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$\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\TM}{\operatorname{TM}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\Cof}{\operatorname{Cof}}$ $\newcommand{\Det}{\operatorname{Det}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\TN}{\operatorname{T\N}}$ $\newcommand{\TstarM}{T^*\M}$

Let $\M,\N$ be $d$-dimensional oriented Riemannian manifolds. Let $\tilde E:C^{\infty}(M,N) \to \R$ be defined by

$$\tilde E(f)=\frac{1}{2}\int_{\M} | \bigwedge^2 df|^2 \text{Vol}_{\M}.$$ $\tilde E(f)$ measures the mean action of $f$ on 2$D$-cubes. (How $f$ affects areas of surfaces, locally).

Note that $\bigwedge^2 df\in \Omega^2\big(\M,\Lambda_2(f^*{\TN})\big)$. Let $\nabla^{\Lambda_2(f^*{\TN})}$ be the induced connection on $\Lambda_2(f^*{\TN})$ and let $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} $ be the adjoint of the covariant exterior derivative $d_{\nabla^{(f^*{\TN})}}$.

Question: Does every critical point of $\tilde E$ satisfy $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$?

Of course, by counting degrees of freedom, this doesn't look reasonable.

The converse direction follows from the Euler-Lagrange equation of $\tilde E$: I proved here that the E-L equation is $$h_{f^*\TN} \big( \tr_{\TM}\big( df \otimes \delta_{\nabla^{\Lambda_2(f^*{\TN})}}(\bigwedge^2 df)\big) \bigg)=0,$$

where $h_{f^*\TN}:f^*\TN \otimes \Lambda_2(f^*\TN) \to f^*\TN$ is the linear extension of

$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2.$$

Thus $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$ implies $f$ is critical.

Edit 1:

A natural way to produce critical points of a functional is to look at its symmetries. $\tilde E$ is conformally-invariant exactly at dimension $4$. So, in dimension $4$ conformal maps are critical. However, I proved that they also satisfy the stronger equation $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$.

(In fact I proved that a conformal map $\M^d \to \N^d$ satisfies $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ if and only if $d=2k$ or it is a homothety).

Edit 2:

As I shall prove below (see "Edit 3"), in dimension $d=2$ the answer is positive. (Note that in this case the "number of constraints" is the same). Focusing on the simplest next case, we consider $d=3,\M=\N=\R^3$ with the flat metrics.

First, let's try to answer a more degenerate version of the question (forgetting $h_{f^*\TN}$):

Is there a smooth map $f:\mathbb{R}^3 \to \mathbb{R}^3$, which satisfy

$$\delta \big( df \wedge df \big) \neq 0, \, \text{and } \, \tr \big( df \otimes \delta(df \wedge df) \big)=0, $$

If there is such an $f$ then it cannot be a local diffeomorphism; its Jacobian must vanish somewhere.

Here $ \delta \big( df \wedge df \big)\in \Omega^1\big(\R^3;\Lambda_2(\R^3)\big)$ is a one-form on $\R^3$ with values in $\Lambda_2(\R^3)$, and

$$ \delta \big( df \wedge df \big)(X)=\sum_i \nabla_{e_i}(df \wedge df)(e_i,X)=\sum_i (\nabla_{e_i} df)(e_i) \wedge df(X) + df(e_i) \wedge (\nabla_{e_i} df)(X)$$ $$ =\Delta f \wedge df(X)+\sum_i df(e_i) \wedge (\nabla_{e_i} df)(X).$$

We can try to look first for harmonic counter-examples, but so far I failed doing even that.

Edit 3:

The answer is positive for $d=2$, so we need to restrict our attention to $d \ge 3$.

Indeed, in this case $E(f)=\int_{\M}| \bigwedge^2 df|^2 \text{Vol}_{\M}=\int_{\M}(\Det df)^2 \text{Vol}_{\M}$. We shall prove a map $f$ is $E$-critical if and only if its determinant is constant.

Since $\Det df$ is constant $\iff$ $\nabla (\bigwedge^2 df)=0 \iff \delta_{\nabla^{\Lambda_2(f^*\TN)}} \big( \bigwedge^2 df \big)=0$, we are done.

The Euler-Lagrange equation of $E$ can be written as

$$ \delta (\Det df \cdot \Cof df) =0, \tag{1}$$ where $\Cof df:\TM \to f^*(\TN)$ is the cofactor map of $df$ defined by $$ \Cof df= \star_{f^*TN}^{d-1} (\wedge^{d-1} df) \star_{TM}^1, $$

and $\delta$ is the adjoint of the pullback connection on $f^*\TN$.

(The Euler-Lagrange equation of the Jacobian functional $E(f)=\int_{\M} \Det df \text{Vol}_{\M}=\int_{\M} f^* \text{Vol}_{\N}$ is $\delta (\Cof df)=0$. Details can be found in lemma 2.9 in my paper here).

Expanding equation $(1)$ we get $$ 0=\delta(\Det df \cdot \Cof df )= \Det df \delta(\Cof df ) - \tr_{g}(d \Det df \otimes \Cof df). $$

We now use the fact the Jacobian functional is a null-Lagrangian, i.e. every smooth map is a critical point, or equivalently $\delta (\Cof df)=0$. (This is essentially Stokes theorem, you can see lemma 2.5 here).

So, the E-L equation $(1)$ reduces to $$\tr_{g}(d \Det df \otimes \Cof df)=0. \tag{2}$$

Let $f$ be a map satisfying equation $(2)$. We shall prove $\Det df$ is constant; suppose that $\Det df_p \neq 0$ for some $p \in \M$. This implies $ \Cof (df_p)$ is invertible, so by equation $(2)$ $d\Det df_p=0$.

Now we observe that any $C^1$ function $g : \M \to \mathbb R$ on a connected manifold, satisfying $g(p) \ne 0 \implies dg_p = 0$ is constant.

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    $\begingroup$ For the local problem, it may be useful to assume that $M,N$ are both flat in which case you can get rid of the geometry and concentrate on the multilinear algebra. $\endgroup$ – Willie Wong Jun 16 '17 at 17:00

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