Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Isomorphism between Clifford algebras CL(4,2) and CL(2,4)

  1. Nov 24, 2014 #1
    Hi,

    I was reading a paragraph of a book (you can find it here) where the author seems to suggest that the Clifford algebras [itex]\mathcal{C}\ell_{2,4}(\mathbb{R})[/itex] and [itex]\mathcal{C}\ell_{4,2}(\mathbb{R})[/itex] are isomorphic. In particular, at the third line after Equation (10.190), when he talks about the algebra with signature (4,2), the authors says:

    "The signature shows that this space is isomorphic to the conformal algebra of spacetime"

    where by "conformal algebra of spacetime" he means the algebra with signature (2,4).

    How did he manage to deduce just from the signatures of these two algebras that they are indeed isomorphic?
    I am not convinced of this statement, and I am wondering if there is a quick way to prove it (that I am missing).

    Thanks!
     
  2. jcsd
  3. Nov 24, 2014 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    The Clifford algebra is generated by matrices ##\Gamma_\mu## satisfying

    $$\{ \Gamma_\mu, \Gamma_\nu \} = 2 \eta_{\mu\nu}$$
    To flip ##\eta_{\mu\nu} \to - \eta_{\mu\nu}##, just send ##\Gamma_\mu \to i \Gamma_\mu##.
     
  4. Nov 24, 2014 #3
    Hi Ben,

    thanks a lot for answering, though I must confess that I didn't understand anything of what you wrote.
    In particular, I don't know what [itex]\Gamma[/itex] and [itex]\eta[/itex] stand for, and I have never seen the notation [itex]\{A,B\}[/itex]. Could you please clarify this?
    Thanks.
     
  5. Nov 26, 2014 #4
    Sorry, but I don't think my original question received an answer.

    Assuming that in Ben's first equation the curly braces denote the anti-commutator, and assuming that η in the right-term denotes the metric tensor (btw the identity matrix next to η is missing), then I don't see how the trick of sending ##\Gamma_\mu \to i \Gamma_\mu## could automatically prove that we have created a Clifford algebra isomorphism.

    It seems to me that we could apply, for instance, the same reasoning to the generating elements of ##\mathcal{C}\ell_{2,0}(\mathbb{R})## and prove that: ##\mathcal{C}\ell_{2,0}(\mathbb{R})\simeq \mathcal{C}\ell_{0,2}(\mathbb{R}) \simeq \mathbb{H}## which is clearly not true.

    I hope someone can answer my original question.
     
  6. Nov 27, 2014 #5

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Good points. Since the map I've written involves multiplication by ##i##, it only establishes the isomorphism of the algebras over ##\mathbb{C}##, not over ##\mathbb{R}##. But I think for any practical purpose, that is enough. I know it makes no physical difference whether you work in a Clifford algebra of signature (p,q) or of (q,p). It should be obvious that ##SO(p,q) \simeq SO(q,p)##, so either Clifford algebra can be used to construct 1/2-integer representations.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Isomorphism between Clifford algebras CL(4,2) and CL(2,4)
Loading...