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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

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By chain rule (the second equality),

$$\delta _{ij} = \frac{\partial q_i}{\partial q_j} = \frac {\partial q_i}{\partial x} \frac {\partial x}{\partial q_j} + \frac {\partial q_i}{\partial y} \frac {\partial y}{\partial q_j} + \frac {\partial q_i}{\partial z} \frac {\partial z}{\partial q_j} = \epsilon^i \cdot \epsilon_j $$

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  • $\begingroup$ 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. $\endgroup$ – levitt Feb 21 '15 at 12:23
  • $\begingroup$ @levitt: For example, if $q_1= y, q_2 = x$. Then $\frac {\partial q_1}{\partial x} \frac {\partial x}{\partial q_1} =0$ $\endgroup$ – user99914 Feb 21 '15 at 22:38

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