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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 $H = H^0 \cdot S[2]$ with $S[2]$ centralizing $H^0$ implies that the center of $H(\mathbf{R})$ meets $H^0(\mathbf{R})$ is exactly $Z_{H^0}(\mathbf{R})$, so the latter is of finite index in the center of $H(\mathbf{R})$. Moreover, $Z_{H^0} = T_I$ by design of $H$, so $Z_{H^0}(\mathbf{R})$ is compact since $T_I$ is $\mathbf{R}$-anisotropic. Thus, the center of $H(\mathbf{R})$ is compact.

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.

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.)

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 $S' := (S \cap \mathbf{D}(G))^0$ a maximal split $\mathbf{R}$-torus in $\mathscr{D}(G)$.

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 $\mathscr{D}(G)$.

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, and $P$ a parabolic $\mathbf{R}$-subgroup of $G$, so $P=L\ltimes U$ for the unipotent radical $U$ of $P$ and Levi subgroup $L \subset P$, so [Edit: what follows is only for minimal $P$; I need to revisit this later.] $L=Z_G(S)$ for a maximal split $\mathbf{R}$-torus $S\subset P$.

Clearly $P(\mathbf{R})=L(\mathbf{R})\ltimes N$ for $N:=U(\mathbf{R})$. Define $A:=S(\mathbf{R})^0$, so $M$ is designed to be a complement to $A$ in $L(\mathbf{R})$. We seek an algebraic construction of $M$. The maximal compact subgroup of $S(\mathbf{R})$ is $S[2](\mathbf{R})$, and it is a complement to $A$ in $S(\mathbf{R})$ since $S$ is a split $\mathbf{R}$-torus and $\mathbf{R}^{\times} = \mathbf{R}^{\times}[2] \times (\mathbf{R}^{\times})^0$.

Let $M \subset L(\mathbf{R})$ be a maximal compact subgroup; typically $M$ is disconnected since $L(\mathbf{R})$ may be disconnected though with finite component group. See Hochschild's book The Structure of Lie Groups for the good theory of maximal compact subgroups of Lie groups with finite component group; this applies to the Lie group $\mathcal{G}(\mathbf{R})$ for any linear algebraic $\mathbf{R}$-group $\mathcal{G}$. By maximality of $M$ and conjugacy/existence results for all such, $M$ contains every central compact subgroup, such as $S[2](\mathbf{R})$.

The "algebraicity" of the theory of compact Lie groups ensures that $M=H(\mathbf{R})$ for a unique closed $\mathbf{R}$-subgroup $H \subset L$ for which $H^0$ is $\mathbf{R}$-anisotropic reductive and $H(\mathbf{R})$ meets every connected component of $H$.

Note that the connected reductive $L/S$ is $\mathbf{R}$-anisotropic, so the group $L(\mathbf{R})/S(\mathbf{R})=(L/S)(\mathbf{R})$ is connected compact. We claim that $H$ is an isogeny complement to $S$ in $L$ as algebraic groups and that $M = H(\mathbf{R})$ is an exact complement to $A$ in $L(\mathbf{R})$; i.e., $M \times A \rightarrow L(\mathbf{R})$ is an isomorphism. Certainly $M \cap A = 1$, and $\dim M + \dim A = \dim L(\mathbf{R})$ since $\mathscr{D}(L)$ is an $\mathbf{R}$-anisotropic subgroup of $L$ that is an isogeny complement to the maximal central torus in $L$ (that in turn is contained in every maximal torus $T$ of $L$, and contains $S$ by design of $L$ with $T/S$ anisotropic).

Thus, the task is entirely about whether $M$ meets every connected component of $L(\mathbf{R})$. But it is a general fact (as proved in Hochschild's book) that maximal compact subgroups of Lie groups with finite component group meet every connected component.

For the good robust theory of maximal compact subgroups of Lie groups with finite component group, the book of Hochschild is the only reference I am aware of which provides a complete treatment of the analytic aspects. For the algebraicity aspects of compact (possibly disconnected) Lie groups, one has to look elsewhere.

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 $S' := (S \cap \mathbf{D}(G))^0$ 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 $H = H^0 \cdot S[2]$ with $S[2]$ centralizing $H^0$ implies that the center of $H(\mathbf{R})$ meets $H^0(\mathbf{R})$ is exactly $Z_{H^0}(\mathbf{R})$, so the latter is of finite index in the center of $H(\mathbf{R})$. Moreover, $Z_{H^0} = T_I$ by design of $H$, so $Z_{H^0}(\mathbf{R})$ is compact since $T_I$ is $\mathbf{R}$-anisotropic. Thus, the center of $H(\mathbf{R})$ is compact.

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.

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 ${\rm{X}}(S/S_I) \subset {\rm{X}}(S)$.