Proving Summation with Kronecker Delta and Levi-Civita

[newton1]

i need help...
prove SUM(k) [E(ijk)E(lmk)]= d(il)d(jm) - d(im)d(jl)
where "d" is Kronecker delta symbol and "E" is permutation symbol or
Levi-Civita density

 

[HallsofIvy]

A clarification: the Kronecker delta, d(ij), is 1 if i= j, 0 otherwise.

The Levi-Civita permutation symbol, E(ijk) {real notation is "epsilon"), is 1 if ijk is an even permutation of 123, -1 if ijk is an odd permutation of 123, and 0 otherwise. While d(ij) is defined for all dimensions, E(ijk) implies that i, j, and k can only be 1, 2 ,3. For higher "dimensions" we would need more indices.

SUM(k) [E(ijk)E(lmk)]= E(ij1)E(lm1)+ E(ij2)E(lm2)+E(ij3)E(lm3)

 

[M98Ranger]

To prove this summation using Kronecker delta and Levi-Civita, we can start by writing out the sum as:

SUM(k) [E(ijk)E(lmk)]

We can then expand the first permutation symbol using its definition:

E(ijk) = d(ij)k - d(ik)j

Substituting this into the original sum, we get:

SUM(k) [d(ij)k - d(ik)j]E(lmk)

Next, we can expand the second permutation symbol in a similar way:

E(lmk) = d(lm)k - d(lk)m

Substituting this into the sum, we get:

SUM(k) [d(ij)k - d(ik)j] [d(lm)k - d(lk)m]

Using the distributive property, we can expand this sum further:

SUM(k) d(ij)kd(lm)k - SUM(k) d(ij)kd(lk)m - SUM(k) d(ik)jd(lm)k + SUM(k) d(ik)jd(lk)m

Now, let's focus on each term separately. The first term is:

SUM(k) d(ij)kd(lm)k

Since we are summing over k, we can treat d(ij) and d(lm) as constants. This means we can pull them out of the sum:

d(ij)d(lm) SUM(k) k

The sum of k from 1 to n is simply n(n+1)/2. Therefore, this term simplifies to:

d(ij)d(lm) n(n+1)/2

Similarly, for the second term, we have:

SUM(k) d(ij)kd(lk)m

Again, we can pull out d(ij) and d(lk) as constants, leaving us with:

d(ij)d(lk) SUM(k) km

The sum of km from 1 to n is simply n(n+1)/2. Therefore, this term simplifies to:

d(ij)d(lk) n(n+1)/2

For the third term, we have:

SUM(k) d(ik)jd(lm)k

Following the same steps as before, we can pull out d(ik) and d(lm) as constants, leaving us with:

d(ik)d(lm) SUM(k)