Lie-Cartan coordinates of the first kind vs Lie-Cartan coordinates of the second kind

Question

Lie-Cartan coordinates of the first kind: $$ R_1 = \exp(\alpha_1w_1 + \alpha_2w_2+\alpha_3w_3) $$ Lie-Cartan coordinates of the second kind: $$ R_2 = \exp(\beta_1w_1) \exp(\beta_2w_2) \exp(\beta_3w_3) $$ My problem is are these two different? For $\exp$ we have $$ b^{a+b} = b^ab^b, $$ does this not apply to matrices?

Answer 1

In the context of Lie groups and Lie algebras, the $w_1$, $w_2$, and $w_3$ quantities you have as arguments to the exponential are usually skew-symmetric matrices, each of which when exponentiated give rise to a rotation action in $SO(3)$. Intuitively, rotations themselves do not commute, and this gives us a good hint as to why your usage of the exponential identity is invalid.

If we expand $\exp{a}\exp{b}$, we get the product of two Taylor series:

$$\left(1 + a + \frac{a^2}{2} + \cdots \right)\left(1 + b + \frac{b^2}{2} + \cdots \right)$$

For comparison, the expansion of $\exp({a + b})$ is:

$$1 + a + b + \frac{(a + b)^2}{2} + \cdots$$

Let's look closely at the expansion of $(a + b)^2$. The result is not in general $a^2 + 2ab + b^2$ as we'll see.

$$ \begin{aligned} (a + b)^2 &= (a + b) (a + b) \\ &= a(a + b) + b(a + b)\\ &= a^2 + ab + ba + b^2 \end{aligned} $$

If $a$ and $b$ do not commute, the expansion of $\exp{(a + b)}$ will have the terms $\frac{ab}{2}$ and $\frac{ba}{2}$, while the expansion of $\exp{a}\exp{b}$ will have the term $ab$. In the case of $\mathfrak{so}(3)$, the group of skew-symmetric matrices you've denoted as $w_1$, $w_2$, and $w_3$, elements in this group do not commute, so $\exp{a}\exp{b} \neq \exp{(a + b)}$.