$Hom(mathbb C^*, mathbb C^*)$?












0












$begingroup$


I want to calculate this group.



So far I noticed that if $h:mathbb C^* to mathbb C^*$ is a hoomorphism, then $h(z)=h(1cdot z)=h(1)cdot h(z)$ for any $zin mathbb C^*$. Thus $h(1)=1$.

Further I know that for any $tin mathbb C$ $h_t: zmapsto z^t$ satisfies $h(xy)=(xy)^t=x^ty^t=h(x)h(y)$. So this is an homomorphism, hence $Hom (mathbb C^*, mathbb C^*)subset mathbb C.$



But is this an equality, or are there more homomorphisms?






abstract-algebra group-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    $z^t$ is not well defined if $t notin Bbb Z$.
    $endgroup$
    – Nicolas Hemelsoet
    Nov 24 '18 at 11:11










  • $begingroup$
    If $tnotinBbb Z$, then $zmapsto z^t$ (whatever that is) is not a homomorphism. If you believe AC then there are lots of homomorphisms.
    $endgroup$
    – Lord Shark the Unknown
    Nov 24 '18 at 11:11










  • $begingroup$
    Complex conjugation is one additional homomorphism you can write down (and of of course all composition of it with the homomorphisms you already mentioned)
    $endgroup$
    – Severin Schraven
    Nov 24 '18 at 11:20
















0












$begingroup$


I want to calculate this group.



So far I noticed that if $h:mathbb C^* to mathbb C^*$ is a hoomorphism, then $h(z)=h(1cdot z)=h(1)cdot h(z)$ for any $zin mathbb C^*$. Thus $h(1)=1$.

Further I know that for any $tin mathbb C$ $h_t: zmapsto z^t$ satisfies $h(xy)=(xy)^t=x^ty^t=h(x)h(y)$. So this is an homomorphism, hence $Hom (mathbb C^*, mathbb C^*)subset mathbb C.$



But is this an equality, or are there more homomorphisms?






abstract-algebra group-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    $z^t$ is not well defined if $t notin Bbb Z$.
    $endgroup$
    – Nicolas Hemelsoet
    Nov 24 '18 at 11:11










  • $begingroup$
    If $tnotinBbb Z$, then $zmapsto z^t$ (whatever that is) is not a homomorphism. If you believe AC then there are lots of homomorphisms.
    $endgroup$
    – Lord Shark the Unknown
    Nov 24 '18 at 11:11










  • $begingroup$
    Complex conjugation is one additional homomorphism you can write down (and of of course all composition of it with the homomorphisms you already mentioned)
    $endgroup$
    – Severin Schraven
    Nov 24 '18 at 11:20














0












0








0





$begingroup$


I want to calculate this group.



So far I noticed that if $h:mathbb C^* to mathbb C^*$ is a hoomorphism, then $h(z)=h(1cdot z)=h(1)cdot h(z)$ for any $zin mathbb C^*$. Thus $h(1)=1$.

Further I know that for any $tin mathbb C$ $h_t: zmapsto z^t$ satisfies $h(xy)=(xy)^t=x^ty^t=h(x)h(y)$. So this is an homomorphism, hence $Hom (mathbb C^*, mathbb C^*)subset mathbb C.$



But is this an equality, or are there more homomorphisms?






abstract-algebra group-theory







share|cite|improve this question









$endgroup$




I want to calculate this group.



So far I noticed that if $h:mathbb C^* to mathbb C^*$ is a hoomorphism, then $h(z)=h(1cdot z)=h(1)cdot h(z)$ for any $zin mathbb C^*$. Thus $h(1)=1$.

Further I know that for any $tin mathbb C$ $h_t: zmapsto z^t$ satisfies $h(xy)=(xy)^t=x^ty^t=h(x)h(y)$. So this is an homomorphism, hence $Hom (mathbb C^*, mathbb C^*)subset mathbb C.$



But is this an equality, or are there more homomorphisms?






