# Is there always a non-trivial homotopy class of maps which minimizes the Dirichlet energy?

Let $$M,N$$ be $$d$$-dimensional oriented Riemannian manifolds, possibly with boundary, $$M$$ compact. Let $$E:C^{\infty}(M,N) \to \mathbb{R}$$ be the Dirichlet energy, i.e.

$$E(f)=\int_M |df|^2 \text{Vol}_M.$$

For any given homotopy class $$\alpha$$ of smooth maps $$M \to N$$, define $$E_{\alpha}=\inf \{ E(f) \, | \,\, f \in \alpha \}.$$

(In some cases, we know $$E_{\alpha}$$ is realized, e.g. when $$N$$ is closed and of negative curvature.)

Now define $$E:=\inf_{\alpha \text{ is not trivial }} E_{\alpha}$$

Is $$E$$ always a minimium? i.e. does there exist a homotopy class $$\alpha$$ such that $$E=E_{\alpha}$$?

Does the answer change if we replace the Dirichlet energy with the $$d$$-energy? (i.e. integrate $$|df|^d$$ instead of $$|df|^2$$?)

In the case $$M=N=\mathbb{S}^1$$, one can prove directly, via Holder's inequality, that the answer is positive:

$$\big(\deg f \cdot \text{Vol}(\mathbb{S}^1)\big)^2=\big(\int_{\mathbb{S}^1} |f'|\big)^2 \le \int_{\mathbb{S}^1} |f'|^2 \int_{\mathbb{S}^1} 1^2=\text{Vol}(\mathbb{S}^1)E(f),$$ so we get

$$E(f) \ge (\deg f)^2 \cdot \text{Vol}(\mathbb{S}^1),$$

and for a map of degree $$n$$, the minimal energy is obtained exactly for $$f(e^{i\theta})=e^{in\theta}$$.

So, the non-trivial homotopy class with minimal infimal energy is the class of degree-one maps.

More generally, if we consider the $$d$$-energy, and assume $$M,N$$ are closed, then

$$E_d(f)=\int_M |df|^d \text{Vol}_M \ge d^{\frac{d}{2}} \int_M \det df \text{Vol}_M=d^{\frac{d}{2}} \int_{M} f^*\text{Vol}_N=d^{\frac{d}{2}} \deg f \int_{N} \text{Vol}_N,$$ so $$E_d(f) \ge d^{\frac{d}{2}} \deg f \cdot \text{Vol}(N),$$ with equality if and only if $$f$$ is conformal. I am not sure this helps to settle the question though.

• I think the case $M=S^1$ gives $E=\ell^2/2\pi$ where $\ell$ is the length of the shortest noncontractible loop i.e. homotopy $1$-systole. In this case the minimizer won't exist if $N$ is for example $\mathbb R^2$ with a disc of radius $\tfrac 1n$ removed from around $(n,0)$ for each $n.$ Even the case $M=S^1$ and $\pi_1(N)$ finitely generated could be interesting perhaps – Dap Mar 20 '18 at 13:58
• Oh, to get it $d$-dimensional: $M=S^1\times [0,1]^{d-1}$ should work, I think. – Dap Mar 20 '18 at 14:26
• Thanks, this is interesting. Do you mean removing open disks from $\mathbb{R}^2$, so the closed non-contractible geodesics you are referring to, are "boundary geodesics" (just the circles, i.e. the boundaries of the removed sets)?. – Asaf Shachar Mar 21 '18 at 6:26
• Since there are no restrictions on $N$ and since there doesn't need to be a minimizing $f,$ you could remove either closed or open discs. You could even get a closed $N$ by gluing something into the holes, like a gluing of the manifold to a copy of itself (using collars). – Dap Mar 21 '18 at 7:10
• Ah, OK you are right. So your example shows the minimizer won't necessarily exist if $N$ is non-compact. Interesting. I wonder what can be said when $N$ is compact. I will think about it... – Asaf Shachar Mar 21 '18 at 12:03