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If a real analytic variety $V$ in $\mathbb{R}^n$ is both bounded and contractible, is it true that $V$ must be a single point?

Here a real analytic variety is the set of zeros of a real analytic function $f: \mathbb{R}^n \rightarrow \mathbb{R}$.

This is certainly true if $V$ is a compact real analytic manifold (without boundary). But what about varieties that are not manifolds? Answers in other settings (complex analytic varieties or real algebraic varieties) would be interesting too.

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  • $\begingroup$ What is your definition of a "real analytic variety in $R^n$?" $\endgroup$
    – Moishe Kohan
    Commented May 9, 2023 at 13:47
  • $\begingroup$ A real analytic variety is the set of zeros of one or more real analytic functions (in this case, real-valued functions that are real analytic on $\mathbb{R}^n$). Of course, a real analytic variety that is defined by more than one real analytic function $f_1, f_2, \ldots, f_k$ can be expressed as the zero set of $f_1^2 + f_2^2 + \ldots f_k^2$. $\endgroup$
    – Brian Lins
    Commented May 9, 2023 at 14:46
  • $\begingroup$ You should update your question accordingly (since there are inequivalent definitions in the literature). $\endgroup$
    – Moishe Kohan
    Commented May 9, 2023 at 14:55
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    $\begingroup$ In your edit, you seem to impose that the function is globally defined, but the usual definition of complex analytic sets is defined locally being a zero set of local analytic functions. $\endgroup$
    – Z. M
    Commented May 9, 2023 at 15:05
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    $\begingroup$ I find it very surprising that somebody can ask a question about contractible spaces and not be familiar with basics of homology. I also have no idea how one can justify the sentence "This is certainly true if $V$ is a compact real analytic manifold (without boundary)" (from your post) without using homology. $\endgroup$
    – Moishe Kohan
    Commented May 13, 2023 at 17:36

2 Answers 2

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It is a consequence of Sullivan's work

Sullivan, D., Combinatorial invariants of analytic spaces, Proc. Liverpool Singularities-Sympos. I, Dept. Pure Math. Univ. Liverpool 1969-1970, Lect. Notes Math. 192, 165-168 (1971). ZBL0227.32005.

that every compact $k$-dimensional real-analytic subset $V\subset {\mathbb R}^n$ is a mod 2 pseudo-manifold: Every $k-1$-dimensional simplex in a triangulation of $V$ is contained in an even number of $k$-simplices. (This is immediate from the main result of Sullivan's paper about local structure of $V$ as a cone over a base of even Euler characteristic.) Now, take the sum of all $k$-simplices in the given triangulation of $V$. This will be a mod 2 cycle. Since there are no simplices of dimension $k+1$, this cycle is not a boundary. Hence, $H_k(V, {\mathbb Z}_2)\ne 0$. In particular, $V$ cannot be contractible. Note that this argument is pretty much the same as in the smooth case, once you have Sullivan's local result.

Sullivan's paper is freely available here.

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  • $\begingroup$ This seems like what I am looking for. Unfortunately, I'm pretty far outside my area of expertise and so I'm not sure how to derive your conclusion ($H_k(V,\mathbb{Z}_2) \ne 0$) from Sullivan's paper. $\endgroup$
    – Brian Lins
    Commented May 10, 2023 at 12:25
  • $\begingroup$ @BrianLins: See the edit. $\endgroup$
    – Moishe Kohan
    Commented May 10, 2023 at 12:57
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I think that the answer is negative for non-compact real analytic manifolds.

Consider the curve $$\gamma \colon (0, \, + \infty) \longrightarrow \mathbb{R}^2, \quad \gamma(t)=(e^{-t} \cos t, \, e^{-t} \sin t).$$

This is an injective map which is also a smooth immersion (since $\gamma'(t)$ is never zero); moreover, it is an open map, hence it is a smooth embedding. This means that the image $M$ of $\gamma$ is homeomorphic to $(0, \, +\infty)$, hence a contractible analytic submanifold of $\mathbb{R}^2$.

Geometrically, $M$ spirals to $(0, \, 0)$ when $t \to + \infty$ and approaches $(1, \, 0)$ as $t \to 0$. Hence $M$ is a bounded subset of $\mathbb{R}^2$, see the picture below.

enter image description here

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  • $\begingroup$ Seemingly something easier is possible: $\arctan\colon\mathbb R\to(-1,1)$ seems to be a homeomorphism and real analytic? $\endgroup$
    – Z. M
    Commented May 9, 2023 at 14:51
  • $\begingroup$ Your manifold $M$ is not closed, but a real analytic variety (as defined belatedly above) would always be closed. $\endgroup$
    – Brian Lins
    Commented May 9, 2023 at 14:52
  • $\begingroup$ @BrianLins: well, when I wrote this example no precise definition was given. Anyway, I will leave the answer, which belongs to an enlarged version of the "other settings" category... $\endgroup$
    – Francesco Polizzi
    Commented May 9, 2023 at 14:55

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