Example for algebraic homomorphism between topological groups which is not continuous

Question

I am quite sure there should be an easy example of:

(algebraic) homomorphism between topological groups which is not continuous.

However, I do not see one immediately.

Answer 1

The identity map from the real numbers with the usual topology to the real numbers with the discrete topology.

Answer 2

Consider $$\begin{align}f\colon\quad\mathbb Q[\sqrt 2]&\to \mathbb Q[\sqrt 2],\\ a+b\sqrt 2&\mapsto a-b\sqrt 2\end{align}$$ where the topology is that inherited from $\Bbb R$.

Define two sequences of integers as follows: Let $a_1=-1,b_1=1$ and recursively $a_{n+1}=a_n^2+2b_n^2, b_{n+1}=2a_nb_n$. Then with $x_n:=a_n+b_n\sqrt 2$ we have $x_{n+1}=x_n^2$ and from $|x_1|=\sqrt 2-1<1$ we see that $x_n\to 0$. On the other hand $x_nf(x_n)=a_n^2-2b_n^2\ne 0$ because $\sqrt 2$ is irrational and $a_n=b_n=0$ (i.e., $x_n=0$) does not occur. We conclude $|f(x_n)|=\frac{|a_n^2-2b_n^2|}{|x_n|}\ge \frac1{|x_n|}\to \infty$. If $f$ were continuous, we'd expect $f(x_n)\to f(0)=0$ instead.

Answer 3

Consider the reals $R=(\mathbb{R}, +)$ as a topological group. Clearly any homomorphism $f: R\rightarrow R$ which sends 1 to 1, continuous or not, must be the identity on $\mathbb{Q}$ (exercise!); this means the only continuous homomorphism from $R$ to $R$ which sends 1 to 1 is the identity. However, algebraically speaking, telling me where $f$ sends 1 gives me absolutely no information about where $f$ sends $\pi$! (Or, more generally, any irrational number.) There is, for instance, a homomorphism $f: R\rightarrow R$ such that

• $f(1)=1$, but

• $f(\pi)=-{17\sqrt{2}\over e}$, so

• $f$ is not continuous.

In general, given a topological group $G$ which has "large" independent subsets, $G$ will admit "many" non-continuous homomorphisms to itself.