A question on hereditary C*-subalgebra.

Question

From: An introduction to the classification of amenable C*-algebras.

Lemma 3.5.8. Let $A$ be a C*-algebra satisfying the condition that every hereditary C*-subalgebra contains at least two mutually orthogonal nonzero positive elements. Then for any nonzero elements $a,b\in A^+$ there are nonzero positive elements $a_1\in \text{ Her}(a)$ and $b_1\in \text{ Her}(b)$ such that $a_1b_1= 0$.

Proof. We may assume that $x = ab \neq 0$. Suppose that $x = v|x|$ is the polar decomposition of $x$ in $A''$. Fix $0 < \epsilon < \|x\|$. By the assumption, there are mutually orthogonal nonzero positive elements $d_1,d_2$ such that $f_\epsilon(|x|)d_i = d_i$, $i = 1,2$. Then $c_i = vd_iv^*\in\text{ Her}(|x^*|)$ and $f_\epsilon(|x^*|)c_i = c_i$,$i = 1,2$ . Since $c_2vd_1d_1v^*C_2 = 0$, $c_2vd_1 = 0$. Let $g(t) = t^{-l}f_\epsilon(t)$. Then $g \in C_0((0, \|x\|])^+$. Set $z_1 = g(|x|)d_1$. Then

$xz_1 = v|x|g(|x|)d_1 = vf_\epsilon(|x|)d_1 = vd_1$.

Therefore $c_2xz_1= c_2vd_1 = 0$. Now set $b_1 = bz_1b$ and $a_1= ac_2a$. Note that $b_1\in\text{Her}(b)^+$, $a_1\in \text{ Her}(a)^+$ and both are nonzero. However, $a_1b_1 = ac_2abz_1b = a(c_2xz_1)b = 0$.

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$f_\epsilon(t)$ is the continuous monotonic function such that $f_\epsilon(t)=0$ on $[0,\epsilon/2]$ and $f_\epsilon(t)=1$ on $[\epsilon,\infty)$.

I see the existence of $d_1,d_2$, but I do not see why $b_1=bz_1b$ is positive.

Also, in the next lemma, lemma 3.5.9

Let $A$ be a unital simple C*-algebra with property (SP). Then for any nonzero positive elements $a,b \in A$ there is $u \in U(A)$ such that $u\text{Her}(a)u^* \cap \text{Her}(b) \neq {0}$.

It says, It follows from the above lemma that we assume that $ab = 0$. I see why we can assume $ab=0$, but if lemma 3.5.8 is not true, how do I prove this without lemma 3.5.8?

Answer 1

There might be tons of ways to argue in favor of the positivity of $b_1$ but, since $b_1=b^*z_1b$ (recall that $b$ is self-adjoint), I suppose the most natural way is to try proving that $z_1$ is positive. However, if one assumes nothing except that $d_1$ is positive, and that $f_\varepsilon (|x|)d_1=d_1$, it is not possible to deduce that $z_1$ is positive.

Here is a counter-example: working within $M_3(\mathbb {C})$, let $x=\text{diag}(1,2,3)$, so that $|x|=x$.

Setting $\varepsilon =2$, we have that $f_\varepsilon (1)=0$, while $f_\varepsilon (2)=f_\varepsilon (3)=1$, so $$ f_\varepsilon (|x|)=\text{diag}(0, 1, 1). $$

On the other hand, with $g(t)=t^{-1}f_\varepsilon (t)$, we have that $g(1)=0$, $g(2)=1/2$, and $g(3)=1/3$, and therefore $$ g(|x|)=\text{diag}(0,1/2,1/3). $$

In order for "$f_\varepsilon (|x|)d_1=d_1$" to hold we may thus take $d_1$ to be any positive matrix of the form $$ d_1 = \pmatrix{ 0 & 0 & 0 \cr 0 & * & * \cr 0 & * & *}. $$

It is therefore easy to choose one which does not commute with $g(|x|)$ (any non-diagonal matrix will do), in which case the product $$ z_1:=g(|x|)d_1 $$ will fail to be self-adjoint, much less positive.