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I want to show that $SU(2)\times SU(2)/\ker(\varphi)\cong SO(4)$ where $\varphi:SU(2)\times SU(2)\rightarrow \rm End(\mathbb{H})$ is the map $\varphi(A,B)=(h\mapsto AhB^{-1})$.

I have shown that $\ker(\phi)={(I,I),(-I,-I)}$, and $\varphi(SU(2)\times SU(2))\subset SO(4)$.

I have tried to use the isomorphism theorem, but I am getting stuck. To use the isomorphism theorem, I want to show that for any $X\in SO(4)$ there is $A, B$ such that $\varphi(A, B)=X$. This gives a rather horrible matrix expression that I cannot solve. Is there a slicker way to approach this?

cqfd
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Emily
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1 Answers1

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Yes. The Cartan-Dieudonne theorem says every rotation in $\mathbb{R}^n$ is a composition of (an even number less than or equal to $2n$) reflections. Notice $F(x)=-\overline{x}$ flips the real component while keeping the imaginary components fixed, so it is a reflection across the $3$D space of purely imaginary quaternions, or equivalently a reflection across the orthogonal complement $1^{\perp}$. To get a reflection $F_q$ across any other orthogonal complement $q^{\perp}$, we may "transport" the one from $1^{\perp}$: first rotate $q$ to $1$, then reflect across $1^{\perp}$, then rotate $1$ back to $q$. The rotation from $1$ to $q$ may be left or right multiplication by $q$ (and rotating back the same but by its inverse $q^{-1}=\overline{q}$, assuming $q$ is a unit quaternion). That is,

$$ F_q(x) ~=~ q\big(-\overline{\overline{q}x}\,\big) ~=~ \big(-\overline{x\overline{q}}\,\big)q ~=~ -q\overline{x}q. $$

Composing two of them gives $(F_p\circ F_q)(x)=(pq)x(qp)$, which is of the form $axb$. Therefore, the group homomorphism $\mathrm{Sp}(1)\times\mathrm{Sp}(1)\to \mathrm{SO}(4)$ must be onto.

(People often use $\mathrm{SU}(2)$ to refer to $\mathrm{Sp}(1)$ since they're isomorphic, for basically the same reason that $SO(2)$ is isomorphic to $U(1)$. But people don't insist on referring to complex numbers as real matrices even though it can be done. The reason for the quaternion-phobia, I expect, is the idea that complex matrices are supposed to be more familiar than quaternions. Here, $\mathrm{Sp}(1)$ refers to the $1\times 1$ quaternionic unitary matrices, which is essentially the group $S^3$ of unit quaternions, just as $U(1)$ is essentially $S^1$.)

To see the kernel is $(-1,-1)$, first suppose $axb=x$ for all $x$, set $x=1$ to see $b=a^{-1}$, then assuming $a$ has nonzero imaginary part let $x$ be a perpendicular imaginary quaternion in which csae $axa^{-1}=x$ simplifies to the expression $a^2x=x$, and $a^2=1$ implies $a=\pm1$.

anon
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  • I don't know how you got $q\big(-\overline{\overline{q}x},\big) = \big(-\overline{x\overline{q}},\big)q$, but another approach is to note that $$q(-\overline{\overline{q}x}) = -q,\overline{x},\overline{\overline{q}} = -q,\overline{x},q$$ using the fact that $\overline{pq} = \overline{q},\overline{p}$. – Michael Albanese Jan 12 '20 at 00:58
  • @MichaelAlbanese Let $L_q(x)=qx$ and $R_q(x)=xq$ and $C(x)=-\overline{x}$. I put both $L_q\circ C\circ L_q^{-1}$ and $R_q\circ C\circ R_q^{-1}$ in (since $L_q$ and $R_q$ are both rotations which send $1\mapsto q$). I wrote them equal to each other since they both simplify to the same thing (the way you did it). I should have been more explicit. – anon Jan 12 '20 at 03:10
  • I see now, thanks. – Michael Albanese Jan 12 '20 at 03:23