Assume that $A$ and $B$ are two unital commutative Banach algebras. Assume that $\phi \in \mathcal{M} (A)$ is an element of the maximal Ideal space. Define $\alpha: A\hat{\otimes} B \to \mathbb{C}\otimes B\simeq B$ with $\alpha=\phi \otimes Id_{B}$. $\;$Put $I=\ker \alpha$. Assume that $J$ is an ideal in $A\hat{\otimes} B$ with $J+ I=A\hat{\otimes} B$.

1.Is there an ideal $K$ in $A$ with $K\hat{\otimes} B \subset J$ and $(K\hat{\otimes} B )+ I= A\hat{\otimes} B$?

2.Consider the same question for two (pure algebraic) complex unital algebras, so we remove $"\hat{}"$ from the above tensor products. and $\phi$ is an algebra morphism from $A$ to $\mathbb{C}$.

This question is motivated by the "tube lemma" in general topology. So the answer to this question is "yes" for commutative $C^{*}$ algebras $A$ and $B$

We can consider the same question for (noncommutative) $C^{*}$ algebras $A$ and $B$, an irreducible representation $\phi: A \to B(H)$ and $\alpha: A\otimes B \to B(H) \otimes B$ and $I=\ker \alpha$.

I explain that why I think that this is a noncommutative version of the tube lemma:

In general, assume that $X$ is a compact Hausdorff space and $F$ and $K$ are two closed sets in $X$, then they are two disjoint set iff $I_{F}+I_{K}=C(X)$ where $I_{F}$ is the ideal in $C(X)$ which consists all $g\in C(X)$ with $g(F)=0$ . Now assume that $F=\{x_{0}\}\times Y$ is a slice in $X\times Y$. Then $I_{F}$ has an algebraic description as the above $I=\ker \alpha$, in my question. If $U$ is an open set containing this slice, then $U^{c}$ and the slice are two disjoint closed set. So the inclusion $\{x_{0}\}\times Y \subset U$ implies that $J+I=C(X)$ where $J=I_{U^{c}}$.

  • $\begingroup$ I removed the "ask WBJ" tag, which if it is to remain used should be used meaningfully. (WBJ is a Banach space theorist, not an operator algebraist) $\endgroup$ – Yemon Choi Apr 24 '14 at 3:33
  • $\begingroup$ The repeated editing of this question seems excessive. $\endgroup$ – Nik Weaver Apr 25 '14 at 0:45
  • $\begingroup$ @NikWeaver I tried to construct a "non unital case" but I need to the following. do you think that the answer to this question is obvious? :math.stackexchange.com/questions/768054/… $\endgroup$ – Ali Taghavi Apr 25 '14 at 1:02
  • $\begingroup$ That is :what is the algebraic analogy of "Two disjoint closed sets" in non unital case? $\endgroup$ – Ali Taghavi Apr 25 '14 at 1:05
  • 1
    $\begingroup$ @AliTaghavi: Given any $f \in C_0(X)$, consider the function on $F \cup K \cup \{\infty\}$ which agrees with $f$ on $K \cup \{\infty\}$ and is zero on $F \cup \{\infty\}$. It is continuous by the gluing lemma and extends to $X \cup \{\infty\}$ by Tietze. Call the extension $g$; then $g + (f - g)$ is the desired decomposition showing $f \in I_F + I_K$. $\endgroup$ – Nik Weaver Apr 25 '14 at 17:06

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