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 '18 at 11:09









J.DoeJ.Doe

214




214












  • $begingroup$
    $z^t$ is not well defined if $t notin Bbb Z$.
    $endgroup$
    – Nicolas Hemelsoet
    Nov 24 '18 at 11:11










  • $begingroup$
    If $tnotinBbb Z$, then $zmapsto z^t$ (whatever that is) is not a homomorphism. If you believe AC then there are lots of homomorphisms.
    $endgroup$
    – Lord Shark the Unknown
    Nov 24 '18 at 11:11










  • $begingroup$
    Complex conjugation is one additional homomorphism you can write down (and of of course all composition of it with the homomorphisms you already mentioned)
    $endgroup$
    – Severin Schraven
    Nov 24 '18 at 11:20


















  • $begingroup$
    $z^t$ is not well defined if $t notin Bbb Z$.
    $endgroup$
    – Nicolas Hemelsoet
    Nov 24 '18 at 11:11










  • $begingroup$
    If $tnotinBbb Z$, then $zmapsto z^t$ (whatever that is) is not a homomorphism. If you believe AC then there are lots of homomorphisms.
    $endgroup$
    – Lord Shark the Unknown
    Nov 24 '18 at 11:11










  • $begingroup$
    Complex conjugation is one additional homomorphism you can write down (and of of course all composition of it with the homomorphisms you already mentioned)
    $endgroup$
    – Severin Schraven
    Nov 24 '18 at 11:20
















$begingroup$
$z^t$ is not well defined if $t notin Bbb Z$.
$endgroup$
– Nicolas Hemelsoet
Nov 24 '18 at 11:11




$begingroup$
$z^t$ is not well defined if $t notin Bbb Z$.
$endgroup$
– Nicolas Hemelsoet
Nov 24 '18 at 11:11












$begingroup$
If $tnotinBbb Z$, then $zmapsto z^t$ (whatever that is) is not a homomorphism. If you believe AC then there are lots of homomorphisms.
$endgroup$
– Lord Shark the Unknown
Nov 24 '18 at 11:11




$begingroup$
If $tnotinBbb Z$, then $zmapsto z^t$ (whatever that is) is not a homomorphism. If you believe AC then there are lots of homomorphisms.
$endgroup$
– Lord Shark the Unknown
Nov 24 '18 at 11:11












$begingroup$
Complex conjugation is one additional homomorphism you can write down (and of of course all composition of it with the homomorphisms you already mentioned)
$endgroup$
– Severin Schraven
Nov 24 '18 at 11:20




$begingroup$
Complex conjugation is one additional homomorphism you can write down (and of of course all composition of it with the homomorphisms you already mentioned)
$endgroup$
– Severin Schraven
Nov 24 '18 at 11:20










2 Answers
2






active

oldest

votes


















1












$begingroup$

One can classify simply the continuous homomorphisms from $Bbb C^*$
to itself. They are the maps
$$zmapsto z^nexp(aln|z|)$$
where $ninBbb Z$ and $ainBbb C$.



But assuming the axiom of choice, one proves that $Bbb C^*$
is isomorphic to the additive $(Bbb Q/Bbb Z)timesbigoplus_IBbb Q$
where $I$ is an uncountable index set. This group has $2^{2^{aleph_0}}$
automorphisms, an awful lot.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I see that the defined map is a continuous homomorphism. But how can we be sure that there are no others?
    $endgroup$
    – J.Doe
    Nov 24 '18 at 13:23



















1












$begingroup$

Multiplicative notation is not common in abelian group theory.



