I'm confused about the Kronecker delta. In the context of four-dimensional spacetime, multiplying the metric tensor by its inverse, I've seen (where the upstairs and downstairs indices are the same):

$$g^{\mu\nu}g_{\mu\nu}=\delta_{\nu}^{\nu}=\delta_{0}^{0}+\delta_{1}^{1}+\delta_{2}^{2}+\delta_{3}^{3}=1+1+1+1=4.$$ But I've also seen (where the upstairs and downstairs indices are different):

$$g^{\mu\nu}g_{\nu\lambda}=\delta_{\lambda}^{\mu}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right).$$

How can there be two different answers to (what appears to me to be) the same operation, ie multiplying the metric tensor by its inverse? Apologies if I've got this completely wrong.

  • 5
    $\begingroup$ They're not the same: your first one has no free indices, your second one has two $\mu$ and $\lambda$ $\endgroup$
    – twistor59
    May 29, 2013 at 18:51
  • $\begingroup$ See also the Einstein notation entry on Wikipedia. $\endgroup$
    – Qmechanic
    May 29, 2013 at 18:52
  • $\begingroup$ Sorry, still can't see it. Don't both equation refer to multiplying a metric by its inverse? If so, why two different answers. Please don't worry about making your answers too simple! $\endgroup$
    – Peter4075
    May 29, 2013 at 19:01
  • 1
    $\begingroup$ The second equation refers to multiplying a metric by its inverse. The first one refers to multiplying a metric by its inverse, and then taking its trace. $\endgroup$
    – Mike
    May 29, 2013 at 19:02
  • $\begingroup$ Or, to paraphrase Mike's comment and Jerry's answer, if you do the second one, you got yourself a matrix. The first one is just got by adding up the diagonal elements of that matrix. The "1"s in your first equation are the same "1"s that are arranged diagonally in the second. $\endgroup$
    – twistor59
    May 29, 2013 at 19:05

2 Answers 2


In terms of your ordinary matrix multiplication, you have, for the case of a 4x4 matrix $M = g_{ab}$:

$M\cdot M^{-1} = I$, which is the same thing as $g_{ab} g^{bc} = \delta_{a}{}^{c}$


$Tr\left(M\cdot M^{-1}\right) = 4$, which is the same thing as $g_{ab}g^{ab} = \delta_{a}{}^{a} = 4$


It's useful to know how matrix multiplication is defined:

For $n \times n$ matrices, $A$ and $B$, denote the entry in the $i$th row and $j$th column by $A^i_j$ and $B^i_j$ respectively. Then for $C = AB$, the entries are given by $$C^i_j = A^i_kB^k_j$$ (summation convention of course), which you can check by working out a few examples.

Now when we have a matrix and it's inverse, multiplying them together yields the identity matrix, or using the definition above:

$$A^i_k (A^{-1})^k_j = \delta^i_j,$$ since the entries of the identity matrix are given by the Kronecker delta symbol.

The trace of a matrix $A$ is simply given by $Tr(A) = A^i_i$. In the case where the matrix $C$ is a product, combining the two formulas (for trace and matrix multiplication), it's trace would be given by $Tr(C) = C^i_i = A^i_kB^k_i$, which is what you're doing in the first case.


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