Mariş’s Theorem

During a literature search (to answer my question concerning symmetries of “ground states” in variational problem, I came across a very nice theorem due to Mihai Mariş. The theorem itself is, more than anything else, a statement about the geometry of Euclidean spaces. I will give a rather elementary write-up of (a special case of) the theorem here. (The proof presented here can equally well be applied to get the full strength of the theorem as presented in Maris’s paper; I give just the special case for clarity of the discussion.)

Notations
Throughout we let $\mathcal{X}$ be a set of real-valued continuously differentiable functions defined over $\mathbf{R}^d$ for some $d \geq 2$. By $S\mathbf{R}^d$ I denote the tangent sphere bundle of Euclidean $d$-space; if you are not familiar with that, just think of it as the Cartesian product $\mathbf{R}^d \times \mathbb{S}^{d-1}$, or ordered pairs $(x,v)$ where $x$ is a point in $\mathbf{R}^d$ and $v$ is a unit vector.

By $\mathfrak{g}$ we denote a function $\mathcal{X}\times S\mathbf{R}^d \to \mathbf{R}$; we put its first argument in square brackets and its second argument in parentheses. We will refer to $\mathfrak{g}$ as the “constraint function”.

By $\Gamma$ we denote the set of all isometries of $\mathbf{R}^d$, which induces also a mapping of $S\mathbf{R}^d$ to itself. We will note a few special elements:

• For $(x,v)\in S\mathbf{R}^d$ we let $r_{x,v}: y\mapsto y - [ 2(y-x)\cdot v] v$ be the reflection over the hyperplane through the point $x$ that is orthogonal to $v$.
• Similarly, if $V = \{v_1,v_2,\ldots,v_k\}$ is a orthonormal set of vectors, we let $r_{x,V} = r_{x,v_1}r_{x,v_2}\cdots r_{x,v_k}$. Notice that the operations on the right hand side commute because the $v_i$‘s are mutually orthogonal.
• And lastly, we use the short hand $r_x: y\mapsto 2x - y$ the complete reflection about the point $x$, which is equivalent to $r_{x,V}$ when $V$ is a set of orthonormal basis vectors.
• For $v,w\in \mathbb{S}^{d-1}$ such that $v\perp w$, and for $\theta\in \mathbf{R}$, we let $\rho^\theta_{x,v,w}$ denote the rotation in the plane spanned by $\{v,w\}$ through $x$ by angle $\theta$. Analytically we can write it as sending $y \mapsto y + (y\cdot v)[ (\cos \theta - 1)v + \sin\theta w] + (y\cdot w)[-\sin\theta v + (\cos\theta -1)w]$.

For each $(x,v)\in S\mathbf{R}^d$ we define three maps on $\mathbf{R}^d$:

• $\pi_{x,v} : y\mapsto y - (y-x)\cdot v$,
• $\pi^+_{x,v}$ which acts as the identity if $(y-x) \cdot v \geq 0$ and as $r_{x,v}$ otherwise,
• $\pi^-_{x,v}$ which acts as the identity if $(y-x)\cdot v \leq 0$ and as $r_{x,v}$ otherwise.

Note that $\pi_{x,v}$ is the projection map to the hyperplane defined by $(x,v)$, and $\pi^+_{x,v}$ is the projection map to the positive half space. Also observe that $\pi^{-}_{x,v} = \pi^+_{x,-v}$.

We extend the above definition to sets $V$ of orthonormal vectors:

• $\pi_{x,V} = \pi_{x,v_1}\pi_{x,v_2} \cdots \pi_{x,v_k}$ and
• $\pi_{x,V}^\perp = 1 - \pi_{x,V}$

are the orthogonal projections to the subspace orthogonal to $V$ and the subspace spanned by $V$ respectively (the subscript denotes the directions to project away, not the ones to keep).

Geometry background
We note a few facts concerning our definitions above.

