# Einstein Notation

1. Evaluate $$\delta^{i}_{j}\delta^{j}_{i}$$when $1\leq i,j \leq n$

2. Simplify $$\delta^{a}_{b}g_{ca}g^{bd}\delta^{c}_{d}$$ when $a,b,c,d\in \{1,2,...,n\}$

1. in Einstein notation a matrix as a linear transformation is written as $$A=a^{i}_{j}$$ So $$\delta^{i}_{j}\delta^{j}_{i}=I$$ when I is the identity matrix. But on the other hand the index $j$ is used for summation so the answer will be $$\delta^{i}_{j}\delta^{j}_{i}=\delta^{i}_{i}+\delta^{i}_{i}+...+\delta^{i}_{i}(\text{n times})=1+1+...+1=n$$

What is the correct answer?

1. $$\delta^{a}_{b}g_{ca}g^{bd}\delta^{c}_{a}=\delta_{b}g_{c}g^{b}\delta^{c}$$

How should I continue?

## 2 Answers

Mainly, the Kronecker delta makes sums collapse, making the two indexes equal everywhere else in the expression. For example: $$\delta_j^i \delta^j_i = \delta_i^i = n,$$and $$\delta^{\color{red}{a}}_{\color{blue}{b}}g_{c\color{red}{a}}g^{bd}\delta^{c}_{d} = g_{c\color{blue}{b}}g^{bd}\delta^c_d.$$I'll use colors again to ilustrate how this computation proceeds: $$g_{\color{red}{c}b}g^{bd}\delta_{\color{blue}{d}}^{\color{red}{c}} = g_{\color{blue}{d}b}g^{bd} \stackrel{(\ast)}{=} \delta_d^d = n,$$where in $(\ast)$ I used the definition of the inverse metric tensor.

• why can we cancel out $\delta$ in $\delta^{i}_{j}\delta^{j}_{i}=\delta^{i}_{i}$? because it is equal to $1$ or $0$? if we had for example $a^{a}_{b}b^{b}_{c}=a^{a}b_{c}$ and we could not simplify more? – gbox May 11 '17 at 18:38
• $\delta$ is $1$ iff indexes assume the same value, $0$ otherwise, so it has the "effect" of replacing the saturated index with the other: $\delta_a^b p^a=p^b$. You can't do what you say, $a$ and $b$ cannot just "lose" an index! – lesath82 May 11 '17 at 18:55
• @gbox you can think of it like in the second example: $\delta_{\color{blue}{j}}^{\color{red}{i}} \delta_{\color{red}{i}}^j = \delta_{\color{blue}{j}}^j$ ("red out, blue in"). – Ivo Terek May 11 '17 at 19:19
• That expression is out of the scope of the Einstein convention, you can't have $j$ appearing twice above – Ivo Terek May 11 '17 at 19:28
• Einstein convention says that if a index appear once above and once below, then you're summing: $$\delta_i^i = \delta_1^1 + \cdots + \delta_n^n = 1+\cdots + 1 = n.$$The identity matrix becomes just ${\rm Id}_n = (\delta^i_{\; j})_{1 \leq i, j \leq n}$. – Ivo Terek May 11 '17 at 19:58
1. Indexes are completely saturated, you end up with a scalar, so $n$ is the only correct answer.

2. $\delta^{a}_{b}g_{ca}g^{bd}\delta^{c}_{d}=g_{cb}g^{bc}=\delta_c^c=n$

$\delta$ is $1$ iff indexes assume the same value, $0$ otherwise, so it has the "effect" of replacing the saturated index with the other. For example: $\delta_a^b p^a = p^b$ (as Ivo says: "$a$ out, $b$ in"). Note that the index is replaced where it was: here it was an upper one at there it remains.

$g$ has a somewhat similar behaviour, but it "moves" indexes up/down: $g^{ab}p_a=p^b$. Moreover $g^{ab}g_{bc}=\delta^a_c$ (in fact you are free to choose what index to saturate first, e.g. you can do $g^{ab}g_{bc}p^c=g^{ab} p_b=p^a$ or $g^{ab}g_{bc}p^c=\delta^a_c p^c=p^a$)