Why is $SO(3)\times SO(3)$ isomorphic to $SO(4)$?

Question

Why is $SO(3)\times SO(3)$ isomorphic to $SO(4)$?

Answer 1

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. $$ \mathfrak{so}(4) \;\cong\; \mathfrak{so}(3) \times \mathfrak{so}(3). $$

Answer 2

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 $Spin(n)$.)

Answer 3

It is well-known that $\mathrm{SL}_2(\mathbb C)\cong\mathrm{Spin}_3(\mathbb C)$. I will give a construction of the map $\mathrm{SL}_2(\mathbb C)\times\mathrm{SL}_2(\mathbb C)\to\mathrm{SO}_4(\mathbb C)$.

The group $\mathrm{SL}_2(\mathbb C)\times\mathrm{SL}_2(\mathbb C)$ acts on $V=\mathbb C^2\otimes\mathbb C^2$ in the obvious way, namely $(g,h)\cdot(v\otimes w):=gv\otimes hw$. This action has determinant $1$, so gives a homomorphism $\mathrm{SL}_2(\mathbb C)\times\mathrm{SL}_2(\mathbb C)\to\mathrm{SL}(V)$.

Also note that $\mathbb C^2$ carries a $\mathrm{SL}_2(\mathbb C)$-equivariant anti-symmetric form $\omega(v,w):=\det(v,w)$. Thus, $V$ carries a bilinear form $\langle v_1\otimes w_1,v_2\otimes w_2\rangle:=\omega(v_1,v_2)\omega(w_1,w_2)$ which is now symmetric. This is $\mathrm{SL}_2(\mathbb C)\times\mathrm{SL}_2(\mathbb C)$-equivariant since $\omega$ is $\mathrm{SL}_2(\mathbb C)$-equivariant. We have thus constructed a homomorphism $$\mathrm{SL}_2(\mathbb C)\times\mathrm{SL}_2(\mathbb C)\to\mathrm{SO}_4(\mathbb C),$$ which has kernel $\{1,(-1,-1)\}$. Surjectivity follows from a dimension-counting argument.

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By the way, the isomorphism $\mathrm{SL}_2(\mathbb C)\cong\mathrm{Spin}_3(\mathbb C)$ can be constructed similarly, by considering the action of $\mathrm{SL}_2(\mathbb C)$ on $\mathfrak{sl}_2(\mathbb C)$, which carries the Killing form: Lie algebra isomorphism between $\mathfrak{sl}(2,{\bf C})$ and $\mathfrak{so}(3,\Bbb C)$.

Answer 4

Here is yet another construction of the double-cover $\mathrm{SL}_2(\mathbb C)\times\mathrm{SL}_2(\mathbb C)\to\mathrm{SO}_4(\mathbb C)$.

The vector space $V=M_2(\mathbb C)$ has an inner product $\langle A,B\rangle:=\frac12\mathrm{tr}(A\cdot JB^tJ^{-1})$, where $J:=\begin{pmatrix}&1\\-1\end{pmatrix}$. (The associated quadratic form is the determinant.) Moreover, $V$ has an action of $\mathrm{SL}_2(\mathbb C)\times\mathrm{SL}_2(\mathbb C)$, where the first copy of $\mathrm{SL}_2(\mathbb C)$ acts by left multiplication, and the second copy acts by right multiplication: $(g,h)\cdot A:=gAh^{-1}$ for $g,h\in\mathrm{SL}_2(\mathbb C)$ and $A\in V=M_2(\mathbb C)$. This action preserves the inner product defined above. Thus, we have produced a homomorphism $\mathrm{SL}_2(\mathbb C)\times\mathrm{SL}_2(\mathbb C)\to\mathrm{SO}_4(\mathbb C)$.

The kernel of this homomorphism consists of pairs $g,h\in\mathrm{SL}_2(\mathbb C)$ such that $gAh^{-1}=A$ for all $A\in M_2(\mathbb C)$, which can only happen when $g=h=\pm I_2$. Thus, we have an injection $\mathrm{SL}_2(\mathbb C)\times\mathrm{SL}_2(\mathbb C)/\{\pm1\}\to\mathrm{SO}_4(\mathbb C)$. This is surjective by dimension-counting.