1. $r_{x,v} \pi^{\pm}_{x,v} = \pi^{\mp}_{x,v}$ and $\pi^{\pm}_{x,v} r_{x,v} = \pi^{\pm}_{x,v}$.
2. If $v\perp w$, we have $r_{x,v}\pi^{\pm}_{x,w} = \pi^{\pm}_{x,w} r_{x,v}$, and $\pi^-_{x,v}\pi^+_{x,w} = \pi^+_{x,w}\pi^-_{x,v}$.
3. In general, given $v\neq w$, let $u\perp v$ be in the span of $\{v,w\}$, then there exists $\theta$ such that $r_{x,v}r_{x,w} = \rho^\theta_{x,v,u}$. Let $h$ be a continuous function on span of $\{v,w\}$, we have three cases
• $\theta / \pi$ is irrational. Then orbit of $v$ under $\left( \rho^\theta_{x,v,u}\right)^k$ is dense on the unit circle. Hence $h$ must be radially symmetric.
• $\theta / \pi = p/q$ where $p$ is odd. Then $r_{x,v}$ and $r_{x,w}$ generate a dihedral group of an even polygon, and $r_{x,\{v,u\}}$ is a symmetry of $h$.
• $\theta/\pi = p/q$ where $p$ is even and $q$ is odd. Then $r_{x,v}$ and $r_{x,w}$ generate a dihedral group of an odd polygon, and hence the two reflections are conjugate by a reflection. (To see this, let $k$ be the order of the element $r_{x,v}r_{x,w}$, by assumption $k$ is odd. Hence $\left(r_{x,v}r_{x,w}\right)^kr_{x,v} = r_{x,v}$. Letting $u'$ be a vector such that $r_{x,u'} = \left(r_{x,v}r_{x,w}\right)^{\frac{k-1}{2}}r_{x,w}$, the LHS can be written as $r_{x,u'}r_{x,w} r_{x,u'}$. Observe that $r_{x,u'}$ is a symmetry of $h$ which swaps the spans of $v$ and $w$.)

Hypotheses
We need the following properties on $\mathfrak{g}$. In the following $f$ will always denote arbitrary functions in $\mathcal{X}$, and $(x,v)$ arbitrary pairs in $S\mathbf{R}^d$.

1. The function $\mathfrak{g}[f](x,\cdot): \mathbb{S}^{d-1} \to \mathbf{R}$ is continuous,
2. $\mathfrak{g}[f](x,v) = - \mathfrak{g}[f](x,-v)$,
3. If $x\cdot v = y\cdot v$ then $\mathfrak{g}[f](x,v) = \mathfrak{g}[f](y,v)$,
4. For every $f, x,v$ there exists a $\lambda \in\mathbf{R}$ such that $\mathfrak{g}[f](x-\lambda v,v) = 0$.

We also will assume certain closure conditions on $\mathcal{X},\mathfrak{g}$:

1. If $f\in \mathcal{X}$ and $\mathfrak{g}[f](x,v) = 0$, then $f\circ \pi^{\pm}_{x,v}\in \mathcal{X}$.
2. If $f\in\mathcal{X}$ satisfies $f = f\circ r_{x,v}$, then $\mathfrak{g}[f](x,v) = 0$.
3. If $\gamma\in\Gamma$ is an isometry, and if $f\in \mathcal{X}$, then $f\circ\gamma\in\mathcal{X}$ and $\mathfrak{g}[f\circ\gamma](\gamma(x,v)) = \mathfrak{g}[f](x,v)$.

Intermediate lemmas

We will use, without proof, the Borsuk-Ulam Theorem

Thm 1 [Borsuk-Ulam]
A continuous odd map $\phi:\mathbb{S}^{d_1} \to \mathbf{R}^{d_2}$ where $d_1 \geq d_2 \geq 1$ admits a vector $v$ such that $\phi(v) = 0$.

Now some technical lemmas.

Lem 2
Let $f,x$ be fixed. If $V$ is a set of orthonormal vectors such that $\mathfrak{g}[f](x,v) = 0$ for all $v\perp V$, we have that $f(y) = f(z)$ whenever $\pi_{x,V}^\perp y = \pi_{x,V}^\perp z$ and $|\pi_{x,V} y| = |\pi_{x,V} z|$. In this case we say $f$ is radially symmetric about $(x,V)$.

