Why is $SO(3)\times SO(3)$ isomorphic to $SO(4)$?
Why is $SO(3)\times SO(3)$ isomorphic to $SO(4)$?
-
1Since $SU(2)/\Bbb{Z}_2 \cong SO(3)$, you might be interested in [More on the Isomorphism $SU(2)\otimes SU(2)\cong SO(4)$](http://arxiv.org/abs/quant-ph/0608186). – 2012-07-14
2 Answers
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:
$SO(3)$ 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.)
$SO(4)$ 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). $
-
0@SRS Yes, in the sense that there are subgroups of SO(4) that are isomorphic to SO(3). For example, the stabilizer of any nonzero vector (i.e. the set of elements of SO(4) that fix the vector) is a subgroup of SO(4) isomorphic to SO(3). – 2018-08-11
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 $Spin(n)$.)