Inspired by comment discussions in this MO post smooth version of splitting principle we ask:
Are there two compact real analytic manifolds $M,N$ of dimension $m>n$ such that there is not any analytic surjection $f:M \to N$?
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2$\begingroup$ Hint: You can approximate any smooth map by real-analytic ones. $\endgroup$– Moishe KohanCommented Jul 23 at 22:16
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1$\begingroup$ @MoisheKohan according to Theorem 1.4 in ALGEBRAIC APPROXIMATION OF SMOOTH MAPS by Bochnak and Kucharz core.ac.uk/download/pdf/222575527.pdf there are almost always a lot of smooth maps between smooth real algebraic varieties that cannot be approximated by algebraic maps. It would be very strange if it was possible in analytic category: by Nash-Tognoli embedd everything into $R^N$, realize an analytic map as a map of $R^N$ and approximate it by algebraic maps obtained by truncating the corresponding series. $\endgroup$– Dmitrii KorshunovCommented Jul 24 at 3:31
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1$\begingroup$ @MoisheKohan Thanks for the reference. But as far as I can see it addresses the approximation of $C^k$-functions by smooth functions. The proof uses Lemma 4.1 which involves partition of unity etc. $\endgroup$– Dmitrii KorshunovCommented Jul 24 at 4:09
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4$\begingroup$ @DmitriiKorshunov: Yes, I realized that it was a wrong reference. The correct one (for compact manifolds) is H.L. Royden: The analytic approximation of differentiable mappings, Math. Ann., 139(1960), 171-179. $\endgroup$– Moishe KohanCommented Jul 24 at 4:11
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2$\begingroup$ @MoisheKohan yes, that works for compact manifolds. although Royden states only the existence of an analytic function in the same homotopy class, the proof actually gives density in the uniform topology. For the benefit of future readers: there is also a textbook reference is Guaraldo, Macri, Tancredi "Topics on Real Analytic Spaces" Chapter VII Theorem 2.25 has all one might wish for -- without compactness assumption and in strong $C^\infty$ topology. $\endgroup$– Dmitrii KorshunovCommented Jul 25 at 22:02
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1 Answer
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- Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Whitney-Grauert-Morrey embedding theorem) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
- Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ again by Whitney-Grauert-Morrey to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
- If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until $fg$ becomes surjective.
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3$\begingroup$ Thank you very much for your very perfect and interesting answer. $\endgroup$ Commented Jul 24 at 8:24