1

The Cuntz algebra is sometimes defined as the (universal) C*-algebra on isometries $s_1,\ldots, s_n$ satisfying $\sum s_is_i^* = 1$. This is for instance the definition on Wikipedia and also in the original paper, which also states that orthogonoality $s_i^*s_j=\delta_{ij}$ follows from this definition.

How does this follow from the definition?

It is easy to see from the concrete C*-algebra construction of $\mathcal O_n$ that orthogonality is satisfied. So once you know this algebra is universal and also simple, it follows that for any such operators one has orthogonality $s_i^*s_j = \delta_{ij}$. Is this what is meant when it is said that it follows from the equation, or is it more obvious? At least in the case of $n=2$ it can easily be proven algebraically.

(Note: This question came up when thinking about find a unitary element in Cuntz algebra $\mathcal{O}_n$)

Ben
  • 7,321

1 Answers1

3

We know that if we have two projections $p,q$, then $p + q$ is a projection if and only if $p$ and $q$ are orthogonal. Thus if you have partial isometries (or just isometries in this case) $v,w$ such that $vv^* + ww^*$ is a projection, then they necessarily have orthogonal ranges ($vv^*ww^* = 0$). Now since $v,w$ are partial isometries, $v = vv^*v, w = ww^*w$, and so

$$ v^*w = v^*vv^*ww^*w = v^*(vv^*)(ww^*)w = 0. $$

In particular if $(s_j)_{j=1}^n$ are $n$ isometries satisfying the Cuntz relation, then they have orthogonal ranges and $s_i^*s_j = \delta_{ij}$.

PStheman
  • 1,711