We can use the fact that $mathbb{C}^*congmathbb{R}oplusmathbb{T}$, where $mathbb{T}=mathbb{R}/mathbb{Z}$, via
$$
(r,u+mathbb{Z})mapsto e^r(cos(2pi u)+isin(2pi u))
$$
Thus the group of endomorphisms of $mathbb{C}^*$ is the direct sum of
$$DeclareMathOperator{Hom}{Hom}
Hom(mathbb{R},mathbb{R})oplusHom(mathbb{R},mathbb{T})oplus
Hom(mathbb{T},mathbb{T})oplusHom(mathbb{T},mathbb{R})
$$



The first two groups are big $mathbb{Q}$-vector spaces, the third one is a wild beast. If we consider the exact sequence $0tomathbb{Z}tomathbb{R}tomathbb{T}to0$, then we have
$$
0toHom(mathbb{T},mathbb{T})toHom(mathbb{R},mathbb{T})toHom(mathbb{Z},mathbb{T})to0
$$
and $Hom(mathbb{Z},mathbb{T})congmathbb{T}$. So $Hom(mathbb{T},mathbb{T})$ sits in a big $mathbb{Q}$-vector space so that the cokernel is $mathbb{T}$.



What's $Hom(mathbb{T},mathbb{R})$? A similar sequence arises:
$$
0toHom(mathbb{T},mathbb{R})toHom(mathbb{R},mathbb{R})toHom(mathbb{Z},mathbb{R})to0
$$



As you see, the group $Hom(mathbb{C}^*,mathbb{C}^*)$ is much bigger than $mathbb{C}$.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    One can classify simply the continuous homomorphisms from $Bbb C^*$
    to itself. They are the maps
    $$zmapsto z^nexp(aln|z|)$$
    where $ninBbb Z$ and $ainBbb C$.



    But assuming the axiom of choice, one proves that $Bbb C^*$
    is isomorphic to the additive $(Bbb Q/Bbb Z)timesbigoplus_IBbb Q$
    where $I$ is an uncountable index set. This group has $2^{2^{aleph_0}}$
    automorphisms, an awful lot.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I see that the defined map is a continuous homomorphism. But how can we be sure that there are no others?
      $endgroup$
      – J.Doe
      Nov 24 '18 at 13:23
















    1












    $begingroup$

    One can classify simply the continuous homomorphisms from $Bbb C^*$
    to itself. They are the maps
    $$zmapsto z^nexp(aln|z|)$$
    where $ninBbb Z$ and $ainBbb C$.



    But assuming the axiom of choice, one proves that $Bbb C^*$
    is isomorphic to the additive $(Bbb Q/Bbb Z)timesbigoplus_IBbb Q$
    where $I$ is an uncountable index set. This group has $2^{2^{aleph_0}}$
    automorphisms, an awful lot.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I see that the defined map is a continuous homomorphism. But how can we be sure that there are no others?
      $endgroup$
      – J.Doe
      Nov 24 '18 at 13:23














    1












    1








    1





    $begingroup$

    One can classify simply the continuous homomorphisms from $Bbb C^*$
    to itself. They are the maps
    $$zmapsto z^nexp(aln|z|)$$
    where $ninBbb Z$ and $ainBbb C$.



    But assuming the axiom of choice, one proves that $Bbb C^*$
    is isomorphic to the additive $(Bbb Q/Bbb Z)timesbigoplus_IBbb Q$
    where $I$ is an uncountable index set. This group has $2^{2^{aleph_0}}$
    automorphisms, an awful lot.






    share|cite|improve this answer









    $endgroup$



    One can classify simply the continuous homomorphisms from $Bbb C^*$
    to itself. They are the maps
    $$zmapsto z^nexp(aln|z|)$$
    where $ninBbb Z$ and $ainBbb C$.



