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Clifford algebra in higher dimensions

  1. Jul 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider gamma matrices ##\gamma^0, \gamma^1, \gamma^2, \gamma^3## in 4-dimension. These gamma matrices satisfy the anti-commutation relation
    $$
    \{ \gamma^\mu , \gamma^\nu \}=2\eta^{\mu \nu}.~~~(\eta^{\mu\nu}=diag(+1,-1,-1,-1))
    $$
    Define ##\Gamma^{0\pm}, \Gamma^{1\pm}## as follows:
    \begin{align}
    \Gamma^{0\pm}:=\dfrac12 (\gamma^0 \pm \gamma^1),\\
    \Gamma^{1\pm}:=\dfrac12 (i\gamma^2 \pm \gamma^3).
    \end{align}

    They satisfy following anti-commutation relations.
    $$
    \{ \Gamma^{a+}, \Gamma^{b-} \}=\delta^{ab},
    \{ \Gamma^{a+}, \Gamma^{b+} \} = \{ \Gamma^{a-}, \Gamma^{b-} \}=0.
    ~~(a,b=0,1)
    $$

    Now, we can construct other Clifford algebra in higher dimensions.

    Question: Construct Clifford algebra for 1+10 dimension and 2+10 dimension as we did in above.

    2. Relevant equations


    3. The attempt at a solution
    The reason this problem is difficult for me is that 1+10 dimension is 11-dimension space-time and they are odd number dimension. If it were even number dimension, then it is easy to construct Clifford algebra; we only have to add new terms such like
    $$
    \Gamma^{2\pm}=\dfrac12 (i\gamma^3 \pm \gamma^4).
    $$
    But in 11-dimension, we can't do this. I tried to think in 3-dimension, but I couldn't figure out how to construct Clifford algebra.

    Thanks.
     
  2. jcsd
  3. Jul 26, 2017 #2

    fresh_42

    Staff: Mentor

    I don't really understand the problem. Firstly, the pairing which leads to the ##\Gamma^a## is simply a basis transformation of the ##\gamma^a##, so we don't have do bother them. Now ##\eta## is all that counts for the definition, since this is the bilinear form. So why isn't ##\eta = \operatorname{diag}(+1,-1,\ldots ,-1)## the solution with as many ##-1## as desired?

    The grading can be done for all Clifford algebras. All you need is an involution. The difference between the given ##(1,3)## and the requested ##(2,10)## example is bigger than between ##(1,3)## and ##(1,10)##.
     
  4. Jul 26, 2017 #3


    I think I should have noted the motivation of this problem. We want to think about a generalization of 4D Clifford algebra. For example, in ##1+3## dimension, we can set ##\Gamma##s as
    \begin{align}
    \Gamma^{0\pm}:=\dfrac12 (\gamma^0 \pm \gamma^1),\\
    \Gamma^{1\pm}:=\dfrac12 (i\gamma^2 \pm \gamma^3).
    \end{align}

    The anti-commutation relations are
    $$
    \{ \Gamma^{a+}, \Gamma^{b-} \}=\delta^{ab},
    \{ \Gamma^{a+}, \Gamma^{b+} \} = \{ \Gamma^{a-}, \Gamma^{b-} \}=0.
    ~~(a,b=0,1)
    $$

    These ##\Gamma^{0+}## and ##\Gamma^{1+}## are kind of raising operators, and ##\Gamma^{0-}## and ##\Gamma^{1-}## are lowering operators. Since the anti-commutation relations are similar to that of spinor fields', we could define a vacuum state ##\Omega## and set
    $$
    \Gamma^{a-}\Omega = 0.
    $$

    Then, states like ##\Gamma^{a+}\Omega## can be thought as a one-particle state.

    If we construct Clifford algebra by using these ##\Gamma##s, we can think of Clifford algebra in various dimensions such like 1+10 or 2+10.
    This is used in string theory.
     
  5. Jul 26, 2017 #4

    fresh_42

    Staff: Mentor

    I know. But it doesn't change my answer. You can extend ##v \longmapsto -v## to an involution ##\varphi## of ##\mathcal{Cl}(V,\eta)## and the grading is then ##\mathcal{Cl}^{(i)}(V,\eta)= \operatorname{im}(\operatorname{id} \pm \varphi)##. With that you can think about an appropriate basis.
     
  6. Jul 29, 2017 #5
    Thanks.
    I will do my best.
     
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