In "Disjointness preserving operators on $C^*$ algebras" by Manfred Wolff, Arch. Math 62, 248-253, 1994 it is presented the concept of zero divisor preserving map as follows:
Let $A$ and $B$ be two $C^*$ algebras. A symmetric bounded linear operator $\phi:A\to B$ is disjointness preserving if $xy=0, x=x^*, y=y^*$ implies that $\phi(x)\phi(y)=0$.
In example 2.2 it is written that every Jordan * homomorphism is a disjointness preserving operator.
What is a proof for this fact?
I tried to prove it via points in the following post but the answer to the post shows that my method does not work.
The equation $ab+ba=0, a=a^*, b=b^*$ in a $C^*$ algebra
I had intention to prove it via the following: Let $\phi:A\to B$ be a symetric Jordan homomorphism. Assume that $xy=0$ for self adjoint $x,y$
Then we have $xy+yx=0$ then we have $\phi (x)\phi(y)+\phi(y)\phi(x)=0$. Moreover $a=\phi(x)$ and $b=\phi (y)$ are self adjoint elements satisfying $ab+ba=0$ but according to the answer to the above post this does not imply that $ab=0$
So how can one prove the assertion of example 2.2 in the Wolff paper?
Note: A symmetric linear operator is a linear map $\phi:A\to B$ with $\phi(x^*)=(\phi(x))^*$. A aJordan Homomorphism is a linear map satisfying one of the following equivalent conditions:
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$\phi(x^2)=(\phi(x))^2$
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$\phi(xy+yx)=\phi(x)\phi(y)+\phi(y)\phi(x)$
Best Answer
For a Jordan homomorphism, $\phi(yxy)=\phi(y)\phi(x)\phi(y)$. Following is a proof from I.N. Herstein, Jordan homomorphisms, Trans. Amer. Math. Soc., 81 (1956), 331–341.
Now, getting back to the problem at hand. Since $xy=0$ then $yxy=xyx=0$ and it follows from the above that $bab=aba=0$. You've already shown that $ab+ba=0$ so $$ \begin{align} (ab-ba)^2 &= abab-abba-baab+baba\\ &= abab - (-ba)ba - (-ab)ab + baba\\ &= abab + baba + abab + baba\\ &= 2a(bab) + 2(bab)a = 0 \end{align} $$ Furthermore, $(ab-ba)^* = b^* a^* - a^* b^* = ba - ab = -(ab-ba)$ so $$(ab-ba)^* (ab-ba) = -(ab-ba)^2 = 0$$ Now, in a $C^*$-algebra, $\|u^*u\| = \|u^*\| \|u\|$, so letting $u=ab-ba$ it follows that $\|ab-ba\| = 0$, so $ab - ba = 0$. You already proved that $ab+ba=0$ so adding the two identities and dividing by two you get $ab=0$.$\Box$