$e^i.e_j=\delta_{ij}$ not true in a two dimensional coordinate system with dual.

Question

Suppose vector space $\mathbb{R}^2$ with standard basis $E=\{e_1,e_2\}$, vector space $V$ with basis $E^\prime=\{e^\prime_1,e^\prime_2\}$, $V^*$ (dual of vector space $V$) with basis $E^*=\{e^1,e^2\}$ and a linear transformation $A^\prime_X:\mathbb{R}^2\rightarrow V$. suppose we have vector $u$ with components $(x,y)$, $(x^\prime,y^\prime)$ and $(x^1,y^1)$ for $\mathbb{R}^2$, $V$ and $V^*$ respectively, and :

$$A^\prime_X:\left[\begin{array}{} x^\prime \\ y^\prime\end{array} \right]= \frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \\ \end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$$

In this figure, the red lines and blue lines represent coordinate system for $V$ and $V^*$ respectively. I found :

$$\left[\begin{array}{} x^\prime \\ y^\prime\end{array} \right]= \frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \\ \end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$$

$$\left[\begin{array}{} x^1 \\ y^1 \end{array} \right]= \frac{1}{\sin\theta}\left[ \begin{array}{} \cos\beta & \sin\beta\\ -\cos\gamma &-\sin\gamma\end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$$

$$\left[\begin{array}{} x^1 \\ y^1 \end{array} \right]= \frac{1}{\sin\theta}\left[ \begin{array}{} 1 & \cos\theta\\ -\cos\theta &-1\end{array} \right]\left[\begin{array}{} x^\prime\\ y^\prime\end{array} \right]$$

$$\left[\begin{array}{} e^\prime_1 \\ e^\prime_2\end{array} \right]= \frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\sin\beta\\ -\cos\gamma & \cos\beta \\ \end{array} \right]\left[\begin{array}{} e_1\\ e_2\end{array} \right]$$

$$\left[\begin{array}{} e^1 \\ e^2\end{array} \right]= \frac{1}{\sin\theta}\left[ \begin{array}{} \cos\beta & -\cos\gamma \\ \sin\beta & -\sin\gamma \\ \end{array} \right]\left[\begin{array}{} e_1\\ e_2\end{array} \right] .$$

My problem: I know $e^i(e^\prime_j)=\delta_{ij}$ but it is not true in this example, I do not know why. where and what are my mistakes?

Figure source

Answer 1

I get $$ \left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right]= \left[ \begin{matrix} \cos\beta & \sin\beta \\ \cos\gamma & \sin\gamma \end{matrix} \right] \left[\begin{matrix} e_1\\ e_2\end{matrix} \right] $$

At least that makes $$ \left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right] \left[\begin{matrix} x' \\ y' \end{matrix} \right] = \left[\begin{matrix} e_1 & e_2\end{matrix} \right] \left[ \begin{matrix} \cos\beta & \cos\gamma \\ \sin\beta & \sin\gamma \end{matrix} \right] \frac{1}{\sin\theta} \left[ \begin{matrix} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \end{matrix} \right] \left[\begin{matrix} x \\ y \end{matrix} \right] \\ = \frac{1}{\sin\theta} \left[\begin{matrix} e_1 & e_2\end{matrix} \right] \left[ \begin{matrix} \cos\beta\sin\gamma - \cos\gamma \sin\beta & -\cos\beta \cos\gamma + \cos\beta \cos\gamma \\ \sin\beta \sin\gamma - \sin\beta \sin\gamma & -\sin\beta \cos\gamma + \sin\gamma \cos\beta \end{matrix} \right] \left[\begin{matrix} x \\ y \end{matrix} \right] \\ = \frac{1}{\sin\theta} \left[\begin{matrix} e_1 & e_2\end{matrix} \right] \left[ \begin{matrix} \sin(\gamma-\beta) & 0 \\ 0 & \sin(\gamma-\beta) \end{matrix} \right] \left[\begin{matrix} x \\ y \end{matrix} \right] \\ = \left[\begin{matrix} e_1 & e_2\end{matrix} \right] \left[\begin{matrix} x \\ y \end{matrix} \right] $$ since $\gamma-\beta = \theta$.

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How did I find the equation at the top of my post?

We wanted $\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right] \left[\begin{matrix} x' \\ y' \end{matrix} \right] = \left[\begin{matrix} e_1 & e_2\end{matrix} \right] \left[\begin{matrix} x \\ y \end{matrix} \right]$.

