Homomorphism between $mathrm{SL}(2, mathbb{C})$ and $mathrm{O}(1, 3)$

Question

By definition if $L$ is an element of the Lorentz group we have $(x,y)=(Lx,Ly)$.
Now $S$ be an element of $mathrm{SL}(2, mathbb{C})$. Usually people defines the homomorphism
$lambda:mathrm{SL}(2, mathbb{C}) longrightarrow mathrm{O}(1, 3)$ by
$$lambda(S)x=SA_xS^dagger$$
Where
$$
A_x:=left(begin{array}{ll}
x_0+x_3 & x_1-ix_2 
x_1+ix_2 & x_0-x_3
end{array}right).  
$$
The proof that $lambda(S)x in mathrm{O}(3, 1)$ is by defining the inner product by $$(lambda(S)x,lambda(S)x)=det[lambda(S)x] tag 1$$
We can show that $det[lambda(S)x]=(x,x)$.
I am not understanding  why  $lambda(S)$ is an element of the Lorentz group. Shouldn't we prove that $$(lambda(S)x,lambda(S)y)=(x,y)~?$$
How to define  $(lambda(S)x,lambda(S)y)$?

Qmechanic · Accepted Answer

OP particular question boils down to the use of the polarization identity, cf. e.g. this related Phys.SE post.

More generally for the group homomorphism, see e.g. this related Phys.SE post.

Mozibur Ullah · Answer

First there is a bijection between Minkowski space, $M$ and Hermetian 2x2 matrices, $Hm[2]$, this is given by $X: M rightarrow Hm[2]$ where $x=(x^mu) rightarrow x^mu.sigma_mu$.
We then see that $det[X(x)] = |x|^2$, where the latter is the square of the Minkowski norm.
Next, we define an action of $SL(2,C)$ on $Hm[2]$ by $S.A:=SAS^dagger$ where $S in SL(2,C)$ and $A in Hm[2]$. This action preserves the determinant, that is $det(S.A)=det(A)$.
Then putting the two together, we see that $det[S.X(x)] = det[X(x)] = |x|^2$. Hence the action acts by orientation preserving isometries on Minkowski space, and so we get a group map $alpha: SL(2,C) rightarrow SO^+(1,3)$. It turns out that this maps is surjective with kernel ${Id, -Id}$, and so is a double cover.

Homomorphism between $mathrm{SL}(2, mathbb{C})$ and $mathrm{O}(1, 3)$

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