Proof. Since $latex$f$is continuously differentiable, it suffices to show that all “angular derivatives” vanish, i.e. that for every $v\perp V$, $v\cdot \nabla f(y) = 0$ when $y\neq x, (y-x)\cdot v = 0$. By hypothesis since $\mathfrak{g}[f](x,v) = 0$, $f\circ \pi^+_{x,v} \in \mathcal{X}$ and so is $C^1$. Hence for $y = \pi_{x,v} y$ we have that $0 = v\cdot \nabla (f\circ \pi^+_{x,v})(y) = v\cdot \nabla f(y)$ as needed. q.e.d. Lem 3 Let $f\in\mathcal{X}$ and $(x,v)\in S\mathbf{R}^d$ such that $f$ is radially symmetric about $(x,v)$. Then there exists $\lambda\in\mathbf{R}$ such that $f$ is radially symmetric about $(x-\lambda v,\{\})$. Proof. By our hypothesis $\lambda$ can be chosen such that $\mathfrak{g}[f](x-\lambda v,v) = 0$. WLOG suppose that $x-\lambda v = 0$. By Lemma 2 it suffices to show that for every $w$, $\mathfrak{g}[f](0,w) = 0$. For $w\neq v$, there exists $u\perp v$ such that $w$ is in the span of $\{u,v\}$. Since $f$ is assumed to be radially symmetric about $(x,v)$ (and hence also $(0,v)$), $f = f\circ r_{0,u}$. Now consider $f^\pm = f\circ \pi^\pm_{0,v}$, this implies that $f^\pm = f^\pm \circ r_{0,\{v,u\}}$. On the other hand, it is clear that $r_{0,\{v,u\}}$ interchanges the range of $\pi^{\pm}_{0,w}$. Since $f^\pm$ is independent of the choice of $u$, this implies that $f^\pm$ are both radially symmetric. However, since $f^+ = f^-$ along the hyperplane $\pi_{0,v}$, we get that $f^+ \equiv f^- \equiv f$. q.e.d. Cor 4 [2D case] If $d = 2$, and $f\in \mathcal{X}$, then there exists $x$ such that $f$ is radially symmetric about $(x,\{\})$. Proof. Choose $y$ arbitrarily. By assumption $\mathfrak{g}[f](y,\cdot)$ is an odd continuous function from the circle to the real numbers, hence it vanishes for some direction $u$. Let $v$ be a direction orthogonal to $u$. By the construction from the proof of Lemma 3, we can find $x = y + \lambda v$ such that $\mathfrak{g}[f](x,v) = 0$. The functions $f^{\pm_1\pm_2} = f\circ \pi^{\pm_1}_{x,u}\pi^{\pm_2}_{x,v}$ satisfy $f^{\pm_1\pm_2} = f^{\pm_1\pm_2}\circ r_{x,\{u,v\}}$ and from the proof of Lemma 3 we see that they are all radially symmetric about $x$. Using that $f^{\pm_1+} = f^{\pm_1-}$ along $\mathbf{R}_{\pm_1}u$, and similarly $f^{+\pm_2} = f^{-\pm_2}$ along $\mathbf{R}_{\pm_2}v$, we have that the four functions all equal and hence $f$ is radially symmetric about $x$. q.e.d. Lem 5 Let $O$ be a set of orthonormal basis. If $f\in\mathcal{X}$ is radially symmetric about $(x,V)$ and $(x,W)$ where $V,W\subseteq O$, and$\latex V^\perp \cap W^\perp \neq \emptyset$, then $f$ is radially symmetric about $V\cap W$. Proof. We can write $O = (V\cap W) \cup (V^\perp \cap W^\perp) \cup (V\cap W^\perp) \cup (W\cap V^\perp)$, and hence every point $y$ in $\mathbf{R}^d$ can be written as a linear combination of $p+q+r+s$ each belonging to one of the subspaces. The two radial symmetries imply that $f(p,q,r,s) = f(p,\sqrt{q^2+r^2},0,s) = f(p,\sqrt{q^2 + r^2 + s^2},0,0)$. This demonstrates the radial symmetry. q.e.d. Lem 6 [Induction hypothesis] If $d \geq 3$, and $f\in \mathcal{X}$. For every $x$ there is a $d-2$ dimensional subspace such that $f$ is radially symmetric about $(x,V)$. Proof. Since $d \geq 3$, using the Borsuk-Ulam theorem we can choose $v$ such that $\mathfrak{g}[f](x,v) = 0$. Let $f^\pm = f\circ \pi^\pm_{x,v}$. Applying the Borsuk-Ulam theorem to the continuous functions $\mathfrak{g}[f^\pm](x,\cdot):\mathbb{S}^{d-1}\cap v^\perp$ we get two vectors $w^\pm$ for which $\mathfrak{g}[f^\pm](x,w^\pm) = 0$. Writing$K^\pm = \{v,w^\pm\}^\perp$, the argument in the proof of Lemma 3 suffices to show that $f^\pm$ are radially symmetric about $(x,K^\pm)$ respectively. We split into several cases depending on how $w^\pm$ compare. Let $K = K^+\cap K^-$. 1. $w^+ = \pm w^-$. Then $K^+ = K^- = K$. Hence $f^\pm$ are two functions radially symmetric about $(x,K)$, and such that they agree on $\pi^\perp_{x,K\cup\{w^+\}}$, so they agree wholly and $f$ is radially symmetric about$latex $(x,K)$.
2. Suppose $r_{x,w^+}r_{x,w^-}$ generate the circle in the plane spanned by $w^\pm$. Then from the definition of radial symmetry we see that $f^\pm$ are both radially symmetric around $(x,K\cup \{v\})$, and hence by Lemma 5 radially symmetric around $(x,K)$. Since they agree on $\pi_{x,v}$, we get $f^\pm = f$ is also radially symmetric..
3. Suppose that $r_{x,w^+}r_{x,w^-}$ generate the dihedral group for an even polygon. Letting $u,u'$ be an orthonormal basis for the span of $w^\pm$, we have then $r_{x,\{u,u'\}}$ is a symmetry of $f^\pm$. Let $u''$ be in the span of $u,u'$, we have that using the invariance property of $\mathfrak{g}$ under isometries, $\mathfrak{g}[f](x,-u'') = \mathfrak{g}[f\circ r_{x,\{u,u'\}}](r_{x,\{u,u'\}}(x,u'')) = \mathfrak{g}[f](x,u'')$. This implies, by Lemma 2, that both $f^\pm$ are radially symmetric around $(x,K\cup\{v\})$. This implies via Lemma 5 that $f = f^\pm$ are radially symmetric around $(x,K)$.
4. Lastly, we have the odd polygon dihedral case. By assumption $f^\pm$ agree on $\pi_{x,v}$, and by radial symmetry is completely determined by the values there. By the discussion previously on geometry, we have that there exists some $u'$ such that $f^\pm |_{\pi_{x,v}(\mathbf{R}^d)}$ is symmetric under $r_{x,u'}$. A direct computation shows that thus $f^\pm \circ r_{x,u'} \circ r_{x,v} = f^\mp$. This implies that $f\circ r_{x,\{u',v\}} = f$. So for any $u$ that is a linear combination of $u',v$ we have that $\mathfrak{g}[f](x,u) = \mathfrak{g}[f](x,-u) = 0$ (again using the action of $\mathfrak{g}$ under isometry). Hence $f$ is radially symmetric about $(x,\{u',v\}^\perp)$.

