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Prove GL(R) is not isomorphic to GL(C)

  1. May 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove GL(R) is not isomorphic to GL(C)

    2. Relevant equations



    3. The attempt at a solution

    Well I don't really have a good grasp of the issue at hand as this is another appendix problem, but my thoughts are I think GL(R) is a subgroup of GL(C) so maybe I could show they don't have the same cardinality? Would that be enough? Been a while since I did algebra... Not sure how I would would show that either.
     
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  3. May 15, 2013 #2

    micromass

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    Do you know why ##\mathbb{R}\setminus \{0\}## is not isomorphic to ##\mathbb{C}\setminus \{0\}##? Try to generalize.
     
  4. May 15, 2013 #3
    Umm [tex] f(I)=f(I^2)=f(I)^2[/tex] implies f(I) is the identity in GL(C). [STRIKE]But [tex]f(-I)^2=f(I)[/tex] so f(-I) is also the identity in GL(C). [/STRIKE]

    Is that how to do it?

    Errr. Flaw in that logic. meh.
     
    Last edited: May 15, 2013
  5. May 20, 2013 #4
    Okay here's what I got. Comments are appreciated.

    Suppose there exists an isomorphism [itex] \phi : \text{GL}(\mathbb{C}) \rightarrow \text{GL}(\mathbb{R})[/itex]. Then [itex]\phi(\text{I})=\text{I}[/itex]. We also have [itex]\phi (-\text{I})^2=\phi((-\text{I})^2)=\phi(\text{I})=\text{I}[/itex]. This implies [itex]\phi(-\text{I})=\pm \text{I}[/itex]. Since [itex]\phi(i\text{I})^2=\phi(-\text{I})[/itex], and [itex]\phi(i\text{I})[/itex] is a matrix of real numbers, it cannot be that [itex]\phi(-\text{I})=-\text{I}[/itex]. Hence, [itex]\phi(-\text{I})=\text{I}[/itex], which shows that \[itex]\phi [/itex] is not injective.
     
  6. May 20, 2013 #5

    micromass

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    Why does ##A^2 = I## imply that ##A=\pm I##? And why can ##A^2 = -I## not occur?
     
  7. May 20, 2013 #6

    Because the square root of I is ±I right?



    Because real numbers squared cannot be negative 1.

    Right?
     
  8. May 20, 2013 #7

    micromass

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    You seem to assume that these things hold for matrices because they hold for real numbers. You need to prove whether they also hold for arbitrary matrices.
     
  9. May 20, 2013 #8

    Office_Shredder

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  10. May 21, 2013 #9
    Alas!

    Well... Do they hold? Seems like they don't based on the link Office_Shredder provided.

    Your original direction was to show it holds for numbers, so if you pretend I wrote 1's instead of I's, that's my logic. Now I just need to get this stuff to generalize. Did you have any hint on how to do that?
     
  11. May 21, 2013 #10

    micromass

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    Maybe you can find the centers of both groups?
     
  12. May 21, 2013 #11
    That would be any scalar (except for 0) times the identity matrix?
     
  13. May 21, 2013 #12

    micromass

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    Yes. Are the centers isomorphic?
     
  14. May 21, 2013 #13
    I would guess not. I would like to say because of my proof, but I know I'm missing something.
     
  15. May 21, 2013 #14

    I like Serena

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    Following up on mm's suggestion, can you think of a group that is isomorphic with a real non-zero scalar times the identity matrix?
    How about a group that is isomorphic with a complex non-zero scalar times the identity matrix?


    I also have an alternative solution method: what you can say about the eigenvalues of the matrix A if ##A^2=I##?


    Btw, I'm somewhat confused about the meaning of ##GL(\mathbb R)##.
    What does it mean?
    I suspect you do not mean what wiki says about the General linear group.
     
  16. May 23, 2013 #15
    Reals minus 0 with multiplication?

    Either nxn invertible matrices or infinite ones, not sure what my professor intended. If I can get it figured out for nxn I'm confident it will generalize.
     
  17. May 23, 2013 #16

    I like Serena

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    Yep.

    Now consider this previous comment of mm...
     
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