# Show that two C*-algebras are morita equivalent

• Oct 3rd 2010, 02:59 AM
Mauritzvdworm
Show that two C*-algebras are morita equivalent
Let $A$ be a C*-algebra, I would like to show that $A$ is Morita equivalent to $M_l(A)$

Here follows a description of my plan to show this:
Define $F(A):=\{(a_1,\dots,a_l):a_i\in A\}$
let $\overline{a}:=(a_1,\dots,a_l)$ be the row matrix, then we can define the following inner products
$_{A}\langle \overline{a},\overline{b}\rangle:=\overline{a}\ove rline{b}^*$
$\langle \overline{a},\overline{b}\rangle_{M_l(A)}:=\overli ne{a}^*\overline{b}$

The above choice of inner product yields the desired result which we need to have an imprimitivity bimodule, namely
$_{A}\langle \overline{a},\overline{b}\rangle \overline{c}=\overline{a}\overline{b}^*\overline{c }=\overline{a}\langle \overline{b},\overline{c}\rangle_{M_l(A)}$

I can show that $F(A)$ in complete in the norm induced by the $A$-inner product. But not for the norm induced by the $M_l(A)$-inner product.

I can also show that $I_{A}:=\text{span}\{_{A}\langle \overline{a},\overline{b}\rangle:\overline{a},\ove rline{b}\in A\}$ is dense in $A$. Is it possible to show that
$I_{M_{l}(A)}:=\text{span}\{\langle \overline{a},\overline{b}\rangle_{M_l(A)}:\overlin e{a},\overline{b}\in A\}$ is dense in $M_l(A)$?

Or is this idea totally wrong?
• Oct 8th 2010, 08:00 AM
Mauritzvdworm
I figured it out, here is an outline for those that might be interested

The idea I had worked perfectly well, we can show that
$I_{M_l(A)}=\{\langle \overline{a},\overline{b}\rangle\}$ is dense in $M_l(A)$ irrespective if $A$ is unital or not. The trick is in noticing that we can create matrices with either 1 (in the unital case) or the approximate identity (non-unital case) at the positions $(i,j)$. The span of these matrices will be dense in $M_l(A)$.

The Cauchy sequences are also easy since we are dealing with a finite dimensional space $F(A)$. The inner products $\langle\cdot,\cdot\rangle_{M_l(A)}$ and $_{A}\langle\cdot,\cdot\rangle$ induce well defined norms. Now there is a theorem which states that every finite dimensional normed space in complete.

All the conditions are then fullfilled and we see that $F(A)$ is in fact an $A-M_{l}(A)$-Imprimitivity bimodule and hence $A$ and $M_l(A)$ are Morita equivalent.