Proving that $A^{[ab]}B_{ab} = A^{ab}B_{[ab]}$

Question

On this problem sheet, I am trying to show that for tensors (given a vector space V and it's dual $V^{*}$) $$A:V^{*} \times V^{*} \to \mathbb{R} \\ B: V \times V \to \mathbb{R}$$ that $A^{[ab]}B_{ab} = A^{ab}B_{[ab]}$ where $A^{[ab]} := \frac{1}{2}(A^{ab} - A^{ba})$. This seems like a straightforward calculation, but I'm not understanding the following $(*)$ line in the solution:

\begin{align*} A^{[ab]}B_{ab} &= \frac{1}{2}(A^{ab}B_{ab}-\color{red}{A^{ba}B_{ab}}) \\ &= \frac{1}{2}(A^{ab}B_{ab}-A^{mn}B_{nm}) \tag{$*$} \\ &= \frac{1}{2}(A^{ab}B_{ab}-A^{ab}B_{ba}) \\ &= A^{ab}B_{[ab]} \end{align*}

The instructor insists that it doesn't matter how we name the indices being summed over, and we can simply rewrite the $\color{red}{red}$ term as $$ A^{ba}B_{ab} = A^{mn}B_{nm} = A^{ab}B_{ba} $$ which ultimately gives us the result we want. Why can we simply swap the indices like this? I feel like I'm not seeing the full picture.

Answer 1

As noted in the comments:

Assume that $\text{dim }V = 3$, the renaming of indices comes from the fact that \begin{align*} A^{ba}B_{ab} &:= A^{11}B_{11} + A^{12}B_{12} + A^{13}B_{13} + A^{21}B_{21} + A^{22}B_{22} + A^{23}B_{23} + A^{31}B_{31} + A^{32}B_{32} + A^{33}B_{33}\\ &= A^{mn}B_{nm} \\ &:= A^{11}B_{11} + A^{12}B_{12} + A^{13}B_{13} + A^{21}B_{21} + A^{22}B_{22} + A^{23}B_{23} + A^{31}B_{31} + A^{32}B_{32} + A^{33}B_{33} \\ &= A^{ab}B_{ba} \end{align*} according to the Einstein summation convention.

Answer 2

Note that $A^{ba}B_{ab} = A^{11}B_{11} + A^{22}B_{22} +...+A^{dd}B_{dd}$ makes it looks like there are $d$ many terms, whereas there is actually $d^2$ many.

Since the indexing sets is finite, we can switch the order of summation and the terms, \begin{align*} A^{ba}B_{ab} &:= \sum_{b=1}^{d}\sum_{a=1}^{d} A^{ba}B_{ab} \\ &= \sum_{a=1}^{d}\sum_{b=1}^{d} A^{ab}B_{ba} \\ &= A^{ab}B_{ba} \end{align*}