$so(4)=su(2)× su(2)$

Question

Modified question There was $sl(2,\mathbb R)$ used instead of $su(2)$ in the previous version. Thanks to MattE for pointing it out.

I have seen it claimed many times that $so(4,\mathbb R)=su(2)\times su(2)$. If $C$ is a Cartan subalgebra of $so(2n)$ then the Weyl group on $so(2n)$ is generated by permutations of $(x_1,...,x_n)\in C$ and by negations of even number of $x_i$s. Hence, for $so(4)$, the orbit of $(x_1,x_2)\in C$ under the Weyl group action is $(x_1,x_2), (-x_1,-x_2), (x_2,x_1), (-x_2,-x_1).$

The Weyl group of $su(2)\times su(2)$ is the product of Weyl groups of its two components. Hence, the orbit of $(x_1,x_2)$ in a Cartan subalgebra of $su(2)\times su(2)$ under its Weyl group action is $(x_1,x_2), (-x_1,x_2), (x_1,-x_2), (-x_1,-x_2).$

That looks like a contradiction! Where do I go wrong? You can of course reformulate this question for $SO(4,\mathbb R)=SU(2)\times SU(2)$ or $SO(4,\mathbb C)=SL(2,\mathbb C)\times SL(2,\mathbb C).$

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Original question

I have seen it claimed many times that $SO(4,\mathbb R)=SL(2,\mathbb R)\times SL(2,\mathbb R),$ However, the orbit of $(x_1,x_2)$ in a max torus of $SO(4)$ under the Weyl group action is $(x_1,x_2), (x_1^{-1},x_2^{-1}), (x_2,x_1), (x_2^{-1},x_1^{-1}),$ while the orbit of $(x_1,x_2)$ in a max torus of $SL(2)\times SL(2)$ under the Weyl group action is $(x_1,x_2), (x_1^{-1},x_2), (x_1,x_2^{-1}), (x_1^{-1},x_2^{-1}).$

That looks like a contradiction! Where do I go wrong?

Answer 1

There is a double cover $SU(2) \times SU(2) \to SO(4)$, which induces an isomorphism of Lie algebras $so(3) \times so(3) \cong so(4)$. One convenient way to compute this double cover is in terms of quaternions:

The unit length quaternions are isomorphic to $SU(2)$. If we define the action of $SU(2) \times SU(2)$ on $\mathbb H = \mathbb R^4$ (the space of quaternions) via $(p,q)\cdot v := p v q^{-1},$ it is not hard to check that this action preserves the norm on $\mathbb H$, and in fact induces a double cover $SU(2) \times SU(2) \to SO(4)$.

In terms of this explicit description of the double cover, it won't be hard to compute what happens on Cartan subgroups of the source and target.

Answer 2

This is quite false because the LHS is compact and the RHS is not. (Where have you seen this claimed?)

The correct statement is an isomorphism of Lie algebras; $\mathfrak{so}(4) \cong \mathfrak{su}(2) \times \mathfrak{su}(2)$. This is induced by an exceptional isogeny between corresponding Lie groups, either

$$\text{SU}(2) \times \text{SU}(2) \to \text{SO}(4)$$

or

$$\text{SO}(4) \to \text{SO}(3) \times \text{SO}(3).$$

These are both double covers. The first can be constructed using quaternions and the second can be constructed using the action of $\text{SO}(4)$ on bivectors.

Answer 3

I think I figured the relation between Cartan subalgebras of $su(2)\times su(2)$ and $so(4)$. The pair $$\left(\begin{matrix} x_1 & 0\\ 0 & -x_1\end{matrix}\right), \left(\begin{matrix} x_2 & 0\\ 0 & -x_2\end{matrix}\right)$$ in the Cartan subalgebra of $su(2)\times su(2)$ is mapped to $$\left(\begin{matrix} 0 & y_1 & 0 & 0\\ -y_1 & 0 & 0 & 0 \\ 0 & 0 & 0 & y_2\\ 0 & 0 & -y2 & 0 \end{matrix}\right),$$ where $y_1=x_1+x_2$ and $y_2=x_1-x_2.$ This map is Weyl group equivariant! Thank you all who responded.