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Let $U\subset \mathbb{R}^n$ be open, and let $f:U\to\mathbb{R}$ be real-analytic. We consider the zero set $Z:=f^{-1}(\{0\})$.

For a paper I am writing, I am looking for the best reference to the following basic fact:

If $Z$ has topological dimension equal to $d$, then $Z$ contains a real-analytic manifold of dimension $d$.

I can get this from Lojasiewicz's theorem or similar results, but that is a slightly unwieldy reference, and something probably needs to be said about how exactly one deduces it. Given that the statement is rather simple, I was wondering if someone knows of a more direct reference to this fact.

And to add a mathematical question: This result is obviously much weaker than Lojasiewicz's theorem. Is there a proof that doesn't require developing the full structure theorem?

Many thanks for any pointers!

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3 Answers 3

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Even to say it must have some topological dimension $d$ requires some background, and it's not really that unreasonable in my opinion to go the extra mile to quote something from stratification theory like Lojasiewicz's theorem.

But you can get real-analyticity in pretty short order from just knowing the stratification is in terms of topological manifolds (I think that's what you're assuming.) Let $D$ be a closed disk in a $d$-dimensional stratum, and let $x$ in $D$ be such that the zero of $f$ at $x$ has minimal order $a$. Then there's some multiindex $\alpha$ of order $a-1$ such that $\nabla(\partial^{\alpha}f)$ is nonzero at $x$ ($\alpha$ may be zero). Thus by the implicit function theorem the zero set of $\partial^{\alpha}f$ is an $n-1$-dimensional real-analytic manifold near $x$. On the other hand, by the minimality of $a$, $D$ is contained in this manifold (assuming the disk is small enough). So now you can look at $f$ as a function on this $n-1$ dimensional manifold, and use induction on the dimension to ultimately get that $D$ is a real-analytic manifold.

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  • $\begingroup$ Could you clarify what you mean by "Even to say it must have some topological dimension $d$ requires some background"? Every topological space has a topological dimension. I agree that knowing there is a stratification by topological manifolds is unlikely to be easier than quoting the full results. Basically a version of the result in the complex setting is contained in Krantz's book on several complex variables (Prop. 7.3.1). Of course here the result is much easier, and the set always has (complex) dimension n-1. I was hoping there might be a similar reference for the real-analytic case. $\endgroup$
    – Lasse Rempe
    Commented Feb 3, 2010 at 11:01
  • $\begingroup$ I do have Krantz's book... ok, the definition given there is not the one I thought you were using. You can try doing what I did up there on a neighborhood of x in f^{-1}(0); taking the zero of minimal order then passing to an n-1 dimensional submanifold via the implicit function theorem and iterating. You'll end out with some manifold... but I don't personally see how to show you can get that the dimension can be the same as the topological dimension given there, which I think he used because he was doing the complex case. $\endgroup$
    – Michael Greenblatt
    Commented Feb 3, 2010 at 15:08
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By now, your paper is probably out, but if you need the result in the future, it can be found in:

B. Malgrange: Ideals of differential functions, Oxford University Press, 1966

In Russian translation (by A. Gabrielov, Mir 1968), this is Proposition VI.3.11:

Let $X_0$ be an analytic germ at $0 \in \mathbb{R}^n$, dim $X_0=k$. Assume that $X_0$ contains a germ $V_0$ of a variety of class $\mathcal{C}^\infty$ of dimension $k$. Then $V_0$ is a germ of an analytic variety (which is an irreducible component of the germ $X_0$).

The proof uses properties of ideals generated by germs of analytic functions. Lojasiewicz' s theorem is not used directly.

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  • $\begingroup$ Thank you. Our paper is submitted, but not yet through the refereeing process, so new references will be helpful. However, before being able to apply this theorem, would we not need to know that, if $Z$ has topological dimension $d$, it contains a $C^{\infty}$-manifold of dimension $d$? Would this not require some kind of stratification result also? Apologies if I am missing something. $\endgroup$
    – Lasse Rempe
    Commented Oct 27, 2011 at 10:09
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    $\begingroup$ Sorry, I misread the assumption that Z is of topological dimension d (rather than being of dimension d as an analytic germ). Stratification results seem necessary, but I do not know what the complete argument should be. (Probably no relation, but there is some interest in complex analytic germs contained in real analytic sets (e.g. arXiv:1006.4190 Title: Tameness of complex dimension in a real analytic set Authors: Janusz Adamus, Serge Randriambololona, Rasul Shafikov)). $\endgroup$
    – Margaret Friedland
    Commented Oct 28, 2011 at 17:20
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This may be a little late, but the following reference seems relevant and also contains many other references.

@article {MR972342, AUTHOR = {Bierstone, Edward and Milman, Pierre D.}, TITLE = {Semianalytic and subanalytic sets}, JOURNAL = {Inst. Hautes \'Etudes Sci. Publ. Math.}, FJOURNAL = {Institut des Hautes \'Etudes Scientifiques. Publications Math\'ematiques}, NUMBER = {67}, YEAR = {1988}, PAGES = {5--42}, URL = {67_5_0">http://www.numdam.org/item?id=PMIHES_1988_67_5_0}, }

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  • $\begingroup$ If I recall correctly, this won't let you bypass Lojasiewicz's theorem, which is what the OP was asking about. $\endgroup$
    – Thierry Zell
    Commented Sep 20, 2011 at 16:39

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