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For a closed manifold the laplacian is almost surjective operator since the index of $\Delta$ is zero and there is no a non constant harmonic function. So the codimension of the image of $\Delta$ is equal to $1$.

Now assume that $M$ is an open manifold(non compact without boundary). What type of results exist for the image of $\Delta$. Is it surjective? Is the codimension of its image finite?

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    $\begingroup$ If $M\cup\partial M$ is compact, the Dirichlet problem is solvable and therefore $\Delta$ is surjective on reasonnable functional spaces. $\endgroup$
    – Denis Serre
    Commented Oct 6, 2014 at 15:04
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    $\begingroup$ You should be more precise about the function spaces. For example, $\Delta:C^\infty(\mathbb R^2) \to C^\infty(\mathbb R^2)$ is surjective but $\Delta:C^2(\mathbb R^2) \to C^0(\mathbb R^2)$ is not. $\endgroup$
    – Jochen Wengenroth
    Commented Oct 6, 2014 at 15:50
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    $\begingroup$ $\Delta$ on $L^2(\mathbb R^n)$ is not surjective either because $0$ is in the spectrum. $\endgroup$
    – Christian Remling
    Commented Oct 6, 2014 at 17:38

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