So, I was chatting with a friend and we noticed something that might be very, very, very stupid, but I found it at least intriguing.

Consider Minkowski spacetime. The trace of a matrix $A$ can be written in terms of the Minkowski metric as $\eta^{\mu \nu} A_{\mu \nu} = \eta_{\mu \nu} A^{\mu \nu} = A^\mu_\mu$.

What about the trace of the metric? Notice that $\eta^\mu_\mu$ cannot be written as $\eta_{\mu \nu} \eta^{\mu \nu}$, because this is equal to $4$, not $-2$. It seemed to us that there is some kind of divine rule that says "You shall not lower nor raise indexes of the metric", because $\eta^{\mu \nu} \eta_{\nu \alpha} = \delta^\mu_\alpha \neq \eta^\mu_\alpha$. Is the metric immune to index manipulations? Is this a notation flaw or am I being ultra-dumb?

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    $\begingroup$ $\delta^\mu{}_\nu=\eta^\mu{}_\nu$ $\endgroup$
    – Ryan Unger
    Oct 13, 2015 at 22:53
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    $\begingroup$ The trace invariant under Lorentz transformations is not the trace of the matrix $(A_{\mu\nu})$, but the trace of the matrix $(A_\mu^\nu)$. Those may both be valid concepts arising here and there, but they will certainly not give the same value in general. (The Lorentz invariant trace is more commonly encountered). $\endgroup$ Oct 13, 2015 at 23:18

2 Answers 2


I know this is an old and already answered question, but I thought I'd elaborate a bit on what's going on "behind the scenes." When dealing with rank-two tensors, it's sometimes better not to think of matrices at all, because that blurs the distinction between something like $A_{\mu \nu}$ and something like $A^\mu_{\ \ \nu}$, which are very qualitatively different beasts. $A_{\mu \nu}$ is a bilinear form - a bilinear map that inputs two vectors and outputs a scalar, or equivalently, a linear map that inputs a vector and outputs a one-form. $A^\mu_{\ \ \nu}$ is a linear operator - a bilinear map that inputs a vector and a one-form and outputs a scalar, or equivalently, a linear map that inputs a vector and outputs another vector. They're very different, because for a bilinear form, the input and output live in two completely different spaces and can't be added or directly compared. (Think of it as being like a non-square matrix.)

If you think back to your linear algebra course, you'll probably remember that the matrices were always considered as linear operators. And in fact, many linear algebra tools, like eigenvectors, eigenvalues, trace, determinant, etc., are only defined for linear operators.

So in fact, you can't take the trace of a bilinear form $A_{\mu\nu}$. Nor can you define its eigenvalues or eigenvectors. Sure, you can write it out as a matrix in some particular basis, pretend it was a linear operator, and then find the eigenvalues of that linear operator, but the result will be basis-dependent and therefore of no physical interest. When you say you can "use the metric to take the trace $\eta^{\mu \nu} A_{\mu \nu}$ of the bilinear form $A_{\mu \nu}$," what you're really doing is (a) using the metric to convert the bilinear form into a linear operator, and then (b) taking the trace of that linear operator without using the metric. In fact, the trace of a linear operator is a metric-independent concept, because the indices are already in the right places and ready to be contracted.

(The determinant is actually a bit of a subtle case, because even though the metric $g_{\mu \nu}$ is a bilinear form, for non-Cartesian coordinate systems, the determinant $\det g$ of the metric is a valid and useful concept. It's not a tensor though, it's a "tensor density." That's another story.)

Anyway, the metric's "job" is to raise and lower indices, so if it's not actually changing index heights then it should act trivially. So the one-index-up, one-index-down form of the metric is always the identity linear operator $\delta^\mu_{\ \ \nu}$. That's true in any coordinate system, curved spacetime, whatever. That's why people almost always use the notation $\delta^\mu_{\ \ \nu}$ instead of $\eta^\mu_{\ \ \nu}$. That's also why the two-upper-index tensor $\eta^{\mu \nu}$ is defined to the inverse of the metric $\eta_{\mu \nu}$. (Again, note the subtlety - the metric inputs a vector and outputs a one-form, so the inverse metric must input a one-form and output a vector.)

