Find $\mathbf{u} \cdot \mathbf{v}$, where $\mathbf{u} = \mathbf{e}^1 - \mathbf{e}^2 + \mathbf{e}^3 - \mathbf{e}^4$.

Question

Please note that Einstein summation notation is used.

Take $B = \{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e_3}, \mathbf{e}_4 \}$ as a basis for $\mathbb{R}^4$, where $\mathbf{e}_1 = (1, 0, 0, 0), \mathbf{e}_2 = (1, 1, 0, 0), \mathbf{e_3} = (1, 1, 1, 0), \mathbf{e_4} = (1, 1, 1, 1)$.

Find the covariant components of $\mathbf{v} = (1, 2, 3, 4)$.

The dual basis is $B^\perp = \{\mathbf{e}^1, \mathbf{e}^2, \mathbf{e}^3, \mathbf{e}^4\}$, where $\mathbf{e}^1 = (1, -1, 0, 0)$, $\mathbf{e}^2 = (0, 1, -1, 0)$, $\mathbf{e}^3 = (0, 0, 1, -1)$, $\mathbf{e}^4 = (0, 0, 0, 1)$.

Find $\mathbf{u} \cdot \mathbf{v}$, where $\mathbf{u} = \mathbf{e}^1 - \mathbf{e}^2 + \mathbf{e}^3 - \mathbf{e}^4$,

I already found the covariant components as follows:

$v_1 = (1, 2, 3, 4) \cdot (1, 0, 0, 0) = 1$

$v_2 = (1, 2, 3, 4) \cdot (1, 1, 0, 0) = 3$

$v_3 = (1, 2, 3, 4) \cdot (1, 1, 1, 0) = 6$

$v_4 = (1, 2, 3, 4) \cdot (1, 1, 1, 1) = 10$

$\mathbf{u} = (1, -1, 0, 0) - (0, 1, -1, 0) + (0, 0, 1, -1) - (0, 0, 0, 1) = (1, -1, 2, -2)$

As I understand it, $(1, -1, 2, -2)$ is in terms of the standard basis?

And we know that $\mathbf{u} \cdot \mathbf{v} = u^i v_i$. Therefore, I think we can calculate the $u^i$ as $\mathbf{u} \cdot \mathbf{e}^i$:

$u^1 = (1, -1, 2, -2) \cdot (1, -1, 0, 0) = 2$

$u^2 = (1, -1, 2, -2) \cdot (0, 1, -1, 0) = -3$

$u^3 = (1, -1, 2, -2) \cdot (0, 0, 1, -1) = 4$

$u^4 = (1, -1, 2, -2) \cdot (0, 0, 0, 1) = -2$

$\mathbf{u} \cdot \mathbf{v} = u^i v_i = (2, -3, 4, -2) \cdot (1, 3, 6, 10) = 2 - 9 + 24 - 20 = -3$

My problem is that the solutions say that $\mathbf{u} \cdot \mathbf{v} = -5$. I am confused as to how this is the case, since I cannot find any flaws in my reasoning? Which part, if any, of my reasoning is erroneous? Why is it erroneous? What is the correct way to solve this?

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

You made a mistake calculating $\mathbf{u}$ in the standard basis. It should be

$$ \mathbf{u} = (1,-\color{red}{2}, 2, -2) $$

If you carry out the computation the same way but with the $-2$ you'll get $\mathbf{u} \cdot \mathbf{v} = -5$.