Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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

2 Answers 2

up vote 19 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. $$ \mathfrak{so}(4) \;\cong\; \mathfrak{so}(3) \times \mathfrak{so}(3). $$

share|improve this answer

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 $Spin(n)$.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.