Why does $\delta^i_j$ $\alpha^j$=$\alpha^i$?

Question

I am currently pursuing a self study in Tensor analysis and Tensor algebra and came across the following statement when exploring the Einstein summation convention:

Why does $\delta^i_j$$\alpha^j$=$\alpha^i$? (1)

My understanding of $\delta^1_j$$\alpha^j$=$\alpha^1$ (A similar example shown in my book) is if we expand the left hand side of the equation we see that the only time $j=1$ is the first term in the sum, therefore we are left with $\alpha^1$. However, I am at a loss as to why the $\alpha$ is raised an $i$ on the right hand side of (1)? I am a beginner so I may be overlooking a simple explanation.

Answer 1

There is an implicit summation here as you mentioned: $$ \sum_{j} \delta^i_j\alpha^j=\alpha^i$$ because $$ \delta^i_j= \begin{cases} 1&i=j\\ 0&i\ne j. \end{cases}$$ So, in the sum when $i\ne j$, the term $\delta^i_j\alpha^j=0.$ In particular, the only nonzero summand is $\delta^i_i\alpha^i=\alpha^i$.

Edit: Also, thanks to the comment below I'll add the following: in differential geometry (or anywhere tensors appear) some indices are written as superscripts: so, $\alpha^i$ is not $\alpha$ raised to the $i^{th}$ power, but rather it is simply an index.

Answer 2

The $i$ in $\alpha^i$ is an index of a the vector $\alpha$, not an exponent. $\alpha^i$ is the $i$-th component of the vector $\alpha$, just like $\delta^a_b$ is the ${}^a_b$-th component of the tensor $\delta$. Specifically, $$\delta^a_b=\begin{cases}1&\text{if }A=b\\ 0&\text{if }a\ne b\end{cases}$$

So, like in the previous case, $$\delta^i_j\alpha^j:=\sum_{j=1}^n\delta^i_j\alpha^j=\alpha^i$$

because $\delta^i_i=1$ and all the other $\delta^i_j$-s are $0$.

Another interpretation of this is that $\delta^\bullet_j\alpha^j$ is the vector $\alpha$ itself.

Answer 3

The explanation you need is that $\delta_j^i=1$ only when $j=i$ and is $0$ otherwise.Therefore $\delta_j^i\alpha^j=\alpha^i$, because the sum over $j$ has only $j=i$ as a non-zero term.