### 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 be a set of real-valued continuously differentiable functions defined over for some . By I denote the tangent sphere bundle of Euclidean -space; if you are not familiar with that, just think of it as the Cartesian product , or ordered pairs where is a point in and is a unit vector.

By we denote a function ; we put its first argument in square brackets and its second argument in parentheses. We will refer to as the “constraint function”.

By we denote the set of all isometries of , which induces also a mapping of to itself. We will note a few special elements:

- For we let be the reflection over the hyperplane through the point that is orthogonal to .
- Similarly, if is a orthonormal set of vectors, we let . Notice that the operations on the right hand side commute because the ‘s are mutually orthogonal.
- And lastly, we use the short hand the complete reflection about the point , which is equivalent to when is a set of orthonormal basis vectors.
- For such that , and for , we let denote the rotation in the plane spanned by through by angle . Analytically we can write it as sending .

For each we define three maps on :

- ,
- which acts as the identity if and as otherwise,
- which acts as the identity if and as otherwise.

Note that is the projection map to the hyperplane defined by , and is the projection map to the positive half space. Also observe that .

We extend the above definition to sets of orthonormal vectors:

- and

are the orthogonal projections to the subspace orthogonal to and the subspace spanned by 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.

- and .
- If , we have , and .
- In general, given , let be in the span of , then there exists such that . Let be a continuous function on span of , we have three cases
- is irrational. Then orbit of under is dense on the unit circle. Hence must be radially symmetric.
- where is odd. Then and generate a dihedral group of an even polygon, and is a symmetry of .
- where is even and is odd. Then and generate a dihedral group of an odd polygon, and hence the two reflections are conjugate
*by a reflection*. (To see this, let be the order of the element , by assumption is odd. Hence . Letting be a vector such that , the LHS can be written as . Observe that is a symmetry of which swaps the spans of and .)

**Hypotheses**

We need the following properties on . In the following will always denote arbitrary functions in , and arbitrary pairs in .

- The function is continuous,
- ,
- If then ,
- For every there exists a such that .

We also will assume certain closure conditions on :

- If and , then .
- If satisfies , then .
- If is an isometry, and if , then and .

**Intermediate lemmas**

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

Thm 1 [Borsuk-Ulam]

A continuous odd map where admits a vector such that .

Now some technical lemmas.

Lem 2

Let be fixed. If is a set of orthonormal vectors such that for all , we have that whenever and . In this case we say isradially symmetric about.

*Proof*. Since $latex $f$ is continuously differentiable, it suffices to show that all “angular derivatives” vanish, i.e. that for every , when . By hypothesis since , and so is . Hence for we have that as needed. q.e.d.

Lem 3

Let and such that is radially symmetric about . Then there exists such that is radially symmetric about .

*Proof*. By our hypothesis can be chosen such that . WLOG suppose that . By Lemma 2 it suffices to show that for every , . For , there exists such that is in the span of . Since is assumed to be radially symmetric about (and hence also ), . Now consider , this implies that . On the other hand, it is clear that interchanges the range of . Since is independent of the choice of , this implies that are both radially symmetric. However, since along the hyperplane , we get that . q.e.d.

Cor 4 [2D case]

If , and , then there exists such that is radially symmetric about .

*Proof*. Choose arbitrarily. By assumption is an odd continuous function from the circle to the real numbers, hence it vanishes for some direction . Let be a direction orthogonal to . By the construction from the proof of Lemma 3, we can find such that . The functions satisfy and from the proof of Lemma 3 we see that they are all radially symmetric about . Using that along , and similarly along , we have that the four functions all equal and hence is radially symmetric about . q.e.d.

Lem 5

Let be a set of orthonormal basis. If is radially symmetric about and where , and $\latex V^\perp \cap W^\perp \neq \emptyset$, then is radially symmetric about .

*Proof*. We can write , and hence every point in can be written as a linear combination of each belonging to one of the subspaces. The two radial symmetries imply that . This demonstrates the radial symmetry. q.e.d.

Lem 6 [Induction hypothesis]

If , and . For every there is a dimensional subspace such that is radially symmetric about .

*Proof*. Since , using the Borsuk-Ulam theorem we can choose such that . Let . Applying the Borsuk-Ulam theorem to the continuous functions we get two vectors for which . Writing $K^\pm = \{v,w^\pm\}^\perp$, the argument in the proof of Lemma 3 suffices to show that are radially symmetric about respectively. We split into several cases depending on how compare. Let .

- . Then . Hence are two functions radially symmetric about , and such that they agree on , so they agree wholly and is radially symmetric about $latex .
- Suppose generate the circle in the plane spanned by . Then from the definition of radial symmetry we see that are both radially symmetric around , and hence by Lemma 5 radially symmetric around . Since they agree on , we get is also radially symmetric..
- Suppose that generate the dihedral group for an even polygon. Letting be an orthonormal basis for the span of , we have then is a symmetry of . Let be in the span of , we have that using the invariance property of under isometries, . This implies, by Lemma 2, that both are radially symmetric around . This implies via Lemma 5 that are radially symmetric around .
- Lastly, we have the odd polygon dihedral case. By assumption 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 such that is symmetric under . A direct computation shows that thus . This implies that . So for any that is a linear combination of we have that (again using the action of under isometry). Hence is radially symmetric about .

q.e.d.

Lem 7 [Induction step]

If is radially symmetric about where is a subspace of dimension at least 2, then there exists such that is radially symmetric about .

*Proof*. Using Borsuk-Ulam theorem again, there exists such that . We can then use the same method as the proof of Lemma 3 to conclude that is radially symmetric about . q.e.d.

**Conclusions**

Putting everything together we get

Thm [Maris]

If and , then is radially symmetric about some point .

In dimension 1, if you let , then we can allow 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 .

As an example application: consider the problem of minimising under the constraint that is fixed. By elliptic regularity (under suitable assumptions on the growth rate of ) we have that any critical point must have regularity . Let denote the set of constrained minimisers. Let . Define . If , then so still solves the constraints. On the other hand, we have that . Since minimises the Dirichlet integral, so must also . The various other properties assumed on and are easily checked: continuity of 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.