Definition of the Lorentz group

Question

Let $x=(x_0,x_1,x_2,x_3) \in {\rm R^4}$. Define de inner product $(x,y)$ by $(x,y)=x_0y_0-x_1y_1-x_2y_2-x_3y_3$. Now define the Lorentz group $L$ by $(Lx,Ly)=(x,y)$. Is sufficient to the define the Lorentz group by $(Lx,Lx)=(x,x)$?

Answer 1

Yes; due to the polarization identity. For convenience, let me denote the (pseudo) inner product by $g$, and the associated quadratic form as $Q(x):=g(x,x)$, then the polarization formula says \begin{align} g(x,y)&=\frac{Q(x+y)-Q(x-y)}{4} \end{align} (just like how $ab=\frac{(a+b)^2-(a-b)^2}{4}$ from basic algebra). So, now if $Q(L(\xi))=Q(\xi)$ for all $\xi$, then for any $x,y$, we have \begin{align} g(L(x),L(y))&=\frac{Q(L(x)+L(y))-Q(L(x)-L(y))}{4}\\ &=\frac{Q(L(x+y))-Q(L(x-y))}{4}\tag{by linearity of $L$}\\ &= \frac{Q(x+y)-Q(x-y)}{4}\tag{by assumption on $L$}\\ &= g(x,y). \end{align}

Answer 2

Actually, the Lorentz group is the group of all invertible linear maps $L\colon\Bbb R^4\longrightarrow\Bbb R^4$ such that$$(\forall x,y\in\Bbb R^4):\bigl(L(x),L(y)\bigr)=(x,y).$$But, yes, it is enough to have$$(\forall x\in\Bbb R^4):\bigl(L(x),L(x)\bigr)=(x,x).$$This follows from the polarization identity:$$(x,y)=\frac12\bigl((x+y,x+y)-(x,x)-(y,y)\bigr).$$