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I am trying to solve Exercise 7.9 of An Introduction to K-theory of C*-algebras, It remains for me to show that the Automorphisms of $\mathbb{Q}\oplus\mathbb{Q}$ which send $ \{(x,y) \in \mathbb{Q}\oplus\mathbb{Q} : x,y>0 \}\cup(0,0) $ to itself and $(1,1)$ to itself are only 2. I was able to find 2 already ,the identity element and the one that sends $(x,y)$ to $(y,x)$ how do I show that these are the only two?

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I don't think what you want is true. Let $$ T=\begin{bmatrix} a&1-a\\ c&1-c\end{bmatrix} $$ with $0<a,c<1$ and $a+a^2\ne c+c^2$. Then $T$ is invertible, $$ T\begin{bmatrix} 1\\1\end{bmatrix} =\begin{bmatrix} a+1-a\\ c+1-c\end{bmatrix}=\begin{bmatrix} 1\\1\end{bmatrix}, $$ and if $x,y>0$ you have $$ T\begin{bmatrix} x\\y\end{bmatrix} =\begin{bmatrix} ax+(1-a)y\\ cx+(1-c)y\end{bmatrix}, $$ with $ax+(1-a)y>0$, $cx+(1-c)y>0$.

The examples you found are the cases $a=1$, $c=0$, and $a=0$, $c=1$, but there are infinitely many others.

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  • $\begingroup$ Yes this is true! Thank you. $\endgroup$ – sirjoe Mar 19 at 17:48
  • $\begingroup$ However, if you insist on the automorphisms to be from $\{(x,y) \in \mathbb{Q}\oplus\mathbb{Q} : x,y>0 \}\cup(0,0)$ onto itself, which is probably what is meant in the exercise, then it appears to be indeed true. $\endgroup$ – Andreas Caranti Mar 19 at 18:26
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$\renewcommand{\epsilon}{\varepsilon}$$\newcommand{\Q}{\mathbb{Q}}$$\DeclareMathOperator{\Aut}{Aut}$Let us show that the statement is true if we assume the automorphism to be from $$\{(x,y) \in \mathbb{Q}\oplus\mathbb{Q} : x,y>0 \}\cup(0,0)$$ onto itself.

If $$ \alpha = \begin{bmatrix} a&b\\ c&d\end{bmatrix} $$ is such an automorphism, then the fact that fixes $\begin{bmatrix} 1\\1\end{bmatrix}$ yields $a + b = 1 = c + d$, and applying $\alpha$ to $\begin{bmatrix} 1\\ \epsilon\end{bmatrix}$, and $\begin{bmatrix} \epsilon\\1\end{bmatrix}$, for $\epsilon \downarrow 0$, we get $a, b, c, d \ge 0$.

Now if $\alpha$ maps $\{(x,y) \in \mathbb{Q}\oplus\mathbb{Q} : x,y>0 \}\cup(0,0)$ onto itself, so does its inverse $$ \alpha^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d&-b\\ -c&a\end{bmatrix}. $$ Therefore for $\epsilon > 0$ $$ \alpha^{-1} \begin{bmatrix} 1\\ \epsilon\end{bmatrix} = \frac{1}{ad - bc} \begin{bmatrix} d - b \epsilon\\-c + a \epsilon\end{bmatrix} \in \{ (x, y) \in \mathbb{Q}\oplus\mathbb{Q} : x,y>0 \}, $$ so that $$ \frac{d - b \epsilon}{ad - bc}, \qquad\frac{-c + a \epsilon}{ad - bc} $$ are both positive. Letting $\epsilon \to 0$, we get that $$\tag{pos-neg} \frac{d}{ad - bc}, \qquad\frac{-c}{ad - bc} $$ are both non-negative. But since $c, d \ge 0$, one of two numbers in (pos-neg) is $\le 0$. Therefore one of $c, d$ is $0$, and then the other is then $1$, as $c + d = 1$. The same holds for $a, b$ (and clearly if $c = 0$, $d = 1$, then $a = 1$, $b = 0$, and conversely), so that we get indeed only the two given automorphisms.

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