    But assuming the axiom of choice, one proves that $Bbb C^*$
    is isomorphic to the additive $(Bbb Q/Bbb Z)timesbigoplus_IBbb Q$
    where $I$ is an uncountable index set. This group has $2^{2^{aleph_0}}$
    automorphisms, an awful lot.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 24 '18 at 11:25









    Lord Shark the UnknownLord Shark the Unknown

    102k960132




    102k960132












    • $begingroup$
      I see that the defined map is a continuous homomorphism. But how can we be sure that there are no others?
      $endgroup$
      – J.Doe
      Nov 24 '18 at 13:23


















    • $begingroup$
      I see that the defined map is a continuous homomorphism. But how can we be sure that there are no others?
      $endgroup$
      – J.Doe
      Nov 24 '18 at 13:23
















    $begingroup$
    I see that the defined map is a continuous homomorphism. But how can we be sure that there are no others?
    $endgroup$
    – J.Doe
    Nov 24 '18 at 13:23




    $begingroup$
    I see that the defined map is a continuous homomorphism. But how can we be sure that there are no others?
    $endgroup$
    – J.Doe
    Nov 24 '18 at 13:23











    1












    $begingroup$

    Multiplicative notation is not common in abelian group theory.



    We can use the fact that $mathbb{C}^*congmathbb{R}oplusmathbb{T}$, where $mathbb{T}=mathbb{R}/mathbb{Z}$, via
    $$
    (r,u+mathbb{Z})mapsto e^r(cos(2pi u)+isin(2pi u))
    $$
    Thus the group of endomorphisms of $mathbb{C}^*$ is the direct sum of
    $$DeclareMathOperator{Hom}{Hom}
    Hom(mathbb{R},mathbb{R})oplusHom(mathbb{R},mathbb{T})oplus
    Hom(mathbb{T},mathbb{T})oplusHom(mathbb{T},mathbb{R})
    $$



    The first two groups are big $mathbb{Q}$-vector spaces, the third one is a wild beast. If we consider the exact sequence $0tomathbb{Z}tomathbb{R}tomathbb{T}to0$, then we have
    $$
    0toHom(mathbb{T},mathbb{T})toHom(mathbb{R},mathbb{T})toHom(mathbb{Z},mathbb{T})to0
    $$
    and $Hom(mathbb{Z},mathbb{T})congmathbb{T}$. So $Hom(mathbb{T},mathbb{T})$ sits in a big $mathbb{Q}$-vector space so that the cokernel is $mathbb{T}$.



    What's $Hom(mathbb{T},mathbb{R})$? A similar sequence arises:
    $$
    0toHom(mathbb{T},mathbb{R})toHom(mathbb{R},mathbb{R})toHom(mathbb{Z},mathbb{R})to0
    $$



    As you see, the group $Hom(mathbb{C}^*,mathbb{C}^*)$ is much bigger than $mathbb{C}$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Multiplicative notation is not common in abelian group theory.



      We can use the fact that $mathbb{C}^*congmathbb{R}oplusmathbb{T}$, where $mathbb{T}=mathbb{R}/mathbb{Z}$, via
      $$
      (r,u+mathbb{Z})mapsto e^r(cos(2pi u)+isin(2pi u))
      $$
      Thus the group of endomorphisms of $mathbb{C}^*$ is the direct sum of
      $$DeclareMathOperator{Hom}{Hom}
      Hom(mathbb{R},mathbb{R})oplusHom(mathbb{R},mathbb{T})oplus
      Hom(mathbb{T},mathbb{T})oplusHom(mathbb{T},mathbb{R})
      $$



      The first two groups are big $mathbb{Q}$-vector spaces, the third one is a wild beast. If we consider the exact sequence $0tomathbb{Z}tomathbb{R}tomathbb{T}to0$, then we have
      $$
      0toHom(mathbb{T},mathbb{T})toHom(mathbb{R},mathbb{T})toHom(mathbb{Z},mathbb{T})to0
      $$
      and $Hom(mathbb{Z},mathbb{T})congmathbb{T}$. So $Hom(mathbb{T},mathbb{T})$ sits in a big $mathbb{Q}$-vector space so that the cokernel is $mathbb{T}$.



