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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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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. – David Speyer May 8 '12 at 14:34
Editing your post so as to remove the question is inappropriate. I've put back the original post. – Zev Chonoles May 8 '12 at 16:17
Thanks, I am sorry! – Aviv May 8 '12 at 16:40
up vote 6 down vote accepted

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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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? – Aviv Mar 9 '12 at 2:00
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 – Yemon Choi Mar 9 '12 at 6:55
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. – Martin Argerami Mar 9 '12 at 16:37
Thanks for your example. Now, I see my mistake. $B=pIp$ is one hereditary $C^*$-subalgebra of $A$, but not necessary to be unital. – Aviv Mar 10 '12 at 1:48

$B$ 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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Oh, Iam sorry, I mean that the projection $p$ is not trivial. – Aviv Mar 8 '12 at 14:38
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? – Ulrich Pennig Mar 8 '12 at 18:36
Thanks for your concern! I see my mistake. – Aviv Mar 10 '12 at 1:51

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