Schaum's Outline to Tensor Calculus — Chapter 1, Solved problem 1.5 — Use the summation convention to write and state the value of $n$ necessary in: $$g^{\LARGE{1}}_{11} + g^{\LARGE{1}}_{12} + g^{\LARGE{1}}_{21} + g^{\LARGE{1}}_{22} + g^{\LARGE{2}}_{11} + g^{\LARGE{2}}_{12} + g^{\LARGE{2}}_{21} + g^{\LARGE{2}}_{22} $$.

My solution. The game plan is just to simplify each of the three indices one at at time in any of the $3 x 2 x 1$ orders. Say I start with the superscript — I'd get $ \forall \, {\LARGE{i}} \in \{1,2\} \, g^i_{11} + g^i_{12} + g^i_{21} + g^i_{22} $.
Then say I pick the second subscript — subsequently $ \forall \, i, k \in \{1,2\} \, g^i_{1k} + g^i_{2k} $.
The ultimate subscript — subsequently $ \forall \, i,j, k \in \{1,2\} \, g^i_{jk} $.

Their solution:

Set $c_i=1$ for each $i$ ($n=2$). Then the expression may be written \begin{align*} g_{11}^i c_i + g_{12}^i c_i + g_{21}^i c_i + g_{22}^i c_i &= (g_{11}^i + g_{12}^i + g_{21}^i + g_{22}^i)\,c_i \\ &= (g_{jk}^i c_j c_k) c_i = g_{jk}^i c_i c_j c_k. \end{align*}

Does their final answer look like mine? But why did they "set $c_i = 1 \, \forall \, i \in \{1,2\}"?$

  • $\begingroup$ The problem with your attempt is that there's no summation going on. You only have one index of each. That's why they introduce those $c_i$. To have a summation. Try to write it down with the conventional summation notation first, i.e. $\sum$. $\endgroup$ – Raskolnikov Jul 5 '13 at 10:03

Your solution is equivalent to

$$\sum_{i,j,k=1}^2g^i_{jk} $$

and no summation convention (I think it is a way to say "Einstein convention on repeated indices") appears, as pointed out by @Raskolnikov.

The aim of the exercise is to arrive at an expression with "repeated indices", i.e. an expression in which you use the summation convention. To do so, one needs to have no free index (like yours $i$, $j$ and $k$) and to contract-or produce pairs of- all indices.

It is clear that the starting expression has 3 indices: so 3 summations, or contractions are needed. The textbook begins to produce a summation considering at first the above index, called $i$. This is done introducing the vector


and realizing the sum $\sum_{i=1}^2 g^i_{j,k}$ (which, once again, uses no summation convention) as

$$g^1_{j,k}+g^2_{j,k}=\sum_{i=1}^2 g^i_{j,k}=g^i_{j,k}c_i,$$

for any $j,k$. On the rightmost r.h.s. of the above expression we use the Einstein convention, summing over $i$, the only repeated index. Please note that the length of $c$ is equal to $2$, i.e. the cardinality of the set $\{1,2\}$ of all the possible values for $i$.

We are left with 2 indices, so 2 more summations are needed. We repeat the above lines to produce the summation w.r.t. to $j$ through

$$g^i_{1,k}c_i+g^i_{2,k}c_i=\sum_{j=1}^2 g^i_{j,k}c_i=g^i_{j,k}c_ic_j,$$

for any $k$. Repeating the same trick with $k$, the only free index remaining, we arrive at



$$\sum_{i,j,k=1}^2g^i_{jk}=g^i_{j,k}c_ic_jc_k.~~~(1) $$

On the r.h.s. of $(1)$ we have all repeated indices and we are done.

  • $\begingroup$ Thanks. I'm addled. You said "my solution $\sum_{ijk = 1}^2 g_ijk$ is equivalent to"$ \sum_{i,j,k=1}^2 g_ijk$ which turns out to be $g^{\LARGE{1}}_{11} + g^{\LARGE{1}}_{12} + g^{\LARGE{1}}_{21} + g^{\LARGE{1}}_{22} + g^{\LARGE{2}}_{11} + g^{\LARGE{2}}_{12} + g^{\LARGE{2}}_{21} + g^{\LARGE{2}}_{22} \text{ = this was what question necessitated }$. What's the problem??? $\endgroup$ – Dwayne E. Pouiller Jul 7 '13 at 9:17
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    $\begingroup$ @DEPouiller The "problem" is that your solution $\sum_{ijk}g^i_{jk}$ uses no summation convention (=there are no contracted/repeated indices), although it gives the expression $g^1_{11}+...+g^2_{22}$,as requested. The exercise asks for an equivalent expression, BUT with repeated/contracted indices. The trick is to introduce the vector $c=(1,1)$ as above. I hope it helps $\endgroup$ – Avitus Jul 7 '13 at 9:51
  • $\begingroup$ Thanks a lot. Last question — Why isn't Einstein Summation Notation defined so that $\large{g_{jk}^i = \sum_{ijk} g_{jk}^i}$? Then I wouldn't have had a problem here? $\endgroup$ – Dwayne E. Pouiller Jul 7 '13 at 11:06
  • $\begingroup$ @DEPouiller you are welcome. Feel free to upvote and flag the answer if it was helpful. Using the notation you introduce, even dealing with matrices would be a bit complicated. For example, it is not clear whether $M_{ii}$ is the $i$-th diagonal entry, or the trace $M_{11}+...+M_{nn}$ of the matrix $M$ :) $\endgroup$ – Avitus Jul 7 '13 at 11:22
  • $\begingroup$ Thanks a lot again. I upvoted and flagged. I'm still bewildered. Even with the current Eistein Summation Notation, $M_ii$ can still mean the $i$th diagonal entry of a matrix or $\sum_i M_{ii}$?. I still don't see the problem with defining Einstein Notation so that $\large{g^i_{jk} = \sum_{i, j, k} {g^i_{jk}}} ?$ $\endgroup$ – Dwayne E. Pouiller Jul 15 '13 at 5:39

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