Characterization for the continuity of Weil representations












2












$begingroup$


Let $K$ be a non-Archimedean local field and $W_K$ be the Weil group of $K$. We consider a representation $rho: W_K to operatorname{GL}_n(mathbb{C})$ between two topological groups. Here, $operatorname{GL}_n(mathbb{C})$ is endowed with the discrete topology and $W_K$ has this weird topology such that
$$
1 to I_K to W_K to mathbb{Z} to 1
$$
becomes a short exact sequence of topological spaces where
$I_K$ is the inertia subgroup of the absolute Galois group $G_K$ and $mathbb{Z}$ denotes the subgroup generated by the Frobenius element $x mapsto x^{|k|}$ of the absolute Galois group $G_k$ of the residue field $k$ of $K$.



I would like to show the equivalence of the following two statements:





  • $rho$ is continuous,


  • $rho(I_K)$ is finite.


If one of these two criteria is satisfied, we will call $rho$ a Weil representation.



Attempts and Ideas:




  • If $rho$ is continuous, then $rho$ maps compact subsets of $W_K$ to compact subsets of $operatorname{GL}_n(mathbb{C})$. If I manage to show that $I_K subseteq W_K$ is compact, then $rho(I_K)$ is finite since $operatorname{GL}_n(mathbb{C})$ has the discrete topology.

  • I have no clue for the other direction. It seems like we need an argument from the theory of topological spaces. However, I have little to no knowledge about it.


Could you please help me with this problem? Thank you!






abstract-algebra representation-theory algebraic-number-theory topological-groups galois-representations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $K$ be a non-Archimedean local field and $W_K$ be the Weil group of $K$. We consider a representation $rho: W_K to operatorname{GL}_n(mathbb{C})$ between two topological groups. Here, $operatorname{GL}_n(mathbb{C})$ is endowed with the discrete topology and $W_K$ has this weird topology such that
    $$
    1 to I_K to W_K to mathbb{Z} to 1
    $$
    becomes a short exact sequence of topological spaces where
    $I_K$ is the inertia subgroup of the absolute Galois group $G_K$ and $mathbb{Z}$ denotes the subgroup generated by the Frobenius element $x mapsto x^{|k|}$ of the absolute Galois group $G_k$ of the residue field $k$ of $K$.



    I would like to show the equivalence of the following two statements:





    • $rho$ is continuous,


    • $rho(I_K)$ is finite.


    If one of these two criteria is satisfied, we will call $rho$ a Weil representation.



    Attempts and Ideas:




    • If $rho$ is continuous, then $rho$ maps compact subsets of $W_K$ to compact subsets of $operatorname{GL}_n(mathbb{C})$. If I manage to show that $I_K subseteq W_K$ is compact, then $rho(I_K)$ is finite since $operatorname{GL}_n(mathbb{C})$ has the discrete topology.

    • I have no clue for the other direction. It seems like we need an argument from the theory of topological spaces. However, I have little to no knowledge about it.


    Could you please help me with this problem? Thank you!






    abstract-algebra representation-theory algebraic-number-theory topological-groups galois-representations







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $K$ be a non-Archimedean local field and $W_K$ be the Weil group of $K$. We consider a representation $rho: W_K to operatorname{GL}_n(mathbb{C})$ between two topological groups. Here, $operatorname{GL}_n(mathbb{C})$ is endowed with the discrete topology and $W_K$ has this weird topology such that
      $$
      1 to I_K to W_K to mathbb{Z} to 1
      $$
      becomes a short exact sequence of topological spaces where
      $I_K$ is the inertia subgroup of the absolute Galois group $G_K$ and $mathbb{Z}$ denotes the subgroup generated by the Frobenius element $x mapsto x^{|k|}$ of the absolute Galois group $G_k$ of the residue field $k$ of $K$.



      I would like to show the equivalence of the following two statements:





      • $rho$ is continuous,


      • $rho(I_K)$ is finite.


      If one of these two criteria is satisfied, we will call $rho$ a Weil representation.



      Attempts and Ideas:




      • If $rho$ is continuous, then $rho$ maps compact subsets of $W_K$ to compact subsets of $operatorname{GL}_n(mathbb{C})$. If I manage to show that $I_K subseteq W_K$ is compact, then $rho(I_K)$ is finite since $operatorname{GL}_n(mathbb{C})$ has the discrete topology.

      • I have no clue for the other direction. It seems like we need an argument from the theory of topological spaces. However, I have little to no knowledge about it.


