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?


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:

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

  2. $\rho_1|_{\text{Gal}(\overline{L}/L)}$ 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:

  1. 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.

  2. 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.

  3. 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.

  4. 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:

  1. 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 $\mathbb{G}_a$.

  2. 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.

  • $\begingroup$ Writing everything as direct sums of Ideals, the question boils down to the case of both summands being simple. Then $[\mathfrak{g},\mathfrak{g}]=\mathfrak g$ and hence $\mathfrak g$ is semisimple. $\endgroup$
    – user473423
    Sep 4, 2022 at 7:35
  • 1
    $\begingroup$ @Echo Dear Echo, for the case $\mathfrak{g}$ is the Lie algebra of standard Borel of $\mathfrak{sl}_2$, $\mathfrak{g}$ don't have simple Lie algebra ideal (since abelian Lie algebra is not), and we can't write it as direct sum of ideals (since it is semi-direct product). $\endgroup$
    – Yu LUO
    Sep 4, 2022 at 13:23
  • $\begingroup$ @Echo The question rather boils down to the ideal being simple and the other subalgebra being abelian 1-dimensional. $\endgroup$
    – YCor
    Sep 4, 2022 at 17:22
  • $\begingroup$ I don't believe there's any algebraic group theory necessary for this. Looking further, it seems that eventually one has to check that if $A,B$ are linear maps (acting on finite-dim spaces $V,W$ in char zero) and the matrix $(v\otimes w)\mapsto Av\otimes w+v\otimes Bw$ is semisimple, then so is $(v\otimes w)\mapsto Av\otimes w$. The latter boils down to the algebraically closed case where it's somewhat immediate. $\endgroup$
    – YCor
    Sep 5, 2022 at 8:07

1 Answer 1


WLOG, we may assume that $\phi$ is injective and identify $\mathfrak{g}$ with its image in $\mathrm{End}(V)$. Our goal is to construct a reductive algebraic subgroup of $\mathrm{GL}(V)$, whose Lie algebra coincides with $\mathfrak{g}$.

We may assume, thanks to Deligne's results, that $\mathfrak{g}^0$ (which comes from Hodge theory) is a semisimple Lie algebra. In addition, we may also assume that $\mathfrak{h}$ is an algebraic Lie subalgebra of $\mathrm{End}(V)$; indeed, since $\mathfrak{h}$ is the Lie algebra attached to the Galois action on the Tate module, its algebraicity is a theorem of Bogomolov.

Let $G_0 \subset \mathrm{GL}(V)$ be the connected semisimple algebraic group, whose Lie algebra coincides with $\mathfrak{g}^0$ and $H\subset \mathrm{GL}(V)$ be the connected reductive algebraic group, whose Lie algebra coincides with $\mathfrak{h}^0$. We know that $H$ normalizes $G_0$.

Let us consider the Lie subalgebra $\overline{\mathfrak{g}^0}\subset \mathrm{End}(V)$ that is the normalizer of $\mathfrak{g}^0$ in $\mathrm{End}(V)$. Clearly, $\overline{\mathfrak{g}^0}$ is an algebraic Lie subalgebra of $\mathrm{End}(V)$; we write $G$ for the connected algebraic subgroup of $\mathrm{GL}(V)$, whose Lie algebra coincides with $\overline{\mathfrak{g}^0}$.

We have $$\mathfrak{g}^0, \mathfrak{h} \subset \overline{\mathfrak{g}^0}\subset \mathrm{End}(V).$$

By definition, $\mathfrak{g}^0$ is an ideal in $\overline{\mathfrak{g}^0}$. This gives rise to a natural Lie algebra homomorphism from $\overline{\mathfrak{g}^0}$ to the Lie algebra $\mathrm{Der}(\mathfrak{g}^0)$ of derivations of $\mathfrak{g}^0$, which is just the restriction of the adjoint representation $$\mathrm{Ad}: \overline{\mathfrak{g}^0} \to \mathrm{Der}(\mathfrak{g}^0)\subset \mathrm{End}(\mathfrak{g}^0).$$

Since $\mathfrak{g}^0$ is semisimple, its every derivation is inner one, i.e., $\mathrm{Der}(\mathfrak{g}^0)=\mathfrak{g}^0$. So, we get the Lie algebra homomorphism $\rho: \overline{\mathfrak{g}^0} \to \mathfrak{g}^0$ that coincides with the identity map on $\mathfrak{g}^0$ (in particular, $\rho$ is surjective) and such that

$[\rho(x),y]=[x,y]$ for all $x \in \overline{\mathfrak{g}^0}, y\in \mathfrak{g}^{0}$.

