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I already created this post on Math Stack Exchange but I was not so sure if this question fits better here. If it is not, I want to apologize in advance and feel free to delete my post.

I want to understand the proof of the following Theorem from "Euler Factors determine Weil Representations" by Tim and Vladimir Dokchitser:

Theorem 1 Every Frobenius-semisimple Weil representation $\rho$ is uniquely determined by its local polynomials $P(\rho/F,T)$ over finite separable extensions $F/K$.

Before we talk about the cyclic case of the proof, let us recall some definitions first:

  • Let $K$ be a local field and $G_K = \operatorname{Gal}(\bar{K}/K)$ be the absolute Galois group of $K$. An (arithmetic) Frobenius element is any element $\operatorname{Frob}_K \in G_K$ that acts as $x \mapsto x^{|k|}$ on $\bar{k}$, the algebraic closure of the residue field $k$ of $K$.
  • The Weil group $W_K$ is the subgroup of $G_K$ of all automorphisms that act as an integral power of Frobenius on the residue field.
  • A Weil representation is a representation $\rho: W_K \to \operatorname{GL}_n(\mathbb{C})$ such that $\rho(I_K)$ is finite. It is called Frobenius-semisimple if the image of some (equivalently, any) Frobenius element is diagonalizable.
  • The local polynomial $P(\rho,T)$ is the inverse characteristic polynomial of $\operatorname{Frob}_K^{-1}$ on the inertia invariants of $\rho$, i.e. $$P(\rho,T) = \det(1-T \cdot \operatorname{Frob_K^{-1}}).$$ Similarly, for a finite extension $F/K$, we write $P(\rho/F)$ for the local polynomial of the restriction of $\rho$ to $W_F$, i.e. $$P(\rho/F,T) = P(\rho|_{W_F},T).$$

Now I would like to talk about the proof of the cyclic case which is given in the paper:

Step 1: Cyclic. Suppose $\rho$ factors through a finite cyclic group $G = \operatorname{Gal}(F/K) \simeq C_n$ and $F/K$ has ramification degree $e$. By Lemma 2 (cf. below), there is a cyclic totally ramified extension $K/K$ of degree $e$ such that $FL/L$ is unramified of degree $n$. The restriction map $\operatorname{Gal}(FL/L) \to \operatorname{Gal}(F/K)$ is an isomorphism, as it is clearly injective and both groups have order $n$. So $\rho/L$ determines $\rho$, and $\rho/L$ can be recovered from its local polynomial $P(\rho/L,T)$.

In our proof we used the following Lemma which we shall take for granted in this post:

Lemma 2 Let $F/K$ be a cyclic extension of degree $n$ and ramification degree $e$. Then there exists a cyclic totally ramified extension $L/K$ of degree $e$ such that $FL/L$ is unramified of degree $n$.

Now I have the following specific questions about the proof above:

  • Could you give me a good argument why the restriction map $\operatorname{Gal}(FL/L) \to \operatorname{Gal}(F/K)$ is injective? Let us say we have $\sigma, \sigma' \in \operatorname{Gal}(FL/L)$ which are not equal. Then there exists an $x \in FL \setminus L$ such that $\sigma(x) \neq \sigma'(x)$. If $x \in F$ then the restrictions $\bar{\sigma}, \bar{\sigma}$ are obviously not equal. But what happens if $x$ is not in $F$?
  • Why does $\rho/L$ determine $\rho$ then? And how can we recover $\rho/L$ from its local polynomial $P(\rho/L,T)$? What do "determining" and "recovering" even mean in these cases?

Additional Remark:

I would also like to add a diagram which my professor drew to explain me something relating to this Theorem (or Lemma?):

$\require{AMScd}\DeclareMathOperator\Gal{Gal}$ \begin{CD} @. \Gal(L F/L) @>{\text{$\chi|_L$, unramified}}>> \Gal(K^{\mathrm{ur}}/K) \\ @. @V{\simeq}VV @| \\ 1 @>>> \Gal(F/K^{\mathrm{ur}}) @>>> \Gal(F/K) @>>> \Gal(K^{\mathrm{ur}}/K) @>>> 1 \\ @. @. @V{\chi}VV \\ @. @. \mathbb C^\times \end{CD}

And next to this diagram, he also wrote $P(\chi|_L,T) = 1 - \operatorname{Frob}_L \cdot T$.

To this remark, I have the following questions (if it is really related to the proof):

  • What is $\chi$ supposed to mean?
  • Is the diagram commutative?
  • Why all of the sudden does the maximal unramified extension (I think?) $K^{ur}$ of $K$ appear here?
  • If or how is this related to the proof of Theorem 1 (resp. Lemma 2)?

