Then this is anticommutative, and not commutative unless $f=0$. So it provides a counterexample.

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  • $\begingroup$ Thank you for your answer. do you think that my question is true for nondegenerate banach algebras, namely for each a, there is b with $ab\neq 0$? Is it true to say each nondegenerate finite D algebra has an approximate identity $\endgroup$
    – Ali Taghavi
    Jan 2, 2014 at 22:04

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