      Could you please help me with this problem? Thank you!






      abstract-algebra representation-theory algebraic-number-theory topological-groups galois-representations







      share|cite|improve this question











      $endgroup$




      Let $K$ be a non-Archimedean local field and $W_K$ be the Weil group of $K$. We consider a representation $rho: W_K to operatorname{GL}_n(mathbb{C})$ between two topological groups. Here, $operatorname{GL}_n(mathbb{C})$ is endowed with the discrete topology and $W_K$ has this weird topology such that
      $$
      1 to I_K to W_K to mathbb{Z} to 1
      $$
      becomes a short exact sequence of topological spaces where
      $I_K$ is the inertia subgroup of the absolute Galois group $G_K$ and $mathbb{Z}$ denotes the subgroup generated by the Frobenius element $x mapsto x^{|k|}$ of the absolute Galois group $G_k$ of the residue field $k$ of $K$.



      I would like to show the equivalence of the following two statements:





      • $rho$ is continuous,


      • $rho(I_K)$ is finite.


      If one of these two criteria is satisfied, we will call $rho$ a Weil representation.



      Attempts and Ideas:




      • If $rho$ is continuous, then $rho$ maps compact subsets of $W_K$ to compact subsets of $operatorname{GL}_n(mathbb{C})$. If I manage to show that $I_K subseteq W_K$ is compact, then $rho(I_K)$ is finite since $operatorname{GL}_n(mathbb{C})$ has the discrete topology.

      • I have no clue for the other direction. It seems like we need an argument from the theory of topological spaces. However, I have little to no knowledge about it.


      Could you please help me with this problem? Thank you!






      abstract-algebra representation-theory algebraic-number-theory topological-groups galois-representations




      abstract-algebra representation-theory algebraic-number-theory topological-groups galois-representations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 9:17







      Diglett

















      asked Jan 4 at 8:53









      DiglettDiglett

      9801521




      9801521






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Let's take a step back for a second and ask: if $H$ is a discrete topological group and $G$ is an arbitrary topological group, what does it mean for a homomorphism $rho:Gto H$ to be continuous?



          Since ${e}$ is open in $H$, we certainly need $ker rho$ to be open in $G$. And this condition is sufficient too: if $H_0subset H$ is a subset (and hence an open subset) of $H$, then
          $$rho^{-1}(H) = bigcup_{hin H}bigcup_{gin rho^{-1}(h)}gker(rho)$$
          is a union of open sets, so is open.



          Now, everything is simpler. Here are some hints:




          • If $rho$ is continuous, then $rho|_{I_K}$ is continuous, so $rho|_{I_K}:I_Kto mathrm{GL}_n(mathbb C)$ has open kernel. But $I_K$ is profinite, so the kernel must be a finite index normal subgroup.


          • If $rho(I_K)$ is finite, then $ker(rho)cap I_K$ is a finite index normal subgroup, and hence is open. The remainder of the Weil group is determined by the Frobenius element.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your response! I understood the first part completely. Could you please explain why $I_K$ being profinite implies that the kernel is a finite index normal subgroup? I still have trouble to understand how the topology of inverse limits work.
            $endgroup$
            – Diglett
            Jan 4 at 12:30










          • $begingroup$
            I would like to gather everything I know: I only know that in a compact topological group (so esp. profinite groups) a subgroup is open if and only if it is closed of finite index. I understand that $operatorname{ker}(rho|_{I_K}) = I_K cap operatorname{ker}(rho)$ has finite index if $rho(I_K)$ is finite, so the kernel must be closed too if I want to convince myself that it is open. In the books I looked up it was never mentioned that a finite index normal subgroup of a profinite group is open. Could you please recommend me a reference where your claim is mentioned?
            $endgroup$
            – Diglett
            Jan 6 at 9:38










          • $begingroup$
            What I said isn't true in general. But it is true for the absolute Galois group, and in particular for the inertia group. Galois theory ensures a bijective correspondence between Galois extensions of $K$ and closed normal subgroups of $G_K$. If $Hsubset I_K$ is finite, then by Galois theory + ramification theory, $H$ corresponds to a finite extension of $K$, and hence $H$ is closed and of finite index, so is open.
            $endgroup$
            – Mathmo123
            Jan 7 at 10:01












