Left-invariant tensor fields
We work with the Lie group defined in the previous section.
We confirm that the Lie algebra of our group G is determined by the commutator
Denote with ∈G the canonical basis vectors for the tangent space G. generates a left-invariant Killing vector field on G. Locally the vector fields induced by and are
Because [,]=, we have just confirmed the entry [,]= in ad commutator table above!
Note, the function GTL transforms a tensor of arbitrary rank to a left-invariant tensor field. The integer r, which denotes the rank of the (r,s)-tensor, is made input to the function GTL as well. So, what is behind GTL?
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GTL is just a change of basis! In order to compute the (r,s)-tensor with respect to the new basis, it uses the command Τα. The basis transformation is determined by GdL and depends the coordinates {,,}.
To obtain a left-invariant metric, i.e. a (0,2)-tensor field with special properties, we apply GΤL to a scalarproduct on G. Recall, a scalar product is a (0,2)-tensor. As a bilinear form, let the scalar product be defined by the symmetric matric B with respect to the canonic basis.
Remember, this is only an example. Any other symmetric matric with det≠0, will do. We show the resulting metric g, and the metric, scaled by the common denominator.
To check, that the metric is indeed left-invariant, we simple input the left-invariant vector field we have yielded above. The outcome is constant and equals the generating vectors scalar producted by B.
What is the left-invariant (1,2)-tensor field induced by the (1,2)-tensor ad? GΤL gives the answer. The Jacobi identity is satisfied at any coordinate point.
Created by Mathematica (September 30, 2006) |