$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$

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:

- Any smooth harmonic map has harmonic differential.
- Thus, any weakly harmonic map should have weakly harmonic differential.
- Since the harmonic equation for forms is linear elliptic, the weak differential of $f$ is smooth.
- 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?

vectorvalued forms- then the target bundle $f^*TN$ also changes with the map. $\endgroup$ – Asaf Shachar May 31 '18 at 12:44