# The splitting of Galois representations

Suppose $$X$$ is a smooth projective variety defined over a number field $$K$$, then the etale cohomology $$H^i_{et}(X,\mathbb{Q}_\ell)$$ defines a continuous representation of the absolute Galois group $$\text{Gal}(\overline{K}/K)$$. Suppose that for every good prime $$\mathfrak{p}$$ of $$K$$, the characteristic polynomial of the Frobenius $$F_{\mathfrak{p}}$$ factors into $$\begin{equation} P_{\mathfrak{p}}(T)=\text{Det}(1-F_{\mathfrak{p}}T)|_{H^i_{et}(X,\mathbb{Q}_\ell)}=f_{\mathfrak{p}}(T) \cdot g_{\mathfrak{p}}(T) \end{equation}$$ where the factorization happens in the ring $$\mathbb{Z}[T]$$. To avoid trivial cases, let us assume $$\text{Deg}\,f_{\mathfrak{p}}>0$$ and $$\text{Deg}\,g_{\mathfrak{p}}>0$$.

Question: is $$H^i_{et}(X,\mathbb{Q}_\ell)$$ the direct sum of two Galois representations, i.e. $$M_1 \oplus M_2$$, such that the characteristic polynomial of the Frobenius acting on $$M_1$$ (resp. $$M_2$$) is $$f_{\mathfrak{p}}$$ (resp. $$g_{\mathfrak{p}}$$)?

P.S. I gather if $$P_{\mathfrak{p}}(T)$$ can be factored further into product of polynomials of lower degree, we should combine correct factors to give the right $$f_{\mathfrak{p}}$$ (resp. $$g_{\mathfrak{p}}$$).

This is false for simple group theoretical reasons. Suppose that $$V$$ is an absolutely irreducible representation of a group $$G$$ which has odd dimension $$d$$ and which is self-dual up to twist, say $$V \simeq V^{\vee} \otimes \chi$$. Then $$\chi = \psi^2$$ is a square (consider determinants), and the characteristic polynomial of an element $$g$$ always has a factor of the form $$(X \pm \psi(g))$$.

As an example of Galois representations with this property, $$V$$ could be $$\mathrm{Sym}^2(W)$$, where $$W = H^1(E,\mathbf{Q}_{\ell})$$ for an elliptic curve $$E$$. Then $$V \simeq V^{\vee} \otimes \varepsilon^2$$, where $$\varepsilon$$ is the cyclotomic character. So if Frobenius at $$p$$ has the characteristic polynomial $$x^2 - a_p x + p$$ acting on $$W$$, then on $$V$$ it will have the characteristic polynomial

$$(x^2 - (a_p^2 - 2p)x + p^2)(x - p),$$

even though (assuming $$E$$ does not have CM) $$V$$ will be irreducible. This example certainly occurs inside etale cohomology, since $$H^2(E \times E,\mathbf{Q}_{\ell}) = V \oplus \mathbf{Q}_{\ell}(-1)^3.$$

Examples like this occur all the time. A single irreducible representation can even "split" on the level of characteristic polynomials into as many different factors as you want; for example $$\mathrm{Sym}^{2n}(W)$$ and $$\mathrm{Sym}^{2n+1}(W)$$ with the same $$W$$ above will exhibit this property where now there are $$n+1$$ factors.

You don't even need to go to positive dimensions to see this, you can already see it in dimension zero. Let $$f(x) \in \mathbf{Q}[x]$$ be any degree four separable polynomial with Galois group $$A_4$$ ($$S_4$$ would work almost exactly the same). If $$X$$ is the underlying set of four points, then

$$H^0(X,\mathbf{Q}_{\ell}) = V_{\ell} \oplus \mathbf{Q}_{\ell},$$

where $$V_{\ell} = V \otimes \mathbf{Q}_{\ell}$$ is the unique $$3$$-dimensional irreducible representation of $$A_4$$, which is also defined over $$\mathbf{Q}$$. Even though there are only two irreducible factors, the characteristic polynomial of Frobenius will always look like $$(X-1)^2 P_g(X)$$ for some quadratic $$P_g(X)$$ depending only in the image of $$g$$.

• I think there's a typo. Shouldn't it be $V\oplus \mathbb{Q}_\ell(-1)^3$? Also, I was trying to think of an example with an actual etale cohomology group, not a subquotient thereof. Do you know of such an example? Apr 7, 2019 at 3:24
• @AlexYoucis, if your question means find an irreducible cohomology group, then sure, you can simply adapt the last example. Let $E/\mathbf{Q}$ be an elliptic curve, and let $K/\mathbf{Q}$ be a degree $4$ extension whose Galois closure $L$ is an $A_4$-extension. Then there is an isogeny $\mathrm{Res}_{K/\mathbf{Q}}(E) = E \times A$, where $A$ is an abelian $3$-fold with $H^1(A) \simeq H^1(E) \otimes V$ for the $3$-dimensional representation of $A_4 =\mathrm{Gal}(L/\mathbf{Q})$. Now $H^1(A)$ is irreducible but the characteristic polynomials on $H^1(A)$ are divisible by those of $H^1(E)$. Apr 7, 2019 at 13:05
• Thanks! I'll think about it. Apr 7, 2019 at 18:25