A square is a critical point of any isometric-invariant permutation-invariant function

Question

$\newcommand{\S}{\mathbb{S}^1}$

This is a self-answered question I have been thinking on quite some time now. Any improvements or cleaner arguments or simplifications are welcome.

Let $E:(\S)^4 \to \mathbb{R}$ be a smooth function. Suppose that $E$ is:

1. Invariant under permutations: For any $\sigma \in S_4$, $E(x_1,x_2,x_3,x_4)=(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)},x_{\sigma(4)})$.
2. Invariant under "square isometries": $E(Qx_1,Qx_2,Qx_3,Qx_4)=E(x_1,x_2,x_3,x_4)$ for any rotation or reflection $Q \in D_4$, where we think on $Q$ as a map $\S \to \S$.

Claim:

Let $p=(p_1,p_2,p_3,p_4)$ be the configuration of a planar square lying on $\S$. Then $p$ is a critical point of $E$.

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Bonus:

Do we really need invariance of $E$ under the full $S_4$ and $D_4$ groups? What is a minimal subgroup which suffice?

Invariance under isometries alone does not suffice for the square to be critical; take e.g. $E=\|x_1-x_2\|^2$.

Answer 1

$\newcommand{\S}{\mathbb{S}^1}$ The idea in a nutshell:

We have two symmetries which fixes $p_1$: one a diagonal reflection/swap $R$, and one "general coordinate-exchange" $P$ ($x_2 \iff x_4$). Since $dR$ acts as minus map on $T_{p_1}\S$ and $dP$ acts as the identity, the relevant derivative should be zero.

Here are the details:

Consider the diagonal reflection $R$ which swaps $p_2$ and $p_4$.

$R$ induces a map $\tilde R:(\S)^4 \to (\S)^4$, given by $$ \tilde R(x_1,x_2,x_3,x_4):=\big(R(x_1),R(x_2),R(x_3),R(x_4)\big). $$ Since $R,\tilde R$ are isometries, we have $E \circ \tilde R = E$, and thus $$ dE_{p}=dE_{\tilde R(p)} \circ \tilde R . $$ Let $v \in T_{p_1}\S$. Then $$ dE_{p}(v,0,0,0)=dE_{\tilde R(p)}(-v,0,0,0). \tag{1} $$ Now, let $P:(\S)^4 \to (\S)^4$ be defined by $$P(x_1,x_2,x_3,x_4)=(x_1,x_4,x_3,x_2).$$ Since $P$ is an isometry, we have $E \circ P=E,$ so $$ dE_{p}(v_1,v_2,v_3,v_4)=dE_{P(p)} \circ P(v_1,v_2,v_3,v_4)=dE_{P(p)}(v_1,v_4,v_3,v_2). $$ In particular, $$ dE_{p}(v,0,0,0)=dE_{P(p)}(v,0,0,0). \tag{2} $$ Since $q:=\tilde R(p)=P(p)$, equations $(1),(2)$ imply $$ dE_{p}(v,0,0,0)=dE_{q}(v,0,0,0)=dE_{q}(-v,0,0,0)=0. \tag{3} $$ for any $v \in T_{p_1}\S$.

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Let $Q$ be counter-clockwise rotation by $\pi/2$, and let $\tilde Q:(\S)^4 \to (\S)^4$ be defined as $\tilde R$ above. Then we have $$ 0=dE_{p}(v,0,0,0)=dE_{\tilde Q(p)}(Qv,0,0,0), $$ so $$ dE_{\tilde Q(p)}(w,0,0,0)=0 \tag{4} $$ for any $w \in T_{p_2}\S$.

Let $S:(\S)^4 \to (\S)^4$ be defined by $$S(x_1,x_2,x_3,x_4)=(x_2,x_3,x_4,x_1).$$ $E \circ S=E,$ implies $$ dE_{p}(0,v,0,0)=dE_{S(p)}(v,0,0,0)=dE_{\tilde Q(p)}(v,0,0,0)=0 $$ by equation $(4)$.

So, we started with equation $(3)$ $dE_{p}(v,0,0,0)=0$ and derived $dE_{p}(0,v,0,0)=0$.