# Summation Convention – Substitution Rule

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1. Feb 16, 2016

### FluidStu

Hello. I'm new to this forum. I'm starting a PhD – it's going to be a big long journey through the jungle that is CFD. I would like to arm myself with some tools before entering. The machete is Cartesian Tensors.

I know the rules regarding free suffix's and dummy suffixes, but I'm having trouble proving the substitution rule:
δi j δj k = δi k

Assuming a range of 3 for each component, I will choose that the free suffix's have the following values: i=3, k=1. Therefore, using the implicit summation rule for the dummy index:
δi j δj k = δ3 1 δ1 1 + δ3 2 δ2 1 + δ3 3 δ3 1 = ?

Is my expansion correct?

Many thanks.

Last edited by a moderator: Feb 17, 2016
2. Feb 16, 2016

### blue_leaf77

You can visualize this equation by remembering that kronecker delta symbol actually represents the matrix elements of an identity matrix.

3. Feb 17, 2016

### HallsofIvy

No, It is not because, since "i" and "k" are NOT repeated your result should have "i" and "k" in it, only "j" is replaced by numbers:
δijδjk= δi1δ1k+ δi2δ2k+ δi3δ3k

4. Feb 24, 2016

### FluidStu

Ok, that makes sense (since the index which appears once simultaneously represents all the possible values of that index).

However, from your expansion, I do not understand how one could end up with δik. This was really the point in my question – I would like to see step-wise how δijδjk = δik.

5. Feb 24, 2016

### blue_leaf77

What do you get when multiplying an identity matrix with another identity matrix?

6. Feb 24, 2016

### FluidStu

I * I = I

7. Feb 24, 2016

### blue_leaf77

Now, imagine you have a 3x3 matrices $A$, $B$, and $C$ such that $AB=C$. How do you write the element $c_{ik}$ of $C$ in terms of the elements of $A$ and $B$?

8. Feb 24, 2016

### FluidStu

cik = aiibik + aijbjk + aikbkk

9. Feb 24, 2016

### blue_leaf77

Almost, just that the index to be summed must be written as numbers, not as indices as you did there.

10. Feb 24, 2016

### FluidStu

Do you mean:

cik = a11b13 + a12b23 + a13b33 ?

11. Feb 24, 2016

### blue_leaf77

No. The left most and right most indices in each term in right hand side should be equal to the indices in the left hand side, which is $ik$. Only the two two middle indices changes due to the summation.
Having fixed this, you should be able to see how that equation becomes of when $A=B=C=I$.

