During my Linear Algebra lecture from the Mathematical Physics class, our professor introduced the Kronecker delta tensor ( but he mainly introduced it as a function ) while explaining inner products between two vectors. Using some tensor tools, things can by way more convenient.
Before starting with the Kronecker delta, we need another tool, the Einstein sum convention, and this will make everything super-light. It's simple : discard the sigma.
\[ \vec{x} = \sum_{i=1}^{n}x_i \hat{e}_i \quad when \quad \vec{x}=\left( x_{1}, x_{2} , \dots , x_{n} \right) \]
And using the Einstein notation,
\[ \vec{x} = x_i \hat{e}_i \]
it is way more simple. There's actually more to talk about subscripts and superscripts, which is important using the Einstein notation, let's us know whether the component is a vector or a covector, but I'll skip this and only use subscripts in this post. I'll explain about vectors and covectors in later posts.
First, let's express inner products between two vectors.
\[ \vec{x} \cdot \vec{y} = ( x_i \hat{e}_i )\cdot ( y_j \hat{e}_j) = x_i y_j(\hat{e}_i \cdot \hat{e}_j) \]
Now, what would $\hat{e}_i \cdot \hat{e}_j$ be...?
We can get some clues from the results of the inner products between $\hat{e}_i$s, or the basis vectors of an orthogonal coordinate system, or to think easily just think of the $R^3$. We know that $\hat{e}_i \cdot \hat{e}_j$ when $i = j$ is 1. But when we inner product two orthogonal vectors, $\hat{e}_i \cdot \hat{e}_j$ when $i \neq j$, we get 0. And this property of basis vectors of an orthogonal coordinate could be easily expressed by using the Kronecker delta tensor. It is defined as,
\[ \delta_{ij} = \begin{Bmatrix} 0 \quad (i \neq j) \\ 1 \quad ( i=j) \end{Bmatrix} \]
and we could apply this to express $\hat{e}_i \cdot \hat{e}_j$.
\[ \hat{e}_i \cdot \hat{e}_j = \delta_{ij} \]
So, we can express the inner product using the Kronecker delta.
\[ \vec{x} \cdot \vec{y} = \delta_{ij} x_i y_j \]
The next step is important. But wait..... isn't i and j already different, and thus it makes the whole thing 0? NO. The indices i and j are like variables, they can become 1 or 3 or etc, thus it means that they are not a constant number or a fixed value we don't know. We need some flexibility dealing with these indices and think of two possibilities, differed by the equality between them. When the indices have different values, they all disappear as $\delta_{ij}=0$. Meanwhile, when the indices are identical, $\delta_{ij}=1$ and it makes $x_i y_j$ 'alive'. Thus, $\delta_{ij} x_i y_j $ could be reduced by,
\[ \delta_{ij} x_i y_j = \delta_{ii} x_i y_i = x_i y_i \]
Thus we obtain an already known definition of inner products,
\[ \vec{x} \cdot \vec{y} = x_i y_i \]
The Kronecker delta can be used as a function as we did, but actually they are tensors. I'll explain about it in detail after some posts.
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