# Does the following $C^{*}$-algebraic result have a purely algebraic proof?

While studying the proof of Bott periodicity for operator $K$-theory in this set of notes, I learned this fact:

Theorem. Let $A$ and $B$ be $C^{*}$-algebras. Let $f,g: A \to B$ be $*$-homomorphisms. Then $f + g$ is also a $*$-homomorphism if and only if the ranges of $f$ and $g$ are orthogonal, i.e., $$f[A] g[A] = g[A] f[A] = \{ 0_{B} \}.$$

In general, $f + g$ is only a $*$-preserving linear map, and multiplication may not be preserved unless further conditions are imposed. I managed to prove the theorem, but my argument is not entirely algebraic in the sense that it uses topological facts about $C^{*}$-algebras.

My proof

The backward implication is trivial enough, so let us prove the forward one only.

Suppose that $f + g$ is a $*$-homomorphism. Then for all $a_{1},a_{2} \in A$, we have \begin{align} (f + g)(a_{1} a_{2}) & = (f + g)(a_{1}) \cdot (f + g)(a_{2}) \\ & = [f(a_{1}) + g(a_{1})] [f(a_{2}) + g(a_{2})] \\ & = f(a_{1}) f(a_{2}) + f(a_{1}) g(a_{2}) + g(a_{1}) f(a_{2}) + g(a_{1}) g(a_{2}), \\ (f + g)(a_{1} a_{2}) & = f(a_{1} a_{2}) + g(a_{1} a_{2}) \\ & = f(a_{1}) f(a_{2}) + g(a_{1}) g(a_{2}). \end{align} It follows immediately that $(\star) ~ f(a_{1}) g(a_{2}) + g(a_{1}) f(a_{2}) = 0_{B}$ for all $a_{1},a_{2} \in A$.

Next, let $a \in A$ be any self-adjoint element. As $(\star)$ implies that $f(a) g(a) = - g(a) f(a)$, we get $$f(a^{2}) g(a^{2}) = f(a) f(a) g(a) g(a) = - f(a) g(a) f(a) g(a) = f(a) g(a) g(a) f(a),$$ and similarly, $$g(a^{2}) f(a^{2}) = g(a) g(a) f(a) f(a) = - g(a) f(a) g(a) f(a) = f(a) g(a) g(a) f(a).$$ We also know from $(\star)$ that $f(a^{2}) g(a^{2}) + g(a^{2}) f(a^{2}) = 0_{B}$, so $f(a) g(a) g(a) f(a) = 0_{B}$. Hence, by the self-adjointness of $a$, we have $$[f(a) g(a)] [f(a) g(a)]^{*} = f(a) g(a) g(a) f(a) = 0_{B}.$$ Therefore, $f(a) g(a) = 0_{B}$, and by interchanging $f$ and $g$, we also obtain $g(a) f(a) = 0_{B}$. As our choice of $a$ was arbitrary, the discussion in this paragraph applies to all self-adjoint elements of $A$.

Finally, let $(e_{i})_{i \in I}$ be any self-adjoint approximate identity in $A$. Then for all $x,y \in A$, we get \begin{align} f(x) g(y) & = \lim_{i \in I} f(x e_{i}) g(e_{i} y) \qquad (\text{$C^{*}$-homomorphisms are automatically continuous.}) \\ & = \lim_{i \in I} f(x) f(e_{i}) g(e_{i}) g(y) \\ & = \lim_{i \in I} f(x) ~ 0_{B} ~ g(y) \qquad (\text{By the previous paragraph.}) \\ & = 0_{B}. \end{align} Similarly, $g(x) f(y) = 0_{B}$ for all $x,y \in A$. This concludes the proof. $\quad \blacksquare$

Question. Can we obtain the same result if we merely assume that $A$ and $B$ are $*$-algebras over $\Bbb{C}$? For convenience, we may suppose that $(a^{*} a = 0_{A}) \Rightarrow (a = 0_{A})$ for all $a \in A$ and likewise for $B$.

Here is a small extension of your idea. You have, for any $a,b\in A$, $$f (a)g (b)+g (a)f (b)=0.$$ Then $$f (ab)g (ba)=f (a)f (b)g (b)g (a)=-f (a)g (b)f (b)g (a)=f (a)g (b)g (b)f( a)$$ and $$g (ab)f (ba)=g (a)g (b)f (b)f (a)=-g (a)f (b)g (b)f (a)=f (a)g (b)g (b)f (a).$$ Now $$0=f (ab)g (ba)+g (ab)f (ba)=2f (a)g (b)g (b)f (a).$$ When $a,b$ are selfadjoint we get $$f (a)g (b)[f (a)g (b)]^*=f (a)g (b)g (b)f (a)=0,$$ and we conclude that $f (a)g (b)=0$ for all selfadjoint $a,b$. But then, as any $x,y\in A$ can be written $x=a+ib$, $y=c+id$, $$f (x)g (y)=f (a+ib)g (c+id)=f (a)g (c)-f (b)g (d)+i [f (b)g (c)+f (a)g (d)]=0.$$
• This is awesome, Martin. Thanks! :) With this, we have the result that if $A$ and $B$ are $*$-algebras over $\Bbb{C}$, and $b^{*} b = 0_{B} \Rightarrow b = 0_{B}$ for every $b \in B$, then the sum of any two $*$-homomorphisms $f,g: A \to B$ is yet again a $*$-homomorphism if and only if the ranges of $f$ and $g$ are orthogonal. – Transcendental Jul 22 '15 at 15:37