Pairwise orthogonality for partitions of unity in a *-algebra

Question

Let $\mathcal{A}$ be a $*$-algebra with unit $1_{\mathcal{A}}$. As in the $\mathrm{C}^*$-setting, a projection is an element $p\in\mathcal{A}$ such that $p=p^2=p^*$. A partition of unity is a finite set of projections $\{p_i\}_{i=1}^N\subset \mathcal{A}$ such that: $$\sum_{i=1}^N p_i=1_{\mathcal{A}}.$$

Is there an example of a partition of unity such that the projections are not pairwise-orthogonal?

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If $\mathcal{A}$ is a $\mathrm{C}^*$-algebra (or presumably, if there exists a *-representation $\pi(\mathcal{A})\subset B(\mathsf{H})$, or something similarly and suitably embedding the partition of unity in a $\mathrm{C}^*$-algebra), then it is well known that the projections in a partition of unity are pairwise-orthogonal. In the theory of quantum permutation groups, there is an approach that uses universal *-algebras generated by various partitions of unity, and because it is believed that the answer to the above question is yes, the orthogonality relations $p_ip_j=\delta_{ij}\,p_i$ are added at that point (where in the universal $\mathrm{C}^*$-algebra approach, such relations are automatic).

I have asked some people for an explicit example/counterexample and thus far nobody knows beyond "person $X$ says there is one".

Answer 1

The following arguments are perhaps not as explicit as you might like, but at least they show that orthogonality is not automatic for $N \ge 5$.

• Consider first the purely algebraic case without involution. For this, Example 4.2 of a paper by Bart, Ehrhardt and Silbermann gives non-orthogonal idempotents $p_1, \ldots, p_4$ summing to $0$. Trivailly, using $p_5 = 1$ in addition results in five idempotents summing to $1$ and violating the orthogonality.

• Now we have a complex algebra with idempotents $p_1, \dots, p_5$ such that $p_1 + \cdots + p_5 = 1$, but such that the orthogonality relation $p_i p_j = 0$ for $i \neq j$ does not hold. This implies that the orthogonality does not hold in the universal algebra $$\langle p_1, \dots, p_5 \mid p_i^2 = p_i, \; p_1 + \cdots + p_5 = 1\rangle$$ either. But this universal algebra is such that the relations only involve palindromes, and therefore we can define an involution $*$ by reversing every word in the generators and extending conjugate linearly. In particular, this makes the generators $p_i$ become self-adjoint, and we therefore have the desired counterexample.

It seems plausible that the construction of Bart, Ehrhardt and Silbermann could be adapted to produce $p_1 + \dots + p_4 = 1$ already, in which case a counterexample with $N = 4$ would follow.