Could you please help me answering my questions? I feel like I lack a lot of background knowledge which is required here, so it would be really nice if you could explain them to me carefully. Thank you in advance!

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  • $\begingroup$ I will just point out that the (abstact-algebra) tag is deprecated on MathOverlow - see the tag-info. (But it is commonly used on Mathematics - which might make things a bit confusing for users of both sites.) Perhaps a more siitable tag can be chosen instead (or the tag could be simply omitted). $\endgroup$ – Martin Sleziak Oct 3 '18 at 17:07
  • $\begingroup$ @MartinSleziak: Thank you for your remark, I did not know this! Actually, I am not often around Math Overflow, so I still have to learn about the rules here. Sorry about that! $\endgroup$ – Diglett Oct 3 '18 at 17:24
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Why is the map ${\rm Gal}(FL/L)\to {\rm Gal}(F/K)$ injective?

Elements of ${\rm Gal}(FL/L)$ are automorphisms of $FL$ that act trivially on $L$. To be in the kernel of the above map means to also act trivially on $F$. A field automorphism that acts trivially on $F$ and $L$ acts trivially on $FL$ (by definition of compositum and of field automorphism).

Why does $\rho|_L$ determine $\rho$? And what does "determine" mean?

Firstly, what does $\rho|_L$ mean? $\rho$ is a representation of the absolute Galois group of $K$, and $\rho|_L$ is really shorthand for the restriction of $\rho$ to the subgroup ${\rm Gal}(\bar{K}/L)$ of ${\rm Gal}(\bar{K}/K)$. "Determine" means that if you have a second representation $\rho'$ satisfying all the hypotheses such that $\rho|_L=\rho'|_L$, then in fact $\rho=\rho'$. So why is this the case? Well, firstly, $\rho$ is trivial on ${\rm Gal}(\bar{K}/LF)$, since this is a subgroup of ${\rm Gal}(\bar{K}/F)$, and you already know that $\rho$ is trivial on the latter (that's what it means to say that $\rho$ factors through $F/K$). So we can just think of $\rho$ as a representation of ${\rm Gal}(LF/K)$ that is trivial on ${\rm Gal}(LF/F)$. So the value of $\rho$ on $\sigma\in {\rm Gal}(LF/K)$ only depends on the image of $\sigma$ under the quotient map ${\rm Gal}(LF/K)\to {\rm Gal}(F/K)$. But we have just observed that this quotient map is an isomorphism on the subgroup ${\rm Gal}(LF/L)$ (i.e. different elements of that subgroup give different cosets in the quotient, and every element of the quotient has a representative in that subgroup), so $\rho$ is determined by its values on ${\rm Gal}(LF/L)$. It's really just a matter of unravelling the definitions, and has not much to do with Galois, it's just a fact about homomorphisms from arbitrary groups to anywhere that factor through a quotient, when the extension of groups splits (the diagram I would draw here is the diamond with $K$, $F$, $L$, and $FL$; just think about what $\rho$ and $\rho'$ would look like on the quotient ${\rm Gal}(F/K)$ if they agreed on ${\rm Gal}(LF/L)$).

How to recover $\rho$ from its local polynomial?

Now you are left with a representation of the Galois group of an unramified extension. That Galois group is generated by Frobenius, let's call him $\phi$ (that's what being unramified has bought you), so the representation $\rho$ is determined by $\rho(\phi)$. By assumption, the representation is Frobenius-semisimple, so $\rho(\phi)$ is just determined by the eigenvalues of $\rho(\phi)$, and these, in turn, can be read off from the characteristic polynomial.

I do not understand your professor's diagram ($K^{\rm ur}$ cannot denote the maximal unramified extension of $K$, since that would be infinite, so certainly not the same degree as $F/K$; it also cannot denote the maximal unramified extension contained within $F$, since that claimed equality would still not hold - $F/K$ was not assumed to be unramified), and I suggest that you ask your professor rather than us.

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  • $\begingroup$ Thank you for your response! Most things became much clearer for me now. I still have a question left: Can you also argue with Galois groups $\operatorname{Gal}(\bar{K}/K)$, $\operatorname{Gal}(\bar{K}/L)$ etc. if we do have Weil representations (which was required) instead of Galois representations (as you did)? I just know that the Weil group is a proper subgroup of the absolute Galois group. $\endgroup$ – Diglett Oct 6 '18 at 22:31
  • $\begingroup$ If a representation factors through a finite quotient, then you can lift it to the Weil group, but you can also lift it to the full Galois group. Moreover, the former lift will be the restriction of the latter. So I guess I showed that the whole lift to the full Galois group will already be determined by the local polynomial. But really, that distinction is empty for representations that factor through a finite quotient. $\endgroup$ – Alex B. Oct 7 '18 at 0:23

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