Is the natural representation of the real group $SO(n)$ on $\Bbb R^n $ unitary?
I am trying to prove that is it not
My definition of unitary representation: A unitary G representaion of a Lie group $G$ is a continuous representation on a Hilbert space $V$ s.t $\pi(g)$ is a unitary operator for all $g \in G$ (namely $<\pi(g)v,\pi(g)v'>=<v,v'>$)
Take $x,y \in \Bbb R^n$, the standart inner product $<x,y>=x^T y$.
Let's say I try to prove that $\pi(g):\Bbb R^n \to \Bbb R^n :v\mapsto gv$ is unitary($g \in SO(n)$):
$<\pi(g)x,\pi(g)y>=<gx,gy>=(gx)^Tgy=x^Tg^Tgy=x^T y=<x,y>$
Isn't this proving that $\pi(g)$ is unitary, contrary to my expectations?
Edit:
So I am kind of confused seems like unitary means is more than preserving the scalar product.
Using $SO(n,\Bbb R)$ and $SO(n,\Bbb C)$ to distinguish the real and complex groups. I have the following representaion , I'd like to determine which of them are or are not unitary, preserve or not the scalar product and/or are unitary representations. I'll'do the preserving part at least, and I hope I can get some help saying with determining if the operator can be called unitary:
i) Representation of $SO(n,\Bbb R)$ on $\Bbb R^n$
Here the standard scalar product $<x,y>=x^T y$.
As above: here $g \in SO(n,\Bbb R)$, so it's a matrix with real coefficients acting on real vectors
$<\pi(g)x,\pi(g)y>=<gx,gy>=(gx)^Tgy=x^Tg^Tgy=x^T y=<x,y>$ the scalar product is conserved
ii) Representation of $SO(n,\Bbb C)$ on $\Bbb R^n$
Here the standard scalar product $<x,y>=x^T y$
$g \in SO(n,\Bbb C)$, so it's a matrix with complex coefficients acting on real vectors, the result will be in general a complex vector so I guess this is not an action anymore, since the codomain has changed
$<\pi(g)x,\pi(g)y>=<gx,gy>=....$ I get complex vectors but my scalar product was defined on $\Bbb R^n$ . I think this makes no sense, or is it a way to make it have sense?
iii) Representation of $SO(n,\Bbb R)$ on $\Bbb C^n$
Here the standard scalar product $<x,y>=x^T \bar y$.
$g \in SO(n,\Bbb R)$, so it's a matrix with real coefficients ( $\overline{g}=g$) acting on complex vectors
$<\pi(g)x,\pi(g)y>=<gx,gy>=(gx)^T\overline{gy}=x^Tg^T\overline{gy}=x^Tg^T\bar{g}\bar{y} =x^T \bar y=<x,y>$
The scalar product is preserved
iv) Representation of $SO(n,\Bbb C)$ on $\Bbb C^n$
Here the standard scalar product $<x,y>=x^T \bar y$.
$g \in SO(n,\Bbb C)$, so it's a matrix with complex coefficients acting on complex vectors, the result will be in general a complex vector
$<\pi(g)x,\pi(g)y>=<gx,gy>=(gx)^T\overline{gy}=x^Tg^T\bar{g}\bar{y} $
Looks like here the scalar product will not be preserved in general since $g^T\overline{g}\neq I$ in general
Is this OK? Which of these operator are unitary, so that I can talk about a unitary representation?