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Question on Clifford Algebra

  1. May 27, 2012 #1
    I was trying to solve the following equation:

    [tex] \bigwedge\limits_{j=1}^{k}\begin{bmatrix}
    a_{1,j}\\
    a_{2,j}\\
    :\\
    .\\
    a_{k+1,j}
    \end{bmatrix} [/tex]

    Does anyone know how I can solve it? Thanks in advance.
     
  2. jcsd
  3. May 28, 2012 #2
    try k=1,2,3 etc.
     
  4. May 28, 2012 #3
    Actually, I already know what that expression results in. I was trying to prove it.
     
  5. May 28, 2012 #4
    There is no equation, just an expression. It is not clear what you are trying to solve.

    Please show us the result. Please tell us exactly what you are trying to prove.
     
  6. May 28, 2012 #5
    Ya I realised that later. Just made a typo error.



    [tex]\sqrt {\sum\limits_{j = 1}^{k + 1} {{{\det }^2}\left( {{{{\mathop{\rm cross}\nolimits} }_j}\left( {\mathop {\bf{A}}\limits_{(k + 1) \times k} } \right)} \right)} } [/tex]

    where A_(k+1)Xk is the matrix formed by augmenting the vectors together and the cross_j function means crossing out the jth row of the matrix A_(k+1)Xk.

    I'm trying to prove that [itex]\sqrt {\sum\limits_{j = 1}^{k + 1} {{{\det }^2}\left( {{{{\mathop{\rm cross}\nolimits} }_j}\left( {\mathop {\bf{A}}\limits_{(k + 1) \times k} } \right)} \right)} } [/itex] is the answer, since I found it by finding the special cases where k=1,2,3.
     
    Last edited: May 28, 2012
  7. May 28, 2012 #6
    I guess I know very little about clifford algebras, I was expecting a wedge not to be a scalar, which I am guessing is what the root of sum of determinants squared would give.
     
  8. May 28, 2012 #7
    Oops! Sorry, you are right. I mean the determinant of the exterior product is equal to [tex]\sqrt {\sum\limits_{j = 1}^{k + 1} {{{\det }^2}\left( {{{{\mathop{\rm cross}\nolimits} }_j}\left( {\mathop {\bf{A}}\limits_{(k + 1) \times k} } \right)} \right)} }[/tex]
     
  9. May 28, 2012 #8
    For k=2 I would have guessed the cross product.

    Ah
     
  10. May 28, 2012 #9
    No... I think it is the hodge dual of the cross product, actually. The wedge alone would be a bivector in that case.
     
  11. May 28, 2012 #10
    Show us the steps for k=2 and/or 3
     
  12. May 28, 2012 #11
    [tex]\begin{array}{l}
    \begin{array}{*{20}{l}}
    {\left\| {\left[ {\begin{array}{*{20}{l}}
    \alpha \\
    \gamma \\
    \varepsilon
    \end{array}} \right] \wedge \left[ {\begin{array}{*{20}{l}}
    \beta \\
    \delta \\
    \zeta
    \end{array}} \right]} \right\| = \left\| {\alpha \delta \left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \alpha \zeta \left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \gamma \beta \left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_1}} \right) + \gamma \zeta \left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right) + \varepsilon \beta \left( {{{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_1}} \right) + \varepsilon \delta \left( {{{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_2}} \right)} \right\|}\\
    { = \left\| {\left( {\alpha \delta - \beta \gamma } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \left( {\alpha \zeta - \beta \varepsilon } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \left( {\gamma \zeta - \delta \varepsilon } \right)\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right)} \right\|}\\
    { = \left\| {\det \left[ {\begin{array}{*{20}{c}}
    \alpha &\beta \\
    \gamma &\delta
    \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \det \left[ {\begin{array}{*{20}{c}}
    \alpha &\beta \\
    \varepsilon &\zeta
    \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \det \left[ {\begin{array}{*{20}{c}}
    \gamma &\delta \\
    \varepsilon &\zeta
    \end{array}} \right]\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right)} \right\|}
    \end{array}\\
    = \sqrt {{{\det }^2}\left[ {\begin{array}{*{20}{c}}
    \alpha &\beta \\
    \gamma &\delta
    \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}
    \alpha &\beta \\
    \varepsilon &\zeta
    \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}
    \gamma &\delta \\
    \varepsilon &\zeta
    \end{array}} \right]}
    \end{array}[/tex]

