Constructing a projection

**Mauritzvdworm** · October 20th 2010, 03:53 AM

Let $B$ be a C*-algebra and let $p$ be a rank-one projection in $K$ , the space of compact operators on some Hilbert space $H$ .

I would like to construct a projection $\mathcal{P}$ in the multiplier algebra $M(B\otimes K)$ such that $\mathcal{P}(B\otimes K)\mathcal{P}=B\otimes p$

here is my current idea:
consider the mapping $\eta:B\otimes K\rightarrow B\otimes p$ such that $b\otimes k\mapsto b\otimes p$ . Here $b\in B,k\in K$
It can easily be shown that $\eta^2=\eta$ , however I'm having some trouble in showing that $\eta^*=\eta$ . Is it possible?

**Opalg** · October 20th 2010, 12:00 PM

Originally Posted by Mauritzvdworm

Let $B$ be a C*-algebra and let $p$ be a rank-one projection in $K$ , the space of compact operators on some Hilbert space $H$ .

I would like to construct a projection $\mathcal{P}$ in the multiplier algebra $M(B\otimes K)$ such that $\mathcal{P}(B\otimes K)\mathcal{P}=B\otimes p$

here is my current idea:
consider the mapping $\eta:B\otimes K\rightarrow B\otimes p$ such that $b\otimes k\mapsto b\otimes p$ . Here $b\in B,k\in K$
It can easily be shown that $\eta^2=\eta$ , however I'm having some trouble in showing that $\eta^*=\eta$ . Is it possible?

The trouble with the mapping $\eta$ is that it is not well defined (for example, $(2b)\otimes k = b\otimes(2k)$ , but $\eta$ takes the first of these to $2b\otimes p$ and the second to $b\otimes p$ ). Try the map $b\otimes k\mapsto b\otimes pkp$ .

**Mauritzvdworm** · October 20th 2010, 10:47 PM

Now I see that my original idea was wrong. For the mapping
$\eta:b\otimes k\mapsto b\otimes pkp$
it is easy to show that $\eta=\eta^2$ , but once again I am missing something in determining $\eta^*$ ?

The way I am thinking about this is to have some kind of inner product in order to determine $\eta$ .

Another thing is if this $\eta$ is in fact the projection I am looking for then
$\mathcal{P}(b\otimes k)\mathcal{P}=(b\otimes pkp)\mathcal{P}$
now if we let $b\otimes pkp:=a$ then
$a\mathcal{P}=\left(\mathcal{P}a^*\right)^*=\left(\ mathcal{P}\left(b^*\otimes pk^*p\right)\right)^*=\left(b^*\otimes pk^*p\right)^*=b\otimes pkp$

but is $pkp=p$ ?

**Opalg** · October 20th 2010, 11:49 PM

Looking at the problem again, I don't see why you need the map $\eta$ at all. Simply define $\mathcal{P} = 1\otimes p$ , where 1 is the identity element in $M(B)$ . The point is that if $p$ is a rank-one projection then $pkp$ is always a scalar multiple of $p$ . In fact, if $\xi$ is a unit vector in the range of $p$ then $pkp = \varphi(k)p$ , where $\varphi$ is the state on $K$ defined by $\varphi(k) = \langle k\xi,\xi\rangle.$

**Mauritzvdworm** · October 20th 2010, 11:51 PM

Aaaaa, I see! Thanks a lot!

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