# Projection and positive elements in unital C*-algebras

If $$a$$ is a positive element in a unital C*-algebra $$\mathcal{A}$$, with norm $$\|a\|=1$$. Let $$P$$ be a non-zero positive projection in $$\mathcal{A}$$. How to prove: $$P\leq a \Longleftrightarrow aP=P ?$$ Also, if for any pure state $$f$$ such that $$f(P)=1$$, we always have $$f(a)=1$$, then it implies $$P\leq a$$.

I'm thinking of the hereditary C*-subalgebra, the closure of $$a\mathcal{A}a$$. But can't see how to proceed.

If $$ap=p$$, taking adjoints you get $$pa=p$$. Then $$ap=pa$$ and $$p=a^{1/2}pa^{1/2}\leq a^{1/2}1a^{1/2}=a.$$ Conversely, if $$p\leq a$$ then $$1-a\leq 1-p$$. So $$0\leq p(1-a)p\leq p(1-p)p=0.$$ Thus $$p(1-a)p=0$$. We can write this as $$[(1-a)^{1/2}p]^*(1-a)^{1/2}p=0$$, and so $$(1-a)^{1/2}p=0$$, from where we get $$(1-a)p=0$$. That is $$ap=p$$.

As for your other question, suppose that $$f(a)=1$$ whenever $$f$$ is a pure state with $$f(p)=1$$. For such a state we have by Cauchy-Schwarz $$\tag1 |f(a(1-p))|≤f(aa^*)^{1/2}\,f(1-p)^{1/2}=0.$$ Therefore $$f(a(1-p))=0$$. Then $$1=f(p)=f(ap)=f(pa)$$.

Let $$g$$ be a pure state of the subalgebra $$pAp$$. We can extend $$g$$ to a state of $$A$$ by $$\tilde g(x)=g(pxp)$$. Suppose that $$\tilde g=th+(1-t)k$$ for states $$h,k$$ and $$t\in[0,1]$$. Since $$th\leq\tilde g$$ and $$\tilde g(1-p)=0$$, we get that $$h(1-p)=0$$. Using Cauchy-Schwarz as above we get $$h((1-p)x)=0$$ for all $$x$$. It follows that $$h(x)=h(pxp)$$ for all $$x$$ and similarly for $$k$$. Then $$h,k$$ can be seen as states on $$pAp$$ and by the purity of $$g$$ we get that $$h=k=g$$. So $$\tilde g$$ is pure.

As we just showed, any pure state $$f$$ of $$pAp$$ can be seen as a pure state of $$A$$ with $$f(p)=1$$. Then $$f(a)=1$$; we have from $$(1)$$ (note that the positivity of $$a$$ was not used there) that $$f(a(1-p))=f((1-p)ap)=0.$$ Then $$1=f(a)=f(ap+a(1-p))=f(ap)=f(pap).$$ This can be done for any pure state of $$pAp$$ and pure states separate points, so we have $$pap=p$$. This we can write as $$p(1-a)p=0$$. As we showed before, this gives us $$ap=p$$ and thus $$p\leq a$$.

• May I ask how to get $p=a^{\frac{1}{2}}p a^{\frac{1}{2}}$? Is it because $p=\left(p^{\frac{1}{2}}\right)^2$, and $p^{\frac{1}{2}}=a^{\frac{1}{2}}p^{\frac{1}{2}}=p^{\frac{1}{2}}a^{\frac{1}{2}}$? Commented Oct 5, 2022 at 5:50
• Also, why is $0\leq p(1-a)p$? Is it because, $0\leq 1-a$, since $\sigma(1-a)=1-\sigma(a)\subset [0,1]$, since $\sigma(a)\subset [0,1]$? Commented Oct 5, 2022 at 9:44
• Once you know that $ap=pa$, it is easy to get that $a^{1/2}p=pa^{1/2}$. Then $$p=ap=a^{1/2}a^{1/2}p=a^{1/2}pa^{1/2}.$$ For your second question, $a\geq0$ and $\|a\|≤1$ is equivalent with $0\leq a\leq 1$, as you say; and conjugating by an element and it's adjoint preserves order ($0\leq b$ $\implies$ $0\leq cbc^*$). Commented Oct 5, 2022 at 11:22
• In the Cauchy-Swcharz, $$|f(a(1-p))|≤f(a^*a)^{1/2}\,f(1-p)=0,$$ may I ask why there isn't the square root for $f(1-p)$? Also, $a$ positive, so $a^* a=a^2$? Also, is there a typo in $f(a(1-p))=f(a(1-p))=0$? And where does the last part in $1=f(p)=f(ap)=f(pa)$ come from? Commented Oct 17, 2022 at 15:38
• Also, can you elaborate on "using Cauchy-Schwarz as before we get that $f(p-pap)=0$? I can't seem to get it work. It seems we have $$\left|f(p(1-ap))\right|^2\leq \left|f(p)\right|\left|f((1-pa)(1-ap)\right|$$ But don't see how to get $0$. Commented Oct 17, 2022 at 15:46