During a literature search (to answer my MathOverflow question concerning symmetries of "ground states" in variational problems, 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.)
Then a minimizer, should it exist, must be radially symmetric about some point $x_0$.
The proof of Mariş's theorem boils down to mostly a statement about the geometry of the Euclidean spaces, using the following properties of minimizers:
- Under suitable assumptions on the growth rate of $V$ near $\infty$, by using Lagrange multipliers we see that minimizers satisfy a semilinear elliptic partial differential equations, and by elliptic regularity the minimizers are always $C^{1,\alpha}$.
- Given any minimizer, if we act on it with a symmetry of $\mathbf{R}^d$ we get another minimizer.
- Given $f$ a minimizer: let $\Pi$ be a hyperplane such that the integral of $V(f)$ on each of the two halves partitioned by $\Pi$ are equal, then the integral of $|\nabla f|^2$ on each of the two halves are also equal. (For if not, replace $f$ on the larger half by its mirror image from the smaller half. This function is still in $W^{1,2}$ (it is continuous along $\Pi$), and this new function would have lower energy than $f$.)
These properties allow us to reduce Mariş's theorem to a statement about sets of $C^1$ functions on $\mathbf{R}^d$ satisfying certain closure properties with respect to a certain functional. We begin by first setting up the notation and formalizing the three properties above in a convenient-to-use way for the geometric argument.
Notations
Throughout we let $\mathcal{Y}$ be a set of functions defined over $\mathbf{R}^d$ for some $d \geq 2$; this $\mathcal{Y}$ will be the domain of our functionals. We will denote by $\mathcal{X}$ a subset of $\mathcal{Y}$ to be described below.
By $S\mathbf{R}^d$ I denote the tangent sphere bundle of Euclidean $d$-space.1 Notice that we can freely identify elements of $S\mathbf{R}^d$ with oriented hyperplanes in $\mathbf{R}^d$; given a position $x$ and a unit vector $v$ based at $x$, we can associate to it the hyperplane through $x$ orthogonal to $v$, with positive orientation given by the direction of $v$. This identification is of course not-unique: given another $y$ on the same hyperplane, $(y,v)$ defines the same hyperplane with the same orientation as $(x,v)$.
By $\mathfrak{g}$ we denote a function $\mathcal{Y}\times S\mathbf{R}^d \to \mathbf{R}$; we put its first argument in square brackets and its second argument in parentheses. We will refer to $\mathfrak{g}$ as the "constraint function(al)".
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 corresponding hyperplane.
- 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 $\theta$, with direction of the rotation determined by the ordering $v,w$ (as opposed to $w,v$). 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 "projection" 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; elements in the negative half space is mapped to their image under reflection across the corresponding hyperplane to $(x,v)$. 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.
- $r_{x,v} \pi^{\pm}_{x,v} = \pi^{\mp}_{x,v}$ and $\pi^{\pm}_{x,v} r_{x,v} = \pi^{\pm}_{x,v}$.
- If $v\perp w$, we have $r_{x,v}\pi^{\pm}_{x,w} = \pi^{\pm}_{x,w} r_{x,v}$, and $\pi^-_{x,v}\pi^+_{x,w} = \pi^+_{x,w}\pi^-_{x,v}$.
- 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 the span of $\{v,w\}$ invariant under this rotation, 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 on this plane.
- $\theta / \pi = p/q$ 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 will state our theorems and lemmas in terms of the following hypotheses imposed on the set $\mathcal{X}$ and the constraint functional $\mathfrak{g}$.
- For fixed $f \in \mathcal{Y}$ and $x\in\mathbf{R}^d$, the function $\mathfrak{g}[f](x,\cdot):\mathbb{S}^{d-1} \to \mathbf{R}$ is continuous.
- The function $\mathfrak{g}$ is odd under reflection: $\mathfrak{g}[f](x,v) = - \mathfrak{g}[f](x,-v)$.
- The function $\mathfrak{g}$ only depends on the hyperplane: if $(x,v)$ and $(y,v)$ define the same hyperplane, then $\mathfrak{g}[f](x,v) = \mathfrak{g}[f](y,v)$.
- For every $f, (x, v)$, there exists a $\lambda \in \mathbf{R}$ such that $\mathfrak{g}[f](x-\lambda v, v) = 0$.
- If $f\in \mathcal{Y}$ is such that $f = f \circ r_{x,v}$ (so that $f$ is symmetric across the corresponding hyperplane), then $\mathfrak{g}[f](x,v) = 0$.
- $\mathfrak{g}$ is symmetric under Euclidean group of motions: if $\gamma \in \Gamma$ is an isometry, then for any $f\in \mathcal{Y}$, we have that
- $f \circ \gamma \in \mathcal{Y}$
- $\mathfrak{g}[f\circ \gamma](\gamma(x,v)) = \mathfrak{g}[f](x,v)$
We assume the following closure conditions on $\mathcal{X}\subset \mathcal{Y}$:
- If $f\in \mathcal{X}$ and $(x,v)\in S\mathbf{R}^d$ is such that $\mathfrak{g}[f](x,v) = 0$, then $f\circ \pi^{\pm}_{x,v} \in \mathcal{X}$.
