Solving Clifford Algebra Equation - Need Help!

dimension10 · May 27, 2012

I was trying to solve the following equation:

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

Does anyone know how I can solve it? Thanks in advance.

algebrat · May 28, 2012

try k=1,2,3 etc.

dimension10 · May 28, 2012

algebrat said:

try k=1,2,3 etc.

Actually, I already know what that expression results in. I was trying to prove it.

algebrat · May 28, 2012

dimension10 said:

I was trying to solve the following equation:

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

Does anyone know how I can solve it? Thanks in advance.

There is no equation, just an expression. It is not clear what you are trying to solve.

dimension10 said:

Actually, I already know what that expression results in. I was trying to prove it.

Please show us the result. Please tell us exactly what you are trying to prove.

dimension10 · May 28, 2012

algebrat said:

There is no equation, just an expression. It is not clear what you are trying to solve.

Ya I realized that later. Just made a typo error.

algebrat said:

Please show us the result. Please tell us exactly what you are trying to prove.

\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)} }

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

I'm trying to prove that \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)} } is the answer, since I found it by finding the special cases where k=1,2,3.

algebrat · May 28, 2012

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.

dimension10 · May 28, 2012

algebrat said:

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.

Oops! Sorry, you are right. I mean the determinant of the exterior product is equal to \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)} }

algebrat · May 28, 2012

For k=2 I would have guessed the cross product.

Ah

dimension10 · May 28, 2012

algebrat said:

For k=2 I would have guessed the cross product.

Ah

No... I think it is the hodge dual of the cross product, actually. The wedge alone would be a bivector in that case.

algebrat · May 28, 2012

Show us the steps for k=2 and/or 3

dimension10 · May 28, 2012

algebrat said:

Show us the steps for k=2 and/or 3

\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}

\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}

But when I tried it for the general case, it was not possible.

algebrat · May 28, 2012

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).

dimension10 · May 28, 2012

algebrat said:

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).

Thanks a lot! I think that you're right! Thanks again!

brombo · Mar 28, 2014

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

Solving Clifford Algebra Equation - Need Help!

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