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.