I think that there are zero degrees of freedom in a flat metric, but I do not know how to count them. I know that any symmetric metric tensor has $n(n-1)/2$ degrees of freedom, where $n$ is the dimension, but if the metric is constrained to flat, I do not know if any degrees of freedom exist.

The Riemann tensor confuses me, because it contains $n^2(n^2-1)/12$ number of independent components for a given metric. Setting all of them to zero yields $n^2(n^2-1)/12$ equations to be solved, yet the metric only has n(n+1)/2 unique tensor elements, which creates an over-determined system of equations.

How many degrees of freedom are in a flat metric and how does one count them?


1 Answer 1


Notice that there is no Riemann tensor for an inner product space. Moreover, at one chosen point $p \in M$ the metric inner product in $T_p M$ can always be made Euclidean. In other words, the condition for a metric to be flat is local, not pointwise.

To see what is going on, one may look at the Taylor expansion of the metric. In special ("geodesic" or "normal") coordinates, the coefficients of this expansion are given in terms of the Riemann tensor, so if the latter vanishes (in a neighborhood of $p$), the metric is locally Euclidean.

Putting this to your perspective, the Riemann tensor gives conditions not only to the components of the metric, but also to their derivatives.


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