Inserting $\left[\begin{array}{} x^\prime \\ y^\prime\end{array} \right]= \frac{1}{\sin\theta}\left[ \begin{array}{} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \\ \end{array} \right]\left[\begin{array}{} x\\ y\end{array} \right]$ gives $$\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right] \left[\begin{matrix} x' \\ y' \end{matrix} \right] = \left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right] \frac{1}{\sin\theta} \left[ \begin{matrix} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \end{matrix} \right] \left[\begin{matrix} x \\ y \end{matrix} \right] $$ so we must have $$\left[\begin{matrix} e_1 & e_2\end{matrix} \right] = \left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right] \frac{1}{\sin\theta} \left[ \begin{matrix} \sin\gamma & -\cos\gamma \\ -\sin\beta & \cos\beta \end{matrix} \right] $$ Solving for $\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]$ gives $$ \left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right] = \left[\begin{matrix} e_1 & e_2\end{matrix} \right] \frac{\sin\theta}{\sin\gamma \cos\beta - \sin\beta \cos\gamma} \left[\begin{matrix} \cos\beta & \cos\gamma \\ \sin\beta & \sin\gamma \end{matrix}\right] = \left[\begin{matrix} e_1 & e_2\end{matrix} \right] \left[\begin{matrix} \cos\beta & \cos\gamma \\ \sin\beta & \sin\gamma \end{matrix}\right] $$

Finally, taking the transpose results in $$ \left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right] = \left[\begin{matrix} \cos\beta & \sin\beta \\ \cos\gamma & \sin\gamma \end{matrix}\right] \left[\begin{matrix} e_1 \\ e_2\end{matrix} \right] $$

One can also see from the picture that we must have $$ \left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right] = \left[\begin{matrix} a \cos\beta & a \sin\beta \\ b \cos\gamma & b \sin\gamma \end{matrix}\right] \left[\begin{matrix} e_1 \\ e_2\end{matrix} \right] $$ for some constants $a$, $b$.

Answer 2

The Einstein notation makes it simple. Here, the vector space is $\mathbb{R}^2$ with standard basis $E = \lbrace \boldsymbol{e}_1,\boldsymbol{e}_2\rbrace$. Now, we consider a basis $E' = \lbrace \boldsymbol{e}'_1,\boldsymbol{e}'_2 \rbrace$ such that $\boldsymbol{e}'_j = A^i_{\; j}\, \boldsymbol{e}_i$ (Einstein notation). Inversely, one has the relation $\boldsymbol{e}_i = B^j_{\; i}\, \boldsymbol{e}'_j$. The coordinates of the transformation matrix $[B^j_{\; i}]$ are given by $A'_X$, i.e. $$ [B^j_{\; i}] = \frac{1}{\sin\theta}\left[ \begin{array}{cc} \sin\gamma & {-\cos\gamma} \\ {-\sin\beta} & \cos\beta \end{array} \right] , $$ where $\theta=\gamma-\beta$ and $\sin\theta\neq 0$. Its matrix inverse is $$ [A^i_{\; j}] = [B^j_{\; i}]^{-1} = \left[ \begin{array}{cc} {\cos\beta} & {\cos\gamma} \\ {\sin\beta} & {\sin\gamma} \end{array} \right] . $$ Therefore, we obtain the following relations between basis vectors: \begin{aligned} &\boldsymbol{e}'_1 = \cos\beta\, \boldsymbol{e}_1 + \sin\beta\, \boldsymbol{e}_2 \, ,\quad & \boldsymbol{e}'_2 = \cos\gamma\, \boldsymbol{e}_1 + \sin\gamma\, \boldsymbol{e}_2 \, . \end{aligned} Now, we introduce the basis $E^* = \lbrace \boldsymbol{e}^1,\boldsymbol{e}^2 \rbrace$, dual of $E'$. By definition, its basis vectors must satisfy $\boldsymbol{e}^k\cdot \boldsymbol{e}'_j = \delta_j^k$, where $\delta_j^k$ is the Kronecker delta. The coordinates $v^i$ of a vector $\boldsymbol{v} = v_k\,\boldsymbol{e}^k = v^i\,\boldsymbol{e}_i$ on the basis $E$ are given by the scalar products \begin{aligned} v^i = \boldsymbol{v}\cdot \boldsymbol{e}_i &= v_k\,\boldsymbol{e}^k \cdot \boldsymbol{e}_i \\ &= v_k\,\boldsymbol{e}^k \cdot (B^j_{\; i}\, \boldsymbol{e}'_j)\\ &= v_k\, \delta^k_j\, B^j_{\; i} \\ &= v_k\, B^k_{\; i} \, . \end{aligned} In particular, we obtain the following relations between basis vectors: \begin{aligned} &\boldsymbol{e}^1 = \frac{1}{\sin\theta} \left(\sin\gamma\, \boldsymbol{e}_1 - \cos\gamma\, \boldsymbol{e}_2\right) ,\quad & \boldsymbol{e}^2 = \frac{1}{\sin\theta} \left( -\sin\beta\, \boldsymbol{e}_1 + \cos\beta\, \boldsymbol{e}_2 \right) . \end{aligned}