          • $begingroup$
            Okay thanks, I think I understand now why $I_K cap operatorname{ker}(rho)$ is open in $I_K$. Because $I_K$ is open in $W_K$, the intersection of $I_K$ and the kernel must be open in $W_K$ too. But I am not able to proceed to show that the kernel is open in $W_K$ which would give me the continuity of $rho$. All I know is that the quotient of $W_K/I_K$ is the cyclic subgroup generated by a Frobenius element. Could you please explain how this is useful here?
            $endgroup$
            – Diglett
            Jan 7 at 15:17










          • $begingroup$
            You can use the fact that $W_K =bigcup _{iinmathbb Z}phi^i I_K$ where $phi$ is the Frobenius.
            $endgroup$
            – Mathmo123
            Jan 7 at 18:44













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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Let's take a step back for a second and ask: if $H$ is a discrete topological group and $G$ is an arbitrary topological group, what does it mean for a homomorphism $rho:Gto H$ to be continuous?



          Since ${e}$ is open in $H$, we certainly need $ker rho$ to be open in $G$. And this condition is sufficient too: if $H_0subset H$ is a subset (and hence an open subset) of $H$, then
          $$rho^{-1}(H) = bigcup_{hin H}bigcup_{gin rho^{-1}(h)}gker(rho)$$
          is a union of open sets, so is open.



          Now, everything is simpler. Here are some hints:




          • If $rho$ is continuous, then $rho|_{I_K}$ is continuous, so $rho|_{I_K}:I_Kto mathrm{GL}_n(mathbb C)$ has open kernel. But $I_K$ is profinite, so the kernel must be a finite index normal subgroup.


          • If $rho(I_K)$ is finite, then $ker(rho)cap I_K$ is a finite index normal subgroup, and hence is open. The remainder of the Weil group is determined by the Frobenius element.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your response! I understood the first part completely. Could you please explain why $I_K$ being profinite implies that the kernel is a finite index normal subgroup? I still have trouble to understand how the topology of inverse limits work.
            $endgroup$
            – Diglett
            Jan 4 at 12:30










          • $begingroup$
            I would like to gather everything I know: I only know that in a compact topological group (so esp. profinite groups) a subgroup is open if and only if it is closed of finite index. I understand that $operatorname{ker}(rho|_{I_K}) = I_K cap operatorname{ker}(rho)$ has finite index if $rho(I_K)$ is finite, so the kernel must be closed too if I want to convince myself that it is open. In the books I looked up it was never mentioned that a finite index normal subgroup of a profinite group is open. Could you please recommend me a reference where your claim is mentioned?
            $endgroup$
            – Diglett
            Jan 6 at 9:38










          • $begingroup$
            What I said isn't true in general. But it is true for the absolute Galois group, and in particular for the inertia group. Galois theory ensures a bijective correspondence between Galois extensions of $K$ and closed normal subgroups of $G_K$. If $Hsubset I_K$ is finite, then by Galois theory + ramification theory, $H$ corresponds to a finite extension of $K$, and hence $H$ is closed and of finite index, so is open.
            $endgroup$
            – Mathmo123
            Jan 7 at 10:01












          • $begingroup$
            Okay thanks, I think I understand now why $I_K cap operatorname{ker}(rho)$ is open in $I_K$. Because $I_K$ is open in $W_K$, the intersection of $I_K$ and the kernel must be open in $W_K$ too. But I am not able to proceed to show that the kernel is open in $W_K$ which would give me the continuity of $rho$. All I know is that the quotient of $W_K/I_K$ is the cyclic subgroup generated by a Frobenius element. Could you please explain how this is useful here?
            $endgroup$
            – Diglett
            Jan 7 at 15:17










          • $begingroup$
            You can use the fact that $W_K =bigcup _{iinmathbb Z}phi^i I_K$ where $phi$ is the Frobenius.
            $endgroup$
            – Mathmo123
            Jan 7 at 18:44


















          3












          $begingroup$

          Let's take a step back for a second and ask: if $H$ is a discrete topological group and $G$ is an arbitrary topological group, what does it mean for a homomorphism $rho:Gto H$ to be continuous?



          Since ${e}$ is open in $H$, we certainly need $ker rho$ to be open in $G$. And this condition is sufficient too: if $H_0subset H$ is a subset (and hence an open subset) of $H$, then
          $$rho^{-1}(H) = bigcup_{hin H}bigcup_{gin rho^{-1}(h)}gker(rho)$$
          is a union of open sets, so is open.