Then $\ker(\rho)$ is an ideal of $\mathfrak{g} $ that meets $\mathfrak{g}^{0}$ precisely at $\{0\}$, because the center of $\mathfrak{g}^{0}$ is $\{0\}$. Hence, $$\overline{\mathfrak{g}}=\mathfrak{g}^{0}\oplus \ker(\rho).$$ In other words, $\ker(\rho)$ is the centralizer $\mathrm{End}_{\mathfrak{g}^{0}}(V)$ of $\mathfrak{g}^{0}$ in $\mathfrak{g}$ in $\mathrm{End}(V)$. Since $\mathfrak{g}^{0}$ is semisimple, the $\mathfrak{g}^{0}$-module $V$ is semisimple and the centralizer $\mathrm{End}_{\mathfrak{g}^{0}}(V)$ is a semisimple associative subalgebra of $\mathrm{End}(V)$. Viewed as the Lie (sub)algebra, $\mathrm{End}_{\mathfrak{g}^{0}}(V)$ is reductive algebraic and coincides with the Lie algebra of the connected reductive algebraic subgroup $\mathrm{Aut}_{\mathfrak{g}^{0}}(V)$ of $\mathrm{GL}(V)$. Clearly, both $G_0$ and $G_1$ are normal closed subgroups of $G$. They mutually commute, and their Lie algebras $\mathfrak{g}^{0}$ and $\mathfrak{g}^{1}=\mathrm{End}_{\mathfrak{g}^{0}}(V)$ meet precisely at $\{0\}$. Hence, the intersection of $G^0$ and $G^1$ (in $\mathrm{GL}(V)$) is a finite central subgroup of both $G^0$ and $G^1$. In addition, $H$ is a closed reductive subgroup of $G$, because its Lie algebra $\mathfrak{h}$ lies in $\overline{\mathfrak{g}}$. So, $$\mathfrak{h}\subset \overline{\mathfrak{g}}=\mathfrak{g}^{0}\oplus \mathfrak{g}^{1}.$$ If $\mathfrak{h}^{1}$ is the image of $\mathfrak{h}$ in $\mathfrak{g}^{1}$ under the projection map then $$\mathfrak{g}=\mathfrak{g}^{0}+\mathfrak{h}=\mathfrak{g}^{0}\oplus \mathfrak{h}^1.$$ We are done if we can prove that $\mathfrak{h}^1$ is the Lie algebra of a reductive algebraic subgroup of $\mathrm{GL}(V)$.

The homomorphism of reductive algebraic group $$\pi: G_0 \times G_1 \to G, (g_0, g_1)=g_0 g_1=g_1 g_0$$ is an isogeny, because its tangent map $$d\pi: \mathfrak{g}^{0}\oplus \mathfrak{g}^{1} \to \overline{\mathfrak{g}}$$ is an isomorphism (the identity map). Let $\tilde{H}\subset G_0 \times G_1$ be the identity component of the preimage $\pi^{-1}(H)$ of $H$. Clearly, $\pi$ induces an isogeny $\tilde{H} \to H$, hence $\tilde{H}$ is a reductive algebraic group. Let us consider the composition of homomorphisms of reductive algebraic groups $$\tilde{H} \stackrel{\pi}{\to} H \subset G_0 \times G_1\to G_1$$ where the last map is the projection map. Let $H_1 \subset G_1$ be the image of $\tilde{H}$. Then the Lie algebra of $H_1$ is precisely $\mathfrak{h}^1$! Since $H_1$ is the image of reductive $\tilde{H}$, it is isomorphic to a quotient of $\tilde{H}$ and therefore is also reductive. Hence, $G_0 \times H_1$ is also reductive and therefore the image $\tilde{G}\subset \mathrm{GL}(V)$ of the homomorphism $$G_0 \times H_1 \to \mathrm{GL}(V), (g_0,h_1)\mapsto g_o h_1=h_1 g_0$$ is a reductive algebraic subgroup, whose Lie algebra is $$\mathfrak{g}^0\oplus \mathfrak{h}^1=\mathfrak{g}.$$ Therefore the $\mathfrak{g}$-module $V$ is semisimple.

  • $\begingroup$ Thank you for the wonderful proof, it is enlightening for me to use the algebraicity of the Lie algebra! $\endgroup$
    – Yu LUO
    Sep 7, 2022 at 2:31
  • $\begingroup$ @YuLuo You are welcome. $\endgroup$ Sep 7, 2022 at 18:54

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