Let $G$ be a a complex reductive algebraic group, together with an $\mathbb{R}$-form. Is it true that any continuous homomorphism $G(\mathbb{R}) \to \mathbb{R}^{\times}$ comes from an algebraic homomorphism $G \to \mathbb{C}^{\times}$?


I ask this because I want to see whether in the Langlands decomposition $P = MAN$, if we start from an algebraic group $G$ the group $M$ is still algebraic ($M$ is defined as the intersection of preimages of $\{ 1,-1\}$ under all such homomorphisms $MA \to \mathbb{R}^{\times}$, where $MA = Z\_G (\mathfrak{a})$ is a reductive algebraic group).


Edit: As was pointed out in the comments, of course the answer is negative. But then I modify the question, asking instead whether the intersection of kernels of $|\chi|$ for "algebraic" $\chi$ the same as for all continuous $\chi$.


Thanks



Elementary question about Langlands decomposition
Now that the motivation for the question has emerged (algebraicity of $M$ inside $G$), here is how to handle it. Let $G$ be a connected reductive $\mathbf{R}$-group, $P$ a parabolic $\mathbf{R}$-subgroup of $G$, and $S$ a maximal split $\mathbf{R}$-torus in $P$ (so $S$ is also maximal as such in $G$). We may and do choose a minimal parabolic $\mathbf{R}$-subgroup $P\_0$ of $G$ contained in $P$ and containing $S$.


The set $\Phi(G,S)$ of nontrivial $S$-weights on ${\rm{Lie}}(G)$ is the (possibly non-reduced) relative root system spanning a finite-index subgroup of the character lattice ${\rm{X}}(S')$ for $S' := (S \cap \mathscr{D}(G))^0$ a maximal split $\mathbf{R}$-torus in $\mathscr{D}(G)$. 


The choice of $P\_0 \supset S$ corresponds to a positive system of roots in the relative root system, or equivalently a basis $\Delta$ of the relative root system, and there is a natural inclusion-preserving bijective correspondence between the set of parabolic $\mathbf{R}$-subgroups $Q$ of $G$ containing $P\_0$ and the set of subsets of $\Delta$. In this way $P$ corresponds to a subset $I \subset \Delta$. Explicitly, $U := \mathscr{R}\_u(P)$ we have $P = L\_I \ltimes U$ for $L\_I := Z\_G(S\_I)$ with $S\_I := (\cap\_{a \in I} \ker a)^0$ a subtorus of $S$. Thus, $P(\mathbf{R}) = L\_I(\mathbf{R}) \ltimes N$ for the group $N:= U(\mathbf{R})$ that is nilpotent (since $U$ is unipotent).


Let $A = S\_I(\mathbf{R})^0$. If I remember correctly, Langlands' definition/construction of $M$ given the above choices is as the unique closed subgroup of $L\_I(\mathbf{R})$ with compact center such that it is complementary to the central closed subgroup $A$ of $L\_I(\mathbf{R})$. Your question is to show that $M = H(\mathbf{R})$ for a unique closed $\mathbf{R}$-subgroup $H \subset L\_I$ with reductive identity component such that $M$ meets every connected component of $H$. Of course, such an $H$ is unique if it exists since its Lie algebra must be that of $M$ (so $H^0$ is uniquely determined in $G$) and it is generated by $H^0$ and any finite subset of $M$ meeting each of its finitely many connected components. The real task is existence of such an $H$.


Define $H = \mathscr{D}(L\_I) \cdot T\_I \cdot S[2]$ where $T\_I$ is the maximal anisotropic central subtorus of $L\_I$. Note that $H^0 = \mathscr{D}(L\_I) \cdot T\_I$, so $H^0$ is reductive, and $H(\mathbf{R})$ meets every connected component of $L\_I(\mathbf{R})$ since $S[2](\mathbf{R})$ does (as $S$ is maximal split in the connected reductive $L\_I$). Since the connected reductive group $H^0$ is unirational, so $H^0(\mathbf{R})$ is Zariski-dense in $H^0$, the equality $H = H^0 \cdot S[2]$ implies that the center of $H(\mathbf{R})$ meets $H^0(\mathbf{R})$ in the subgroup of $Z\_{H^0}(\mathbf{R})$ centralizing $S[2]$. Thus, to show $H(\mathbf{R})$ has compact center it suffices to show that $Z\_{H^0}(\mathbf{R})$ is compact. But $Z\_{H^0} = T\_I$ by design of $H$, so $Z\_{H^0}(\mathbf{R})$ is compact since $T\_I$ is $\mathbf{R}$-anisotropic. 


By design $H \cap S\_I$ is finite, so $H(\mathbf{R}) \cap A=1$ since $A$ has no nontrivial finite subgroup. Thus, the multiplication map $H(\mathbf{R}) \times A \rightarrow L\_I(\mathbf{R})$ is a closed embedding meeting every connected component, and hence this is an isomorphism if and only if the dimensions agree, or equivalently $\dim H + \dim S\_I = \dim L\_I$. Once this is shown, it follows that $H(\mathbf{R})$ satisfies all of the properties that uniquely characterize Langlands' construction $M$, so this $H$ would do the job. 
(Not only is the Lie group $M$ often disconnected, but $H$ is generally not connected as an $\mathbf{R}$-group, which is to say $S[2]$ is not contained $\mathscr{D}(L\_I) \cdot T\_I$; one sees this already when $L\_I$ is a direct product of ${\rm{GL}}\_{n\_j}$'s.)


Since $L\_I$ is the isogenous central quotient of the direct product of $\mathscr{D}(L\_I)$ and the maximal central torus in $L\_I$, in view of how $H$ was built it is equivalent to show that the split central torus $S\_I$ in $L\_I$ and the maximal anisotropic central torus $T\_I$ in $L\_I$ together generate the maximal central torus of $L\_I$. In other words, is $S\_I$ actually maximal as a central split torus in $L\_I$? 


Our task has now been reduced to something in the Borel-Tits structure theory for connected reductive groups over arbitrary fields $k$ as follows. Let $G$ be a connected reductive $k$-group, $S \subset G$ a maximal split $k$-torus, and $\Delta$ a basis of the relative root system $\Phi(G,S)$ of nontrivial $S$-weights on ${\rm{Lie}}(G)$. (This set of weights spans a finite-index subgroup of the character lattice of the maximal split $k$-torus $S' := (S \cap \mathscr{D}(G))^0\_{\rm{red}}$ in $\mathscr{D}(G)$.) For a subset $I \subset \Delta$, define the $k$-subtorus $S\_I = (\cap\_{a \in I} \ker a)^0\_{\rm{red}} \subset S$ and let $L\_I := Z\_G(S\_I)$. The task is show that $S\_I$ maximal as a central split $k$-torus in $L\_I$. 


By the centrality of $S\_I$ in $L\_I$ and the $L\_I(k)$-conjugacy of all maximal split $k$-tori in $L\_I$ (of which $S$ is one such), it suffices to show that no larger $k$-subtorus of $S$ is central in $L\_I$. But $\Delta$ is linearly independent in ${\rm{X}}(S)$, so $S\_I$ has codimension $\#I$ in $S$ and hence it suffices to show that the adjoint action on $S/S\_I$ on ${\rm{Lie}}(L\_I)$ supports $\#I$ linearly independent weights. But $I$ itself is such a set of weights in the subset ${\rm{X}}(S/S\_I) \subset {\rm{X}}(S)$.