          Now, everything is simpler. Here are some hints:




          • If $rho$ is continuous, then $rho|_{I_K}$ is continuous, so $rho|_{I_K}:I_Kto mathrm{GL}_n(mathbb C)$ has open kernel. But $I_K$ is profinite, so the kernel must be a finite index normal subgroup.


          • If $rho(I_K)$ is finite, then $ker(rho)cap I_K$ is a finite index normal subgroup, and hence is open. The remainder of the Weil group is determined by the Frobenius element.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your response! I understood the first part completely. Could you please explain why $I_K$ being profinite implies that the kernel is a finite index normal subgroup? I still have trouble to understand how the topology of inverse limits work.
            $endgroup$
            – Diglett
            Jan 4 at 12:30










          • $begingroup$
            I would like to gather everything I know: I only know that in a compact topological group (so esp. profinite groups) a subgroup is open if and only if it is closed of finite index. I understand that $operatorname{ker}(rho|_{I_K}) = I_K cap operatorname{ker}(rho)$ has finite index if $rho(I_K)$ is finite, so the kernel must be closed too if I want to convince myself that it is open. In the books I looked up it was never mentioned that a finite index normal subgroup of a profinite group is open. Could you please recommend me a reference where your claim is mentioned?
            $endgroup$
            – Diglett
            Jan 6 at 9:38










          • $begingroup$
            What I said isn't true in general. But it is true for the absolute Galois group, and in particular for the inertia group. Galois theory ensures a bijective correspondence between Galois extensions of $K$ and closed normal subgroups of $G_K$. If $Hsubset I_K$ is finite, then by Galois theory + ramification theory, $H$ corresponds to a finite extension of $K$, and hence $H$ is closed and of finite index, so is open.
            $endgroup$
            – Mathmo123
            Jan 7 at 10:01












          • $begingroup$
            Okay thanks, I think I understand now why $I_K cap operatorname{ker}(rho)$ is open in $I_K$. Because $I_K$ is open in $W_K$, the intersection of $I_K$ and the kernel must be open in $W_K$ too. But I am not able to proceed to show that the kernel is open in $W_K$ which would give me the continuity of $rho$. All I know is that the quotient of $W_K/I_K$ is the cyclic subgroup generated by a Frobenius element. Could you please explain how this is useful here?
            $endgroup$
            – Diglett
            Jan 7 at 15:17










          • $begingroup$
            You can use the fact that $W_K =bigcup _{iinmathbb Z}phi^i I_K$ where $phi$ is the Frobenius.
            $endgroup$
            – Mathmo123
            Jan 7 at 18:44
















          3












          3








          3





          $begingroup$

          Let's take a step back for a second and ask: if $H$ is a discrete topological group and $G$ is an arbitrary topological group, what does it mean for a homomorphism $rho:Gto H$ to be continuous?



          Since ${e}$ is open in $H$, we certainly need $ker rho$ to be open in $G$. And this condition is sufficient too: if $H_0subset H$ is a subset (and hence an open subset) of $H$, then
          $$rho^{-1}(H) = bigcup_{hin H}bigcup_{gin rho^{-1}(h)}gker(rho)$$
          is a union of open sets, so is open.



          Now, everything is simpler. Here are some hints:




          • If $rho$ is continuous, then $rho|_{I_K}$ is continuous, so $rho|_{I_K}:I_Kto mathrm{GL}_n(mathbb C)$ has open kernel. But $I_K$ is profinite, so the kernel must be a finite index normal subgroup.


          • If $rho(I_K)$ is finite, then $ker(rho)cap I_K$ is a finite index normal subgroup, and hence is open. The remainder of the Weil group is determined by the Frobenius element.







          share|cite|improve this answer









          $endgroup$



          Let's take a step back for a second and ask: if $H$ is a discrete topological group and $G$ is an arbitrary topological group, what does it mean for a homomorphism $rho:Gto H$ to be continuous?



          Since ${e}$ is open in $H$, we certainly need $ker rho$ to be open in $G$. And this condition is sufficient too: if $H_0subset H$ is a subset (and hence an open subset) of $H$, then
          $$rho^{-1}(H) = bigcup_{hin H}bigcup_{gin rho^{-1}(h)}gker(rho)$$
          is a union of open sets, so is open.



          Now, everything is simpler. Here are some hints:




          • If $rho$ is continuous, then $rho|_{I_K}$ is continuous, so $rho|_{I_K}:I_Kto mathrm{GL}_n(mathbb C)$ has open kernel. But $I_K$ is profinite, so the kernel must be a finite index normal subgroup.


