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DifferentialGeometry

 DualBasis
 calculate the dual basis to a given basis of vectors or 1-forms

 Calling Sequence DualBasis(S, T)

Parameters

 S - a list of independent vectors or 1-forms T - (optional) a list of independent 1-forms if S is a list of vectors; a list of independent vectors if S is a list of 1-forms

Description

 • Let S = [X_1, X_2, ..., X_n] be a list of vectors, defined on a manifold M, which define a basis for the tangent space at a point p.  Then the dual basis for the cotangent space at p is the list of 1-forms B = [alpha_1, alpha_2, ..., alpha_n] such that alpha_i(X_j) = delta_ij = {0 if i <> j and 1 if i = j}. The command DualBasis(S) will return the list of 1-forms B.
 • Let S = [alpha_1, alpha_2, ..., alpha_n] be a list of 1-forms, defined on a manifold M, which define a basis for the cotangent space at a point p.  Then the dual basis for the tangent space at p is the list of vectors B = [X_1, X_2, ..., X_n] such that alpha_i(X_j) = delta_ij. The command DualBasis(S) will return the list of 1-forms B.
 • More generally, let S = [X_1, X_2, ..., X_k] be a list of independent vectors defined on a manifold M and let T = [theta_1, theta_2, ..., theta_k] be a list of independent 1-forms which are transverse to S in the sense that the k x k matrix A_ij = alpha_i(X_j) is non-singular.  In this case DualBasis(S, T) returns a list of 1-forms B = [alpha_1, alpha_2, ..., alpha_k] such that span(B) = span(T) and alpha_i(X_j)  = delta_ij.
 • This command is part of the DifferentialGeometry package, and so can be used in the form DualBasis(...) only after executing the command with(DifferentialGeometry).  It can always be used in the long form DifferentialGeometry:-DualBasis.

Examples

 > $\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):$

Initialize a 3-dimensional manifold M with coordinates [x, y, z].

 > $\mathrm{DGsetup}\left(\left[x,y,z\right],M\right):$

Example 1.

 > $\mathrm{S1}≔\left[\mathrm{D_x},\mathrm{D_y},\mathrm{D_z}\right]$
 ${\mathrm{S1}}{≔}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{\mathrm{D_z}}\right]$ (1)
 > $\mathrm{B1}≔\mathrm{DualBasis}\left(\mathrm{S1}\right)$
 ${\mathrm{B1}}{≔}\left[{\mathrm{dx}}{,}{\mathrm{dy}}{,}{\mathrm{dz}}\right]$ (2)

Example 2.

 > $\mathrm{S2}≔\mathrm{evalDG}\left(\left[\mathrm{D_x}+\mathrm{D_y}-\mathrm{D_z},\mathrm{D_y}-\mathrm{D_z},\mathrm{D_x}+\mathrm{D_y}+3\mathrm{D_z}\right]\right)$
 ${\mathrm{S2}}{≔}\left[{\mathrm{D_x}}{+}{\mathrm{D_y}}{-}{\mathrm{D_z}}{,}{\mathrm{D_y}}{-}{\mathrm{D_z}}{,}{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{3}{}{\mathrm{D_z}}\right]$ (3)
 > $\mathrm{B2}≔\mathrm{DualBasis}\left(\mathrm{S2}\right)$
 ${\mathrm{B2}}{≔}\left[{\mathrm{dx}}{-}\frac{{\mathrm{dy}}}{{4}}{-}\frac{{\mathrm{dz}}}{{4}}{,}{-}{\mathrm{dx}}{+}{\mathrm{dy}}{,}\frac{{\mathrm{dy}}}{{4}}{+}\frac{{\mathrm{dz}}}{{4}}\right]$ (4)

We check the answer by computing the interior products of S2[i] with B2[j].

 > $\mathrm{Matrix}\left(3,3,\left(i,j\right)→\mathrm{Hook}\left({\mathrm{S2}}_{i},{\mathrm{B2}}_{j}\right)\right)$
 $\left[\begin{array}{rrr}{1}& {0}& {0}\\ {0}& {1}& {0}\\ {0}& {0}& {1}\end{array}\right]$ (5)

Example 3.

The dual basis for the forms B2 from Example 2 are the vectors S2.

 > $\mathrm{B3}≔\mathrm{DualBasis}\left(\mathrm{B2}\right)$
 ${\mathrm{B3}}{≔}\left[{\mathrm{D_x}}{+}{\mathrm{D_y}}{-}{\mathrm{D_z}}{,}{\mathrm{D_y}}{-}{\mathrm{D_z}}{,}{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{3}{}{\mathrm{D_z}}\right]$ (6)

Example 4.

Calculate the dual basis to the vectors S3 relative to the subspace of 1-forms T3.

 > $\mathrm{S4}≔\mathrm{evalDG}\left(\left[\mathrm{D_x}+\mathrm{D_y}+\mathrm{D_z},\mathrm{D_x}-2\mathrm{D_y}+\mathrm{D_z}\right]\right)$
 ${\mathrm{S4}}{≔}\left[{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{\mathrm{D_x}}{-}\left({2}{}{\mathrm{D_y}}\right){+}{\mathrm{D_z}}\right]$ (7)
 > $\mathrm{T4}≔\mathrm{evalDG}\left(\left[\mathrm{dx}+\mathrm{dy}+\mathrm{dz},\mathrm{dx}-2\mathrm{dy}+3\mathrm{dz}\right]\right)$
 ${\mathrm{T4}}{≔}\left[{\mathrm{dx}}{+}{\mathrm{dy}}{+}{\mathrm{dz}}{,}{\mathrm{dx}}{-}\left({2}{}{\mathrm{dy}}\right){+}{3}{}{\mathrm{dz}}\right]$ (8)
 > $\mathrm{B4}≔\mathrm{DualBasis}\left(\mathrm{S4},\mathrm{T4}\right)$
 ${\mathrm{B4}}{≔}\left[\frac{{\mathrm{dx}}}{{3}}{+}\frac{{\mathrm{dy}}}{{3}}{+}\frac{{\mathrm{dz}}}{{3}}{,}\frac{{\mathrm{dx}}}{{24}}{-}\frac{{\mathrm{dy}}}{{3}}{+}\frac{{7}{}{\mathrm{dz}}}{{24}}\right]$ (9)
 > $\mathrm{Matrix}\left(2,2,\left(i,j\right)→\mathrm{Hook}\left({\mathrm{S4}}_{i},{\mathrm{B4}}_{j}\right)\right)$
 $\left[\begin{array}{rr}{1}& {0}\\ {0}& {1}\end{array}\right]$ (10)
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