Let $G$ be a semisimple Lie group and $Lie(G)$ be its Lie algebra. I guess the following proposition should be true by the theory of smooth manifolds (Let me know if not):

(1) If $G \cong G_1\times \cdots \times G_n$ ($\cong$ means isomorphic as Lie groups), then

$$Lie(G)=Lie(G_1)\oplus \cdots \oplus Lie(G_n).$$

But I am not sure if the following converse is also true:

(2) If $Lie(G)=\mathfrak{g_1} \oplus \cdots \oplus \mathfrak{g_n}$, then there exists Lie groups $G_1,\cdots G_n$, with $Lie(G_i)=\mathfrak{g_i},\forall i$, such that (as isomorphism of Lie groups)

$$G \cong G_1\times \cdots \times G_n.$$

I have noticed that

If the Lie algebra is a direct sum, then the Lie group is a direct product?

But here I am not assuming $G$ to be simply connected


No, in the non-simply connected case this is not true, and the fundamental group is exactly the obstruction. If $\text{Lie}(G) = \mathfrak{g}_1 \times \dots \times \mathfrak{g}_n$ and for each index $i$ we take $G_i$ to be the simply connected Lie group with Lie algebra $\mathfrak{g}_i$, then $G_1 \times \dots \times G_n$ is necessarily the universal cover of $G$, so there's a covering map

$$G_1 \times \dots \times G_n \to G$$

and it's an isomorphism iff $G$ is simply connected.

In the non-simply connected case here is the kind of thing that can go wrong. Let $G = SO(4)$. Its Lie algebra satisfies $\mathfrak{so}(4) \cong \mathfrak{su}(2) \times \mathfrak{su}(2)$ so its universal cover is $SU(2) \times SU(2)$, but the covering map

$$SU(2) \times SU(2) \to SO(4)$$

is nontrivial, and in fact has diagonal kernel $(-1, -1)$. The Lie groups with Lie algebra $\mathfrak{so}(4)$ which can be expressed as a nontrivial product must be expressible as a nontrivial product of two Lie groups with Lie algebra $\mathfrak{su}(2)$, so the possibilities are $SU(2) \times SU(2), SO(3) \times SU(2), SU(2) \times SO(3), SO(3) \times SO(3)$ and $SO(4)$ is none of these (this can be seen by inspecting the kernels of the corresponding covering maps).

  • $\begingroup$ Thanks, although it would take me a while to digest your example $\endgroup$ – No One Aug 31 '20 at 19:54

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