A question about orthogonality Let $\mathcal{A}$ be a unital $*$-algebra over $\mathbb{C}$ and let $a,b\in\mathcal{A}$ be projections, that is, $a=a^*=a^2$ and $b=b^*=b^2$. If $a+b=1$, then $ab=0$. This follows from -  \begin{align*}
ab = a(a+b)b = a(ab+b^2)=a(ab+b)=a^2b+ab=ab+ab,
\end{align*} which implies that $ab=0$.
Question: Does this result extend to $n$ elements? That is, if $x_1,...,x_n$ are projections in $\mathcal{A}$ such that $x_1+...+x_n=1$, then is $x_ix_j=0$ for $i\neq j$ and $1\leq i,j\leq n$?
 A: I just want to add a couple of observations to Jose Brox's answer.
Firstly your question had the additional assumption that $A$ has an anti-involution $*$ fixing each of the $x_i$, but this doesn't make a difference. Indeed consider unital $\mathbb C$-algebra with presentation
$$
  F=\mathbb C\langle x_1,\ldots,x_n \mid x_i^2=x_i,\;x_1+\ldots+x_n=1\rangle.
$$
Since the relations are unchanged when the order of multiplication is reversed, there is a unique anti-involution $*$ on $F$ fixing each $x_i$. If $A$ is a counterexample without an anti-involution, then it admits a homomorphism from $F$ whose kernel doesn't contain $x_ix_j$ for some $i\neq j$. Hence $x_ix_j\neq0$ in $F$, so $F$ is a counterexample with an anti-involution.
Secondly the statement does hold if $A$ is finite dimensional. Left multiplication by $x_i$ defines a linear transformation $y_i\in\mathrm{End}_{\mathbb C}(A)$, and these satisfy the same relations: $y_i^2=y_i$ and $\sum_iy_i=\mathrm{id}_A$. Each $y_i$ can be written as a composite of a surjection and injection:
$$
  y_i=p_iq_i,\hspace{5mm}
  p_i:V_i\hookrightarrow A,\hspace{5mm}
  q_i:A\twoheadrightarrow V_i
$$
for some vector space $V_i$. Consider the (external) direct sum
$$
  V=\bigoplus_i V_i
$$
and the maps $p:V\rightarrow A$, $q:A\rightarrow V$ obtained from $p_i$ and $q_i$. Then
$$
  pq=\sum_ip_iq_i=\sum_i y_i=\mathrm{id}_A.
$$
Also
$$
  \dim A=\mathrm{tr}\;\mathrm{id}_A=\sum_i\mathrm{tr}\;y_i
    =\sum_i\dim V_i=\dim V.
$$
Hence $qp=\mathrm{id}_V$. Let $\pi_i:V\rightarrow V_i$ and $\iota_i:V_i\rightarrow V$ denote the natural projection and injection respectively. If $i\neq j$ then
$$
  q_ip_j=\pi_iqp\iota_j=\pi_i\iota_j=0
$$
so $y_iy_j=p_iq_ip_jq_j=0$. Hence
$$
  x_ix_j=(y_iy_j)(1)=0.
$$
A: Let $R$ be a ring free of all torsion, with unit and idempotents $e_1,\ldots,e_n$ such that $e_1+\ldots+e_n=1.$ If $n\leq 3$ then necessarily $e_ie_j=0$ for $i\neq j$, but if $n\geq 4$ the claim is false. This second part is proved in Nonorthogonal idempotents whose sums is idempotent (Mauldon, 1964):
Nonorthogonal idempotents whose sums is idempotent
Let us give a proof for $n=3$:
First we show (lemma) that if $e,f$ are idempotents with idempotent sum then they are orthogonal:
$(e+f)^2=e+f\Rightarrow e^2+f^2+ef+fe=e+f\Rightarrow ef+fe=0$. Multiplying by $e$ on the left we get $ef=-efe$, if on the right, $fe=-efe$, so $ef=fe$. Hence from $ef+fe=0$ we deduce $2ef=0$, which implies $ef=0=fe$.
Now consider idempotents $e,f,g$ such that $e+f+g=1$. Then $e=1-(f+g)$ is idempotent, so $(1-(f+g))^2 = 1-(f+g)\Rightarrow 1+(f+g)^2-2(f+g) = 1-(f+g)\Rightarrow (f+g)^2=f+g$. Hence $f,g$ are idempotents with idempotent sum and therefore orthogonal by our lemma.
Finally, note that $f=(e+f+g)f=ef+f+gf=ef+f\Rightarrow ef=0$ and analogously $fe=0=eg=ge$.

Additional remark:
If you have four idempotents $e,f,g,h$ such that $e+f+g+h=1$ then there is something you can say:
Since $1-(f+g+h)=e$ is idempotent we get as before $(f+g+h)^2=f+g+h$ and $fg+gf+fh+hf+gh+hg=0$.   (1)
On the other hand, since $e+f+g+h=1$, multiplying by $f$ on the left we get $fe+fg+fh=0,$ and on the right $ef+gf+hf=0$, so $ef+fe+fg+gf+fh+hf=0$. (2)
Substracting (1) from (2) we get $$ef+fe=-(gh+hg).$$
A: To answer the new request from the OP (expressed in a commentary), let us see that if $R$ is a ring and $e_i\in R$ are nonzero idempotent and commutative such that $e_1+\ldots+e_n=1$ then they are in fact orthogonal, provided that $R$ is free of torsion up to $n-1$:
I have thought of a proof by backwards induction. First observe that
$$e_1\ldots e_n = e_1\ldots e_n\cdot 1 = e_1\ldots e_n(e_1+\dots + e_n) = n\cdot e_1\ldots e_n,$$
so that
$(n-1)e_1\ldots e_n = 0$. Since $R$ is free of $n-1$ torsion (and $n>1$) we get $e_1\ldots e_n=0$.
We have shown the base case. Now pick $2\leq m\leq n$, suppose that every product of exactly $m$ different idempotents from $\{e_i\}$ is $0$ and let us prove that the same is true for products of $m-1$ different idempotents. Suppose without loss of generality that the $m-1$ idempotents are $e_1,\ldots,e_{m-1}$. Then 
$$e_1\ldots e_{m-1} = e_1\ldots e_{m-1}(e_1+\ldots +e_n) = e_1\ldots e_{m-1}(e_1+\dots + e_{m-1} + e_m + \ldots + e_n) = (m-1)e_1\ldots e_{m-1},$$
since the products of $m$ different idempotents are $0$.
Therefore $$(m-2)e_1\ldots e_{m-1}=0$$ and, since $R$ is free of $m-2$ torsion, we get $e_1\ldots e_{m-1}=0$.
Finally, taking $m=3$ we get that every pair of different idempotents is orthogonal.
Additional remark: If one of the idempotents happens to be $1$, then a posteriori the others are $0$ (by the same proof), so if there are at least two different nonzero idempotents then none of them can be $1$.
