(This is a follow-up to this question of mine.)

Is there an example of a connected reductive algebraic group $G$ over $\mathbb{R}$ such that:

  • $G$ is not isomorphic to a product $G_1 \times G_2$ of smaller groups (isogenous to a product is OK)
  • $G$ is not a torus,
  • the quotient of $G$ by a maximal compact-mod-centre subgroup has a complex structure,
  • $Z_G(\mathbb{R})$ is not contained in the identity component of $G(\mathbb{R})$?

The condition $Z_G(\mathbb{R}) \subseteq G(\mathbb{R})^\circ$ is vacuously satisfied if $G$ is adjoint, because then $Z_G = \{1\}$; but it is also vacuously satisfied if $G$ is semisimple and simply-connected, because then $G(\mathbb{R})$ is connected as a Lie group by a theorem of Cartan. So any example would have to lie somewhere in between the two (which makes me wonder if there are any examples at all).

PS: Of course $GL_3$ is an example if the "complex structure" condition is dropped.

  • $\begingroup$ Consider $G=SO(m,n)$ with $m\ge 1$, $n\ge 3$ both odd. It's connected, and its center is $\{\pm 1\}$. The group $G(\mathbf{R})$ has 2 components; a maximal compact subgroup is $S(O(m)\times O(n))$ in which we see that $-1$ does not belong to the unit component. $\endgroup$
    – YCor
    Jun 14 '19 at 10:28
  • $\begingroup$ $G=GL_{3}$ over $\mathbb R$. $\endgroup$ Jun 14 '19 at 10:33
  • $\begingroup$ Oops. Of course those are both perfectly good examples. However, I stupidly left out an important (crucial) hypothesis: that $G^{\mathrm{ad}}(\mathbb{R})^\circ$ mod its max compact should have a complex structure. That rules out $GL_3$, and I believe it also rules out YCor's example as well. $\endgroup$ Jun 14 '19 at 11:50
  • 2
    $\begingroup$ A silly observation from someone concerned with discrete series: if $G$ has a compact Cartan subgroup $T$, then $Z_G(\mathbb R)$ is contained in $T(\mathbb R)$, which is a product of circle groups and thus connected. $\endgroup$
    – LSpice
    Jun 14 '19 at 15:14
  • 2
    $\begingroup$ It seems that there is no such example. If $G$ is of Hermitian type and $Z_G(\Bbb R)$ is not contained in the identity component of $G(\Bbb R)$, then $G$ is a product. $\endgroup$ Jun 15 '19 at 1:46

There is NO such example.

Note that any semisimple algebraic ${\mathbb{R}}$-group $H$ of Hermitian type has a compact (anisotropic) maximal torus. Indeed, by a definition of a group of Hermitian type (see, e.g., Deligne, Travaux de Shimura, condition (1.5.3) on page 128), $H$ is an inner form of a compact algebraic $\Bbb R$-group $K$, namely, $H=\,_\sigma K$, where $\sigma={\rm inn}(x)\in {\rm Aut}(K)$, $x^2=1$, $x\in K^{\rm ad}(\Bbb R)$, $K^{\rm ad}=K/Z_K$. Let $T_K\subset K$ be a maximal $\Bbb R$-torus such that $T_K^{\rm ad}(\Bbb R)$ contains $x$, where $T_K^{\rm ad}=T_K/Z_K$. Then $T_K=\,_\sigma T_K\subset \,_\sigma K=H$ is a compact maximal torus of $H$.

Theorem. Let $G$ be a (connected) reductive ${\mathbb{R}}$-group. Write $G^{\rm der}=[G,G]$. Assume that $G^{\rm der}$ has a compact maximal torus $T^{\rm der}$. If the image of $Z_G({\mathbb{R}})$ in $\pi_0(G({\mathbb{R}}))$ is nontrivial, then there exists a nontrivial split ${\mathbb{R}}$-subtorus $T'\subset Z_G$ and a reductive ${\mathbb{R}}$-subgroup $G''\subset G$ such that $G=T'\times_{\mathbb{R}} G''$.

Proof. Write $T=Z_G\cdot T^{\rm der}$; then $T$ is a maximal torus of $G$. We have maps $$ Z_G({\mathbb{R}})\to \pi_0(T({\mathbb{R}}))\to \pi_0(G({\mathbb{R}})).$$ Since the image of $Z_G({\mathbb{R}})$ in $\pi_0(G({\mathbb{R}}))$ is nontrivial, we have $ \pi_0(T({\mathbb{R}}))\neq 1$.

Write $T=T_0\times_{\mathbb{R}} T_1\times_{\mathbb{R}} T_2$, where $T_0$ is a compact ${\mathbb{R}}$-torus, $T_1$ is a split ${\mathbb{R}}$-torus, and $T_2\simeq (R_{{\mathbb{C}}/{\mathbb{R}}}{\mathbb G}_{m,{\mathbb{C}}})^{n}$. We have $\pi_0(T_0({\mathbb{R}}))=1$, and $\pi_0(T_2({\mathbb{R}}))=1$. Since $\pi_0(T({\mathbb{R}}))\neq 1$, we conclude that $T_1\neq 1$. Note that the ${\mathbb{R}}$-torus $T/Z_G$ is isogenous to $T^{\rm der}$, and hence, compact. It follows that $T_1\subset Z_G$.

Set $T'=T_1$ and $T''=T_0\times_{\mathbb{R}} T_2$; then clearly $T=T'\times_{\mathbb{R}} T''$. Set $G''=T''\cdot G^{\rm der}$. Then $$T'\cdot G''=T'\cdot T''\cdot G^{\rm der}=T\cdot G^{\rm der}=G\quad \text{and}\quad T'\cap G''=T'\cap T''=1.$$ Since $T'\subset Z_G$, it commutes with $G''$. Thus $G=T'\times_{\mathbb{R}} G''$, as required.

  • $\begingroup$ Great! Thank you. $\endgroup$ Jun 17 '19 at 2:51

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