# Connected components of real Lie groups

(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.

• 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.
– YCor
Jun 14 '19 at 10:28
• $G=GL_{3}$ over $\mathbb R$. Jun 14 '19 at 10:33
• 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. Jun 14 '19 at 11:50
• 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. Jun 14 '19 at 15:14
• 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. 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.

• Great! Thank you. Jun 17 '19 at 2:51