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Could you please explain me the reason why they are isomorphic? Thanks, bye!

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Since $SU(2)/\Bbb{Z}_2 \cong SO(3)$, you might be interested in More on the Isomorphism $SU(2)\otimes SU(2)\cong SO(4)$. – draks ... Jul 14 '12 at 19:18
up vote 21 down vote accepted

This isn't quite true: $SO(3) \times SO(3)$ is isomorphic to $SO(4) / \mathbb{Z}_2$, where $\mathbb{Z}_2 = \{1,-1\}$. (Topologically speaking, $SO(4)$ is a double cover of $SO(3)\times SO(3)$.)

The simple explanation for this is the following:

  1. $SO(3)$ is isomorphic to $U(\mathbb{H})/\mathbb{Z}_2$, where $U(\mathbb{H})$ is the group of unit quaternions and $\mathbb{Z}_2 = \{1,-1\}$. Specifically, the action of $U(\mathbb{H})$ on $\mathbb{R}^3$ is by conjugation, where $\mathbb{R}^3$ is identified with the set of quaternions of the form $ai+bj+ck$ for $a,b,c\in\mathbb{R}$. (See the Wikipedia article on quaternions and spatial rotation for more information on this action.)

  2. $SO(4)$ is isomorphic to $\bigl(U(\mathbb{H})\times U(\mathbb{H})\bigr)/\mathbb{Z}_2$, where $\mathbb{Z}_2 = \{(1,1),(-1,-1)\}$. In particular, any rotation of $\mathbb{R}^4$ can be defined by an equation of the form $$ R(x) \;=\; axb $$ where $a$ and $b$ are quaternions and the input vector $x\in\mathbb{R}^4$ is interpreted as a quaternion.

One consequence of this is that the spin group $\text{Spin}(3)$ is isomorphic to $U(\mathbb{H})$, while $\text{Spin}(4)$ is isomorphic to $U(\mathbb{H}) \times U(\mathbb{H})$. Thus, $$ \text{Spin}(4) \;\cong\; \text{Spin}(3) \times \text{Spin}(3). $$ The statement you gave is also true on the level of Lie algebras, i.e. $$ \mathfrak{so}(4) \;\cong\; \mathfrak{so}(3) \times \mathfrak{so}(3). $$

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On the level of Lie algebras we have that ${\mathfrak {so}}(n)$ are just antisymmetric matrices $n \times n$. It turns out that the six-dimensional space of such $4\times 4$ matrices decomposes into two three-dimensional subspaces that are each closed under taking commutators and each of them satisfies precisely the commutation relations of $\mathfrak{so}(3)$.

Because exponentiation defines an isomorphism between a neighborhood of the identity and a Lie algebra, we have that the two groups are locally isomorphic. It only remains to check global properties, like simple-connectedness, number of components, etc., to be sure the groups are really isomorphic (and not just an universal cover of each other, say, as in the case of $SO(n)$ and $Spin(n)$.)

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