Why is the Kronecker Delta crucial in Einstein summation for vector products?

[MadMax]

Using the Einstein summation convention...

Why is

$$\mathbf{a}^2 \mathbf{b}^2$$

not the same as

$$3 a_i a_j b_j b_i = 3(\mathbf{a} \cdot \mathbf{b})^2$$

given that

$$\mathbf{a}^2 = a_i \cdot a_i = a_i a_j \delta_{ij}$$

$$\mathbf{b}^2 = b_i \cdot b_i = b_i b_j \delta_{ij}$$

and

$$\delta_{ij} \delta_{ji} = 3$$

-> $$\mathbf{a}^2 \mathbf{b}^2 = a_i a_j \delta_{ij} b_i b_j \delta_{ij} = 3(\mathbf{a} \cdot \mathbf{b})^2$$

??

Any help would be much appreciated.

 

[MadMax]

Hmm I think i just figured it out maybe...

Is it 'cos you can't split the a^2 and b^2 up?

 

[Hurkyl]

$$\mathbf{a}^2 \mathbf{b}^2 = a_i a_j \delta_{ij} b_i b_j \delta_{ij} = 3(\mathbf{a} \cdot \mathbf{b})^2$$

You labelled two different dummy variables with the letter i. By doing so, you tricked yourself into thinking they were the same dummy variable.


P.S. your first clue that something is horribly wrong should have been when you had the same letter appear four times in that term as an index.

 

[MadMax]

Ahh yeah, good point :P Forgot about that. Thanks Hurkyl.