# How to construct a harmonic function with non-zero gradient on manifold with two nonparabolic ends?

We know that if a complete noncompact manifold M has two nonparabolic ends, then we can construct a nonconstant bounded harmonic function with finite Dirichlet integral defined on the whole $M$.

More over, can we construct a harmonic function $f$ which not only satisfies all the above properties, but also has $\nabla f \ne 0$ on $M$? If not, please, give counterexamples.

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There are counter examples .Consider any compact Riemann surface of genus at least one and remove two disjoint closed discs ,call it X.Let f be the harmonic function .In this case f is proper onto its image .If gradf is never zero then X is diffeomorphic to an annulus which contradicts the genus assumption .

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The standard (and probably the simplest) example of a manifold with two non-parabolic ends is obtained by taking two disjoint copies of $\mathbb R^3$ and joining them with a "bridge" (or a "tube"). Then the space of bounded harmonic functions on the resulting manifold $M$ is two-dimensional, which means that the gradient vector fields corresponding to bounded harmonic functions form a one-dimensional space, i.e., all of them are proportional to a certain non-zero field $v$.

Now, instead of just one tube one can take two (or more) tubes, and make sure that the whole construction is symmetric. The field $v$ above must also be symmetric, so that it has to vanish at any fixed point of such a symmetry.

More rigorously, let $g$ be an isometry of $M$ which fixes its ends. Then $g$ also fixes the field $v$. It leads to a contradiction if $g$ also fixes a point $p\in M$ without fixing any non-zero vector in $T_p M$.

For a concrete example take in $\mathbb R^4$ two parallel copies $A$ and $A'$ of $\mathbb R^3$, which we shall identify by the means of the corresponding orthogonal projection. Let $D_1, D_2\subset A$ be two disjoint disks of the same radius, and let $D'_1,D'_2$ be the corresponding disks in $A'$. Remove all of them, join the corresponding boundaries with cylindrical tubes, and make the surgery smooth in the same rotationally invariant way on all boundaries. Then the central symmetry of $A$ whose center $p$ is halfway between the centers of $D_1$ and $D_2$ obviously extends to an isometry of the resulting manifold $M$ which satisfies the above conditions.

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Hi, for someone less familiar with the theory, do you mean indicating (a reference perhaps) why the space of bounded harmonic functions on $M$ is two dimensional, and why this also carries over in the case of more than one tube? – Willie Wong Oct 9 '13 at 7:59
Also, if I understand correctly, the harmonic functions would be asymptotically constant, right? Do we have any statements about the shape of the level set $f^{-1}(1-\epsilon)$, assuming $f\to 1$ on one of the ends (in particular is it topologically $\mathbb{S}^2$? If so in the case where there are two cylinders the assertion $\nabla f \neq 0$ is ruled out topologically via some version of mountain-pass, and we don't even need the isometry. (Come to think of it, this is basically Mohan's answer below, right?) – Willie Wong Oct 9 '13 at 8:03
The reason why the space of bounded harmonic functions is 2-dimensional is that there are two types of behaviour at infinity of sample paths of the Brownian motion: namely, convergence to one of the ends. As for the level sets, they should indeed be topological spheres - however, this may require some justification. – R W Oct 9 '13 at 12:58