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I hope my question isn't too vague:

Let $M$ be a real-analytic compact manifold (say surface for simplicity). How big is $\Omega^k(M)$ (space of global real-analytic $k$-forms)?

Let me explain better by comparing with the complex case. If $X$ is a compact Riemann surface of genus $g$ then the $\mathbb C$-vector space of global holomorphic $1$-forms is $g$-dimensional.

This does not work for real-analytic manifolds, however there is still the rigidity given by real-analytic continuation, which reduces the number of possibilities. This gives a related question:

When does a real-analytic form $\omega$ defined on a region $U\subset M$ extend to the whole $M$?

Are there some tools to deal with such kind of problems? Any reference would be appreciated. Thank you.

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    $\begingroup$ Your $M$ embeds analytically into Euclidean space by Nash-Tognoli, I think, and then you get an infinite dimensional space of real analytic differential forms. $\endgroup$
    – Ben McKay
    Commented Apr 29, 2016 at 13:38
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    $\begingroup$ Definitely, it would be infinite dim in general. Take $M=S^1=\mathbb{R}/2\pi \mathbb{Z}$, then consider $\{\sin (nx) dx\}$. $\endgroup$
    – Donu Arapura
    Commented Apr 29, 2016 at 13:59
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    $\begingroup$ More amazingly, for any real-analytic vector bundle $E$ on a real-analytic manifold $M$ (e.g., $E = \wedge^k(T^{\ast}M)$ in your case), the real-analytic sections are dense (in a suitable sense) inside the space of $C^{\infty}$-sections; this is a theorem of Grauert and Morrey, and doesn't even have compactness assumptions on $M$. The key tool is a relation with Stein spaces. To generalize Arapura's example, there are many smooth affine algebraic varieties $X$ over $\mathbf{R}$ such that $M=X(\mathbf{R})$ is compact (and non-empty), so then algebraic differential forms do the job. $\endgroup$
    – nfdc23
    Commented Apr 29, 2016 at 14:01

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