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I Groups of Automorphisms - Aut(C) ...

  1. Jul 2, 2017 #1
    I am reading Anderson and Feil - A First Course in Abstract Algebra.

    I am currently focused on Ch. 24: Abstract Groups ... ...

    I need some help in understanding some claims in Chapter 24 by Anderson and Feil ... ...


    Anderson and Feil claim that ##\text{Aut} ( \mathbb{C} )## is a group with only two elements ##\{i, f \}## ... ... where ##i## is the identity automorphism and ##f## is the complex conjugation map defined by ##f(a + bi) = a - bi## ... ...

    ... can someone please help me to prove the assertion that ##\text{Aut} ( \mathbb{C} )## is a group with only two elements ##\{i, f \}## ...

    Peter
     
  2. jcsd
  3. Jul 2, 2017 #2

    Orodruin

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    For a general automorphism ##h##, you need to preserve the structure of ##\mathbb C## so ##h(zw) = h(z) h(w)## and ##h(z+w) = h(z)+h(w)##. From the second of those relations follows that ##0 = h(0) = h(1-1) = h(1)-h(-1) = 1-h(-1)## and so ##h(-1) = -1##. From the first of the relations follows that ##h(-1) = h(i^2) = h(i)^2## and so there are two possibilities, either ##h(i) = i## or ##h(i) = -i##. In general, taking a complex number ##z = x + iy##, you would now have ##h(z) = h(x+iy) = h(x) \pm i h(y) = x \pm iy##. The choice ##+## gives you the trivial automorphism and the choice ##-## gives you the complex conjugation.
     
  4. Jul 2, 2017 #3
    @Orodruin's proof works if you also require your automorphisms to restrict to the identity on ##\mathbb{R}## (used in the step ##h(x)=x,h(y)=y##) but in general there are other automorphisms of ##\mathbb{C}##. The construction requires a bit of field theory but I'll give a link: http://math.ucr.edu/~res/progeom/oldversions/pgnotesappd.pdf
     
  5. Jul 2, 2017 #4
    Thanks Orodruin and Infrared ... really appreciate your help in this matter ...

    But I still need further help ... Orodruin has produces a convincing proof that there are only two automorphisms of ##\mathbb{C}## ... but the link supplied by Infrared implies that this is the case only when the mappings are also continuous ... but Anderson and Feil say nothing of the need for the automorphisms to be continuous ... but the author of the link goes on to argue that if we do not demand that the automorphisms are continuous then the automorphism group of ##\mathbb{C}## is huge ....


    Can someone please clarify this situation ...



    The key passage from the link supplied by Infrared is as follows:

    upload_2017-7-3_13-31-53.png


    Peter
     

    Attached Files:

  6. Jul 2, 2017 #5
    As I remarked in my last post, Oroduin's proof doesn't work since there is no reason to assume ##h(x)=x,h(y)=y##.

    Any automorphism ##f:\mathbb{C}\to\mathbb{C}## must restrict to the identity on ##\mathbb{Q}## and take ##i## to either ##i## or ##-i##. Hence, the restriction of ##f## to ##\mathbb{Q}(i)## is either the identity function or complex conjugation. In the case ##f## is continuous, this determines [itex]f[/itex] since [itex]\mathbb{Q}(i)[/itex] is dense in ##\mathbb{C}## and so your ##f,i## are the only two continuous automorphisms of ##\mathbb{C}##. However, if we don't require ##f## to be continuous, then it is not necessarily determined by its behavior on ##\mathbb{Q}(i)## and in fact there are many possible automorphisms as my link shows.
     
  7. Jul 2, 2017 #6
    HI Orodruin and Infrared ...

    I thought I would try to clarify exactly what Anderson and Feil had to say about this matter ...

    I am beginning to understand that whether the automorphisms of ##\Bbb C## fix ##\Bbb R## may be an/the issue ... but from my reading of Anderson and Feil they seem to imply that the the automorphisms of ##\Bbb C## actually fix ##\mathbb{R}## anyway ... but you seem to be implying that this is not necessarily the case ...

    The relevant text from Anderson and Feil is as follows:


    First the example that started me thinking:


    ?temp_hash=074f1eb25039629f658ef13a88ba37ed.png




    Now the above example involves a Galois Group ... and in this group ##\mathbb{R}## is certainly fixed under Anderson and Feil's definition of a Galois Group ... which reads as follows:



    ?temp_hash=074f1eb25039629f658ef13a88ba37ed.png
    ?temp_hash=074f1eb25039629f658ef13a88ba37ed.png



    Now, Example 47.3 refers to Exercise 24.14 which asserts that ##\text{Aut} ( \mathbb{C} )## has only two elements ... note the exercise does not mandate the fixing of ## \ \mathbb{R} ## ... Exercise 24.14 reads as follows:



    ?temp_hash=074f1eb25039629f658ef13a88ba37ed.png



    Hope that clarifies what Anderson and Feil have to say about the issue ... ...

