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?

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    $\begingroup$ Because in matrix notation identical indices go next to each other. That's how matrix multiplication works. $\endgroup$
    – Javier
    Commented Jan 10 at 19:42
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    $\begingroup$ You can write the matrix multiplication as $\tilde A = \eta \cdot A \cdot \eta^T$. Your second expression would be wrong without the transpose, except that $\eta$ happens to be symmetric. $\endgroup$
    – hft
    Commented Jan 10 at 20:01
  • $\begingroup$ @Javier is there any source which clearly states this? $\endgroup$
    – Tiago
    Commented Jan 11 at 9:20
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    $\begingroup$ Similar question: Confusion with Lorentz indices notation $\endgroup$ Commented Jan 11 at 11:33
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    $\begingroup$ Any linear algebra book that writes out matrix multiplication in components: $(AB)_{ij} = \sum_k A_{ik} B_{kj}$. $\endgroup$
    – Javier
    Commented Jan 11 at 14:15

3 Answers 3


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


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


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.

  • $\begingroup$ I understand that order does not matter in Einstein notation, and that I can swap the order of the indices $\beta$ and $\nu$ because of symmetry. But why is the correct order in matrix form $ (A_{\alpha\beta}) = (\eta_{\alpha\mu})(A^{\mu\nu})(\eta_{\nu\beta}) $ and not another one? $\endgroup$
    – Tiago
    Commented Jan 11 at 9:16
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    $\begingroup$ @Tiago In matrix notation we would write $C=AB$ and $D=BA$ and $C$ and $D$ are in general not the same matrix. In Einstein notation we write $C^{\alpha}_{\beta} = A^{\alpha}_{\mu}B^{\mu}_{\beta}$ and $D^{\alpha}_{\beta} = B^{\alpha}_{\mu}A^{\mu}_{\beta} = A^{\mu}_{\beta}B^{\alpha}_{\mu}$. Notice that the order of terms on the right hand side does not matter, but the order of indices and the choice of which indices we are contracting over does matter. $\endgroup$
    – gandalf61
    Commented Jan 11 at 11:44

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}

  • $\begingroup$ In the answer above, the vectors $\mathbf{b}^j$ are vectors of the reciprocal base of the base $\mathbf{b}_i$, defined as $\mathbf{b}^j \cdot \mathbf{b}_i = \delta_i^j$, so that it's easier to evaluate the dot product. $\endgroup$
    – basics
    Commented Jan 10 at 20:08
  • $\begingroup$ This does not explain why the right order is $(A_{\alpha\beta}) = (\eta_{\alpha\mu})(A^{\mu\nu})(\eta_{\nu\beta})$. $\endgroup$
    – Tiago
    Commented Jan 11 at 9:22
  • $\begingroup$ Try to compute $\mathbb{B} \cdot \mathbb{A}$ with the notation I used above. And then try to compute $\eta \hspace{-4pt} \eta \cdot \mathbb{A} \cdot \eta \hspace{-4pt} \eta$ and compare it with $\eta \hspace{-4pt} \eta\cdot \eta \hspace{-4pt} \eta \cdot \mathbb{A} $ or any other combination you like $\endgroup$
    – basics
    Commented Jan 11 at 10:38
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    $\begingroup$ I performed the product you need in the edit at the end of the answer. Now, you should easily realize that the "lq" component of the 2-nd order tensor $\eta \hspace{-4pt} \eta \cdot \mathbb{A} \cdot \eta \hspace{-4pt} \eta $ is $ \eta_{li} A^{ik} \eta_{kq}$, and thus this should solve your doubts. $\endgroup$
    – basics
    Commented Jan 11 at 10:49

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.


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