standard picture of the index map

102 Views Asked by At

enter image description here

In the above proposition,there is a remark,which states that the right-hand side of (9.6) does not depend on the choice of $v$.

For diffenrent $v$s,there may exist different $p$-s and $q$-s,how to show that the index map $\delta_1$ is well defined?

1

There are 1 best solutions below

1
On BEST ANSWER

The right hand side does not depend on $v$ because $\delta_1$ is already a well-defined map (see the previous section). The proposition gives that, for any $v$ satisfying the hypothesis, with $p,q$ defined as in the statement of the proposition, $\delta_1([u]_1) = [p]_0 - [q]_0$. So if $v_1$ also satisfy the hypothesis, with $p_1,q_1$ being the corresponding projections, then $$ [p_1]_0 - [q_1]_0 = \delta_1([u]_1) = [p]_0 - [q]_0. $$ Also, if you are concerned with different $p,q$'s for the same partial isometry $v$, remember that $\tilde{\phi}$ is injective (by 4.3.1).