# $so(4)=su(2)× su(2)$ contradiction

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 $SO(4,\mathbb C)=SL(2,\mathbb C)\times SL(2,\mathbb C).$

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?

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What are your $(x_1, x_2)$ for the $SO(4)$ representation? –  Steven Stadnicki Aug 7 '12 at 16:10
@student One should never edit the question in such a way that renders answers extended to you look irrelevant. This wastes people's effort. I have restored the original question alongside the updated version. –  Sasha Aug 7 '12 at 18:13
Dear student, I think you should complexify your Lie algebras in order for your claimed isomorphism to be true. Regards, –  Matt E Aug 7 '12 at 18:16
Also, there is a question as to whether one's choice of coordinates is necessarily a/the one which makes an isomorphism visible. Sometimes one proves that the "obvious" choice isn't it! –  paul garrett Aug 7 '12 at 20:57
Dear student, I dont' understand your lack of agreement. If it were true that $so(4)$ were the product of $sl_2$ with itself, then by the Lie group/Lie algebra correspondence, the Lie group $SL_2(\mathbb R) \times SL_2(\mathbb R)$ would be a finite cover of $SO(4)$, which it's not. Alternatively, $sl_2(\mathbb R) \times sl_2(\mathbb R)$ is a split Lie algebra, while $so(4)$ is not. Am I blundering? Regards, –  Matt E Aug 7 '12 at 21:37
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## 3 Answers

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.

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Dear Matt: how do you arrive from $SU(2)\times SU(2)\rightarrow SO(4)$ to $so(3)\times so(3)\cong so(4)$? Is there a typo here? –  user32240 Aug 7 '12 at 22:22
@user32240: Dear user, there is a double cover $SU(2) \to SO(3)$, which is an isomorphism on Lie algebras. So I wrote $so(3)$ rather than $su(2)$, just because I thought the former notation fitted in better to the context of the question. You are free to write $su(2) \times su(2) \cong so(4)$ instead, if you prefer. Regards, –  Matt E Aug 7 '12 at 22:34
Dear Matt: Thanks. I was confused by using type $D$ in case for type $C$. Now this is making a lot sense. –  user32240 Aug 8 '12 at 0:02
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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.

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FYI, The OP has updated the question based on your answer. Please see the updated question. –  Sasha Aug 7 '12 at 18:15
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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.

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I think the $-y_1$ in the 2nd row, 2nd column should be a $0$, but other wise I think you're exactly right. –  Jason DeVito Aug 8 '12 at 1:21
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