# Einstein Summation Convention Question 2

• Athenian
In summary: B_{ij}## as follows: ##A_{ij} B_{ij} = (A_{ji} - B_{ji}) \, A_{ij}##.Now suppose that ##A## is symmetric and that ##B## is antisymmetric. Using the properties I give above, rewrite ##A_{ij}...B_{ij}## as follows: ##A_{ij} B_{ij} = (A_{ji} - B_{ji}) \, A_{ij}##.This is correct. Asymmetric matrices always have a product of negative terms, and symmetric matrices always have a product of positive terms.
Athenian
Homework Statement
Solve for ##\epsilon_{ij \ell} \, \epsilon_{km \ell} \, \epsilon_{ijm} \, a_k##
Relevant Equations
Refer Below ##\longrightarrow##
Below is my attempted solution:

$$\epsilon_{ij \ell} \, \epsilon_{km \ell} \, \epsilon_{ijm} \, a_k$$
$$\Rightarrow (\delta_{ik} \, \delta_{jm} - \delta_{im} \, \delta_{jk}) \epsilon_{ijm} \, a_k$$
$$\Rightarrow \delta_{ik} \, \delta_{jm} \, \epsilon_{ijm} \, a_k - \delta_{im} \, \delta_{jk} \epsilon_{ijm} \, a_k$$
$$\Rightarrow a_i \, \delta_{jm} \, \epsilon_{ijm} - a_j \delta_{im} \, \epsilon_{ijm}$$

From here on out, I am not sure what I am doing. Therefore, any guidance or assistance would be greatly appreciated.

$$\Rightarrow a_m \epsilon_{ijm} - a_m \epsilon_{ijm} = 0$$

Perhaps, my biggest question is would eliminating the Kronecker delta and getting ##a_m## be mathematically legal?

Also, I have heard from someone else that the above equation results to zero because of anti-symmetry. How is one able to determine that?

Thank you for the help!

Athenian said:
Also, I have heard from someone else that the above equation results to zero because of anti-symmetry. How is one able to determine that?

If ##A## and ##B## are general, is it always true that

$$A_{ij} B_{ij} = A_{ji} B_{ji}?$$

If so, why? If not, why not?

Unfortunately, I am not quite sure what you mean by "general" as I have only seen the Kronecker delta possessing two subscripts thus far in my studies. However, if I am to apply the above equation using "Kronecker delta" logic, I would say yes as they will both ultimately equate to 0.

I apologize if this doesn't properly answer your question. Nonetheless, I really appreciate the helping hand.

Athenian said:
Unfortunately, I am not quite sure what you mean by "general" as I have only seen the Kronecker delta possessing two subscripts thus far in my studies.

Think of ##A## and ##B## as being matrices (##3\times3##, if you like). The Kronecker delta is the component version of the identity matrix. By "general", I mean that ##A## and ##B## do not possesses any special symmetry or antisymmetry properties. I am trying to illustrate and import aspect of the Einstein summation convention.

George Jones said:
If ##A## and ##B## are general, is it always true that

$$A_{ij} B_{ij} = A_{ji} B_{ji}?$$

If so, why? If not, why not?
Hmmm, despite thinking for a bit to find a better answer, I still only came up with an assumption of "no" since matrix multiplication is not communicable.

While I did try to prove my guess by writing something along the lines of ##A_{ij} \, B_{ik} \, \delta_{jk} = A_{ij} \, B_{ij}##, I didn't think that was going very well in the end ...

Athenian said:
Hmmm, despite thinking for a bit to find a better answer, I still only came up with an assumption of "no" since matrix multiplication is not communicable.

While I did try to prove my guess by writing something along the lines of ##A_{ij} \, B_{ik} \, \delta_{jk} = A_{ij} \, B_{ij}##, I didn't think that was going very well in the end ...
Try to write the sum explicitly for ##i,j=1,2,3##.

Wrote the sum explicitly down and that definitely helped me better visualize what's going on.

Anyway, I take back my answer. They are equal to each other. In short, ##A_{ij}## and ##A_{ji}## are the same because one is simply moving the ##\Sigma## (i.e. sum) for ##i## and ##y## back and forth (or in different orders). However, in the end, though, the resulting sum would be the same.

Would this logic, I can thus say that the below equation is correct:
George Jones said:
##A_{ij} B_{ij} =A_{ji} B_{ji}##

PeroK
Athenian said:
Wrote the sum explicitly down and that definitely helped me better visualize what's going on.

Anyway, I take back my answer. They are equal to each other. In short, ##A_{ij}## and ##A_{ji}## are the same because one is simply moving the ##\Sigma## (i.e. sum) for ##i## and ##y## back and forth (or in different orders). However, in the end, though, the resulting sum would be the same.

