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Let $A$ be a $C^*$-algebra and $I$ be a two side closed (essential) ideal of $A$. Suppose that $p \in A\backslash I$ is a non trivial projection. Let $B=pIp$. My questions are:

(1) Is $B$ a $C^*$-subalgebra of $I$?

(2) If (1) is correct, then, is $B$ unital?

(3) If both (1) and (2) are right, then, what is the "unit" of $B$? is it the projection "p"?

Special case of this may be: $A=M(I)$, the multiplier algebra of $C^*$-algebra $I$. Hope any comments for these.

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    $\begingroup$ I'm glad that you now understand your question well enough that it seems trivial to you, but it would be more useful to everyone to leave the question visible to give a context to the answers. $\endgroup$
    – David E Speyer
    Commented May 8, 2012 at 14:34
  • $\begingroup$ Editing your post so as to remove the question is inappropriate. I've put back the original post. $\endgroup$
    – Zev Chonoles
    Commented May 8, 2012 at 16:17

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It is true that $B$ is a C$^*$-subalgebra. But it doesn't have to be unital. Consider for example $A=M_2(\ell^\infty(\mathbb{N}))$, $I=M_2(c_0(\mathbb{N}))$, and $$ p=\begin{bmatrix}1&0 \\\\ 0&0\end{bmatrix}. $$ Then $pIp$ is $c_0(\mathbb{N})$, which is not unital.

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  • $\begingroup$ Thanks Argerami! But every element $b$ in $B=pIp$ has this form $b=pip$, where $i \in I$, then $pb=ppip=b=pipp=bp$ since $p$ is a projection. How do we explain this? $\endgroup$
    – Aviv
    Commented Mar 9, 2012 at 2:00
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    $\begingroup$ Aviv, what is there to explain? $p$ does not belong to $B$, so there's no contradiction with the fact that $B$ is not unital $\endgroup$
    – Yemon Choi
    Commented Mar 9, 2012 at 6:55
  • $\begingroup$ Aviv, the double commutant of a C$^*$-algebra (i.e. the closure on the weak operator topology) is always a von Neumann algebra, so in particular it is unital. Or, seen from another point of view, any C$^*$-algebra has an approximate unit, and this net converges to the identity in the double commutant. If this is not clear to you, you may want to try to show that the $I$ in my example is unital. $\endgroup$
    – Martin Argerami
    Commented Mar 9, 2012 at 16:37
  • $\begingroup$ Thanks for your example. Now, I see my mistake. $B=pIp$ is one hereditary $C^*$-subalgebra of $A$, but not necessary to be unital. $\endgroup$
    – Aviv
    Commented Mar 10, 2012 at 1:48
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    $\begingroup$ Done. But it's a losing battle, unfortunately. Those people get their highs from their power trips, and because of that they are way more active than anybody else. One of them has twice been suspended for a year, and they are still very active within the clique. $\endgroup$
    – Martin Argerami
    Commented Nov 14, 2021 at 19:35
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$B$ does not have to be unital. Think of the case $A = M(I)$. Then $p =1$ is a reasonable projection in $A \backslash I$. In this case $B= I$. Since a unit $1$ in $B$ has to satisfy $1 = p\cdot 1\cdot p = p^2 = p$, $p$ is the only choice you have. Therefore $B$ is never unital for $p \in A \backslash I$.

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  • $\begingroup$ Oh, Iam sorry, I mean that the projection $p$ is not trivial. $\endgroup$
    – Aviv
    Commented Mar 8, 2012 at 14:38
  • $\begingroup$ Okay, but still. If you have $1 \in B$, then $1 = p \cdot 1 \cdot p = p^2 = p$, so $B$ can not be unital. Or am I misinterpreting something? $\endgroup$
    – Ulrich Pennig
    Commented Mar 8, 2012 at 18:36

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