      What's $Hom(mathbb{T},mathbb{R})$? A similar sequence arises:
      $$
      0toHom(mathbb{T},mathbb{R})toHom(mathbb{R},mathbb{R})toHom(mathbb{Z},mathbb{R})to0
      $$



      As you see, the group $Hom(mathbb{C}^*,mathbb{C}^*)$ is much bigger than $mathbb{C}$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Multiplicative notation is not common in abelian group theory.



        We can use the fact that $mathbb{C}^*congmathbb{R}oplusmathbb{T}$, where $mathbb{T}=mathbb{R}/mathbb{Z}$, via
        $$
        (r,u+mathbb{Z})mapsto e^r(cos(2pi u)+isin(2pi u))
        $$
        Thus the group of endomorphisms of $mathbb{C}^*$ is the direct sum of
        $$DeclareMathOperator{Hom}{Hom}
        Hom(mathbb{R},mathbb{R})oplusHom(mathbb{R},mathbb{T})oplus
        Hom(mathbb{T},mathbb{T})oplusHom(mathbb{T},mathbb{R})
        $$



        The first two groups are big $mathbb{Q}$-vector spaces, the third one is a wild beast. If we consider the exact sequence $0tomathbb{Z}tomathbb{R}tomathbb{T}to0$, then we have
        $$
        0toHom(mathbb{T},mathbb{T})toHom(mathbb{R},mathbb{T})toHom(mathbb{Z},mathbb{T})to0
        $$
        and $Hom(mathbb{Z},mathbb{T})congmathbb{T}$. So $Hom(mathbb{T},mathbb{T})$ sits in a big $mathbb{Q}$-vector space so that the cokernel is $mathbb{T}$.



        What's $Hom(mathbb{T},mathbb{R})$? A similar sequence arises:
        $$
        0toHom(mathbb{T},mathbb{R})toHom(mathbb{R},mathbb{R})toHom(mathbb{Z},mathbb{R})to0
        $$



        As you see, the group $Hom(mathbb{C}^*,mathbb{C}^*)$ is much bigger than $mathbb{C}$.






        share|cite|improve this answer











        $endgroup$



        Multiplicative notation is not common in abelian group theory.



        We can use the fact that $mathbb{C}^*congmathbb{R}oplusmathbb{T}$, where $mathbb{T}=mathbb{R}/mathbb{Z}$, via
        $$
        (r,u+mathbb{Z})mapsto e^r(cos(2pi u)+isin(2pi u))
        $$
        Thus the group of endomorphisms of $mathbb{C}^*$ is the direct sum of
        $$DeclareMathOperator{Hom}{Hom}
        Hom(mathbb{R},mathbb{R})oplusHom(mathbb{R},mathbb{T})oplus
        Hom(mathbb{T},mathbb{T})oplusHom(mathbb{T},mathbb{R})
        $$



        The first two groups are big $mathbb{Q}$-vector spaces, the third one is a wild beast. If we consider the exact sequence $0tomathbb{Z}tomathbb{R}tomathbb{T}to0$, then we have
        $$
        0toHom(mathbb{T},mathbb{T})toHom(mathbb{R},mathbb{T})toHom(mathbb{Z},mathbb{T})to0
        $$
        and $Hom(mathbb{Z},mathbb{T})congmathbb{T}$. So $Hom(mathbb{T},mathbb{T})$ sits in a big $mathbb{Q}$-vector space so that the cokernel is $mathbb{T}$.



        What's $Hom(mathbb{T},mathbb{R})$? A similar sequence arises:
        $$
        0toHom(mathbb{T},mathbb{R})toHom(mathbb{R},mathbb{R})toHom(mathbb{Z},mathbb{R})to0
        $$



        As you see, the group $Hom(mathbb{C}^*,mathbb{C}^*)$ is much bigger than $mathbb{C}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 24 '18 at 12:20

























        answered Nov 24 '18 at 12:15









        egregegreg

        180k1485202




        180k1485202






























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