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. $g_{\mathfrak{p}}$).


1 Answer 1


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 $g$.

  • $\begingroup$ 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? $\endgroup$ Apr 7, 2019 at 3:24
  • $\begingroup$ @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 $H^1(E)$. $\endgroup$
    – Furlo Roth
    Apr 7, 2019 at 13:05
  • $\begingroup$ Thanks! I'll think about it. $\endgroup$ Apr 7, 2019 at 18:25

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