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Recently in more technical settings (I was learning algebraic QFT), I encountered the term "real analytic" manifolds (Lorentzian manifolds, to be precise). This is in contrast to smooth manifolds, the same way real analytic functions are not the same as smooth ($C^\infty$) functions.

What I have trouble wrapping my head around are "simple" examples of analytic Lorentzian manifolds that differ from standard smooth Lorentzian manifolds. This issue is related to another post here, but I guess I am looking for something more concrete (and possibly simpler). I can frame this in two ways:

  1. Are the standard examples in general relativity considered real-analytic manifolds? The ones I encounter most of the time---Minkowski, Schwarzschild, FRW spacetimes, seem to be the case: the Schwarzschild geometry has $f(r)=1-2M/r$. The FRW with good enough scale factors (like $a(\eta)=-1/(H\eta)$ for de Sitter case) is also real-analytic on $\eta<0$. I believe the same goes for the entire Kerr-Newman family and its maximal extension. In the Stackexchange post above, the argument seems to be a matter of talking about "the" maximal analytic extension, but if I just want to talk about the exterior of Schwarzschild or exterior of a star, for example, is real-analyticity really too strong? Another example I can probably think of is "gravitational shockwaves" (e.g., Aichelburg-Sexl metric) where the metric is not real-analytic (it is distributional, I guess). Does that mean all spacetimes whose metric can be written in terms of elementary real functions automatically real-analytic then?

  2. If the spacetime is real-analytic, does the test fields on top of it have to be? For example, must hyperbolic wave equation be "real-analytic" (e.g., with real-analytic potentials) if the metric is real analytic? It looks to me that usually the fields living on some spaces does not need to inherit the properties of the background geometry, but I don't know.

Qmechanic
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Evangeline A. K. McDowell
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1 Answers1

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We only impose $C^{\infty}$ when setting up the general theory, because real-analyticity would be too strong a condition to impose. However, when trying to find specific solutions (Schwarzschild, Reissner Nordstrom etc etc) it turns out that we end up with a set that can be covered by charts whose transitions are real-analytic; so it is by a miracle that we end up with a real-analytic structure. Also, the metric when expressed in any coordinate system in these examples has real-analytic components. Therefore, it is a very fortunate matter that having solved Einstein's equations in one chart automatically means it is satisfied on the entire manifold.

So once again, we do not require that our manifolds/fields be real-analytic simply because real-analyticity would be too strong a condition to demand a-priori (because of uniqueness of analytic continuation). If it turns out after our symmetry ansatz's and calculations that we end up with something analytic, we just thank our lucky stars and be happy with it. Hopefully this answers your second question as well: no we do not have to (nor do we) demand that fields be analytic; if we have an analytic manifold, then we're perfectly within our right to consider fields which are analytic but we can also consider $C^{\infty}$ fields, or if we want, just $C^2$ fields.

Edit:

@BenceRacskó has kindly pointed out that every $C^k$ manifold ($k\geq 1$) actually has an analytic sub-atlas, which I didn't realize. But still, having an analytic fields (in particular the metric) is non-trivial because of analytic continuation: its values on some non-empty open set (no matter how small) would determine it uniquely on the entire manifold (provided such an extension exists).

peek-a-boo
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  • and I suspect it's only for such 'simple' examples like Schwarzschild, Kerr etc where we can write things down explicitly (i.e in terms of 'elementary functions' or some special functions such as the W Lambert function or Gamma fuction) that we end up with real-analytic manifolds and metric tensors. But there's no reason to expect this to be always possible, nor do we demand this. – peek-a-boo May 15 '22 at 01:02
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    I mean look at $(\Bbb{R}^4,\eta)$ the Minkowski space. This has the structure of a vector space (the simplest possible thing), the structure of a $C^k$ manifold for all $0\leq k\leq \infty$, and also the structure of a $C^{\omega}=$ real-analytic manifold. But usually we only think of it as a vector space/smooth manifold, because we may be interested in solving equations with only smooth functions (e.g imagine a source which is concentrated in one region of space, and vanishing elsewhere; this can't be modelled using analytic functions). – peek-a-boo May 15 '22 at 01:05
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    Small comment: Every $C^k$ manifold (for $k\ge 1$ ) has a unique analytic structure, so having an analytic manifold structure is not at all miraculous. In fact you can consider analytic manifolds exclusively from the get-go. The metric being analytic is the nontrivial thing. – Bence Racskó May 17 '22 at 08:13
  • @BenceRacskó hmm I thought the statement was that every $C^k$ manifold has a $C^{\infty}$ structure... guess I misremembered the theorem – peek-a-boo May 17 '22 at 09:32
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    Apparantly it is true for analytical structure as well and the proof is probably found in Munkres' Elementary Differential Topology. I mainly know this from a remark in Hawking/Ellis on page 58 where they refer to a 1936 paper by Whitney and Munkres' book. I never did check the proof though so it is also possible that Hawking/Ellis wrote it down wrong though. – Bence Racskó May 17 '22 at 10:14
  • Ok, I checked the sources. Munkres only proves the $C^\infty$ case, but Whitney's paper does contain the proof for analytic manifolds. – Bence Racskó May 17 '22 at 10:24
  • @BenceRacskó I am not aware of analytic structure either, I know the $C^\infty$ one. Thanks! – Evangeline A. K. McDowell May 18 '22 at 15:03
  • @peek-a-boo I thought your write-up is quite convincing and sufficient for my purposes. Thanks! – Evangeline A. K. McDowell May 18 '22 at 15:03