12. Feb 24, 2016

### HallsofIvy

I presume this is the "Kronecker delta"- $\delta_{ii}= 1$, $\delta_{ij}= 0$ if $i\ne j$.
Then $\delta_{ij}\delta_{jk}= \delta_{i1}\delta_{1k}+ \delta_{i2}\delta_{2j}+ \delta_{i3}\delta_{3j}$
If i= 1, k= 1, then only the first term is non-zero and it is 1: $\delta_{11}= 1+ 0+ 0= 1[/quote]. If i= 1, k= 2, then at least one of the factors in every term is 0: [itex]\delta_{12}= 0+ 0+ 0= 0[/quote]. If i= 1, k= 3, then at least one of the factors in every term is 0: [itex]\delta_{13}= 0+ 0+ 0= 0. If i= 2, k= 2, then only the second term is non-zero and it is 1: [itex]\delta_{22}= 0+ 1+ 0= 1. etc. 13. Feb 24, 2016 ### FluidStu Ok, so then cik = ai1b1k + ai2b2k + ai3b3k? In this way I can see that A=B=C=I. However, I still don't see how this leads to the proof that δijδjk= δik. If you want to expand on this for others visiting the thread, feel free. However, I can see it clearly from this: In all the cases where i = k, we end up with 1, while in all the cases where i ≠ k, we end up with 0. In other words, we end up with δik. Very good, thank you both for your help. 14. Feb 24, 2016 ### blue_leaf77 That the Kronecker delta is defined as $\delta_{ik}=1$ for $i=k$ and zero for $i\neq k$, doesn't it remind you of the elements of the identity matrix? 15. Feb 24, 2016 ### FluidStu Yes, it does. I can see that the Kronecker Delta in matrix form is simply an identity matrix. But how does this help me to prove: ? 16. Feb 24, 2016 ### blue_leaf77 \delta_{ij}\delta_{jk} = \delta_{i1}\delta_{1k} + \delta_{i2}\delta_{2k} + \delta_{i3}\delta_{3k} = \delta_{ik} Now you already have for the matrix equation $AB=C$. If A=B=C=I, you will only need to replace $a$, $b$, and $c$ with $\delta$ in that quoted equation. 17. Feb 24, 2016 ### FluidStu I see that replacing each with δ gives the required proof. But is it "ok" to replace a, b and c with δ? Aren't a, b and c simply elements of matrices? How can we just replace them with δ? 18. Feb 24, 2016 ### zinq At the risk of repetition: To put the essential point in one place: Let A = (aij), B = (bij), and C = (cij) be n-dimensional matrices such that AB = C.​ Then you know from the definition of matrix multiplication that cij = Σ aik bkj for any i, j in the range 1 ≤ i, j ≤ n, where the sum is over k = 1,...,n. Specializing to the case of n = 3, for the case of A = B = C = I3 (the 3D identity matrix), and then using the repeated-index summation convention, you get δik δkj = δij for any i, j in the range 1 ≤ i, j ≤ 3. 19. Feb 24, 2016 ### blue_leaf77 Note that the last sentence in post #16 is a conditional statement, look what the condition is in order for the conclusion to be true. 20. Feb 24, 2016 ### PeroK Here's an alternative proof (for which I'll abandon the summation convention): $\Sigma_{j} \delta_{ij} \delta_{jk} = \delta_{ii} \delta_{ik} = \delta_{ik} \ \$ (as $\delta_{ij} = 0 \ (i \ne j)$) Or, more laboriously: $\Sigma_{j} \delta_{ij} \delta_{jk} = \delta_{ii} \delta_{ik} + \Sigma_{j \ne i} \delta_{ij} \delta_{jk} = \delta_{ik} + 0 \ \$ (as $\delta_{ii} = 1, \ \delta_{ij} = 0 \ (i \ne j)$) 21. Feb 25, 2016 ### Edgardo Try to get an intuitive feeling for what the Kronecker-Delta [itex]\delta_{ij}$ does.
Consider this sum:
$\sum_{j=1}^{n} c_j = c_1 + c_2 + \dots + c_n$

Now observe what happens if we multiply with the Kronecker-Delta:
$\sum_{j=1}^{n} \delta_{ij} c_j = \delta_{i1} c_1 + \delta_{i2}c_2 + \dots + \delta_{ii}c_i + \dots + \delta_{in}c_n$

There will be exactly one summand with $j=i$, that is $\delta_{ii} c_i = 1 \cdot c_i = c_i$.
For the remaining summands with $j \neq i$ we have $\delta_{ij} c_j = 0 \cdot c_j = 0$
(by the definition of the Kronecker-Delta).

We get:
$\sum_{j=1}^{n} \delta_{ij} c_j$
$= \delta_{i1} c_1 + \delta_{i2}c_2 + \dots + \delta_{ii}c_i + \dots + \delta_{in}c_n$
$= 0 \cdot c_1 + 0 \cdot c_2 + \dots + 1 \cdot c_i + \dots + 0 \cdot c_n$
$= c_i$

With the Einstein summation convention we have $\delta_{ij} c_j = c_i$.
Intuitively, $\delta_{ij}$ evaluates the expression $c_j$ at $j=i$.

Now, similarly, $\delta_{ij} \delta_{jk} = \delta_{ik}$ since
intuitively $\delta_{ij}$ evaluates the expression $\delta_{jk}$ at $j=i$.

22. Feb 25, 2016

### FluidStu

The condition being that A=B=C=I≡ δij? Then we may replace the a, b and c with δ and the following two are equivalent:

Having this then makes it easy to see how both post #12 and #21 explain the required proof. Thanks.