**Qeustion:**

Given a Lie algebra $\mathfrak{g}$ over $\mathbb{Q}_\ell$ with an ideal $\mathfrak{g}^O$ and a subalgebra $\mathfrak{h}$, such that $\mathfrak{g}=\mathfrak{g}^O+\mathfrak{h}$.

Now given a faithful representation

$$\varphi:\mathfrak{g}\hookrightarrow \mathfrak{gl}(V)$$

such that the restrictions on $\mathfrak{g}^O$ and $\mathfrak{h}$ are semisimple, Faltings claim that $\varphi$ is semisimple.

My question is: why this is true?

**Background:**

In Faltings' book "Rational Points" Chapter VI "Complements", he generalized his result about Tate conjecture for abelian varieties over number field into finitely generated field over $\mathbb{Q}$. The main idea is to combine the complex Hodge theory and Tate conjecture over number field.

For example, consider the case $K$ is the function field of the (smooth geometric irreducible) scheme $X$ over a number field $L$ with a rational point $p\in X(L)$, we have a split exact sequence $$ e\rightarrow \widehat{\pi_1(X_{\mathbb{C}})}\rightarrow \pi_1^{\text{ét}}(X)\rightarrow \text{Gal}(\overline{L}/L)\rightarrow e $$ where $X_\mathbb{C}$ is the base change from $L$ to $\mathbb{C}$, $V_\ell(A)$ is the Tate module tensor with $\mathbb{Q}_\ell$.

It terms out that the Galois representation $\rho:\text{Gal}(\overline{K}/K)\rightarrow \text{Aut}(V_\ell(A))$ will factor through the étale fundamental group.

Hence to show the $\rho$ is semisimple, we reduce to show that $\rho_1:\pi_1^{\text{ét}}(X)\rightarrow \text{Aut}(V_\ell(A))$ is semisimple.

Now we have:

$\rho_1|_{\widehat{\pi_1(X_{\mathbb{C}})}}$ is semisimple, from the complex Hodge theory by Deligne.

$\rho_1|_{\text{Gal}(\overline{L}/L)}$ is semisimple by Tate conjecture over number field by Faltings, here we taking the restriction via the splitting by the rational point $p\in X(L)$.

Faltings claim that, therefore the representation $\rho$ is semisimple. To do that, he taking $\mathfrak{g},\mathfrak{g}^O,\mathfrak{h}$ to be the Lie algebra of the complex $\ell$-adic group $\rho_1(\pi_1^{\text{ét}}(X)),\rho_1(\widehat{\pi_1(X_{\mathbb{C}})})$ and $\rho_1(\text{Gal}(\overline{L}/L))$. We want to show that $\mathfrak{g}$ is completely reducible in $V_\ell(A)$.

We know that this already holds for $\mathfrak{g}^O$ and $\mathfrak{h}$, and $\mathfrak{g}^O$ is an ideal in $\mathfrak{g}$, and $\mathfrak{g}=\mathfrak{g}^O+\mathfrak{h}$, hence completely reducible for $\mathfrak{g}$.

**What I tried:**

From the definition, a faithful representation $\rho:\mathfrak{g}\rightarrow\mathfrak{gl}(V)$ is semisimple if $\mathfrak{g}=\mathfrak{c}\times\mathfrak{l}$ is reductive, and $\mathfrak{c}$ acting on $V$ is semisimple, where $\mathfrak{c}$ is the radical and is abelian, and $\mathfrak{l}$ is Levi factor. We need at least show that $\mathfrak{g}$ is reductive. But consider the standard Borel of $\mathfrak{sl}_2$, it is the extension of trivial Lie algebra with a trivial Lie algebra, one is the Lie algebra of $\mathbb{G}_m$ and another the Lie algebra of $\mathbb{G}_a$, they are reductive, but the lie algebra of Borel is not since it is solvable and non-abelian. The trouble here is that, we can't distinct Lie algebra of $\mathbb{G}_a$ and $\mathbb{G}_m$, we do need to use the fact that the radical acting on $V$ is semisimple to get reductive. I tried different attempts but failed. Even the case $\mathfrak{g}^O$ is semisimple and $\mathfrak{h}$ is abelian is still hard to prove.

I tried to apply the Hochschild-Serre spectral sequence to get the completely reducible: for an exact sequence $$0\rightarrow \mathfrak{h}\rightarrow\mathfrak{g}\rightarrow\mathfrak{g}/\mathfrak{h}\rightarrow 0$$ we have $$ H^p(\mathfrak{g}/\mathfrak{h},H^q(\mathfrak{h},V))\Rightarrow H^{p+q}(\mathfrak{g},V). $$ We want to show that $H^1(\mathfrak{g},\text{Hom}_k(V/W,W))=0$ for all sub representation $W$, what we have is $H^1(\mathfrak{g}^O,\text{Hom}_k(V/W,W))=0$, $H^1(\mathfrak{h},\text{Hom}_k(V/W,W))=0$. But to make spectral sequence works, we need $H^1(\mathfrak{g}/\mathfrak{h},\text{Hom}_k(V/W,W)^\mathfrak{h})=0$. Again, I can't find a good way to fix it.

I tried to use universal enveloping algebra, and reduce to an algebra representation question. But we don't have a nice formula even for the universal enveloping algebra of semi-direct product of two Lie algebras.

I also tried to prove the result without using any Lie algebra. For example, taking the algebra generated by the image in $\text{End}(V_\ell(A))$, or consider the representation of $\ell$-adic Lie groups, but does not help.

**Why I think it is a research level problem:**

In the note, Faltings used the terminology $\mathfrak{g}$ is reductive in $M$ to say that $M$ is a semisimple $\mathfrak{g}$-module. His statements seems to be more natural if we have algebraic group in mind, and the claim is easy in the algebraic group setting. So I think what he really thought is the algebraic group. But there are crucial difference between algebraic group and lie algebra: we can't distinct $\mathbb{G}_m$ and $\mathbb{G}_a$.

In Lei Fu's paper On the semisimplicity of pure sheaves, he uses more several pages to prove the same question over finite fields, and crucially using that $\text{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)\simeq \mathbb{Z}$ in the proof. If we adopt Faltings' argument, we can greatly simplify Fu's paper, by replacing complex Hodge theory by Weil conjecture.