Tensor product of basis and dual basis

Question

In my textbook is stated the following expression: $$\hat{\delta}=\delta^i_j(\vec{a}_i\otimes\vec{a}^j)=\vec{a}_i\otimes\vec{a}^i$$ And I just can't understand what the logic here is. First of all which coordinate system are we using because this tensor product is gonna have different components depending on our basis vectors. Does it play a role somewhere that the dot product between the two is equal to one? I'm just so confused here.

Answer 1

The Kronecker delta symbol indicates when the indices match: $$ \delta^i_j = \begin{cases} 1 & \text{if } i = j, \\ 0 & \text{else.} \end{cases} $$

Thus, for each pair of indices $(i, j)$, $$ \delta^i_j (\vec{a}_i \otimes \vec{a}^j) = \begin{cases} \vec{a}_i \otimes \vec{a}^i & \text{if } i = j, \\ 0 & \text{else.} \end{cases} $$

For instance, in $3$-dimensions, we have $3 \times 3 = 9$ pure tensors $$ \bigl\{ \vec{a}_i^j \bigr\} = \left\{ \begin{array}{ccc} \vec{a}_1 \otimes \vec{a}^1 & \vec{a}_1 \otimes \vec{a}^2 & \vec{a}_1 \otimes \vec{a}^3 \\ \vec{a}_2 \otimes \vec{a}^1 & \vec{a}_2 \otimes \vec{a}^2 & \vec{a}_2 \otimes \vec{a}^3 \\ \vec{a}_3 \otimes \vec{a}^1 & \vec{a}_3 \otimes \vec{a}^2 & \vec{a}_3 \otimes \vec{a}^3 \end{array} \right\}, $$ but only $3$ survive (nonzero) after pairing with $\delta_j^i$: $$ \begin{align} \delta_j^i\vec{a}_i^j &= \delta_1^1(\vec{a}_1 \otimes \vec{a}^1) + \delta_2^1(\vec{a}_1 \otimes \vec{a}^2) + \delta_3^1(\vec{a}_1 \otimes \vec{a}^3) \\ &\quad{}+ \delta_1^2(\vec{a}_2 \otimes \vec{a}^1) + \delta_2^2(\vec{a}_2 \otimes \vec{a}^2) + \delta_3^2(\vec{a}_2 \otimes \vec{a}^3) \\ &\qquad{}+ \delta_1^3(\vec{a}_3 \otimes \vec{a}^1) {}+ \delta_2^3(\vec{a}_3 \otimes \vec{a}^2) {}+ \delta_3^3(\vec{a}_3 \otimes \vec{a}^3) \\[4pt] &= \vec{a}_1 \otimes \vec{a}^1 + \vec{a}_2 \otimes \vec{a}^2 + \vec{a}_3 \otimes \vec{a}^3 \end{align} $$