Suppose that $M$ is a differentiable manifold which represents spacetime. Located at ever point $p$ in spacetime there exists a tanget space at that point. For any tanget space there also exists a metric that is a member of that tangent space. The metric follows as $$g_{\mu\nu}(x) dx^\mu \otimes dx^\nu,$$ where $x$ is really $$x=x^\alpha=(x^0,x^1,x^2,x^3),$$ so the metric is a function of the spacetime. This is commonly summerized as just $g_{\mu\nu}$. It is common in general relativity that something called einstein summation convention is used. For example, with the christoffel symbol $$\Gamma^{\rho}_{\alpha\beta}A_\rho \implies \sum_{\rho=0}^3\Gamma^{\rho}_{\alpha\beta}A_\rho.$$ This is known as the einstein convention, where the $\rho$ index is refered to as a "dummy index". Question is does this same logic apply for the metric $$g_{\mu\nu}(x) dx^\mu \otimes dx^\nu,$$ or do the indicies here not contract?

  • 2
    $\begingroup$ What would that expression mean if there wasn't the summation convention in effect? $\endgroup$
    – ACuriousMind
    Dec 19, 2021 at 23:15

2 Answers 2


In short, yes, the convention does apply, but let me make a few remarks before I get to the main point:

  • the tangent space at a point is not an element of the manifold, nor a subset. More specifically, the expression $T_p \in M$ is wrong. $T_p$ can be seen, from an extrinsic point of view, as the hyperplane tangent to the manifold at $p$
  • the metric is not an element of the tangent space. It is a $(0,2)$-type tensor-field defined on the tangent space, meaning that for each point $p \in M$ it is an element of $T_p^* \otimes T_p^*$. Equivalently, it is a bilinear map $g \colon T_p \times T_p \to \mathbb{R}$. It does depend on spacetime by means of its dependency on the point $p$
  • $x = x^{\alpha} = (x^0, x^1, x^2, x^3)$ in a coordinate chart. In general, notice that the metric depends on the spacetime point, which can't be split in this manner without a choice of coordinates. This is a subtle remark, but it is important to recall that coordinates are a choice, not a fundamental structure on the manifold

Finally, the expression $g_{\mu\nu}(x) \textrm{d}x^\mu \otimes \textrm{d}x^\nu$ does employ Einstein's convention. The object $g = g_{\mu\nu}(x) \textrm{d}x^\mu \otimes \textrm{d}x^\nu$ is the metric itself, the tensor, the bilinear map I described above. $g_{\mu\nu}(x)$ are its components on the coordinate chart $x^\alpha$. The idea is pretty similar to defining the vectors $$e_1 = \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}, \quad e_2 = \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}, \quad e_3 = \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$$ and then writing the vector $$v = \begin{pmatrix}v^1 \\ v^2 \\ v^3\end{pmatrix}$$ as $$v = v^1 e_1 + v^2 e_2 + v^3 e_3 = \sum_{i=1}^{3} v^i e_i = v^i e_i,$$ the only difference being that the basis for the metric is provided by the objects $\textrm{d}x^\mu \otimes \textrm{d}x^\nu$ instead of the $e_i$. The $\textrm{d}x^\mu$ are the basis for the cotangent space at $p$, $T_p^*$, and hence $\textrm{d}x^\mu \otimes \textrm{d}x^\nu$ provide the basis for $T_p^* \otimes T_p^*$ which, as mentioned above, is the space where the metric "lives".


The metric, written invariantly is:

$g: XM \otimes_M XM \rightarrow M\mathbb{R}$

Here, $XM$, is the sheaf of tangent fields, that is the sheaf of sections of the tangent bundle, $TM$. The metric, locally, and which means in any chart, is expressed as you have written in it. And this is normally how it's written in the physical literature. And this means that the Einstein convention is used. It's used less often in the math literature.


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