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While trying to understand some regularity results, I thought about the following "naive" approach for establishing regularity of weakly harmonic maps between Riemannian manifolds. I would like to know if my approach makes sense. (can it be made valid under suitable technical adjustments?).

Here is the sketch:

First, we begin with the smooth case: A map $f:\M \to \N$ is harmonic if and only if $df \in \Omega^1(\M,f^*T\N)$ is a harmonic form.

Indeed, let $\nabla$ be the pullback connection of the Levi-Civita connection of $\N$. Let $d_{\nabla}:\Omega^k(\M,f^*T\N ) \to \Omega^{k+1}(\M,f^*T\N )$ be the associated exterior derivative and $\delta_{\nabla}$ its adjoint. Then $$ d_{\nabla} df=0 \tag{1}$$ follows from the symmetry of the connection on $\N$. (Equation $(1)$ holds for every smooth map).

Harmonicity is $ \delta_{\nabla} df =0$; combining this with equation $(1)$ we obtain the equivalence.

My idea is the following:

  1. Any smooth harmonic map has harmonic differential.
  2. Thus, any weakly harmonic map should have weakly harmonic differential.
  3. Since the harmonic equation for forms is linear elliptic, the weak differential of $f$ is smooth.
  4. This implies $f$ is smooth.

I am trying to use rather standard linear elliptic regularity results; the classical regularity proofs for harmonic maps are more involved.

Somewhere in these steps there is a failure, since there are weakly harmonic maps which are not continuous. Of course, the devil must be somewhere in the details, which I did not really specify yet.

Morally, I am imagining "intrinsice weak formulations" as follows:

The weak version of $(1)$ is $$ \int_{\M} \langle df , \delta_{\nabla} \sigma \rangle =0, \tag{1'}$$ for all $\sigma \in \Omega^2(\M,f^*T\N )$ which are compactly supported in the interior of $\M$.

Of course, this doesn't make sense as stated, since if $f$ is a Sobolev map, $f^*T\N$ is not a smooth vector bundle. Perhaps we should assume here that $f$ is continuous, and restrict attention to "continuous/weak Sobolev sections" $\sigma$ (which vanish on $\partial \M$ in the trace sense).

Anyway, I think there should be an extrinsic version of $(1')$, by embedding $N$ in a higher dimensional Euclidean space.

(Although this doesn't seem trivial. Any suggestions are welcome).


$\delta(df)=0$ has a well-known extrinsic version, which is "morally equivalent" to $$ \int_{\M} \langle df , \nabla \sigma \rangle =0, \tag{2'} $$ for all compactly supported $\sigma \in \Gamma(f^*T\N )$.

(The extrinsic version uses the second fundamental form of $\N$ inside the Euclidean space).


If we assume $ f \in W^{1,p}(\M,\N)$ where $p \ge \dim \M$, then I think this might work. Sobolev maps can then be approximated by smooth maps (If $p<\dim \M$ we need to restrict the topology of the manifolds to get density.)

In that case, $(1')$ should hold for all Sobolev maps, and so $(1'),(2')$ together should hold for all weakly harmonic maps.

So, assuming Sobolev maps can be approximated by smooth maps, does the suggested approach have a chance?

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    $\begingroup$ Your proposed equation $\delta_\nabla df=0$ is not actually linear, since $d_\nabla$ itself depends on $f$ (it acts on sections of a pullback bundle via $f$), and so does $\delta_\nabla$. $\endgroup$ – YangMills May 29 '18 at 12:25
  • $\begingroup$ Thanks, you are right of course. Indeed, my approach was quite naive; Moreover, if we are thinking in terms of vector valued forms- then the target bundle $f^*TN$ also changes with the map. $\endgroup$ – Asaf Shachar May 31 '18 at 12:44
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Regularity of harmonic map is true in dimension with $f\in W^{1,2}$ but it is subtle see Helein, Harmonic maps, conservation laws and moving frames.

It is false in higher dimension, even if f\in W^{1,n}$, see Rivière, Everywhere discontinuous harmonic maps into spheres.

Whet you need is to control the morrey norm, see Rivière and Struwe, Partial regularity for harmonic maps and related problems.

You can also look to the survey by Hélein, Frédéric and Wood, John C. ,Harmonic maps.

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  • $\begingroup$ Thanks for the references. I was aware of some of them. Actually, the point of my question was whether a more naive approach for proving regularity could work. (I think now that this is unlikely). $\endgroup$ – Asaf Shachar Mar 30 '18 at 10:11

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