Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that the complex plane is a Lie group with +, but is it also a Lie group with the usual complex multiplication?

This would give us a nice geometrical interpretation of the famous Euler formula:

We have $exp:T_1\mathbb{C} \to \mathbb{C}$ and $T_1\mathbb{C} \cong \mathbb{C}$ because the latter is a linear space.

Also the complex multiplication is linear, so its differential is itself. Hence any left invariant vector field $X$ on $\mathbb{C}$ can be obtained by choosing a vector $X_0$ in $T_1\mathbb{C} \cong \mathbb{C}$ and using the formula

$$ X(z) = z X_0 $$

Now looking at the curve

$$ t \mapsto exp(tX_0) $$

in the case $X_0 = i$ give us the nice geometrical interpretation of the Euler Formula!

share|cite|improve this question
What does this have to do with the Euler formula? – Chris Eagle Mar 11 '12 at 0:06
$\mathbb{C}$ isn't a group under multiplication (no $0^{-1}$). The unit circle is a Lie group though. – anon Mar 11 '12 at 0:10
Aren't multiplication and inversion actually homolorphic , making $\mathbb{C}^*$ a complex Lie group? – you Mar 11 '12 at 0:50
@Chris Eagle: look at the edit – Abramo Mar 11 '12 at 17:05
up vote 8 down vote accepted

The full $\mathbb{C}$ isn't a group under multiplication, but there is an isomorphism

$$\mathbb{C}^\times \xrightarrow{\sim}\, \mathbb{S}^1\times\mathbb{R} \;:\; w\mapsto (\arg w,\, \log|w|)$$

Both $\mathbb{S}^1$ and $\mathbb{R}$ are Lie groups, and thus so is $\mathbb{C}^\times$.

share|cite|improve this answer

The group of complex units, $\mathbb{C}^*$, is indeed a Lie group since by basic complex analysis, $w\mapsto zw$ is smooth. And, indeed, the Euler formula $e^{it} = \cos{\theta} + i\sin{\theta}$ describes a homomorphism from $\mathbb{R}$ to $\mathbb{C}^*$. (It is also the universal covering map of the compact Lie group $U(1)$, which is a one-dimensional subgroup of $\mathbb{C}^*$.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.