Mariş’s Theorem

by Willie Wong

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.

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