Given a C*-algebra with unit $1\in\mathcal{A}$.

Define positive elements as: $$A\geq0\iff\sigma(A)\geq0\quad(A=A^*)$$

Positive elements can be characterized by: $$A\geq0\iff A=B^*B$$


One direction easily follows from the continuous calculus: $$A\geq0\implies A=\sqrt{A}\sqrt{A}$$

For operator algebras the numerical range becomes accessible: $$A=B^*B\implies\mathcal{W}(A)\geq0\implies\sigma(A)\geq0$$

For general C*-algebras one has a faithful representation : $$\pi:\mathcal{A}\to\mathcal{B}(\mathcal{H}):\quad\ker\pi=(0)$$

But how to show the above without exploiting advanced tools like Gelfand Naimark?

  • $\begingroup$ To quote Arveson concerning the Gelfand theory applied to the algebra generated by $1$, $x$: "... Significantly, this argument does not imply $z^{\star}z$ has nonnegative spectrum for nonnormal elements $z\in A$, and in fact, the proof that $z^{\star}z \ge 0$ in general is the cornerstone of the Gelfand-Naimark theorem." It seems to me that you are chasing your tail by assuming the operator representation in order to prove that $z^{\star}z$ has non-negative spectrum because the proof of such is the cornerstone of the representation theory. pg 137 Arveson's "A Short Course on Spectral Theory" $\endgroup$ – DisintegratingByParts Jul 1 '14 at 23:41
  • $\begingroup$ No I didn't assume an operator representation (if you meant this) its just that people switch back and forth between positive numerical range and positiv spectrum and that might be not good - you remember the problem with the two definitions of the resolvent set where in the end it came out that there is a dramatic misconjecture arising for nonclosable operators... $\endgroup$ – C-Star-W-Star Jul 1 '14 at 23:53
  • $\begingroup$ Numerical range makes sense only when the algebra is considered to be an algebra of operators. If you start with a $C^{\star}$ algebra, then you don't assume that algebra is an algebra of operators, which doesn't give you a way to define numerical range for an element of the algebra. It is only after establishing the GNS construction that you can show a $C^{\star}$ algebra is faithfully represented as $\mathcal{B}(H)$. But part--indeed, a cornerstone--of proving the representation uses the positivity of the spectrum of $x^{\star}x$ assuming only a $C^{\star}$ algebra. $\endgroup$ – DisintegratingByParts Jul 2 '14 at 1:01
  • $\begingroup$ I know I know remember you reminded me on that in the other thread ;) ...yeah I know the GNS construction uses as the crucial part that $\omega(a^*a)\geq 0$ so proving positivity of $a^*a$ by choosing a representation is biting into the tail - luckily there is a proof for $a^*a$ before digressing into representations (just found it in Bratelli-Robinson) but unfortunately I had no time yet to work through the proof $\endgroup$ – C-Star-W-Star Jul 2 '14 at 1:26
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    $\begingroup$ Sorry, page 127 (not 137) is where the Theorem is located. The discussion from which I quoted is on page 126 of "A Short Course on Spectral Theory" by William Arveson. Read the discussion, and look at the proofs on the next page. Very nice, compact and elegant. $\endgroup$ – DisintegratingByParts Jul 2 '14 at 4:12

Every C$^*$- algebra has a faithful representation into $B(H)$ for some $H$. So the question can be reduced to $\mathcal A\subset B(H)$.

  • $\begingroup$ My question was what if we don't have $\mathcal{A}\subseteq\mathcal{B}(\mathcal{H})$... $\endgroup$ – C-Star-W-Star Jul 1 '14 at 14:59
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    $\begingroup$ Usually the definition of "positive" in a C$^*$-algebra is that the element is of the form $A^*A$. What is your definition of positive? $\endgroup$ – Martin Argerami Jul 1 '14 at 15:16
  • $\begingroup$ I'm mainly interested in having positiv spectrum (plus being selfadjoint) - I edited that in the question. $\endgroup$ – C-Star-W-Star Jul 1 '14 at 15:50
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    $\begingroup$ @MartinArgerami I think the embedding of $C^{*}$ algebras is not necessarily spectrum preserving: For example $\mathbb{C}\to M_{2}(\mathbb{C})$ via $a\rightarrow \begin{pmatrix} a&0\\0&0\end{pmatrix}$. $\endgroup$ – Ali Taghavi Jul 1 '14 at 19:23
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    $\begingroup$ @AliTaghavi: of course. That's why only non-degenerate representations need to be considered. $\endgroup$ – Martin Argerami Jul 1 '14 at 19:46

Ok, I guess I have it...

First, let's do some recapitulation...


By the C*-property one has: $$A=A^*:\quad\|A\|=r(A)$$ Remember also that in a ring: $$\sigma(AB)\cup\{0\}=\sigma(BA)\cup\{0\}$$


Every element has a decomposition into selfadjoints: $$A=\Re A+\imath\Im A\quad(\Re A=\Re A^*,\,\Im A=\Im A^*)$$ and every selfadjoint has a decomposition into positive ones: $$A=A^*:\quad A=A_+-A_-\quad(A_\pm\geq0)$$ Especially they cancel each other: $A_+A_-=A_-A_+=0$

Now, let's tackle the problem...


Observe that the order is definit: $$0\leq A\leq0\implies\|A\|=r(A)=0\implies A=0$$ Also, one has the noncommutative complex analog: $$0\leq\Re A^2+\Im A^2=A^*A+AA^*$$ Thus, combos cannot be negative: $A^*A\leq0\implies A^*A=0$

Exploiting this one deduces: $$\left(A\sqrt{(A^*A)_-}\right)^*\left(A\sqrt{(A^*A)_-}\right)=(A^*A)_-(A^*A)=-(A^*A)_-^2\leq0\\\implies-(A^*A)_-^2=\left(A\sqrt{(A^*A)_-}\right)^*\left(A\sqrt{(A^*A)_-}\right)=0\implies A^*A=(A^*A)_+\geq0$$

Concluding, combos are positive: $A^*A,AA^*\geq0$


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