# Decomposition of semisimple Lie groups (lie algebras) into direct products (direct sums) and the correspondence.

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).

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