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 $\varphi$?



Lifting projective Galois representation with condition
$\newcommand\A{\widetilde{A}}$
$\newcommand\Z{\mathbf{Z}}$
$\newcommand\Q{\mathbf{Q}}$
$\newcommand\rhobar{\overline{\rho}}$


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.