Would this logic, I can thus say that the below equation is correct:

Now suppose that ##A## is symmetric and that ##B## is antisymmetic.

I see. Then, using my logic from earlier, ##B_{ij}## would be symmetric while ##-B_{ji}## would be antisymmetric.

Therefore, ##A_{ij} \, B_{ij} = - A_{ij} \, B_{ji}##, right?

Provided that I am correct, I still don't exactly see how this solution method helps me solve my above equation. Are there dots I am supposed to be connecting here?

Athenian said:
I see. Then, using my logic from earlier, ##B_{ij}## would be symmetric while ##-B_{ji}## would be antisymmetric.

I am not sure what what you mean, here. ##A## symmetric means ##A_{ij} = A_{ji}## always, i.e,, both inside and outside sums. The Kronecker delta is symmetric. ##B## antisymmetric means ##B_{ij} = -B_{ji}## always, i.e,, both inside and outside sums. The Levi-Civita symbol is antisymmetric with respect to any pair of indices.

Where we stand: for any ##A## and ##B##, ##A_{ij} B_{ij} = A_{ji} B_{ji}##, because of the implied sums.

Now suppose that ##A## is symmetric and that ##B## is antisymmetic. Using the properties I give above, rewrite ##A_{ij} B_{ij}##.

Athenian
Athenian said:
Homework Statement:: Solve for ##\epsilon_{ij \ell} \, \epsilon_{km \ell} \, \epsilon_{ijm} \, a_k##
Relevant Equations:: Refer Below ##\longrightarrow##

Below is my attempted solution:

$$\epsilon_{ij \ell} \, \epsilon_{km \ell} \, \epsilon_{ijm} \, a_k$$
$$\Rightarrow (\delta_{ik} \, \delta_{jm} - \delta_{im} \, \delta_{jk}) \epsilon_{ijm} \, a_k$$
$$\Rightarrow \delta_{ik} \, \delta_{jm} \, \epsilon_{ijm} \, a_k - \delta_{im} \, \delta_{jk} \epsilon_{ijm} \, a_k$$
$$\Rightarrow a_i \, \delta_{jm} \, \epsilon_{ijm} - a_j \delta_{im} \, \epsilon_{ijm}$$
As I said in another thread, when you see a Kronecker delta, use it to eliminate one of the indices that appears in it. For example here what do you get if you do the sum over m in both terms?

Athenian
George Jones said:
I am not sure what you mean, here. ##A## symmetric means ##A_{ij} = A_{ji}## always, i.e,, both inside and outside sums. The Kronecker delta is symmetric. ##B## antisymmetric means ##B_{ij} = -B_{ji}## always, i.e,, both inside and outside sums. The Levi-Civita symbol is antisymmetric with respect to any pair of indices.

Where we stand: for any ##A## and ##B##, ##A_{ij} B_{ij} = A_{ji} B_{ji}##, because of the implied sums.

Now suppose that ##A## is symmetric and that ##B## is antisymmetic. Using the properties I give above, rewrite ##A_{ij} B_{ij}##.

My mistake here. In that case, to answer the question ##\Longrightarrow A_{ij} B_{ij} = -A_{ji} B_{ji}## assuming that ##A_{ij}## is symmetric and ##B_{ij}## is anti-symmetric.

nrqed said:
As I said in another thread, when you see a Kronecker delta, use it to eliminate one of the indices that appears in it. For example here what do you get if you do the sum over m in both terms?

Thanks a lot for the help. I figured out the logic behind solving the question. Essentially, if I assume ##j \neq m##, then both Kronecker deltas will go to zero - making the entire solution ##0##. However, if I make ## j = m##, then both Kronecker deltas will go to one. However, the Levi-Civita will then become ##\epsilon_{ijj}## or ##\epsilon_{imm}##. And, in this case, the Levi-Civita would also go to zero. Therefore, regardless whether ##j = m## or ##j \neq m##, the entire solution will go to ##0##.

Is this reasoning accurate?

nrqed
Athenian said:
Thanks a lot for the help. I figured out the logic behind solving the question. Essentially, if I assume ##j \neq m##, then both Kronecker deltas will go to zero - making the entire solution ##0##. However, if I make ## j = m##, then both Kronecker deltas will go to one. However, the Levi-Civita will then become ##\epsilon_{ijj}## or ##\epsilon_{imm}##. And, in this case, the Levi-Civita would also go to zero. Therefore, regardless whether ##j = m## or ##j \neq m##, the entire solution will go to ##0##.

Is this reasoning accurate?
That's completely correct.

