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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!

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What does this have to do with the Euler formula? –  Chris Eagle Mar 11 '12 at 0:06
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$\mathbb{C}$ isn't a group under multiplication (no $0^{-1}$). The unit circle is a Lie group though. –  anon Mar 11 '12 at 0:10
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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

2 Answers 2

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 $\mathbb{C}^\times$.

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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 $\mathbb{C}^*$.)

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