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I am trying to calculate:$$\epsilon_{ijk}n^iM1^j\epsilon^{lmk}n_lM2_m$$

Defining the vectors as:

M1={M1x,M1y,M1z}
M2={M2x,M2y,M2z}
n={nx,ny,nz}

and using

FullSimplify@ TensorContract[  TensorProduct[LeviCivitaTensor[3, List], n, M1, LeviCivitaTensor[3, List], n, M2], {{1, 4}, {2, 5}, {3, 8}, {6, 9}, {7, 10}}]

I get the result:

M1z M2z (nx^2 + ny^2) - M1z (M2x nx + M2y ny) nz +  M1y (-ny (M2x nx + M2z nz) + M2y (nx^2 + nz^2)) +  M1x (-M2y nx ny - M2z nx nz + M2x (ny^2 + nz^2))

This is correct, but I would rather that the result would be displayed as:

$$(M1\cdot M2)-(M1\cdot n)(M2\cdot n)$$

($n$ has norm one).

How do I get this simplification?

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  • $\begingroup$ Thanks. It works as shown now. $\endgroup$ – Michael E2 Jul 21 at 18:17
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In order to simplify the quadruple product, I would do like this

r=Sum[LeviCivitaTensor[3][[i,j,k]] n[i]M[1][j] 
      LeviCivitaTensor[3][[l,m,k]] n[l]M[2][m],
      {i,3},{j,3},{l,3},{m,3},{k,3}];

Simplify[r, Assumptions->n[1]^2+n[2]^2+n[3]^2==1
            && n[1] M[1][1]+n[2] M[1][2]+n[3] M[1][3]==Dot[n.M[1]]
            && n[1] M[2][1]+n[2] M[2][2]+n[3] M[2][3]==Dot[n.M[2]]
            && M[1][1] M[2][1]+M[1][2] M[2][2]+M[1][3] M[2][3]==Dot[M[1].M[2]]]

Out[1]= -n.M[1] n.M[2] + M[1].M[2]

Of course, all the assumptions can be generated programmatically.

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