- If $f\in \mathcal{X}$ and $\gamma \in \Gamma$, then $f\circ \gamma \in \mathcal{X}$.
Additionally, we assume that $\mathcal{X} \subset C^1(\mathbf{R}^d)$.
We our hypotheses stated, we can now state the main geometric result.
We can reduce Theorem 1 to this theorem by setting $\mathcal{Y} = W^{1,2}$, letting $\mathcal{X}$ be the set of minimizers, and setting the functional \[ \mathfrak{g}[f](x,v) := \int_{\pi^+_{x,v}\mathbf{R}^d} V(f) ~\mathrm{d}x - \int_{\pi^-_{x,v}\mathbf{R}^d} V(f) ~\mathrm{d}x,\] the difference between the integral of $V(f)$ between the two half spaces demarcated by $(x,v)$.
That the items of Hypothesis 1 is satisfied follows mainly from the properties of Lebesgue integration. That the items of Hypothesis 2 is satisfied follow from the discussion in the introductory paragraphs of this post.
The dimensional restriction is necessary. In dimension 1, if you let $\mathfrak{g}[f](x,v) = v\cdot f'$, then we can allow $\mathcal{X} = \mathcal{Y}$ to be the space of all continuously differentiable functions with compact support, 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 $\mathcal{X}$.
Intermediate lemmas and proof of the theorem
We will use, without proof, the Borsuk-Ulam Theorem
Now some technical lemmas.
By our hypothesis, there exists some $\lambda$ such that $\mathfrak{g}[f](x-\lambda v,v) = 0$. Denote by $z = x-\lambda v$ for convenience.
Consider the function $f^+ = f \circ \pi^+_{z,v}$, which by definition is symmetric under the reflection $r_{z,v}$. Clearly $f^+$ is still radially symmetric about $(z,v)$. So now let $w$ be an arbitrary direction $\neq v$. There exists a unique direction $u\perp v$ such that $w$ is in the span of $\{u,v\}$. Since $u\perp v$ and $f^+$ is radially symmetric about $(z,v)$, we must also have $f^+$ being symmetric under the reflection $r_{z,u}$. This implies that $f^+ = f^+\circ \rho^{\pi}_{z,u,v}$ using that $\rho^\pi_{z,u,v} = r_{z,\{u,v\}}$. By our hypothesis then \[ \mathfrak{g}[f^+](z,w) = \mathfrak{g}[f^+](z,-w) \] using that the isometry $\rho^{\pi}_{z,u,v}(z,w) = (z,-w)$. This implies therefore \[ \mathfrak{g}[f^+](z,w) = 0.\] Since $w\neq v$ is arbitrary, this implies that $\mathfrak{g}[f^+](z,w) = 0$ for any $w$. Combined with the fact that $z$ is chosen so that $\mathfrak{g}[f^+](z,v) = 0$, we have, by the previous Lemma, that $f^+$ must be radially symmetric about $(z,\{\})$.
The same argument also shows that $f^-$ is radially symmetric. As $f^+$ and $f^-$ agree along the hyperplane defined by $(x,v)$, we conclude that $f = f^+ = f^-$ everywhere and is radially symmetric.
Choose $y$ arbitrarily. By our hypotheses $\mathfrak{g}[f](y,\cdot)$ is an odd continuous function from the circle to the reals. Hence it must vanish for some direction $u$. Let $v$ be orthogonal to $u$. Then we can choose $x = y + \lambda v$ such that $\mathfrak{g}[f](x,v) = \mathfrak{g}[f](x,u) = 0$.
Now, $f^{++}:= f \circ \pi^+_{x,u} \pi^+_{x,v}$ is invariant under rotation by angle $\pi$, and by construction belongs to $\mathcal{X}$. Hence the same argument as the previous Lemma shows that $f^{++}$ is radially symmetric. Similarly we can construction $f^{+-}, f^{-+}, f^{--}$ and conclude they are all radially symmetric about $x$. The overlaps along the $\{u,v\}$ axes then allow us to conclude that $f = f^{++} = f^{--}= f^{+-} = f^{-+}$ and is radially symmetric.
With this set up, we can prove the theorem by induction on dimension.
- $w^+ = \pm w^-$. 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 $(x,K)$.
- 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 11 radially symmetric around $(x,K)$. Since they agree on $\pi_{x,v}$, we get $f^\pm = f$ is also radially symmetric..
- 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 8, that both $f^\pm$ are radially symmetric around $(x,K\cup\{v\})$. This implies via Lemma 11 that $f = f^\pm$ are radially symmetric around $(x,K)$.
- 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 $(x,\{u',v\}^\perp)$.
After the induction brings $V$ down to only 1 dimensional, the theorem follows by applying Lemma 9.
- If you are not familiar with the tangent sphere bundle, 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. ^
