The metric tensor $g$ is a symmetric matrix. It occurred to me that one way to create symmetric matrices is to symmetrize any other matrix: $g = A^T A$. Presumably we would also get $\det A = i \sqrt{- \det g}$, which might give a more explicit meaning to the $\sqrt{- \det g}$ term that appears in many integrals.

So I'm wondering:

  • is there a name for a matrix like $A$ such that $A^T A = g$ gives a metric tensor for a particular choice of metric? I've been trying to google for it but it's hard to look it up without knowing its name, and 'square root of the metric' has lots of hits for the determinant that get in the way.
  • is there an obvious reason I'm missing as to why this isn't a useful construct?
  • $\begingroup$ See related discussion about "square roots" of curved space time metric here: physics.stackexchange.com/questions/514592/is-gravity-a-tensor/… $\endgroup$
    – MadMax
    Feb 4, 2021 at 3:36
  • $\begingroup$ the name for this matrix in 3D space is Jacobi matrix, which is die derivative of the position vector to the generalized coordinates $\endgroup$
    – Eli
    Feb 4, 2021 at 8:20

1 Answer 1


What you're asking is already well known: it's the vierbein, or tetrad field, a set of four four-vectors of components $e_{\mu}^a$ (actually its dual field) such that $$\tag{1} g_{\mu \nu}(x) = e_{\mu}^a(x) \, \eta_{ab} \, e_{\nu}^b(x). $$ So the set of components $e_{\mu}^a$ defines a matrix $A$ (your notation) that diagonalises the metric. The tetrad field is such a "square root" of the metric. Just check this page:



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