# A possible gap in Faltings note to prove the Tate conjecture for finitely generated field over $\mathbb{Q}$

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:

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.

• 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.
– user473423
Sep 4, 2022 at 7:35
• @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). Sep 4, 2022 at 13:23
• @Echo The question rather boils down to the ideal being simple and the other subalgebra being abelian 1-dimensional.
– YCor
Sep 4, 2022 at 17:22
• 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.
– YCor
Sep 5, 2022 at 8:07

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.

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