In general, when you have a Kronecker delta, it is very nice because you can do the sum over one of its indices trivially, so you should always do that and get rid of all the Kronecker deltas this way. In this problem, if you had not realized the point you just made, you would have ended up with Levi-Civita symbols with two indices equal, which would have told you the result is zero.

Athenian
Thank you for confirming my explanation. It really helped and I sincerely appreciate it!

As @nrqed has explianed, it is very important to know how to eliminate sum using the Konecker delta. It also can be useful to know how to use symmetry properties.

Athenian said:
I can thus say that the below equation is correct:

George Jones said:
$$A_{ij} B_{ij} = A_{ji} B_{ji}$$
Athenian said:
##A_{ij} B_{ij} = -A_{ji} B_{ji}## assuming that ##A_{ij}## is symmetric and ##B_{ij}## is anti-symmetric.

Putting these two things together gives
\begin{align} A_{ij} B_{ij} &= -A_{ij} B_{ij} \nonumber \\ 2A_{ij} B_{ij} &= 0 \nonumber \\ A_{ij} B_{ij} &= 0 \nonumber \end{align}

for any symmetric ##A## and antisymmetric ##B## (or vice versa). In particular, it can be used in both terms of ##a_i \, \delta_{jm} \, \epsilon_{ijm} - a_j \delta_{im} \, \epsilon_{ijm}##.

The first term contains ##\delta_{jm} \, \epsilon_{ijm}##, with ##\delta_{jm}## symmetric in ##j## and ##m##, and ##\epsilon_{ijm}## antisymmetric in ##j## and ##m##, so ##\delta_{jm} \, \epsilon_{ijm} = 0##.

The second term contains ##\delta_{im} \, \epsilon_{ijm}##, with ##\delta_{im}## symmetric in ##i## and ##m##, and ##\epsilon_{ijm}## antisymmetric in ##i## and ##m##, so ##\delta_{im} \, \epsilon_{ijm} = 0##.

Athenian and nrqed
George Jones said:
As @nrqed has explianed, it is very important to know how to eliminate sum using the Konecker delta. It also can be useful to know how to use symmetry properties.Putting these two things together gives
\begin{align} A_{ij} B_{ij} &= -A_{ij} B_{ij} \nonumber \\ 2A_{ij} B_{ij} &= 0 \nonumber \\ A_{ij} B_{ij} &= 0 \nonumber \end{align}

for any symmetric ##A## and antisymmetric ##B## (or vice versa). In particular, it can be used in both terms of ##a_i \, \delta_{jm} \, \epsilon_{ijm} - a_j \delta_{im} \, \epsilon_{ijm}##.

The first term contains ##\delta_{jm} \, \epsilon_{ijm}##, with ##\delta_{jm}## symmetric in ##j## and ##m##, and ##\epsilon_{ijm}## antisymmetric in ##j## and ##m##, so ##\delta_{jm} \, \epsilon_{ijm} = 0##.

The second term contains ##\delta_{im} \, \epsilon_{ijm}##, with ##\delta_{im}## symmetric in ##i## and ##m##, and ##\epsilon_{ijm}## antisymmetric in ##i## and ##m##, so ##\delta_{im} \, \epsilon_{ijm} = 0##.

To the OP: make sure you understand George's example with an the product of arbitrary symmetric ##A## and antisymmetric ##B##, it can be very useful.

Athenian
Wow, thank you so much for the explanation, @George Jones! This was incredibly insightful and I have managed to connect all the conceptual dots to understanding the symmetry properties.

I really appreciate the help and I'll be sure to practice more similar problems.

berkeman

## 1. What is the Einstein summation convention?

The Einstein summation convention is a mathematical notation used to simplify the writing and understanding of equations involving summation. It was introduced by physicist Albert Einstein and is commonly used in fields such as physics, engineering, and mathematics.

## 2. How does the Einstein summation convention work?

The convention states that when an index appears twice in a single term of an equation, it is implicitly summed over all possible values. This means that instead of writing out each term in a summation, the index is omitted and the summation is understood to be taken over all possible values of the index.

## 3. Why is the Einstein summation convention useful?

The convention helps to simplify and condense complicated equations, making them easier to read and understand. It also allows for the use of shorter and more concise notation, which can save time and space when writing out equations.

## 4. Are there any limitations to the Einstein summation convention?

Yes, the convention can only be applied to equations involving repeated indices. It cannot be used for equations with non-repeated indices or for equations with more than two repeated indices in a single term.

## 5. Can the Einstein summation convention be used in all branches of science?

While the convention is most commonly used in fields such as physics and engineering, it can also be applied in other branches of science where summations are used, such as statistics and computer science.

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