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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 $\varphi$?

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

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.

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  • $\begingroup$ 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?'. $\endgroup$
    – Bear
    Jun 14, 2015 at 13:25
  • $\begingroup$ There is no need to be mean. You are free not to answer the questions. In any case I would be very happy to have hints about how to approach this problem. $\endgroup$
    – Bear
    Jun 14, 2015 at 21:32
  • $\begingroup$ Be mean? I answered the first three versions of your question. You haven't shown any indication that you read those answers in any detail and tried to understand them, nor did you really indicate any gratitude for the effort that went in to writing those answers. $\endgroup$
    – user74597
    Jun 14, 2015 at 22:23
  • $\begingroup$ Of course, you are free to post questions anonymously without making any effort to indicate whether you have thought about the questions in advance, and you are free to change the question if it turns out you meant something else, and you are free to ignore any answers to previous versions of the question that people in their free time have written for your benefit, and you are free to sit back and wait for someone to write a complete answer to the latest version of your question. And yes, I am free not to answer your question (for the fourth time), which is what I shall do. $\endgroup$
    – user74597
    Jun 14, 2015 at 22:23
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$\newcommand\rhobar{\overline{\rho}}$

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 $x^3 + \sqrt{2} \cdot x + 1$.

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  • $\begingroup$ Why is $\ker(\rho)$ a characteristic subgroup? (Also, there's an extra right parenthesis in the definition of your subgroup $Z$, which is driving me crazy. :-) ) $\endgroup$
    – LSpice
    Jun 12, 2015 at 21:31
  • $\begingroup$ What I wrote was linguistically ambiguous and thus confusing: the claim was that $Z$ (the center) was a characteristic subgroup of $\mathrm{im}(\rho)$; the group $\ker(\rho)$ is preserved by $\varphi$ by assumption, so $\varphi$ acts on $\mathrm{im}(\rho)$. I also fixed the parenthesis! $\endgroup$
    – user74913
    Jun 12, 2015 at 21:52
  • $\begingroup$ Thank you for your answer, but I don't understand one point. Do you mean that the stability of $\ker{\bar{\rho}}$ implies the stability of the kernel of any lift? $\endgroup$
    – Bear
    Jun 12, 2015 at 23:08
  • $\begingroup$ As far as I can see you are only showing that the stability of $\ker\bar{\rho}$ is a necessary condition for having a lift with stable kernel, but not a sufficient one. $\endgroup$
    – Bear
    Jun 14, 2015 at 1:04

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