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If you take, say, set of real points of the group-scheme $O(n)$, i.e., $O(n, {\mathbb R})$, then you recover the usual orthogonal (real Lie) group, which you know as $O(n)$. Same applies to $SL(n)$, etc. There is one case when this does not work well, namely when you deal with character varieties. For instance, take $\pi$, say, the free group on two generators, and try to form the quotient $Q=Hom(\pi, SU(2))/SU(2)$. The standard way to do this is to consider the corresponding character variety (or, rather, affine scheme) $X$ and take its set of real points. However, the result will contain both equivalence classes of representations of $\pi$ to $SU(2)$ (as you expected), but also equivalence classes of representations to $SL(2, {\mathbb R})$! The easiest way to see this is to realize that the coordinate ring of $X$ is generated by traces of the elements $A, B, C=AB$ of $\pi$ (where $A, B$ are the free generators). To get the set of real points, you need to use points with real traces, so you end up with the elements of both real Lie groups $SU(2)$ and $SL(2, {\mathbb R})$. This is rather annoying, but one can learn to live with this problem. Namely, in order to isolate $Q=Hom(\pi, SU(2))/SU(2)$ inside $X({\mathbb R})$, you impose also some inequalities, so $Q$ becomes a real semi-algebraic subset. Same problem appears if you consider $Hom(\pi, SL(2, {\mathbb R}))$: Character variety will give you unitary representations as well. The standard way to deal with this problem (in Teichmuller theory) is to consider not all representations to $SL(2, {\mathbb R})$, but only discrete and faithful ones, so that the commutator $[A,B]$ maps to elements of the fixed trace. Then you can form the (topological) quotient by $SL(2, {\mathbb R})$ by taking slice, i.e., restricting to representations $\rho$ so that the (attractive, repulsive) fixed points of $\rho(A)$ are $0, \infty$ and the attractive fixed point of $\rho(B)$ is $1$.