I have the following function:
$$ \alpha = \alpha_0 + \frac{1}{2} V_{ij}V_{ij} + \frac{1}{3} V_{ij}V_{jk}V_{ki}+\frac{1}{8}V_{ij}V_{ij}V_{kl}V_{kl} $$
which is a scalar and depends on the $V$ matrix:
$$ V_{ij} = \beta( S_{ij} – \mathbb{I}_{ij})$$
where $\mathbb{I}_{ij}$ is the identity matrix. At first glance I thought the term $V_{ij}V_{ij}$ was a simple matrix multiplication, but it turns out that we can write it as:
$$ V_{ij}V_{ij} = \beta^2\textrm{Tr}( S_{ij} – \mathbb{I}_{ij})^2 \space\space\space\space\space\space\space\space\space\space\space\space\space(1)$$
The third term can be written as:
$$ V_{ij}V_{jk}V_{ki} = \beta^3\textrm{Tr}( S_{ij} – \mathbb{I}_{ij})^3 \space\space\space\space\space\space\space\space\space\space\space\space\space(2)$$
and the last one:
$$ V_{ij}V_{ij}V_{kl}V_{kl} = \beta^4 \left[\textrm{Tr}( S_{ij} – \mathbb{I}_{ij})^2 \right]^2 \space\space\space\space\space\space\space\space\space\space\space\space\space(3)$$
Using the Einstein summation convention, I know that when we have repeated indicies, there's an implied summation over that index but I dont see how the traces arise from here.
How can I obtain equations $(1)$,$(2)$ and $(3)$?
Best Answer
If we interpret the Einstein notation: $$V_{ij}V_{ij} := \sum_i\sum_j V_{ij}^2$$ Thus $V_{ij}V_{ij}$ is the sum of the squares of all the $V_{ij}$. Note that if you treat the $V_{ij}$ as the components of a matrix $$V = \begin{bmatrix} V_{11}&V_{12}&\dots&V_{1n}\\ V_{21}&V_{22}&\dots&V_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ V_{n1}&V_{n2}&\dots&V_{nn}\end{bmatrix}$$
Then $$V^TV = \begin{bmatrix} V_{11}^2 + V_{12}^2 + \dots + V_{1n}^2&&&\\ &V_{21}^2 + V_{22}^2 + \dots + V_{2n}^2&&\\ &&\ddots&\\ &&&V_{n1}^2 + V_{n2}^2 + \dots + V_{nn}^2\end{bmatrix}$$ where I've only bothered to calculate the diagonal entries. So $\text{Tr}(V^TV)$ is also the sum of the squares of all the $V_{ij}$. That is: $$V_{ij}V_{ij} =\text{Tr}(V^TV)$$
Now $\beta^2\text{Tr}(S_{ij} - \Bbb I_{ij})^2$ is an unholy mashup of matrix and tensor notations, and one can hardly be blamed for having difficulty in interpreting it. But since $V_{ij} =\beta(S_{ij} - \Bbb I_{ij})$, you should be able to see where equation $(1)$ comes from, or almost. There is no transposition in your equation $(1)$. Therefore $(1)$ only follows if $S$, and therefore $V$, is symmetric: $S_{ij} = S_{ji}, V_{ij} = V_{ji}$ for all $i,j$. Presumably you either overlooked this information, or at least forgot to include it in your question.
Skipping over equation $(2)$ for the moment, let's look at $(3)$. $$\begin{align}V_{ij}V_{ij} V_{kl}V_{kl}&:= \sum_i\sum_j \sum_k\sum_lV_{ij}^2V_{kl}^2 \\&= \left(\sum_i\sum_jV_{ij}^2\right)\left(\sum_k\sum_lV_{kl}^2\right)\\&=\text{Tr}(V^2)\text{Tr}(V^2)\\&=\left[\text{Tr}(V^2)\right]^2\end{align}$$ (since we now know $V^T = V$). And sure enough equation $(3)$ is just the square of equation $(1)$.
For equation $(2)$, $V_{ij}V_{jk}$ sums over the $j$, which is exactly the rule for matrix multiplication. That is, $$V_{ij}V_{jk} = \left[V^2\right]_{ik}$$ If we multiply this by $V_{kl}$, it is the rule for matrix multiplication again: $$V_{ij}V_{jk}V_{kl} = \left[V^2\right]_{ik}V_{kl} = \left[V^3\right]_{il}$$ But in $(2)$ the $l$ is replaced by $i$, and as $i$ is now a repeated index, it gets summed over as well: $$V_{ij}V_{jk}V_{ki} := \sum_i \left[V^3\right]_{ii}$$ That is the definition of the trace: $$V_{ij}V_{jk}V_{ki} = \text{Tr}(V^3)$$ which holds even if $V$ is not symmetric. And if you plug in the formula for $V$ in terms of $S$ and $\Bbb I$, this gives equation $(2)$.