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This is a follow up question to my answer here How do you define the Euler Characteristic of a scheme?

A real analytic space is a ringed space locally isomorphic to $(X,O/I)$ where $X$ is the zero locus of some number of real analytic functions $f_1,\ldots, f_k$ on an open set $U$ of $\mathbf{R}^n$, $O$ is the sheaf of germs of real analytic functions on $U$ and $I$ is the ideal sheaf generated by $f_1,\ldots, f_k$ (see e.g. http://eom.springer.de/a/a012430.htm) I would like to ask if it is true that each real analytic space with a countable base can be embedded as a closed analytic subset of some Euclidean space.

The motivation behind this comes from the triangulation theorem for complex algebraic varieties: the only proof of that that I know of (Hironaka's 1974 notes) is based on triangulating analytic subvarieties of Euclidean spaces. So to apply this one must embed a complex algebraic variety as a real subvariety of a Euclidean space. This is easy for projective varieties and is probably possible in general, but I don't know a reference for the general case. (I'm mainly interested in the complex algebraic case, but I don't see why it should be any easier that embedding arbitrary real analytic spaces; however if it is easier, I'd be interested to know.)

A related question: is it possible to prove the triangulation theorem (for complex algebraic varieties or in general) without using embeddings in Euclidean spaces?

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  • $\begingroup$ I imagine Hironoka's triangulation theorem is a mild tweak of Whitehead's theorem that smooth manifolds admit triangulations. The idea of the proof is to take any embedding of the manifold $M$ into Euclidean space, subdivide a triangulation of Euclidean space sufficiently and keep $M$ transverse to the strata of the triangulation, then pull back the stratifications back to $M$, giving a smooth polyhedral decompostion of $M$. Subdivide to a triangulation of $M$. So you really only need an embedding of $M$ in some object that admits a triangulation, and enough flexibility to get transversality. $\endgroup$
    – Ryan Budney
    Aug 19, 2010 at 6:09
  • $\begingroup$ Are you asking if a real analytic space is a real affine algebraic variety? What kind of embedding are you happy with? There's a characterisation of closed subsets of Euclidean space as something like all spaces that have finite Lebesgue covering dimension, are Hausdorff and are 2nd countable. Perhaps I'm forgetting a criterion, but it's something that appears in many point-set topology texts. $\endgroup$
    – Ryan Budney
    Aug 19, 2010 at 6:30
  • $\begingroup$ Ryan -- Hironaka's proof (based on an earlier proof by Lojasiewicz) is by projecting and using induction on the dimension. The idea you mention probably works as well, but some modifications will be necessary: eg the fact that a triangulation is transverse to each stratum does not guarantee a polyhedral decomposition: take a simplex in the 3-space that contains the vertex of a quadratic cone and intersects transversally each stratum. In general the local structure of real analytic spaces can be pretty messy. $\endgroup$
    – algori
    Aug 19, 2010 at 12:44
  • $\begingroup$ Re what kind of embedding I'm looking for: as a closed analytic subset. Will clarify that in the posting. I'm not sure point set topology does the trick since it is about continuous embeddings and the image can be a complete mess to which the triangulation theorem does not apply. $\endgroup$
    – algori
    Aug 19, 2010 at 12:50
  • $\begingroup$ @algori:There is a proof of triangulation of real analytic spaces by B Giesecke Math Zeutschrift vol 83 ,pages 177-213 yr 1964 $\endgroup$
    – Mohan Ramachandran
    Jun 13, 2011 at 16:08

2 Answers 2

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If you just want a proper 1-1 real analytic map whose image is a real analytic variety then the result is theorem 2 page 593 of a paper of Tognoli and Tomassini in Ann.Scuola.Norm.Pisa (3) vol 21 yr 1967 pages 575-598. This means there is no control over the differential of the map.I am assuming that the real analytic space has finite dimension.If the dimension of the Zariski tangent spaces of a connected reduced real analytic space is bounded then it can be real analytically embedded in euclidean space, see paper by Aquistapace Broglia and Tognoli Ann.Scuola.Norm.Pisa (4) vol 6 yr 1979 p 415-426.

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  • $\begingroup$ The second reference I mention answers question 1 for complex algebraic varieties.Since all these proofs eventually use embedding theorem for stein spaces or manifolds one needs the bound on the dimension of Zariski tangent spaces.This is true for abstract real algebraic varieties. $\endgroup$
    – Mohan Ramachandran
    Aug 23, 2010 at 15:51
  • $\begingroup$ Remark:A necessary condition to embed a real analytic space as a real analytic subspace of Euclidean space is that the dimensions of the Zariski tangent spaces is bounded. $\endgroup$
    – Mohan Ramachandran
    Aug 23, 2010 at 17:44
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I do not know an answer to the embedability question, but the triangulation can be deduced from the existence of a (Thom-Mather) stratification by [Johnson "On the triangulation of stratified sets and singular varieties", Trans. Amer. Math. Soc. 275 (1983), no. 1, p. 333–343] or [Goresky "Triangulation of stratified objects", Proc. Amer. Math. Soc. 72 (1978), no. 1, p. 193–200]; moreover it is known that analytic subvarieties of Euclidean space are Whitney stratified [Whitney "Local properties of analytic varieties", Differential and Combinatorial Topology (A Symposium in Honor of Marston Morse), Princeton Univ. Press, Princeton, N. J., 1965, p. 205–244 & "Tangents to an analytic variety", Ann. of Math. (2) 81 (1965), p. 496–549].

Moreover in [Mather "Notes on topological stability", 1970, Harvard University & "Stratifications and mappings", Dynamical systems (Proc. Sympos., Univ. Bahia, Salvador, 1971), Academic Press, New York, 1973, p. 195–232] you can find the result that Whitney stratified sets are Mather stratified. Finally, Mather stratification are given by local conditions (there might be an issue gluing the local strata but I do not think so), so these result should imply that analytic varieties are triangulable. I do not think it is the most efficient to do so.

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  • $\begingroup$ Benoit -- thanks. Indeed, it sounds very plausible that one can glue a stratification of an analytic manifold out of stratifications of the neighborhoods. $\endgroup$
    – algori
    Aug 19, 2010 at 13:36

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