Lifting projective Galois representation with condition

Question

Let $\bar{\rho}: G_K\to PGL_n(\mathbb{C})$ be projective representation of the absolute Galois group of a number field $K$ and $\varphi\in Aut(G_K)$.

A theorem of Tate tells us that we can always lift $\bar{\rho}$ to some $\rho: G_K \to GL_n(\mathbb{C})$. I am wondering if there is a lift $\rho$ whose kernel is preserved by $\varphi$, i.e. $\varphi(\ker\rho)=\ker\rho$.

Edit. A better question would be: Do you have any idea about how to determine necessary and sufficient conditions for the existence of a lift with kernel stable under the automorphism $\varphi$?

Answer 1

$\newcommand\A{\widetilde{A}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\Q{\mathbf{Q}}$ $\newcommand\rhobar{\overline{\rho}}$

This is in response to the comment of the OP: As far as I can see you are only showing that the stability of $\ker(\rho)$ is a necessary condition for having a lift with stable kernel, but not a sufficient one.

Yes, that is correct. And as far as I can see, your question was "[is] there a lift $\rho$ whose kernel is preserved by $\varphi$" and the above shows that the answer is "no," and moreover will always be no if the kernel of $\rhobar$ is not preserved by $\varphi$. I made no claim that $\ker(\rhobar)$ being preserved by automorphisms implies that there exists a lift with this property. Perhaps this is your actual question?

Question: Suppose that $\ker(\rhobar)$ is preserved by automorphisms of $G_K$. Then can one find a lift $\rho$ such that $\ker(\rho)$ is preserved by automorphisms of $G_K$?

The answer to this question is no, in general.

Let $\A_5$ denote the perfect central extension of $A_5$ by $\Z/2\Z$. Let $G$ be the semi-direct product of $\Z/2\Z$ by $H = (\A_5)^2$, where the action of $\Z/2\Z$ on $H$ is given by permuting the coordinates. Now suppose that $M/\Q$ is a Galois extension with Galois group $G$. Let $K$ be the corresponding quadratic subfield with fixed field $H$. There is a representation:

$$\rho: G_K \rightarrow \mathrm{Gal}(M/\Q) = H \rightarrow \mathrm{GL}_{6}(\mathbf{C})$$

which factors through the quotient $\A_5 \times A_5$ of $H$ (the latter group has a faithful representation of dimension $2 \times 3$). On one hand, the projective representation $\rhobar$ factors through $A_5 \times A_5$, and so the kernel of $\rhobar$ is Galois over $\Q$. On the other hand, the fixed field of $\ker(\rho)$ is not Galois over $\Q$. Since $\rhobar$ is irreducible, all other lifts are of the form $\rho \otimes \chi$ for some character $\chi$ of $G_K$ (Schur's Lemma). Because $H$ is perfect, the image of such a representation will be $\A_5 \times A_5 \times C$ for some cyclic group, and the same argument as above will apply to show that $\ker(\rho)$ is not normal in $G_{\Q}$.

(To make this complete, one should actually produce an extension of $\Q$ with Galois group $G$, but this should be easily enough to find by taking a generic $\A_5$ extension of some quadratic field.)

There may well be simpler examples, but the motivation behind this particular example is to find a group $H$ with a non-cyclic Schur multiplier with the property that $\mathrm{Out}(H)$ acts non-trivially on $H^2(H,\mathbf{C}^{\times})$.

Edit This is in response to follow up comment of the OP: I'm sorry for not being clear. My question was a 'short hand' for 'is there a lift stable under the automorphism? If the answer is not in general, can we find conditions to ensure the existence of such a lift?'

As noted before, a necessary condition is that the fixed field of $\ker(\rhobar)$ is Galois over $E$, where $E$ is the fixed field of $\Aut(K)$. If $K = E$ this is a sufficient condition. If $K \ne E$, one additional condition that ensures a lift with suitable properties is that the Schur multiplier of the image of $\rhobar$ is trivial, because then there will exist a lift with $\ker(\rhobar) = \ker(\rho)$. So this answers the third version of your question.

Wait, perhaps your question is actually (version 4): give complete necessary and sufficient conditions for the existence of a lift in all circumstances. Indeed I certainly have some ideas about this, although at this point, I am no longer inclined to work out the details.

Answer 2

$\newcommand\rhobar{\overline{\rho}}$

Assume that the projective image of $\rho$ is irreducible.

Suppose that $\varphi \ker(\rho) = \ker(\rho)$. I claim that $\varphi \ker(\rhobar) = \ker(\rhobar)$. The assumption $\varphi \ker(\rho) = \ker(\rho)$ implies that $\varphi$ acts via automorphisms on the image of $\rho$. The irreducibility assumption implies that the subgroup

$$Z = \ker \left(\mathrm{im}(\rho) \rightarrow \mathrm{im}(\rhobar) \right)$$

is precisely the centre of $\ker(\rho)$, and so $Z$ is a characteristic subgroup of $\mathrm{im}(\rho)$, and is thus preserved by $\varphi$, and hence $\varphi$ must also act on $\mathrm{im}(\rhobar)$ and hence fix $\ker(\rhobar)$. So the question is independent of which lift you choose.

For a number field $K$, there is an isomorphism (Neukirsch, 12.2.2):

$$\mathrm{Aut}(G_K)/\mathrm{Inn}(G_K) = \mathrm{Aut}(K)$$

Let us consider two cases:

1. $\mathrm{Aut}(K)$ is trivial. In this case, any automorphism is inner, and so preserves normal subgroups, and so all lifts have the required property.

2. $\mathrm{Aut}(K)$ is non-trivial. Let $E \subset K$ denote the (proper) field field of $\mathrm{Aut}(K)$, so $K/E$ is Galois. In this case, automorphisms of $G_K$ come from inner automorhpisms of $G_E$, and hence the requirement is that the fixed field of $\ker(\rhobar)$ (or of $\ker(\rho)$) is Galois over $E$. However, if $K \ne E$, there will always be irreducible $n$-dimensional representations of $G_K$ whose fixed field is not normal over $E$, and indeed this will be the generic situation.

For a very explicit example, take $K = \mathbf{Q}(\sqrt{2})$, take $n = 2$, and let $\rhobar$ be the representation with image $S_3$ coming from the splitting field of $x^3 + \sqrt{2} \cdot x + 1$.