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Math Help - Show that two C*-algebras are morita equivalent

  1. #1
    Member Mauritzvdworm's Avatar
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    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?
    Last edited by Mauritzvdworm; October 3rd 2010 at 02:00 AM. Reason: typo in description
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  2. #2
    Member Mauritzvdworm's Avatar
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    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.
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