# Complete reducibility of a Lie algebra $\mathfrak{g}$ and splitting of short exact sequence of $\mathfrak{g}$-modules.

Suppose that $$\mathfrak{g}$$ is a semisimple Lie algebra. By Weyl's theorem on complete reducibility, $$\mathfrak{g}$$ is completely reducible. Now my book says that the following is equivalent: For every submodule $$\mathfrak{a}$$ with complementary submodule $$\mathfrak{b}$$, the short exact sequence $$0\to \mathfrak{a}\to\mathfrak{b}\underbrace{\to}_{p}\mathfrak{c}\to0$$ splits, so there exists a $$\mathfrak{g}$$-module homomorphism $$\varphi:\mathfrak{c}\to\mathfrak{b}$$ such that $$p\circ\phi=I_{\mathfrak{c}}$$. They do not state a reason why this holds. How can we see it?

• I do not understand "the following is equivalent". Which two statements, precisely, are supposed to be equivalent to each other? – Torsten Schoeneberg Dec 3 '19 at 19:40
• The condition of $\mathfrak{g}$ being completely reducible and the splitting of every short exact sequence of the form in the question. – Lucas Smits Dec 3 '19 at 19:51

Complete reducibility says that any module is a direct sum of irreducible submodules. This is equivalent to that statement that any submodule has a complement. [Why? Let $$M$$ be a module. Suppose every submodule has a complement, take any proper submodule $$N$$. It has a complement $$N'$$. Inductively decompose $$N$$ and $$N'$$ into irreducibles. Now put these decompositions together and you've decomposed $$M$$ into irreducibles. Conversely, suppose reducibility of your module. Take a submodule $$N$$. Every irreducible submodule of the whole module is either completely contained in $$N$$ or not. The irreducibles contained in $$N$$ form its decomposition. The irreducibles not contained in $$N$$ form a complement for $$N$$.]
Suppose that every short exact sequence is split. Say $$0\rightarrow\mathfrak{a}\stackrel{f}{\rightarrow}\mathfrak{b}\stackrel{p}{\rightarrow}\mathfrak{c}\rightarrow 0$$ is split by $$\mathfrak{c}\stackrel{\varphi}{\rightarrow}\mathfrak{b}$$. Then $$f$$ is injective and $$p$$ is surjective. $$f(\mathfrak{a}) \oplus \varphi(\mathfrak{c}) = \mathfrak{b}$$. Why? Suppose $$x \in \mathfrak{b}$$ and consider $$y=b-\varphi(p(b))$$. Notice that $$p(y)=p(b)-p(\varphi(p(b)))=p(b)-p(b)=0$$ since $$p\circ\varphi$$ is the identity. Thus $$y$$ is in the kernel of $$p$$ hence in the image of $$f$$. Thus $$y=f(a)$$ for some $$a \in \mathfrak{a}$$. This means that $$b=y+\varphi(p(b))=f(a)+\varphi(p(b)) \in f(\mathfrak{a})+\varphi(\mathfrak{c})$$. Suppose $$b\in f(\mathfrak{a}) \cap \varphi(\mathfrak{c})$$. Then $$b=f(a)$$ for some $$a \in\mathfrak{a}$$ and $$b=\varphi(c)$$ for some $$c \in \mathfrak{c}$$. But then $$c = p(\varphi(c))=p(b)=p(f(a))=0$$ (because $$p\circ\varphi$$ is the identity and our sequence is exact at $$\mathfrak{b}$$). Thus $$b=\varphi(c)=\varphi(0)=0$$. So the sum is direct.
Conversely suppose every submodule has a complement. Consider an exact sequence $$0\rightarrow\mathfrak{a}\stackrel{f}{\rightarrow}\mathfrak{b}\stackrel{p}{\rightarrow}\mathfrak{c}\rightarrow 0$$. Let $$\mathfrak{n}=f(\mathfrak{a})=\mathrm{ker}(p)$$. This submodule of $$\mathfrak{b}$$ has a complement, call it $$\mathfrak{m}$$. Notice that $$p$$ restricted to $$\mathfrak{m}$$ is an isomorphism. Why? If $$p(m)=0$$, then $$m \in \mathrm{ker}(p)=\mathfrak{n}$$ so $$m=0$$ (because $$\mathfrak{n}$$ and $$\mathfrak{m}$$ are complementary). Thus $$p$$ is injective (when restricted to $$\mathfrak{m}$$). Next, $$p$$ is surjective because our sequence is exact at $$\mathfrak{c}$$. Take any $$y \in \mathfrak{c}$$ there is some $$x \in \mathfrak{b}$$ such that $$p(x)=y$$. But $$x=x'+x''$$ for some $$x'\in\mathfrak{n}$$ and $$x''\in\mathfrak{m}$$. However, $$p(x'')=0+p(x'')=p(x')+p(x'')=p(x)=y$$ since $$x' \in \mathfrak{n}=\mathrm{ker}(p)$$. Thus $$p$$'s restriction to $$\mathfrak{m}$$ is onto. This restriction to $$\mathfrak{m}$$'s inverse is the splitting $$\varphi$$ that we want. Thus our sequence splits.