          • If $rho(I_K)$ is finite, then $ker(rho)cap I_K$ is a finite index normal subgroup, and hence is open. The remainder of the Weil group is determined by the Frobenius element.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 9:55









          Mathmo123Mathmo123

          17.9k33166




          17.9k33166












          • $begingroup$
            Thank you for your response! I understood the first part completely. Could you please explain why $I_K$ being profinite implies that the kernel is a finite index normal subgroup? I still have trouble to understand how the topology of inverse limits work.
            $endgroup$
            – Diglett
            Jan 4 at 12:30










          • $begingroup$
            I would like to gather everything I know: I only know that in a compact topological group (so esp. profinite groups) a subgroup is open if and only if it is closed of finite index. I understand that $operatorname{ker}(rho|_{I_K}) = I_K cap operatorname{ker}(rho)$ has finite index if $rho(I_K)$ is finite, so the kernel must be closed too if I want to convince myself that it is open. In the books I looked up it was never mentioned that a finite index normal subgroup of a profinite group is open. Could you please recommend me a reference where your claim is mentioned?
            $endgroup$
            – Diglett
            Jan 6 at 9:38










          • $begingroup$
            What I said isn't true in general. But it is true for the absolute Galois group, and in particular for the inertia group. Galois theory ensures a bijective correspondence between Galois extensions of $K$ and closed normal subgroups of $G_K$. If $Hsubset I_K$ is finite, then by Galois theory + ramification theory, $H$ corresponds to a finite extension of $K$, and hence $H$ is closed and of finite index, so is open.
            $endgroup$
            – Mathmo123
            Jan 7 at 10:01












          • $begingroup$
            Okay thanks, I think I understand now why $I_K cap operatorname{ker}(rho)$ is open in $I_K$. Because $I_K$ is open in $W_K$, the intersection of $I_K$ and the kernel must be open in $W_K$ too. But I am not able to proceed to show that the kernel is open in $W_K$ which would give me the continuity of $rho$. All I know is that the quotient of $W_K/I_K$ is the cyclic subgroup generated by a Frobenius element. Could you please explain how this is useful here?
            $endgroup$
            – Diglett
            Jan 7 at 15:17










          • $begingroup$
            You can use the fact that $W_K =bigcup _{iinmathbb Z}phi^i I_K$ where $phi$ is the Frobenius.
            $endgroup$
            – Mathmo123
            Jan 7 at 18:44




















          • $begingroup$
            Thank you for your response! I understood the first part completely. Could you please explain why $I_K$ being profinite implies that the kernel is a finite index normal subgroup? I still have trouble to understand how the topology of inverse limits work.
            $endgroup$
            – Diglett
            Jan 4 at 12:30










          • $begingroup$
            I would like to gather everything I know: I only know that in a compact topological group (so esp. profinite groups) a subgroup is open if and only if it is closed of finite index. I understand that $operatorname{ker}(rho|_{I_K}) = I_K cap operatorname{ker}(rho)$ has finite index if $rho(I_K)$ is finite, so the kernel must be closed too if I want to convince myself that it is open. In the books I looked up it was never mentioned that a finite index normal subgroup of a profinite group is open. Could you please recommend me a reference where your claim is mentioned?
            $endgroup$
            – Diglett
            Jan 6 at 9:38










          • $begingroup$
            What I said isn't true in general. But it is true for the absolute Galois group, and in particular for the inertia group. Galois theory ensures a bijective correspondence between Galois extensions of $K$ and closed normal subgroups of $G_K$. If $Hsubset I_K$ is finite, then by Galois theory + ramification theory, $H$ corresponds to a finite extension of $K$, and hence $H$ is closed and of finite index, so is open.
            $endgroup$
            – Mathmo123
            Jan 7 at 10:01












          • $begingroup$
            Okay thanks, I think I understand now why $I_K cap operatorname{ker}(rho)$ is open in $I_K$. Because $I_K$ is open in $W_K$, the intersection of $I_K$ and the kernel must be open in $W_K$ too. But I am not able to proceed to show that the kernel is open in $W_K$ which would give me the continuity of $rho$. All I know is that the quotient of $W_K/I_K$ is the cyclic subgroup generated by a Frobenius element. Could you please explain how this is useful here?
            $endgroup$
            – Diglett
            Jan 7 at 15:17










