Today I Learned

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November 28th

Today I learned a proof that the only automorphism of $\RR$ is the identity. Let $\sigma$ be an automorphism so that we want to show $\sigma=\op{id}.$

We start by getting the easy stuff out of the way. Because $\sigma$ is a bijection (automorphism), there exists some $k\in\RR$ such that $\sigma(k)=1.$ However,\[1=\sigma(k)=\sigma(k\cdot1)=\sigma(k)\sigma(1)=\sigma(1).\]So we see $\sigma(1)=1.$ We can show that all positive integers are fixed by induction. Quickly, $\sigma(1)=1$ is our base case, and if $\sigma(k)=k,$ then $\sigma(k+1)=\sigma(k)+\sigma(1)=k+1$ provides our inductive step. We can also note that\[\sigma(-1)^2=\sigma\left((-1)^2\right)=\sigma(1)=1\]implies $\sigma(-1)=\pm1.$ However, $1$ has already been taken by $\sigma(1),$ so $\sigma(-1)=-1.$ This implies that for any negative integer $-k,$ we have $\sigma(-k)=\sigma(-1)\sigma(k)=-k.$ Finally, $\sigma(0)=\sigma(1+(-1))=1+(-1)=0.$ It follows that all integers are fixed. Further, for any rational $\frac ab\in\QQ$ for $a,b\in\ZZ$ with $b\ne0,$ we have that\[b\sigma\left(\frac ab\right)=\sigma(b)\sigma\left(\frac ab\right)=\sigma(a)=a.\]Because $b\ne0,$ we may say $\sigma\left(\frac ab\right)=\frac ab.$ It follows that all of $\QQ$ is fixed by $\sigma.$

So far we have been treating $\RR$ as an entirely algebraic quantity, which is why I called the above "the easy stuff.'' However, an example pathology of doing so is that there are non-linear functions which satisfy $f(x+y)=f(x)+f(y)$ for all real $x,y.$ The reason we don't fall into this trap is that the graphs of all these functions are dense in $\RR^2.$ Why isn't the graph of $\sigma$ dense in $\RR^2$? Well, for any positive $r\in\RR^+,$ we see\[\sigma(r)=\sigma\left(\left(\sqrt r\right)^2\right)=\sigma\left(\sqrt r\right)^2\in\RR^+,\]so the fourth quadrant of the graph of $\sigma$ is guaranteed to be empty. This does tell us that $\sigma$ is linear automatically, but we would like to avoid this machinery. Regardless, we intuitively know that we're going to have to use some topology of $\RR$ to finish this.

With this in mind, we show that $\sigma$ satisfies a weak form of continuity—strictly increasing. (Any strictly increasing function over $\RR$ can only have jump discontinuities and therefore has only countably many discontinuities.) We show that for any $r,s\in\RR,$ with $r \lt s$ implies\[\sigma(r) \lt \sigma(s).\]This inequality is equivalent to $\sigma(s)-\sigma(r)=\sigma(s-r) \gt 0.$ However, $s-r \gt 0$ by hypothesis, $\sigma(s-r) \gt 0$ because $\sigma$ sends positives to positives from above.

We are now ready to finish the proof. In particular, we roughly know that $\sigma$ is continuous, and we know that $\sigma$ is the identity on a dense set of $\RR$ (named $\QQ$), so we should be able to push this through. Suppose for the sake of contradiction that $\sigma(r)\ne r$ for some real $r\in\RR.$ Because $\QQ$ is dense in $\RR,$ we may fit some rational $q\in\QQ$ between $r$ and $\sigma(r).$ (Say, $q=\floor{\frac{r+\sigma(r)}2\ceil{\frac1{|r-\sigma(r)|}}+0.5}/\ceil{\frac1{|r-\sigma(r)|}}.$) We now have two cases.

  • If $\sigma(r) \gt r,$ then note $\sigma(r) \gt q \gt r$ implies $\sigma(r) \gt q=\sigma(q) \gt \sigma(r),$ a contradiction.

  • If $\sigma(r) \lt r,$ then note $\sigma(r) \lt q \lt r$ implies $\sigma(r) \lt q=\sigma(q) \lt \sigma(r),$ a contradiction.

Having reached contradiction in both cases, we must have $\sigma(r)=r$ for each $r\in\RR,$ so we are done here.

As an aside, we can use this result to also classify all automorphisms of $\CC,$ with little pain. Fix $\sigma$ an automorphism of $\CC.$ We claim that either $\sigma$ is the identity or the conjugation map, both of which can be quickly checked to be automorphisms. Because $\RR\subseteq\CC,$ we can restrict $\sigma$ to be an automorphism of $\RR,$ but we know the automorphisms of $\RR,$ so $\sigma$ must be the identity over $\RR.$ To expand out to $\CC,$ we have to compute $\sigma(i).$ Note that\[\sigma(i)^2=\sigma\left(i^2\right)=\sigma(-1)=-1,\]so $\sigma(i)=\pm i.$ This gives us two cases.

  • If $\sigma(i)=i,$ then for any $a+bi\in\CC,$ we have $\sigma(a+bi)=\sigma(a)+\sigma(b)\sigma(i)=a+bi,$ so $\sigma$ is the identity.

  • If $\sigma(i)=-i,$ then for any $a+bi\in\CC,$ we have $\sigma(a+bi)=\sigma(a)+\sigma(b)\sigma(i)=a-bi,$ so $\sigma$ is conjugation.

It follows $\sigma$ is either the identity or the conjugation mapping, which is what we wanted.