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Ivancevic_Applied-Diff-Geom

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<strong>Applied</strong> Bundle <strong>Geom</strong>etry 673form Q. Then any vector v ∈ V may be written as v = v k e k and theClifford algebra isC(R 4 ) = span{ c n11 cn2 2 cn3 3 cn4 4 | n i ∈ {0, 1} }, where c i = q( e i ).For convenience, we may denote c i by e i without ambiguity. Especially,C 2 (R 4 ) = span{ e 1 e 2 , e 1 e 3 , e 1 e 4 , e 2 e 3 , e 2 e 4 , e 3 e 4 }Now consider the isomorphism τ : C 2 (R 4 ) −→ so(4) = so(R 4 ). And thecorresponding τ : Spin(4) −→ so(4). We haveτ(e i e j ) · v = [ e i e j ,==4∑v k e k ] =k=14∑v k [ e i e j , e k ] =k=14∑v k ( e i e j e k − e k e i e j )k=14∑v k ( e i e j e k + e i e k e j − e i e k e j − e k e i e j )k=14∑v k ( −2 Q( e j , e k ) e i + 2 Q( e k , e i ) e j ) = −2 v j e i + 2 v i e j .k=1Hence τ ( e i e j ) corresponds to 2 · m(i, j) ∈ so(4) where m(i, j) =(m(i, j) αβ ) is a matrix with entriesExplicitly, we have⎧⎪⎨ 1, if α = j and β = i,m(i, j) αβ = −1, if α = i and β = j,⎪⎩ 0, otherwise.⎛ ⎞⎛ ⎞0 −1 0 00 0 −1 0τ( e 1 e 2 ) = 2 ⎜1 0 0 0⎟⎝0 0 0 0⎠ τ( e 1 e 3 ) = 2 ⎜0 0 0 0⎟⎝1 0 0 0⎠0 0 0 00 0 0 0⎛ ⎞⎛ ⎞0 0 0 −10 0 0 0τ( e 1 e 4 ) = 2 ⎜0 0 0 0⎟⎝0 0 0 0 ⎠ τ( e 2 e 3 ) = 2 ⎜0 0 −1 0⎟⎝0 1 0 0⎠1 0 0 00 0 0 0

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