So now the answer to your question is clear: the trace of the metric is always just $\delta^\mu_{\ \ \mu} = d$, the number of spacetime dimensions. Again, true in any coordinate system, any metric signature, curved spacetime, what have you. That fact that the trace of the matrix representation of $\eta_{\mu \nu}$ is 2 has no physical significance. (Well, okay, fine, in Cartesian coordinates on flat spacetime it's the summed metric signature, which is physically significant. But in general it has no physical significance.)

  • $\begingroup$ If eigenvalues are only defined for linear operator, why does the metric have eigenvalues when its not a linear operator $\endgroup$
    – Shashaank
    Apr 30, 2021 at 19:42
  • $\begingroup$ @Shashaank Strictly speaking, the metric tensor does not have eigenvalues, exactly because it isn't a linear operator. What it does have is a metric signature. This is sometimes loosely defined as "the number of positive, negative, and zero metric eigenvalues", but this is not quite correct. The eigenvalues aren't actually the eigenvalues of the metric itself, but of the formal linear operator that arises if you essentially ignore the difference between forms and vectors. Look up "Sylvester's law of inertia" for a more rigorous discussion. $\endgroup$
    – tparker
    May 1, 2021 at 0:59
  • $\begingroup$ thanks, I will have a look. But just one clarification more. For eg: in Schwarzschild metric you can calculate the signature easily but if you write it in Eddington Finkelstein coordinates the metric is not diagonal. So I will need to calculate the eigenvalues. I haven't done it yet for 4 by 4 matrix, but would the eigenvalues I get be exactly the same as in the Schwarzschild case ( I know I should get the same signature in the end) ..and is the matrix that I write for $g$ a linear operator or a bilinear form. Let me know these 2 things before I move to the theorem os Sylvester $\endgroup$
    – Shashaank
    May 1, 2021 at 4:22
  • $\begingroup$ @Shashaank No, the eigenvalues will not be the same as in Schwartzchild coordinates (although the signs will be the same). The metric $g$ is a bilinear form. Strictly speaking, a matrix itself is neither a linear operator nor a bilinear form, but merely a representation of one or the other. In the case of the metric $g$, the matrix always represents a bilinear form and not a linear operator. $\endgroup$
    – tparker
    May 1, 2021 at 14:45
  • $\begingroup$ thanks, I was expecting the eigenvalues to be different since the metric is a bilinear form for which eigenvalues kind of thing should be a basis dependent thing, but in your 1st comment, you say “ but of a formal linear operator that arises....” ...is the matrix representation of the formal linear operator different from the matrix representation of the metric and Will the eigenvalues of that formal linear operator the same in any basis. $\endgroup$
    – Shashaank
    May 1, 2021 at 14:52

The mistake you made is this: $\eta^{\mu}_{\nu} \neq \eta_{\mu\nu} $. When you raise index $\mu$ from downstairs to upstairs, the matrix elements change. $\eta^{0}_{0} = 1$, $\eta_{00} = -1$. That is why if you take the trace of $\eta_{\mu\nu}$, you get 2, but if you take the trace of $\eta^{\mu}_{\nu}$ you get 4.

  • $\begingroup$ Thanks for the answer. I think it solved my problem. This would mean that $\eta^\mu_\mu = \eta^{\mu \nu} \eta_{\mu \nu} = 4$ and $\eta^{\mu \mu} = \eta^\mu_\nu \eta^{\nu \mu} = -2$, right? $\endgroup$ Oct 13, 2015 at 23:33
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    $\begingroup$ if you are using the (1,-1,-1,-1) convention (the particle physics convention)... I was using the (-1,1,1,1) convention which is more common for gravitational physicists. $\endgroup$ Oct 13, 2015 at 23:36
  • $\begingroup$ actually your notation is a bit confusing... Einstein's convention is that summation is only performed when the same index appears both upstairs and downstairs. So Einstein's convention only take traces for (1,1) tensors. For (0,2) or (2,0) tensors, the convention doesn't work... so in order to be consistent, you should try to avoid writing $\eta^{\mu\mu}$ $\endgroup$ Oct 13, 2015 at 23:39
  • $\begingroup$ So how would I express this trace? Since $\eta^\mu_\mu \neq \eta^{\mu \mu}$ I cannot interchange them and use the proper notation. $\endgroup$ Oct 13, 2015 at 23:44
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    $\begingroup$ well. The point of Einstein Summation Convention is that any sum performed by this convention is an INVARIANT quantity. So it is very important! But if you want some other quantities, you have to resort to regular summation $\endgroup$ Oct 13, 2015 at 23:48

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