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I want to do a proof that $SL(2,\mathbb R)\times SL(2, \mathbb R) \cong SO^+(2,2)$. My idea was to use the same Argument as in this Question. So I wanted to begin with the Basis of the Lie algebra $\mathfrak{sl(2,\mathbb R)} \times \mathfrak{sl(2, \mathbb R)}$, but i am not sure how to do it. I know that the basis of $\mathfrak{sl(2, \mathbb R)}$ is $$ \mathfrak{sl(2,\mathbb R)} = span \left\{ \left(\begin{matrix} 0 & 1\\ 0 & 0 \end{matrix}\right), \left(\begin{matrix} 0 & 0\\ 1 & 0 \end{matrix}\right), \left(\begin{matrix} 1 & 0\\ 0 & -1 \end{matrix}\right) \right\}. $$ But i dont know what to do with this result. I read, that $$ \left(\begin{matrix} 1 & 0\\ 0 & 1 \end{matrix}\right), \left(\begin{matrix} 1 & 0\\ 0 & -1 \end{matrix}\right), \left(\begin{matrix} 0 & 1\\ -1 & 0 \end{matrix}\right), \left(\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix}\right) (*) $$ Is a basis of $\mathfrak{sl(2,\mathbb R)} \times \mathfrak{sl(2, \mathbb R)}$ wich gives for the following bilinear form: $$ \langle x, y \rangle := tr(x\cdot wy^Tw^{-1}) \qquad \qquad w := \left(\begin{matrix} 0 & -1\\ 1 & 0 \end{matrix} \right) $$ the desired signature $2,-2,2,-2$. On the other hand, this question, $\dim_{\mathbb{R}}(\mathfrak{so(2,2)}) = 6$. Therefore $\dim_{\mathbb R}(\mathfrak{sl(2, \mathbb R) \times sl(2, \mathbb R)})$ should also be $6$. Was my conclusion in this question wrong, and the statement, $\dim_{\mathbb{R}}(\mathfrak{so(2,2)}) = 6$ is wrong, or is $(*)$ not a basis of $\mathfrak{sl(\mathbb R, 2)\times sl(\mathbb R,2)}$?

What am i missing, and can someone help me, to write this proof in a more explicit way?

The Phenotype
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iqopi
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    http://www.math.umn.edu/~garrett/m/v/sporadic_isogenies.pdf treats this and other similar. – paul garrett Feb 19 '16 at 19:16
  • @paulgarrett yeah i already read this one (even linked it in the question), but still don't really get it. – iqopi Feb 19 '16 at 19:55
  • Ah. The repn on the whole Lie algebra of two copies of $SL_2$ is indeed six-dimensional, so, one way or another, we need to identify a decomposition into a four-dimensional piece and a two-dimensional piece. The four-dimensional piece will give the orthogonal group. That's what's happening, explicitly or implicitly. – paul garrett Feb 19 '16 at 20:49
  • @paulgarrett but whats the basis of $\mathfrak{sl(2, \mathbb R)\times sl(2, \mathbb R)}$ than? – iqopi Feb 19 '16 at 23:18

1 Answers1

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Let us consider the usual sporadic isogenies related to the one you ask. They are associated with quadratic forms of different signatures.

The case of the Lorentz group and the $2:1$ isogeny from its universal cover: $$SL_2(\mathbb{C})\twoheadrightarrow SO^{\uparrow}_+(1,3)$$

The case of the quaternion algebra, the double cover $SU(2)\to SO(3)$ and the total group of proper isometries of the Euclidean quaternion algebra (with its quaternionic norm as the associated quadratic form):

$$\Psi:SU(2)\times SU(2)/\{\pm 1\}\cong SO(\mathbb{H})$$ given by $\Psi(A,B)(X)=AXB^{\dagger}$, where all quaternions are written in their standard complex $2$-dimensional representation (i.e. general unitary matrices when nonzero).

In the case of signature $(2,2)$ we have a nice way of presenting this quadratic space which suits us very well. That is, just write a $2 \times 2$-matrix and compute its determinant; this indeed corresponds to a neutral quadratic form on $\mathbb{R}^4$.

Now, it suffices to write down the action of $SL_2(\mathbb{R})^2$ on $M_2(\mathbb{R})$ similarly to the one above, that is:

$\Psi(A,B)(X):=AXB^{-1}$ defines a left action on $M_2(\mathbb{R})$ via proper isometries with respect to the quadratic form $A\mapsto \det A$, which in turn furnishes a double cover, which is the one you were looking for.

Note that both double covers have isomorphic complexifications. Hope this helps.

$$\Psi:SL_2(\mathbb{R})\times SL_2(\mathbb{R})\to SO^+(2,2).$$

Theon Alexander
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