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hi,

I have a general question. Assume we have a real analytic $n-$dim. manifold $X$ and $M$ a real analytic compact submanifold of $X$ (of dimension less that the dimension of $X$, say $k < n$). Furthermore assume that on $M$ we have a real analytic $l-$form $\Omega$ ($l < k$). Let $\{U_{i}\}$ be a finite covering of $M$ by open sets in $X$ and on each $U_{i}$ we have a real analytic $l-$form $\alpha_{i}$ which restricted on $M$ is the form $\Omega$. Is it possible to use analycity to glue all forms $\alpha_{i}$ together to get one form $\alpha$ on a neighbourhood of the real manifold $M$?

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The question should be posed in this way: given a real analytic $l$-form $\alpha$ on $M$, is there a real analytic extension $\widetilde \alpha$ defined in an open neighborhood of $M$ in $X$? The answer is yes. In fact there is a tubular neighborhood $U$ of $M$ in $X$ and a real analytic bundle map $p : U \to M$ such that $p_{|M} = \text{id}_M$ (this can be constructed by considering the exponential map of the normal bundle of $M$ in $X$ with respect to a real analytic metric on $X$, see Morrey). Then put $\widetilde \alpha = p^*(\alpha)$.

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  • $\begingroup$ is the extension then unique? can one use analicity for uniqueness ? $\endgroup$
    – pascal
    May 9, 2012 at 13:09
  • $\begingroup$ this is actually where I am interested in . $\endgroup$
    – pascal
    May 9, 2012 at 13:10
  • $\begingroup$ The extension is not strictly unique, perhaps is so up to real analytic homotopy rel. M (depending on which analytic metric you pick). May be, is unique the real analytic homotopy class rel. M of the germ of such forms (but is a very "weak" uniqueness). $\endgroup$
    – Daniele Zuddas
    May 9, 2012 at 13:19
  • $\begingroup$ what do you mean by "real analytic homotopoy rel. M" ? $\endgroup$
    – pascal
    May 9, 2012 at 13:54
  • $\begingroup$ The fact that there is a real analytic family $\widetilde\alpha_t$, $t \in [0,1]$, connecting any two such extensions (possibly up to passing to a smaller neighborhood), such that $\widetilde\alpha_t = \alpha$ in $M$ for all $t$. $\endgroup$
    – Daniele Zuddas
    May 9, 2012 at 14:38

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