I'm having trouble understanding the covariance and contravariance of a vector. So I watched this video and after the first 5 mins, this is how I understand it: contravariance and covariance are just two ways of identifying a vector.
- Contravariance means that the components change inversely to the change of the basis (i.e. if the basis vectors $e_1, e_2, e_3$ increase to $(2,0,0),(0,2,0),(0,0,2)$ then a vector $\vec{v}=(2,4,6)$ will now be decreasely displayed as $(1,2,3)$)
- Covariance means the covariant components change alike to the change of the basis. In the video, the covariant components is obtained by taking the dot product of $\vec{v}$ with the basis:$\vec{v}=(\vec{v} \cdot e_1,\vec{v} \cdot e_2,\vec{v} \cdot e_3)$. So if the basis vectors $e_1, e_2, e_3$ increase to $(2,0,0),(0,2,0),(0,0,2)$ then a vector $\vec{v}=(2,4,6)$ will now be increased to $\vec{v}=(4,8,12)$
However, if I look at this example in the Wikipedia page, the dual basis for the covariance of a vector $\vec{v}$ is obtained by:
$$\mathbf { e } ^ { 1 } = \frac { 1 } { 2 } \mathbf { e } _ { 1 } - \frac { 1 } { \sqrt { 2 } } \mathbf { e } _ { 2 } \\ \mathbf { e } ^ { 2 } = - \frac { 1 } { \sqrt { 2 } } \mathbf { e } _ { 1 } + 2 \mathbf { e } _ { 2 }$$
and the covariant components is obtained by:
The covariant components are obtained by equating the two expressions for the vector $v$: $$v = v _ { 1 } \mathbf { e } ^ { 1 } + v _ { 2 } \mathbf { e } ^ { 2 } = v ^ { 1 } \mathbf { e } _ { 1 } + v ^ { 2 } \mathbf { e } _ { 2 }$$
Doesn't this mean that if $\mathbf { e } _ { 1 }, \mathbf { e } _ { 2 }$ increase, $\mathbf { e } ^ { 1 }, \mathbf { e } ^ { 2 }$ will also increase and thus both the contravariant $v^1,v^2$and covariant $v_1,v_2$ components will have to decrease? What am I understanding wrong here? Thank you!
*Note: In the example, I found
$$\mathbf { e } _ { 1 }=(2,0), \mathbf { e } _ { 2 }=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}) \\ \Rightarrow \mathbf { e } ^ { 1 }=(\frac{1}{2},\frac{-1}{2}), \mathbf { e } ^ { 2 }=(0,\sqrt{2})$$