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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 it; 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 Weyl Theorem.

Second, 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, this kind 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).

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).

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 {\it $G(\mathbb C)$ 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 $G_0=SL_3(\mathbb R)\subset G_0(\mathbb C)= SL_3(\mathbb C)/scalars$. 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 it; 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 Weyl Theorem.

Second, 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, this kind 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).

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 it; 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 Weyl Theorem.

Second, 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, this kind 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).

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, 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 complex simple algebraic group, viewed as the group of real points by restriction of scalars.

[Edit] Since @jerry has asked for details, I give it; 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 Weyl Theorem.

Second, if $G_0$ is an algebraic semisimple group defined over $\mathbb R$, 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 $G$ covers $G_0(\mathbb C)$ it follows that this complex representation is an algebraic representation of $G_0(\mathbb C)$.

Incidentally, this kind 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).

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 {\it $G(\mathbb C)$ 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 $G_0=SL_3(\mathbb R)\subset G_0(\mathbb C)= SL_3(\mathbb C)/scalars$. 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 it; 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 Weyl Theorem.

Second, 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, this kind 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).