    I have been alerted that a new post has come in from Infrared ... will now study that post ...

    Peter
     

    Attached Files:

  8. Jul 2, 2017 #7

    Hi Infrared ...

    Still thinking over your post ...


    BUT ... just a worry ...

    You write:

    " ... ... However, if we don't require ##f## to be continuous, then it is not necessarily determined by its behavior on ##\mathbb{Q}(i)## and in fact there are many possible automorphisms as my link shows. ... ... "

    Now, Anderson and Feil in Exercise 24.14 state there are only two automorphisms of ##\mathbb{C}## ... but they do not say that these automorphisms have to be continuous ... ...

    Exercise 24.14 reads as follows:



    ?temp_hash=8a6bdbd982b283e3f1dd7a618b24dc9d.png


    Are Anderson and Feil wrong or, at least, missing a condition on the automorphisms they are referring to ... ..


    Peter
     

    Attached Files:

  9. Jul 3, 2017 #8
    Yes, this looks wrong to me.
     
  10. Jul 3, 2017 #9

    Hi Orodruin and Infrared ...

    Can now see ... thanks to Infrared that there seems to be a problem with asserting that :

    ##h(z) = h(x+iy) = h(x) \pm i h(y) = x \pm iy##.

    since it assumes that for all ##x, y \in \mathbb{R}## we have that ##h(x) = x## and ##h(y) = y##


    So maybe to get the result that ##\text{Aut} ( \mathbb{C} )## has only/exactly two elements ... we need to assume that the automorphisms are fixed on ##\mathbb{R}## ... or alternatively that the automorphisms are continuous ... ...

    Peter
     
    Last edited: Jul 3, 2017
  11. Jul 3, 2017 #10
    Thanks Infrared ...

    Peter
     
  12. Jul 3, 2017 #11
    I think this isn't what you mean to say. We know from a previous thread of yours that ##\text{Aut}(\mathbb{R})## is trivial. But here an automorphism of ##\mathbb{C}## might take real values to non-real values.

    At any rate, I'm going to bed now.
     
  13. Jul 3, 2017 #12

    Oh sorry, Infrared ... it is a typo ...

    Meant to say:

    "... ... So maybe to get the result that ##\text{Aut} ( \mathbb{C} )## has only/exactly two elements ... we need to assume that the automorphisms are fixed on ##\mathbb{R}## ... or alternatively that the automorphisms are continuous ... ... "

    Peter

    *** EDIT ***

    (1) Thanks for all your help om this matter ...

    (2) I have altered the previous post so it is (should be ...) now correct ...
     
  14. Jul 3, 2017 #13

    WWGD

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    I think @Orodruin addressed this in his post, didn't he?
     
  15. Jul 3, 2017 #14

    Orodruin

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    Addressed what? I willingly admit that I took the physicist approach of sweeping the continuity assumption under the carpet. :rolleyes:
     
  16. Jul 3, 2017 #15

    WWGD

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    Ah, sorry about that :(.
     
  17. Jul 3, 2017 #16

    Orodruin

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    You are sorry that I was sloppy? Don't be, it's on me. :smile:
     
  18. Jul 3, 2017 #17

    WWGD

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    All I remember vaguely here is using ultraproducts and ultrafilters. " Ultraproducts, America's new Supermodels"..
     
  19. Jul 3, 2017 #18
    Hi Orodruin ...

    Just wondered if you agree that an alternative to the continuity assumption, is assuming that the automorphisms we are concerned with are fixed on ##\mathbb{R}## ... is that correct? ... (no idea why this assumption might be equivalent ... but it seems to be ...)

    Peter
     
  20. Jul 3, 2017 #19
    I addressed this in my post #5, but to be more explicit: we know that any automorphism ##\phi:\mathbb{C}\to\mathbb{C}## fixes ##\mathbb{Q}##. If in addition ##\phi## is continuous, then ##\phi## must fix ##\mathbb{R}## since ##\mathbb{Q}## is dense in ##\mathbb{R}##.

    Conversely, if ##\phi## fixes ##\mathbb{R}##, then it is either the identity or complex conjugation (by @Orodruin's first post), both of which are continuous.
     
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