If $\sum_{j=1}^n v_j^v_j=1_A=\sum_{j=1}^n v_jv_j^$ in a C* algebra A, then $\sum_{j=1}^n v_j$ is unitary.

Question

Let $v_1,v_2, \cdot v_n$ be partial isometries in a unital $C^*$-algebra $A$ and suppose that $$\sum_{j=1}^n v_j^*v_j=1_A=\sum_{j=1}^n v_jv_j*$$ Show that $\sum_{j=1}^n v_j$ is unitary.

This a exercise from the book "An Introduction to K-theory for $C^*$-algebras". If we let $p_i=v_i^*v_i$, $p_i$ is a projection as $v_i$ is partial isometry. Since $\sum_{j=1}^n p_j=1$ we have $p_ip_j=0$ for each $i \neq j$. If I can show $v_i^*v_j=0$, then clearly $\sum_{j=1}^n v_j$ is unitary. I could not proceed from here.

Can you state which exercise this is? I have a pdf of exercises I completed from this book, and this one might be in there. — Aweygan, Jun 24 '19 at 15:38

score 3 · Accepted Answer · answered Jun 24 '19 at 18:24

The equality $p_kp_j=0$ is $$ v_k^*v_kv_j^*v_j=0. $$ Multiply by $v_k$ on the left and $v_j$ on the right to get $$ (v_kv_k^*)v_k\,v_j^*(v_jv_j^*)=0. $$ Now you use that $$ (v_kv_k^*)v_k=v_k. $$ This works for any partial isometry: if you have $v^*v=p$ and $vv^*=q$ with $p,q$ projections, then $$ 0=p-p^2=v^*v-v^*vv^*v=v^*(1-q)v=v^*(1-q)^*(1-q)v. $$ Then $(1-q)v=0$, which is $v=qv$.

If $\sum_{j=1}^n v_j^*v_j=1_A=\sum_{j=1}^n v_jv_j^*$ in a C* algebra A, then $\sum_{j=1}^n v_j$ is unitary.

1 Answers1

If $\sum_{j=1}^n v_j^v_j=1_A=\sum_{j=1}^n v_jv_j^$ in a C* algebra A, then $\sum_{j=1}^n v_j$ is unitary.