- Vectors and Dual Vectors

1. Deriving Orthonormality of Basis and Dual Basis Through the Definition of Vector and Dual Vector Properties #

In the textbook “An Introduction to Tensors and Group Theory for Physicists” that I am reading, the following definitions are given: $$ e^i(v) = v^i, \quad f(e_i) = f_i $$

Through this definition, the text explains that the following property is satisfied: $$ e^j(e_i) = \delta_i^j $$

2. Linearity of Functions #

However, using the following property (definition) of functions, we can simplify this explanation: $$ (c\ g_1 + g_2)(x) = c\ g_1(x) + g_2(x) $$

3. Deriving Vector and Dual Vector Properties Through the Orthonormality of Basis and Dual Basis #

When we denote a vector in $V^{\ast}$ as $f$, it is a linear functional. Therefore, if we denote the basis vectors of $V^{\ast}$ as $e^j$ and define: $$ e^j(e_i) = \delta_i^j $$

Then we get: $$ f = f_j e^j, \quad f(e_i) = f_j e^j(e_i) = f_j \delta_i^j = f_i $$

Additionally, for a vector $v = v^j e_j$ in $V$: $$ e^i(v) = e^i(v^j e_j) $$

Since the dual vector $e^i$ is a linear functional: $$ e^i(v) = e^i(v^j e_j) = v^j e^i(e_j) = v^j \delta_j^i = v^i $$ This relation can be demonstrated.

Furthermore, when we input $e_i$ into the dual vector $f$: $$ f(e_i) = (f_j e^j)(e_i) = f_j e^j(e_i) = f_j \delta_i^j = f_i $$ In other words, it is sufficient to define only the basis of the dual space.