I am looking for examples of amenable Banach algebras which have non-amenable subalgebra

I know

1: Each amenable Banach algebra has a bounded approximate identity

2: If $I$ be a closed ideal in an amenable Banach algebra, then

$I$ amenable if and only if $I$ has a bounded approximate identity

Does anybody have an example?

Thank you for any suggestions!


2 Answers 2


Mateusz's answer mentions lots of good mathematics but I feel obliged to point out that the fundamental example which answers your original question in the negative is $M_2({\bf C})$. (Banach algebras behave very differently from ${\rm C}^*$-algebras and $L^1$-group algebras.)

The point is that the algebra $$ {\bf C}[x] / (x^2) \cong \left\{ \begin{pmatrix} a & b \cr 0 & a \end{pmatrix} \colon a,b \in {\bf C} \right\} \subset M_2({\bf C}) $$ is not amenable, and should not be "generalized amenable" in any sensible version of "generalized amenability".

Non-amenability can be seen in many ways, but the most direct one -- if you use the definition of amenability in terms of derivations -- is to note that the mapping $$ \begin{pmatrix} a & b \cr 0 & a \end{pmatrix} \to b $$ is a non-zero derivation. In some sense this is the philosophical idea behind derivations on associative algebras, they are the 1st-order terms that arise when one perturbs a homomorphism.

  • 1
    $\begingroup$ This is a better example. I guess I'm too attached to $\ast$-stuff. $\endgroup$ Jul 16, 2020 at 15:01
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    $\begingroup$ @MateuszWasilewski In general it does seem that $\ast$-stuff is the way to go :) $\endgroup$
    – Yemon Choi
    Jul 16, 2020 at 17:26
  • $\begingroup$ @Yemon Choi, thank you this is very helpful information $\endgroup$
    – user62498
    Jul 17, 2020 at 14:18

There are plenty of examples, even for $C^{\ast}$-algebras. By the results of Connes and Haagerup, for $C^{\ast}$-algebras amenability is equivalent to nuclearity, so I will work with nuclearity, which is more familiar for operator algebraists.

For group $C^{\ast}$-algebras of discrete groups nuclearity is equivalent to amenability of the group. So, for example, the group $C^{\ast}$-algebra of the free group $F_n$ is not nuclear. On the other hand, it can be embedded in a nuclear $C^{\ast}$-algebra, as I will show below.

Whenever we have an action of a group $\Gamma$ on a compact space $X$, we can construct the (reduced) crossed product $C(X) \rtimes \Gamma$, which contains the group $C^{\ast}$-algebra $C^{\ast}_{r}(\Gamma)$ as a subalgebra. We now need an example of an action of a free group such that the crossed product is nuclear. We can use, for example, the action of the free group on its boundary. You can visualise the free group as a tree, using its Cayley graph, and the boundary in this case will be the boundary of this tree, i.e. space of infinite paths up to natural equivalence. This action is amenable, which is exactly a property needed for proving nuclearity of the crossed product. To sum up, the inclusion $C_{r}^{\ast}(F_n) \subset C(\partial F_n) \rtimes F_n$ gives you an example of a nonamenable subalgebra of an amenable algebra.

  • $\begingroup$ Mateusz Wasilewski,Thank you. This is very helpful. $\endgroup$
    – user62498
    Jul 16, 2020 at 11:12

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