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I have the following exercise:

Let $\pi:\mathcal A \rightarrow \mathcal B$ be a *-homomorphism between two unital $C^*$ algebras $\mathcal A$ and $\mathcal B$ which maps the unit to the unit. Assume $\forall A \in \mathcal A: A > 0 \Rightarrow \pi(A) > 0$. Proof, that $\pi$ is isometric.

My approach: I know $\pi \text{ isometric} \Leftrightarrow \pi \text{ injective}$, so I tried to prove, that there is an $A>0$ with $\pi(A) \not> 0$, if $\pi$ is not injective.

If $\pi$ is not injective, $\mathrm{ker}\ \pi \ne \{0\}$, so that I can take an $B\in \mathcal A$ with $B\ne 0$ and $\pi(B) = 0$. So $\tilde B = BB^*$ is self-adjoint and $\tilde B^2 \ge 0$. If $\tilde B$ is invertible, one has $\tilde B^2 > 0$ and $\pi(\tilde B^2) = 0 \not > 0$, which would be a contradiction to $\forall A \in \mathcal A: A > 0 \Rightarrow \pi(A) > 0$.

But want can I do, if $\tilde B^2$ is not invertible. Is there a way to construct an invertible element from $\tilde B^2$ with which I can do the proof? Thanks for any help!

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  • $\begingroup$ Does $A>0$ mean $A$ self-adjoint with spectrum in $(0,+\infty)$? $\endgroup$ – Julien May 15 '13 at 20:56
  • $\begingroup$ Yes, I think so (In our lecture we defined $A\ge 0$ iff the spectrum is in $[0,\infty)$, so I guess, this would be the right definition) $\endgroup$ – Stephan Kulla May 15 '13 at 21:01
  • $\begingroup$ Given that $A\leq B$ is a partial order on the self-adjoint elements, $A>0$ could simply mean $A\geq 0$ and $A\neq 0$. In which case you are done. $\endgroup$ – Julien May 15 '13 at 21:07
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    $\begingroup$ Take $\mathcal A = C([0,1]):= \{ f: [0,1]\rightarrow \mathbb{C}\}$. The function $f\in C([0,1])$ with $f(x)=x$ would be an element in $\mathcal A$ with $f\ge 0$ and $f \ne 0$. ($0$ is in the spectrum of $f$ because $f(0)=0$ and the sprectrum of $f$ is the image $f([0,1])$) $\endgroup$ – Stephan Kulla May 15 '13 at 21:13
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    $\begingroup$ And since you mentioned that example, the unital $*$-homomorphism $\pi:f\longmapsto f(0)$ from $C([0,1])$ to $\mathbb{C}$ does verify your interpretation of $f>0 \Rightarrow \pi(f)>0$, and yet, it is not injective, nor isometric. So I believe my second interpretation is the correct one here: $A>0$ means $A$ positive nonzero element, i.e. self-adjoint, nonzero, spectrum in $[0,+\infty)$. And then you can check injectivity trivially. $\endgroup$ – Julien May 15 '13 at 21:26
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I know you solved the problem years ago, nevertheless one can justify things slightly more thoroughly: by absurd you supposed $\pi$ non injective and want to show it is not strictly positive (the def. of strictly positive for an element is indeed that of user 1015).

The disjunction $\overline{B}$ invertible or not (and taking the square) is not relevant: what could make the argument fail is if $\overline{B}=0$. But here you use the $C^*$ property of the norm $$\left\lVert B B^* \right\rVert = \left\lVert B \right\rVert^2 $$ A norm is definite, i.e. $\left\lVert A \right\rVert = 0\quad \Leftrightarrow \quad A =0$, so that $B\neq 0 \quad \Leftrightarrow \quad \overline{B}\neq0$

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  • $\begingroup$ I have attended to the lecture a long time ago and thus I have to admit that I do not really understand this topic any more... ;-) Before accepting your answer: Why is it enough to exclude $\overline B = 0$? $\endgroup$ – Stephan Kulla Jan 20 '16 at 18:05
  • $\begingroup$ The statement "$\pi$ is not strictly positive" reads: $\exists\ A\in \mathcal{A},\ A> 0$ (i.e. $A\geq 0$ and $A\neq 0$) but $\pi(A) = 0$. In our proof by contradiction we want to exhibit such an element (which we finally called $\tilde{B}$) in case $\pi$ is not injective: the candidate is to take $B\in \mathrm{Ker}(\pi)$ and then $\tilde{B}:=BB^*$ ($B^*B$ would also work) and it seems to work but it could be that $BB^*$ were actually $0$, in which case our proof is void. $\endgroup$ – Noix07 Jan 21 '16 at 17:12
  • $\begingroup$ funnily my tilde{} doesn't appear as such... $\endgroup$ – Noix07 Jan 21 '16 at 17:13

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