Describe the continuous homomorphisms of $(\mathbb{R}, +) \to (\mathbb{C} - \{0\}, \cdot)$

Question

The maps in question all have the form:

(1) $x \mapsto e^{ax}$ with $a$ any complex number.

These maps include the

(1) The real-valued exponential maps, $x \mapsto b^x$

(2) The unit circle-valued 'wrap around' maps, $x \mapsto e^{ipx}$

(3) The 'spiral' maps, like $x \mapsto e^{(1+i)x}$

Note that you can always use the complex conjugate automorphism after the initial mapping, getting another such map. But the result can still be put in the form of (1).

Using the proof argument from here on isomorphisms and cyclicality, and the fact that the kernel of continuous homomorphisms are closed, I think this is it.

Also, do the images of these maps give us all the locally compact and connected proper subgroups of $(\mathbb{C} - \{0\}, \cdot)$?

Let me know if I am missing any maps and if there are any 'go to' spots in the literature focusing on this question.

Thanks - looks like I missed the 'spiral' subgroups! Will edit them into the question. — CopyPasteIt, May 23 '17 at 16:11
All of these are of the form $x \mapsto e^{ax}$ for some complex constant $a$. That's it. — GEdgar, May 23 '17 at 17:15
As for the last question consider the set $Z_n$ of the $n$th roots of unit of $\Bbb C$. Each is a proper subgroup but (for $n > 1$) is not connected. — Travis Willse, May 23 '17 at 17:27
And for motivation for a unified expression, note that the maps in (2) can be written as $x \mapsto e^{cx}$, where respectively $c = \log b$. — Travis Willse, May 23 '17 at 17:32

Travis Willse · Answer 1 · 2017-05-23T17:22:21.327

1

Hint The fact that classes (2) and (3) really are homomorphisms relies on the property that $$\exp (ax + ay) = \exp (ax) \exp (ay),$$ respectively for real and imaginary $a$. On the other hand, the identity holds for all complex $a$.

Edit The question has been edited since my original answer. In particular, the 'spiral' maps of the added class (4) are exactly the maps $x \mapsto \exp(ax)$, where $\textrm{Re}\, a \neq 0$ and $\textrm{Im}\, a \neq 0$.

edited May 23 '17 at 17:22

answered May 23 '17 at 15:14

Travis Willse

99,363

1

I think you goofed on $\exp (ax + ay) = \exp a \exp (x + y)$. – GEdgar May 23 '17 at 17:17
It did look fishy, and so I went with (1 + i) comment. – CopyPasteIt May 23 '17 at 17:49

score 0 · Answer 2 · answered May 28 '17 at 20:34

Sketch:

Recall the results and techniques found in $f:\mathbb{R}\rightarrow S^1$. We also have the natural 'projection':

$P: \mathbb{C} / \{0\} \rightarrow S^1$, which is a continuous homomorphism.

So if $f: (\mathbb{R}, +) \to (\mathbb{C} - \{0\}, \cdot)$ is given we can analyze $P \circ f$ and use polar coordinates. By so doing we can show that, once we know $f(1)$ there is only one possible continuous homomorphism. So, indeed, the exponential functiions,

$x \mapsto e^{ax}$ are precisely the answer to this question.

Describe the continuous homomorphisms of $(\mathbb{R}, +) \to (\mathbb{C} - \{0\}, \cdot)$

2 Answers2