q.e.d.

Lem 7 [Induction step]
If $f\in\mathcal{X}$ is radially symmetric about $(x,V)$ where $V$ is a subspace of dimension at least 2, then there exists $w\in V$ such that $f$ is radially symmetric about $(x,V\cap\{w\}^\perp)$.

Proof. Using Borsuk-Ulam theorem again, there exists $w\in V$ such that $\mathfrak{g}[f](x,w) = 0$. We can then use the same method as the proof of Lemma 3 to conclude that $f$ is radially symmetric about $V\cap \{w\}^\perp$. q.e.d.

Conclusions

Putting everything together we get

Thm [Maris]
If $f\in\mathcal{X}$ and $d\geq 2$, then $f$ is radially symmetric about some point $x$.

In dimension 1, if you let $\mathfrak{g}[f](x,v) = v\cdot f'$, then we can allow $\mathcal{X}$ to be the space of all continuously differentiable functions, and the theorem is clearly false in that case.

One should think of the theorem as providing a very stringent constraint on how many elements there can be in $\mathcal{X}$.

As an example application: consider the problem of minimising $\int |\nabla f|^2 dx$ under the constraint that $\int V(f) dx$ is fixed. By elliptic regularity (under suitable assumptions on the growth rate of $V:\mathbf{R}\to\mathbf{R}$) we have that any critical point must have regularity $C^{1,\alpha}$. Let $\mathcal{X}$ denote the set of constrained minimisers. Let $\mathfrak{g}[f](x,v) = \int_{\pi^+_{x,v}\mathbf{R}^d} - \int_{\pi^-_{x,v}\mathbf{R}^d} V(f) dx$. Define $f^\pm = f\circ \pi^\pm_{x,v}$. If $\mathfrak{g}[f](x,v) = 0$, then $\int V(f^\pm) dx = \int V(f) dx$ so $f^\pm$ still solves the constraints. On the other hand, we have that $\int |\nabla f^+|^2 + |\nabla f^-|^2 dx = 2 \int |\nabla f|^2 dx$. Since $f$ minimises the Dirichlet integral, so must also $f^\pm$. The various other properties assumed on $\mathcal{X}$ and $\mathfrak{g}$ are easily checked: continuity of $\mathfrak{g}$ follows in particular from the Lebesgue dominated convergence theorem. Hence we conclude

Cor
For the constrained minimisation problem described, any minimiser must be radially symmetric.