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Quaternions in Clifford algebras

  1. Sep 7, 2010 #1
    Hello,
    it is known quaternions are isomorphic to [tex]\mathcal{C}\ell^{+}_{3,0}[/tex], which is the even subalgebra of [tex]\mathcal{C}\ell_{3,0}[/tex]

    Is it possible to find an isomorphism between [tex]\mathcal{C}\ell_{2,0}[/tex] and [tex]\mathbb{H} \cong \mathcal{C}\ell^{+}_{3,0}[/tex] ?

    *** my attempt was: ***

    Let's consider [tex]\{1,e_1, e_2, e_{12}\}[/tex] and the morphism f defined as follows:

    [tex]f(1)=1[/tex]

    [tex]f(e_{32})=e_1[/tex]

    [tex]f(e_{13})=e_2[/tex]

    [tex]f(e_{21})=e_{21}[/tex]

    This almost works, in fact:
    [itex]f(xy)=f(x)f(y)[/itex] always holds with one exception:

    [tex]f(e_1 e_2) = -e_{21} = f(e_2)f(e_1)[/tex]
    So, in this case, the morphism f does not preserve well the geometric product.
    Is it possible to make the whole thing work?
     
  2. jcsd
  3. Sep 7, 2010 #2
    What you have written doesn't quite make sense. Since you have not defined [tex]f(e_i)[/tex] for [tex]i=1,2[/tex].

    I think what you wanted was
    [tex]
    f(e_{12}) = -f(e_{21}) = -e_{21} = -e_2 e_1 = -f(e_{13})f(e_{32})
    =-f(e_{13}e_{32})=f(e_{12})
    [/tex]
    which is all as it should be.
     
  4. Sep 7, 2010 #3
    You got all right squares and all right anticommutators. So it works.
    And indeed , there is something wrong with your last formula. It is not clear what you were trying to do.
     
  5. Sep 7, 2010 #4
    thanks for both your replies.
    Maybe I now understood the mistake.

    Now, please tell me if this is the correct way to proceed:
    I want to find an isomorphism between the subalgebra of quaternions [tex]\mathcal{C}\ell^{+}_{3,0}[/tex] and the algebra [tex]\mathcal{C}\ell_{2,0}[/tex]

    I define [tex]f:\mathcal{C}\ell^{1}_{3,0} \rightarrow \mathcal{C}\ell^{1}_{2,0}[/tex] as:

    [tex]f(1)=1[/tex]
    [tex]f(e_1)=e_1[/tex]
    [tex]f(e_2)=e_2[/tex]
    [tex]f(e_3)=e_{12}[/tex]

    Now, I simply verify that the quaternion rules [tex]i^2 = j^2 = k^2 = ijk = -1[/tex], where [itex]i=e_{32}[/itex] [itex]j=e_{13}[/itex] [itex]k=e_{21}[/itex] correctly hold as follows [tex]f(i^2) = f(j^2) = f(k^2) = f(ijk) = f(-1)[/tex]

    Does this make more sense?
     
  6. Sep 7, 2010 #5
    Well, you need to show that [tex]f(e_i e_j)=f(e_i)f(e_j)[/tex].
    If you have that [tex]f(e_1 e_2 e_3)=f(e_1)f(e_2)f(e_3)[/tex] then [tex]f(e_1e_2)=f(e_3)[/tex]
    [tex]f(e_1)f(e_2)f(e_3)=-1,[/tex] therefore (multiply on the right by [tex]f(e_3)[/tex])
    [tex]f(e_1)f(e_2)=f(e_3)=f(e_1 e_2),[/tex]
    and you are home.
    If this is what you mean, then you are alright.
     
    Last edited: Sep 7, 2010
  7. Sep 7, 2010 #6
    What you did is a particular case of a general theorem:

    If the quadratic form [tex]q[/tex] on [tex]E[/tex] is non-zero, the even subalgebra [tex]C_+[/tex] of [tex]C(E;q)[/tex] can be, in a natural way, considered as the Clifford algebra [tex]C(E_1;q_1)[/tex] of the subspace [tex]E_1=e_1^{\perp}[/tex]
    of [tex]E[/tex], orthogonal to a regular vector [tex]e_1[/tex], where [tex]q_1=-q(e_1)q[/tex].

    For this you define [tex]f(y)=e_1 y,[/tex] [tex]y\in e_1^{\perp}[/tex].

    Then [tex]f(y)^2 = e_1 y e_1 y = -e_1^2 y^2 = -q(e_1)q(y).[/tex]

    P.S. I am using the convention [tex]C(E;q)= \mathcal{C} l (E,-q)[/tex]
     
    Last edited: Sep 7, 2010
  8. Sep 8, 2010 #7
    Sorry,
    I am a bit confused again now.
    After sleeping over it, What I wrote in my previous post doesn't sound correct anymore, because the quaternion units do not square to -1 in [tex]\mathcal{C}\ell_{2,0}[/tex].

    For example, let's take one quaternion unit in [tex]\mathcal{C}\ell_{3,0}[/tex]:

    [tex]i = e_{32}[/tex]

    We have that:

    [tex]f(e_{32}e_{32}) = f(-1) = -1 \neq f(e_{32})f(e_{32})[/tex]

    so f is not a morphism because it does not preserve the geometric-product.


    PS: @arkajad: could you please give me a reference for that general theorem (if possible)?
     
    Last edited: Sep 8, 2010
  9. Sep 8, 2010 #8
    Rene Deheuvels, "Tenseurs et Spineurs", Presses Universitaires de France, (1993), pp. 247-248 (Very good book. Good to have in the library.)

    I have scanned for you the Theorem, its proof, and and example
     

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  10. Sep 8, 2010 #9
    ... it seems to me that it is not possible to create an isomorphism between [tex]\mathcal{C}\ell_{2,0}[/tex] and the quaternion subalgebra [tex]\mathcal{C}\ell^{+}_{3,0}[/tex], unless one reverses the metric signature into:

    [tex]\mathcal{C}\ell_{0,2}[/tex]

    This is probably the only way to obtain [tex]e_i e_i = -1[/tex].
     
  11. Sep 8, 2010 #10
    Which convention are you using for Cl(q)?

    xy+yx = 2(x,y) or xy+yx = - 2(x,y)

    I should have asked this question first.
     
  12. Sep 8, 2010 #11
    I was assuming the first one: xy+yx = 2(x,y).
    Sorry for having skipped that.
     
  13. Sep 8, 2010 #12
    Then indeed you have an isomorphism of [tex]Cl^+(3,0)[/tex] with [tex]Cl(0,2)[/tex]

    P.S. I was too quick in my first replies!
     
    Last edited: Sep 8, 2010
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