# contravariant and covariant basis

My text book defines covariant (1) and contravariant (2) basis as follows. $$\epsilon_i=\frac {\partial x}{\partial q_i} \hat e_x + \frac {\partial y}{\partial q_i} \hat e_y + \frac {\partial z}{\partial q_i} \hat e_z --(1)$$ $$\epsilon^i=\frac {\partial q_i}{\partial x} \hat e_x + \frac {\partial q_i}{\partial y} \hat e_y + \frac {\partial q_i}{\partial z} \hat e_z--(2)$$ Then it says, using chain rule $$\epsilon^i.\epsilon_j = \frac {\partial q_i}{\partial x} \frac {\partial x}{\partial q_j} + \frac {\partial q_i}{\partial y} \frac {\partial y}{\partial q_j} + \frac {\partial q_i}{\partial z} \frac {\partial z}{\partial q_j} = \delta^i_j$$ When I see the last equation I think there should be a $3$ factor coming making it $3\delta^i_j$. Please help me to see what's wrong with my understanding. Thanks

$$\delta _{ij} = \frac{\partial q_i}{\partial q_j} = \frac {\partial q_i}{\partial x} \frac {\partial x}{\partial q_j} + \frac {\partial q_i}{\partial y} \frac {\partial y}{\partial q_j} + \frac {\partial q_i}{\partial z} \frac {\partial z}{\partial q_j} = \epsilon^i \cdot \epsilon_j$$
• Thanks, now this looks intuitive to me, could you explain a little why $\frac {\partial q_i}{\partial x} \frac {\partial x}{\partial q_j} = \delta^i_j$ is not correct. – levitt Feb 21 '15 at 12:23
• @levitt: For example, if $q_1= y, q_2 = x$. Then $\frac {\partial q_1}{\partial x} \frac {\partial x}{\partial q_1} =0$ – user99914 Feb 21 '15 at 22:38