    [tex]\begin{array}{l}
    \left\| {\left[ \begin{array}{l}
    \alpha \\
    \delta \\
    \eta \\
    \kappa
    \end{array} \right] \wedge \left[ \begin{array}{l}
    \beta \\
    \varepsilon \\
    \theta \\
    \lambda
    \end{array} \right] \wedge \left[ \begin{array}{l}
    \gamma \\
    \zeta \\
    \iota \\
    \mu
    \end{array} \right]} \right\| = \left\| {\left( {\alpha {{{\bf{\hat e}}}_1} + \delta {{{\bf{\hat e}}}_2} + \eta {{{\bf{\hat e}}}_3} + \kappa {{{\bf{\hat e}}}_4}} \right) \wedge \left( {\beta {{{\bf{\hat e}}}_1} + \varepsilon {{{\bf{\hat e}}}_2} + \theta {{{\bf{\hat e}}}_3} + \lambda {{{\bf{\hat e}}}_4}} \right) \wedge \left( {\gamma {{{\bf{\hat e}}}_1} + \zeta {{{\bf{\hat e}}}_2} + \iota {{{\bf{\hat e}}}_3} + \mu {{{\bf{\hat e}}}_4}} \right)} \right\|\\
    {\rm{ }} = \left| {\left| {\left( {\alpha \varepsilon \iota - \delta \beta \iota - \alpha \theta \zeta + \eta \beta \zeta + \delta \theta \gamma - \eta \varepsilon \gamma } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right) + } \right.} \right.\\
    {\rm{ }}\left( {\alpha \varepsilon \mu - \delta \beta \mu - \alpha \lambda \zeta + \kappa \beta \zeta - \delta \lambda \gamma + \kappa \varepsilon \gamma } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_4}} \right) + \\
    {\rm{ }}\left( {\alpha \theta \mu - \eta \beta \mu - \alpha \lambda \iota + \kappa \beta \iota - \eta \lambda \gamma + \kappa \theta \gamma } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_4}} \right) + \\
    \left. {\left. {{\rm{ }}\left( {\delta \theta \mu - \eta \varepsilon \mu - \delta \lambda \iota + \kappa \varepsilon \iota - \eta \lambda \zeta + \kappa \theta \zeta } \right)\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_4}} \right)} \right|} \right|\\
    {\rm{ }} = \left| {\left| {\det \left[ {\begin{array}{*{20}{c}}
    \alpha &\beta &\gamma \\
    \delta &\varepsilon &\zeta \\
    \eta &\theta &\iota
    \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right) + } \right.} \right.\\
    {\rm{ }}\det \left[ {\begin{array}{*{20}{c}}
    \alpha &\beta &\gamma \\
    \delta &\varepsilon &\zeta \\
    \kappa &\lambda &\mu
    \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_4}} \right) + \\
    {\rm{ }}\det \left[ {\begin{array}{*{20}{c}}
    \alpha &\beta &\gamma \\
    \eta &\theta &\iota \\
    \kappa &\lambda &\mu
    \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_4}} \right) + \\
    \left. {\left. {{\rm{ }}\det \left[ {\begin{array}{*{20}{c}}
    \delta &\varepsilon &\zeta \\
    \eta &\theta &\iota \\
    \kappa &\lambda &\mu
    \end{array}} \right]\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_4}} \right)} \right|} \right|\\
    {\rm{ }} = \sqrt {{{\det }^2}\left[ {\begin{array}{*{20}{c}}
    \alpha &\beta &\gamma \\
    \delta &\varepsilon &\zeta \\
    \eta &\theta &\iota
    \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}
    \alpha &\beta &\gamma \\
    \delta &\varepsilon &\zeta \\
    \kappa &\lambda &\mu
    \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}
    \alpha &\beta &\gamma \\
    \eta &\theta &\iota \\
    \kappa &\lambda &\mu
    \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}
    \delta &\varepsilon &\zeta \\
    \eta &\theta &\iota \\
    \kappa &\lambda &\mu
    \end{array}} \right]}
    \end{array}[/tex]

    But when I tried it for the general case, it was not possible.
     
  13. May 28, 2012 #12
    maybe you could somehow argue that the coefficient of e_1^...^e_{j-1}^e_{j+1}^...^e_{k+1} would be the det of cross_j(A).
     
  14. May 28, 2012 #13
    Thanks a lot! I think that you're right! Thanks again!
     
  15. Mar 28, 2014 #14
    Basic Geometric Algebra

    See attached file DET.pdf. The thing to note is that in the case of an orthogonal basis e1,...,en we have e1^...^en = e1...en (wedge product is the same as the geometric/Clifford product).

    Detailed note based on "Geometric Algebra for Physicists" by Doran and Laseby is at

    https://github.com/brombo/GA [Broken]
     

    Attached Files:

    • DET.pdf
      DET.pdf
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    Last edited by a moderator: May 6, 2017
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