Predrag,
let me try to explain
V is a finite =d dimensional vector space over a field R or C or Q or ..
its elements are vectors and given a choice of basis vectors
{\bfe_1 ... e_d} every vector {\bf X} has components X^1 X^d
Note my choice of indices
upstairs for coordinates and downstairs for basis vectors
{\bf X} is the geometrical thing it is basis independent upon a linear GLd
change of basis if the column of X^i's transforms by left multiplication
as X^i--> X'^i'=L^i'_j X^j or X'=LX then the row of basis
vectors transforms as {\bf e'}=L^{-1}{\bf e} or better the column of basis
vectors
transforms as {\bf e'^T}=L^{-1T} {\bf e^T}. The matrix =L^{-1T} ie the
transpose of
the inverse of L is called the contragredient of L (actually an old
reference I believe is the book by Weyl on quantum mechanics and group
theory)
If V is called the vector representation space the covector
representation space is the dual space V* namely the set of linear forms
on V over the field of choice R say. In other words {\bf f} a basis vector
in V*
like any element of V* is a linear map from V to R. It is given by its
values on the {\bf e_i}'s and the dual basis (f) of the basis (e) is
defined by{\bf f^i(e_j)} =\delta^i_j
Note I have been extra careful with indices. the Kronecker symbol is an
invariant of GL(d,R) no metric there
In coordinates you might prefer to use Y_i so called covariant coordinates
if the X's are called contravariant. The Y's transform with the
contragredient representation L^{-1T}.
So * here has no complex conjugation implied You may prefer to use
Cech instead
Finally the Kronecker delta is the invariant one dimensional
representation inside the (V tensor product V*) representation
namely the trace. It is indeed invariant under the action of L on the
left and L^{-1} on the right
hope it helps
Bernard L. JULIA ENS-CNRS/LPT
24 rue Lhomond
75005 Paris FRANCE
Tel. (01)47077146. FAX. (01)43367666.