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Assume that $A$ is a Banach algebra with two closed two sided ideals $I$ and $J$ such that $I$ and $J$ are commutative and $A=I+J$. Does this implies that $A$ is commutative? For the $C^{*}$ algebra, the answer is "Yes".

The motivation: A (noncommutative) compact n dimensional topological manifold could be defined as follows:

A (non commutative) $C^{*}$ or Banach algebra $A$ such that there are ideals $I_{k}$, $k=1,2,\ldots,n$ such that $A=I_{1}+I_{2}+\ldots I_{k}$ and each $I_{j}$ is isomorphic to $C_{0}(\mathbb{R}^{n})$.

But the above answer in MSE shows that, in the context of $C^{*}$ algebras, this definition does not give any non commutative example,.

So we search for a non commutative example in Banach algebras.

Note: According to the comment on my MSE question: To what extent Banach or $C^{*}$ algebras whose underline Lie algebras are metabelian are studied and classified?

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  • $\begingroup$ I just rolled back an edit which made one unnecessary change and which made one change which alters the meaning of one of the sentences $\endgroup$
    – Yemon Choi
    Commented Nov 8, 2014 at 19:38
  • $\begingroup$ By the way, now you have changed this to Banach algebras I think this is not a good way to try to define a NC manifold. Recall that the Gelfand-Naimark correspondence only works for commutative C*-algebras. Also please read about general Banach algebras to see that they behave very very very very very very very very differently from the C*-case and to call a Banach algebra a noncommutative space is IMHO extremely tendentious $\endgroup$
    – Yemon Choi
    Commented Nov 8, 2014 at 19:48
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    $\begingroup$ @YemonChoi "Dense subalgebra" you mean smooth function? So what about the algebraic picture of topological manifold without smooth structure? $\endgroup$
    – Ali Taghavi
    Commented Nov 8, 2014 at 20:07
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    $\begingroup$ Good point. OK, what you really need to do is work out a NC version of affine space - within the class of abelian $C^*$-algebras, what properties single out $C_0({\bf R}^k)$; and then what non-abelian $C^*$-algebras also satisfy those properties. Then you can start to ask about gluing, as in your original question. $\endgroup$
    – Yemon Choi
    Commented Nov 8, 2014 at 20:09
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    $\begingroup$ Well, a manifold is a topological space that looks locally like ${\bf R}^k$. So a NC manifold should be something that behaves "locally" (whatever that means) like a NC version of ${\bf R}^k$. Andreas Thom's answer shows that at least in the unital (i.e. compact) case there is nothing gained by trying to glue commutative pieces, so first work out what kinds of NC algebras would be good pieces $\endgroup$
    – Yemon Choi
    Commented Nov 8, 2014 at 20:26

1 Answer 1

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Theorem Let $A$ be a unital ring and $I_1,\dots,I_n \subset A$ be 2-sided commutative ideals such that $A=I_1+\dots + I_n$. Then, $A$ is commutative.

Proof: If $A=I_1+\dots+I_n$, then $1 = x_1+\dots+x_n$ for $x_i \in I_i$. But then, $$1 = (x_1+\dots+x_n)^{n+1} \in I_1^2 +\dots+ I_n^2$$ and we conclude that $A=I_1^2 + \dots + I_n^2$.

Now, if $I \subset A$ is any abelian 2-sided ideal, then $I^2$ is central, since $$a(bc) = (ab)c= c(ab) = (ca)b = b(ca) = (bc)a,$$ for any $a \in A$ and $b,c \in I$. Since $A = I_1^2 + \dots + I_n^2$, $A$ is commutative. q.e.d.

If $A$ is not unital, then the same argument still shows that $A^{n+1} \subset A$ is central. If $A^2 \neq A$, then there are counterexamples - even for Banach algebras. For example, one can consider the non-abelian algebra $$A = \left( \begin{matrix} 0 & * & * \\ 0 & 0 & *\\0 & 0 & 0 \end{matrix}\right)$$ of strictly upper-triangular $3\times 3$-matrices (with entries in a field), which is is a sum of two abelian ideals $$I = \left( \begin{matrix} 0 & 0 & * \\ 0 & 0 & *\\0 & 0 & 0 \end{matrix}\right), \quad J = \left( \begin{matrix} 0 & * & * \\ 0 & 0 & 0\\0 & 0 & 0 \end{matrix}\right).$$

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  • $\begingroup$ Thank you very much for your interesting answer. So We conclude that every non commutative topological manifold, with the above definition, is necessarily "noncompact",i.e nonunital. Your example is very interesting, however it gives the empty manifold since the multplication of each ideal is identically zero. So the next questions could be:What is an example of a non commutative manifold(non unital non commutative banach algebra)? Is it true to say that the tensor product of two manifold is again a manifold? Are there universal NC manifolds $A_{n}$ such that.... $\endgroup$
    – Ali Taghavi
    Commented Nov 8, 2014 at 18:46
  • $\begingroup$ ...such that every NC manifold is the image of some $A_{N}$, for some $N\in \mathbb{N}$?(The analogy for Withney theorem). Thanks again for your answer. $\endgroup$
    – Ali Taghavi
    Commented Nov 8, 2014 at 18:48
  • $\begingroup$ So a pre-question:Is the proper analogy of the Withney embedding theorem, true?Every manifold is properly embedded in some $\mathbb{R}^{n}$? $\endgroup$
    – Ali Taghavi
    Commented Nov 8, 2014 at 19:16

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