# Can every compact connected Lie group be decomposed as $G=Z_G^0\times G_{ss}$?

Let $G$ be a compact connected Lie group with Lie algebra $\mathfrak{g}$. Then, $\mathfrak{g}=Z_{\mathfrak{g}}\oplus[\mathfrak{g},\mathfrak{g}]$ where $Z_{\mathfrak{g}}$ is the center of $\mathfrak{g}$ and $[\mathfrak{g},\mathfrak{g}]$ is semisimple. Let $Z^0_G$ be the identity component of the center $Z_G$ of $G$ and let $G_{ss}$ be the unique connected subgroup of $G$ with Lie algebra $[\mathfrak{g},\mathfrak{g}]$.

Question. Is $G$ always isomorphic to the direct product $Z_G^0\times G_{ss}$?

Clearly, $Z_G^0\times G_{ss}$ is a Lie group with Lie algebra $\mathfrak{g}$. Moreover, according to Knapp's Lie groups beyond an introduction (Theorem 4.29), $G_{ss}$ and $Z_G^0$ are closed subgroups of $G$, and $G$ is the commuting product $G=Z_G^0G_{ss}$. Hence, the product map $$\varphi:Z_G^0\times G_{ss}\to G$$ is a surjective Lie group homomorphism. Moreover, $G_{ss}\cap Z_G^0$ is a connected closed subgroup with Lie algebra $[\mathfrak{g},\mathfrak{g}]\cap Z_{\mathfrak{g}}=\{0\}$. So $G_{ss}\cap Z_G^0$ is finite and connected, hence trivial. Therefore $\varphi$ is injective and hence an isomorphism. Is this argument correct?

• This seems unlikely. You could always take a product of the simply connected groups and then quotient this by a discrete action not respecting the product structure. Of course one has to make sure it's still the center. – Mikhail Katz Aug 9 '16 at 14:10
• As a complement to Ben Webster's answer below: for such a question, it helps to consider examples, in addition to trying to construct arguments. In order that the example, in this case, be interesting, we should consider a group whose semisimple part is non-trivial, and whose centre is not finite. One of the first examples that comes to mind is $U(2)$: in this case, you are asking if it is the direct product of $U(1)$ (the centre) and $SU(2)$. Is it? – tracing Aug 9 '16 at 17:32
• A more general remark: there are lots of examples in which $Z_G = Z_G^0$; e.g. $U(n)$ satisfies this for any $n$. If this is the case, then since $G = Z_G^0 G_{ss}$ (as you observe), it's easy to see that $G_{ss} \cap Z_G^0$ equals the centre of $G_{ss}$. The centre of a semisimple Lie group is finite (as your argument with Lie algebras shows). Thus in this case, i.e. if $Z_G$ is connected, your argument carries through exactly when $G_{ss}$ has trivial centre. So any $G$ whose centre is connected, and for which $G_{ss}$ has non-trivial centre, gives a counterexample to your statement. – tracing Aug 9 '16 at 17:38

The intersection $G_{ss}\cap Z^0_G$ is not necessarily connected. For example, in $U(n,\mathbb{C})$, we have that $G_{ss}=SU(n,\mathbb{C})$ and $Z^0_G$ is unit length complex numbers times the identity. The intersection between these is the $n$ matrices of the form $e^{2\pi ik/n}I$. So $U(n,\mathbb{C})$ is not a product of this form.
• For a compact example, just take the quotient of $\mathbb{S}^1\times \mathrm{SO}(2n)$ by the involution $(z,A)\mapsto (-z,-A)$. – abx Aug 9 '16 at 14:10
• Ah. $\ \ \!\!\!$ – Allen Knutson Aug 10 '16 at 3:00