Correct order for tensor operations, when expressed in matrix form

Question

When using Einstein convention, with explicit indices, we usually write $$ A_{\alpha\beta} = \eta_{\alpha\mu}\eta_{\beta\nu}A^{\mu\nu} $$ But in matrix form, the order of the operations matters, and we often see the tensor $A$ being put in the middle, like here, when it was stated that the correct order is: $$ (A_{\alpha\beta}) = (\eta_{\alpha\mu})(A^{\mu\nu})(\eta_{\nu\beta}) $$

As matrix multiplication is not commutative, there is clearly a right order. But why is it this exact ordering and not another one?

Answer 1

One of the advantages of Einstein notation is that the order of the terms does not matter. When we write

$\eta_{\alpha\mu}\eta_{\beta\nu}A^{\mu\nu}$

we are implicitly summing over all values of the dummy indices $\mu$ and $\nu$. Each of the terms in the implicit sum is the product of real (or complex) numbers, which commute, so we could write this as

$\eta_{\alpha\mu}A^{\mu\nu}\eta_{\beta\nu}$

without changing the value of the expression.

Note, however, that the order of indices does matter. It is not in general true that

$\eta_{\alpha\mu}\eta_{\beta\nu}A^{\mu\nu} = \eta_{\alpha\mu}A^{\mu\nu}\eta_{\nu\beta}$

where we have reversed the order of indices from $\eta_{\beta\nu}$ to $\eta_{\nu\beta}$. The only circumstance in which we can reverse the order of indices without changing the value of the expression is if $\eta$ is symmetric i.e.

$\eta_{\beta\nu} = \eta_{\nu\beta} \space \forall \nu, \beta$

Fortunately, metric tensors are usually symmetric, so if $\eta$ is a metric tensor you can get away with this.

Answer 2

With Einstein notation you're writing scalar or components of vectors or tensor in general (referred to a base that you should always care about, even if you don't explicitly write it) of tensors as the sum of products of the components of other vectors/tensors

Commutation holds for product, so order is not important when using Einstein notation.

As an example, evaluating the dot product $\mathbb{A} \cdot \mathbb{B}$ of two 2nd-order tensors,

\begin{equation} \mathbb{A} = a^{11} \mathbf{b}_1\mathbf{b}_1 + a^{12} \mathbf{b}_1\mathbf{b}_2 + a^{21} \mathbf{b}_2\mathbf{b}_1 + a^{22} \mathbf{b}_2\mathbf{b}_2 = \sum_{i,j = 1}^{2} a^{ij} \mathbf{b}_i \mathbf{b}_j \\ \mathbb{B} = b_{11} \mathbf{b}^1\mathbf{b}^1 + b_{12} \mathbf{b}^1\mathbf{b}^2 + b_{21} \mathbf{b}^2\mathbf{b}^1 + b_{22} \mathbf{b}^2\mathbf{b}^2 = \sum_{i,j = 1}^{2} b_{ij} \mathbf{b}^i \mathbf{b}^j \end{equation}

you get the 2-nd order tensor

\begin{equation} \begin{aligned} \mathbb{A} \cdot \mathbb{B} = & \left( a^{11} b_{11} + a^{12} b_{21} \right) \mathbf{b}_1 \mathbf{b}^1 + \\ & \left( a^{11} b_{12} + a^{12} b_{22} \right) \mathbf{b}_1 \mathbf{b}^2 + \\ & \left( a^{21} b_{11} + a^{22} b_{21} \right) \mathbf{b}_2 \mathbf{b}^1 + \\ & \left( a^{21} b_{12} + a^{22} b_{22} \right) \mathbf{b}_2 \mathbf{b}^2 = \\ = & \sum_{i,j=1}^{2} \sum_{k=1}^{2} a^{ik} b_{kj} \mathbf{b}_i \mathbf{b}^j \ , \end{aligned} \end{equation} or, shortly and keeping in mind the base you're referring to, using Einstein notation \begin{equation} = a^{ik} b_{kj} \ . \end{equation} As you can see, each component is the sum of products of 2 (here) scalars, and the result doesn't change if you switch the order of the factors.

Edit: explicit evaluation of the product with explicitly writing the base vectors. This edit may complete the answer.

Let's evaluate the product $\eta \hspace{-4pt} \eta \cdot \mathbb{A} \cdot \eta \hspace{-4pt} \eta$, where \begin{equation} \mathbb{A} = A^{ik} \mathbf{b}_i \mathbf{b}_k \ , \ \ \eta \hspace{-4pt} \eta = \eta_{lm} \mathbf{b}^l \mathbf{b}^m \ . \end{equation}

The product reads \begin{equation} \begin{aligned} \eta \hspace{-4pt} \eta \cdot \mathbb{A} \cdot \eta \hspace{-4pt} \eta & = \left( \eta_{lm} \mathbf{b}^l \mathbf{b}^m \right) \cdot \left( A^{ik} \mathbf{b}_i \mathbf{b}_k \right) \cdot \left( \eta_{pq} \mathbf{b}^p \mathbf{b}^q \right) = \\ & = \eta_{lm} A^{ik} \eta_{pq} \mathbf{b}^l \underbrace{\mathbf{b}^m \cdot \mathbf{b}_i}_{\delta^m_i} \underbrace{\mathbf{b}_k \cdot \mathbf{b}^p}_{\delta_k^p} \mathbf{b}^q = \\ & = \eta_{li} A^{ik} \eta_{kq} \mathbf{b}^l \mathbf{b}^q \ . \end{aligned} \end{equation}

Answer 3

But why is it this exact ordering and not another one?

When using the Einstein convention the order of the factors do not matter in an expression.

However, it does have the disadvantage that the indices are explicit. We can keep the indices implicit by moving to matrix notation. This has the disadvantage that we can only use at most two indices in any factor - they are either a vector or a matrix. A further disadvantage is that we must fix a specific ordering in that indices that are summed over must be next to each other.