# $SL_2 (\Bbb R) × SL_2 (\Bbb R)/ ± (I_2 , I_2 ) → (SO_{2,2})^\circ$

I went through this problem in Lie groups:

i) Prove that $$SL_2 (\Bbb R) × SL_2 (\Bbb R)$$ is a linear Lie group.

I identified $$SL_2 (\Bbb R) × SL_2 (\Bbb R)$$ with $$\{\begin{pmatrix} A & 0 \\ 0 & D \end{pmatrix} \in GL_4(\Bbb R), (A,D)\in SL_2 (\Bbb R) × SL_2 (\Bbb R) \}$$

ii) Prove that $$(M, N ) · A = M AN^{−1}$$ is a topological group action of $$SL_2 (\Bbb R) × SL_2 (\Bbb R)$$ on $$M_2 (\Bbb R)$$ ,that preserves the déterminant.

I applied definitions...

iii) prove that the déterminant on $$M_2(\Bbb R)$$ is a quadratic form of signature $$(2,2)$$.

Done.

iv) Deduce a un morphism of Lie groups: $$φ : SL_2 (\Bbb R) × SL_2 (\Bbb R)/ ± (I_2 , I_2 ) → (SO_{2,2})^\circ$$ where $$SO_{2,2}^\circ$$ is the connected component of the neutral element of $$SO_{2,2}$$.

This is where I stuck and I need some help. How to use the questions above to solve this one? I have little knowledge about quadratic forms