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Consider Gibbons and Hawkings paper wherein a Riemannian metric $\overset{\mathcal{R}}{g}_{\mu\nu}$ and everywhere well defined normalized line field $l_{\mu}$ on spacetime $M$ may be used to construct a Lorentzian metric $\overset{\mathcal{L}}{g}_{\mu\nu}$ on M.

$$\overset{\mathcal{L}}{g}_{\mu\nu}=\overset{\mathcal{R}}{g}_{\mu\nu}-2l_{\mu}l_{\nu}$$

Suppose we choose everwhere orthonormal tetrads for both our Riemannian and Lorentzian metrics.

$$\begin{array}{ccc} \overset{\mathcal{L}}{g}_{\mu\nu}=e_{\mu}^{a}\eta_{ab}e_{\nu}^{b} & , & \overset{\mathcal{R}}{g}_{\mu\nu}=E_{\mu}^{a}\mathbb{I}_{ab}E_{\nu}^{b}\end{array}$$

Where $\eta_{ab}$ and $\mathbb{I}_{ab}$ are the Minkowskian and Euclidean metric components respectively. The problem I'm working on is: I would like to find an expression relating a local Lorentzian coframe to it's Euclidean counterpart $e^{a}$ and $E^{a}$ respectively.

We could also choose to represent our line field in terms of the flat space basis:

$$l_{\mu}=E_{\mu}^{a}l_{a}$$

Then we have for expression [eq:1]:

$$e_{\mu}^{a}\eta_{ab}e_{\nu}^{b}=E_{\mu}^{a}\left(\mathbb{I}_{ab}-2l_{a}l_{b}\right)E_{\nu}^{b}$$

But in flat space basis, locally we can also use Gibbons and Hawkings formula and write:

$$\eta_{ab}=\left(\mathbb{I}_{ab}-2l_{a}l_{b}\right)$$

Implying that we can always choose $e_{\mu}^{a}=E_{\mu}^{a}$ locally. Then the problem seems to boil down to the Clifford algebra underlying the flat space metric components? Clearly in an adapted frame, three of the four components are unaffected by the change. For the fourth, I keep getting spinor like quantities..

Since multiplying a clifford matrix by $i$ changes the diagonal metric component sign, this seems to involve complexifying the $\gamma^0$ component or in a general coordinate system, complexifying the Whole Clifford algebra itself which I suppose makes sense since then the Lorentzian and Euclidean Clifford algebras are isomorphic. I'm guessing we're then working with complexifications of the clifford algebra (and hence any structure groups of frame bundles we're looking at) Any help is appreciated.

Another viewpoint is that we might possibly consider the line field as a complex $su(2)$ field so that in making metric components we get a negative sign??? We might then view the field as deforming the Euclidean frame to a lorentzian one?

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  • $\begingroup$ Maybe I am being pedantic, but the title of your question should be “… on the same manifold”, since “spacetime” is a term of Lorentzian geometry. $\endgroup$
    – A.V.S.
    Sep 10, 2023 at 17:50
  • $\begingroup$ @A.V.S. took your suggestion, I just wanted to avoid it getting tagged as a pure math question. Thanks $\endgroup$
    – R. Rankin
    Sep 11, 2023 at 0:10
  • $\begingroup$ @A.V.S. mostly I'm interested in writing EH action in terms of Riemannian metric/tetrads and a field deforming it to Lorentzian. $\endgroup$
    – R. Rankin
    Sep 11, 2023 at 0:16
  • $\begingroup$ @A.V.S. there might be a problem with setting lorentzian and Riemannian verbeins equal.. since any two lorentzian verbeins are related by a local $SO(1,3)$ transform and we'd expect the $SO(4)$ equivalent for Riemannian verbeins? $\endgroup$
    – R. Rankin
    Sep 11, 2023 at 0:22
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    $\begingroup$ Have you seen recent work by Visser? At the very least bibliography might contain other works of note. $\endgroup$
    – A.V.S.
    Sep 11, 2023 at 6:21

1 Answer 1

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Why are you assuming that $\eta_{ab} = (\mathbb{I}_{ab}-2l_{a}l_{b})$? The quantity $\mathbb{I}_{ab}-2l_{a}l_{b}$ gives you a Lorentzian metric, but could be different from Minkowski. For example, you may have non-diagonal elements due to the vector field.

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  • $\begingroup$ Note in my question that $l$ is a normalized line field. Then a locally Euclidean basis can always be chosen parallel to $l$ resulting in the minkowskian metric. The normalization is important for this to occur of course. $\endgroup$
    – R. Rankin
    Sep 13, 2023 at 5:58
  • $\begingroup$ For the general case, I guess you could say that it's always within an $SO(4)$ rotation of being the minkowskian metric. However it is easier to just choose basis such that they're aligned $\endgroup$
    – R. Rankin
    Sep 13, 2023 at 6:15

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