If $G$ is the real points of a semisimple algebraic group and $\rho:G\to GL(n,\mathbb R)$ is continuous representation. Is $\rho$ an algebraic morphism?



Do representations of real semisimple algebraic group have to be algebraic?
If $G$ is a real simple algebraic group and $\rho$ is a finite dimensional continuous representation, and if $G$ is not the group of complex points and *$G(\mathbb C)$ is simply connected*, then $\rho$ is indeed algebraic. This follows, for example, by going to the Lie algebra level and using classification of lie algebra representations. 


If $G=SL\_2(\mathbb C)$, for example, you can have $\rho\otimes {\overline \rho}$ which is not quite algebraic, but can be viewed to be algebraic, if we view the complex group $SL\_2(\mathbb C)$ as the group of real points of the Weil restriction of scalars $R\_{\mathbb C/\mathbb R}(SL\_2)$. 


The same sort of thing holds for any simply connected complex simple algebraic group, viewed as the group of real points by restriction of scalars. 


If $G(\mathbb C)$ is not simply connected, then there is some (small) ambiguity. Take 
$$G\_0=SL\_3(\mathbb R)\subset G\_0(\mathbb C)= SL\_3(\mathbb C)/\text{scalars}.$$ We can view $SL\_3(\mathbb R)$ as an algebraic group in $G\_0(\mathbb C)$; then the standard representation of $SL\_3(\mathbb R)$ is not algebraic with this algebraic structure, but algebraic, if we view $SL\_3(\mathbb R)$ as real points of $SL\_3(\mathbb C)$. 


[Edit] Since @jerry has asked for details, I give them; I am sure that all this is explained in some textbook, but not having one at hand, I cannot give references. 


The first point is that given a complex semi-simple Lie algebra $\mathfrak g$ , there is a simply connected complex algebraic group $G$ with Lie Algebra $\mathfrak g$ (this is perhaps due to Hermann Weyl). So any representation of the Lie algebra $\mathfrak g$ integrates to a holomorphic representation of the complex group $G$. That holomorphic finite dimensional representations of $G$ are algebraic is also well known, perhaps by using the Borel Weil Theorem. 


Secondly, if $G\_0$ is an algebraic semisimple group defined over $\mathbb R$ with $G(\mathbb C)$ simply connected, and $\rho:G(\mathbb R) \rightarrow SL\_n(\mathbb C)$ is a representation, it is known (as Dimitrov has remarked) that $\rho $ is smooth and hence yields a complex representation of the Lie algebra $\mathfrak g\_0$, and therefore of its complexification $\mathfrak g$. Since $G$ is simply connected, this gives an algebraic representation of $G$. Since $G= G\_0(\mathbb C)$ it follows that this complex representation is an algebraic representation of $G\_0(\mathbb C)$. 


Incidentally, these kinds of algebraicity results are proved in great generality by Borel and Tits in an Annals paper "abstract homomorphisms of algebraic groups" (it is in French). 