          • $begingroup$
            You can use the fact that $W_K =bigcup _{iinmathbb Z}phi^i I_K$ where $phi$ is the Frobenius.
            $endgroup$
            – Mathmo123
            Jan 7 at 18:44


















          $begingroup$
          Thank you for your response! I understood the first part completely. Could you please explain why $I_K$ being profinite implies that the kernel is a finite index normal subgroup? I still have trouble to understand how the topology of inverse limits work.
          $endgroup$
          – Diglett
          Jan 4 at 12:30




          $begingroup$
          Thank you for your response! I understood the first part completely. Could you please explain why $I_K$ being profinite implies that the kernel is a finite index normal subgroup? I still have trouble to understand how the topology of inverse limits work.
          $endgroup$
          – Diglett
          Jan 4 at 12:30












          $begingroup$
          I would like to gather everything I know: I only know that in a compact topological group (so esp. profinite groups) a subgroup is open if and only if it is closed of finite index. I understand that $operatorname{ker}(rho|_{I_K}) = I_K cap operatorname{ker}(rho)$ has finite index if $rho(I_K)$ is finite, so the kernel must be closed too if I want to convince myself that it is open. In the books I looked up it was never mentioned that a finite index normal subgroup of a profinite group is open. Could you please recommend me a reference where your claim is mentioned?
          $endgroup$
          – Diglett
          Jan 6 at 9:38




          $begingroup$
          I would like to gather everything I know: I only know that in a compact topological group (so esp. profinite groups) a subgroup is open if and only if it is closed of finite index. I understand that $operatorname{ker}(rho|_{I_K}) = I_K cap operatorname{ker}(rho)$ has finite index if $rho(I_K)$ is finite, so the kernel must be closed too if I want to convince myself that it is open. In the books I looked up it was never mentioned that a finite index normal subgroup of a profinite group is open. Could you please recommend me a reference where your claim is mentioned?
          $endgroup$
          – Diglett
          Jan 6 at 9:38












          $begingroup$
          What I said isn't true in general. But it is true for the absolute Galois group, and in particular for the inertia group. Galois theory ensures a bijective correspondence between Galois extensions of $K$ and closed normal subgroups of $G_K$. If $Hsubset I_K$ is finite, then by Galois theory + ramification theory, $H$ corresponds to a finite extension of $K$, and hence $H$ is closed and of finite index, so is open.
          $endgroup$
          – Mathmo123
          Jan 7 at 10:01






          $begingroup$
          What I said isn't true in general. But it is true for the absolute Galois group, and in particular for the inertia group. Galois theory ensures a bijective correspondence between Galois extensions of $K$ and closed normal subgroups of $G_K$. If $Hsubset I_K$ is finite, then by Galois theory + ramification theory, $H$ corresponds to a finite extension of $K$, and hence $H$ is closed and of finite index, so is open.
          $endgroup$
          – Mathmo123
          Jan 7 at 10:01














          $begingroup$
          Okay thanks, I think I understand now why $I_K cap operatorname{ker}(rho)$ is open in $I_K$. Because $I_K$ is open in $W_K$, the intersection of $I_K$ and the kernel must be open in $W_K$ too. But I am not able to proceed to show that the kernel is open in $W_K$ which would give me the continuity of $rho$. All I know is that the quotient of $W_K/I_K$ is the cyclic subgroup generated by a Frobenius element. Could you please explain how this is useful here?
          $endgroup$
          – Diglett
          Jan 7 at 15:17




          $begingroup$
          Okay thanks, I think I understand now why $I_K cap operatorname{ker}(rho)$ is open in $I_K$. Because $I_K$ is open in $W_K$, the intersection of $I_K$ and the kernel must be open in $W_K$ too. But I am not able to proceed to show that the kernel is open in $W_K$ which would give me the continuity of $rho$. All I know is that the quotient of $W_K/I_K$ is the cyclic subgroup generated by a Frobenius element. Could you please explain how this is useful here?
          $endgroup$
          – Diglett
          Jan 7 at 15:17












          $begingroup$
          You can use the fact that $W_K =bigcup _{iinmathbb Z}phi^i I_K$ where $phi$ is the Frobenius.
          $endgroup$
          – Mathmo123
          Jan 7 at 18:44






          $begingroup$
          You can use the fact that $W_K =bigcup _{iinmathbb Z}phi^i I_K$ where $phi$ is the Frobenius.
          $endgroup$
          – Mathmo123
          Jan 7 at 18:44




















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