20

No. There exist closed manifolds which do not admit any compact group actions, in particular, no $Z_2$ actions. For example, Shultz showed in "Group actions on hypertoral manifolds. II." that in dimensions $\ge 4$ every oriented cobordism class contains such a manifold.


17

No. Let $G$ be a finite subgroup of $GL_n(\mathbb R)$ containing $W(E_8)$ as a subgroup. Because $G$ is compact, $G$ must preserve a symmetric positive definite form on $\mathbb R^8$. Since $W(E_8)$ preserves a unique such form, it must be that one. Let $H$ be the largest subgroup of $G$ generated by reflections. Then $H$ contains $W(E_8)$ and thus is an ...


12

If I understand correctly, you mean the definition of an object’s {symmetries} as its stabilizer under some group action on an ambient space. While a clearcut “first” might not exist, I’d say Jordan has early approximations to that, provided you are willing to replace “function” by “transformation”, “substitution”, or “motion”. E.g. in his Traité des ...


11

Plotting things on a circle like this focuses attention on dihedral symmetry, which is implemented on the integers $\bmod n$ by translation $t \mapsto t+1$ and negation $t \mapsto -t$. Addition $t \mapsto t + k$ is itself a translation, so of course it has translational symmetry (it commutes with translations). Scaling $t \mapsto kt$ always at least commutes ...


10

Outside of quantum physics, both SU(2) and SU(3) appear in the context of polarization optics. For a plane wave the rotation of the two components of the polarization (perpendicular to the propagation direction) are represented by an element of SU(2). To describe the more general case of an arbitrary wave field (without a fixed propagation direction), three ...


9

One of the simplest mechanical systems having an SU(3) dynamical symmetry is the three dimensional isotropic harmonic oscillator. Its phase space is $\mathbb{R}^6$ parametrized by 3 position coordinates $x_i$ and 3 momenta $p_i$, $i=1,2,3$. A fixed energy hypersurface in phase space is given by the constraint (in units where the natural frequency is 1): $$\...


8

From the introduction of Legendre's Revolution (1794): The Definition of Symmetry in Solid Geometry, Giora Hon and Bernard R. Goldstein, Archive for History of Exact Sciences, Vol. 59, No. 2 (January 2005), pp. 107-155. The modern scientific concept of symmetry is, in the first place, a mathematical idea. It refers to an intrinsic property of a ...


8

$A_1^*$ is maximal. $A_2 = I_3$ and $A_2^* = I_6 = G_2$ (for the first one, consider the Dynkin diagram, for the second one, any group containing it as an index $2$ subgroup must normalize it, hence must be the normalizer) so $A_2^*$ is not maximal. $A_3 = D_3$ and $A_3^* = B_3$. $B_3$ has order $48$, a multiple of $16$, while the only larger $3$-dimensional ...


8

The answer is negative. Already in dimension 4 there are fake real-projective spaces, which are smooth 4-manifolds homotopy-equivalent but not homeomorphic to $RP^4$. These correspond to smooth free involutions $\sigma: S^4\to S^4$ which are not topologically conjugate to orthogonal transformations. Similar examples exist in higher dimensions. See manifold ...


7

Noether's theorem was successfully generalized to the setting of quantum field theory in the fifties. This is called Ward-Takahashi identity.


7

"Can there be made mathematical sense out of it?" Perhaps as follows: given a finite set $X$ labeled with bits $y\in\{0,1\}^X$, indicating whether the corresponding element is present or missing, decide whether $\sum_{x\in X}y(x)=|X|$ or $<|X|$. This is a strictly weaker operation than counting. Certainly a counter that evaluates $\sum_{x\in X}y(x)$ can ...


6

A very nice survey is given by Francesco Oliveri. Oliveri, Francesco, Lie symmetries of differential equations: classical results and recent contributions, Symmetry 2, No. 2, 658-706 (2010). ZBL1284.22014. He has plenty of references, in particular to computer algebra system implementations of the algorithm (which goes back to Lie). See, e.g. reference 15. ...


6

The $\nabla$-compatible metrics on $E$ are the positive-definite $\nabla'$-parallel sections of $S^2(E^*)$, where $\nabla'$ is the connection on $S^2(E^*)$ induced by $\nabla$. When $M$ is connected and $x\in M$ is fixed, the space of $\nabla'$-parallel sections of $S^2(E^*)$ is isomorphic to the set $Q_x(\nabla)\subset S^2(E^*_x)$ of quadratic forms on $...


6

Yes, this is true for all polyhedral graphs. See the following reference: P. Mani, Automorphismen von polyedrischen Graphen, Mathematische Annalen Volume 192, Issue 4, pp 279-303, August 1971 My translation: Theorem. For every graph $\mathfrak{Q}$ there is a three dimensional convex polyhedron $P$ with the properties: (a) $\mathfrak{Q}$ is isomorphic to ...


6

$\DeclareMathOperator{\SO}{SO}$$\DeclareMathOperator{\RP}{RP}$$\DeclareMathOperator{\Stab}{Stab}$There is a difference between the sets $C = [-1,1]^3$ and $E = [0,1]^3$. Note that $\Stab(C)$, the stabilizer of $C$ inside of $\SO(3)$, is called the cube group and it has 24 elements. Note that $\Stab(E)$ has three elements. Are you sure you want $E$ and not ...


6

One way you can use conservation laws of a PDE is in numerics; you check at each moment in time the value of the conserved quantity coming from the conservation law, to see if it is still being conserved approximately, and as soon as it is clearly not conserved (and not nearly conserved), you should no longer believe that the numerical approximation to the ...


6

The answer given by Robert Israel seems worth elaborating on. Knowing if everyone is present at a large family gathering is like going through the family history, or family tree: grandmother is here, her daughters A, B, and son C are here. Of course, son D isn't here, since he died in childhood. Daughter A's son E and his wife F are here with their baby son ...


6

You don't need parallelism: a symmetric tensor of type $\binom{0}{2}$ is positive definite if and only if it is a Riemannian metric, and has signature $(1,n-1)$ (where $n$ is the dimension of your manifold) if and only if it is a Lorentzian metric, and has signature $(p,q)$ if and only if it is a $(p,q)$ pseudo-Riemannian metric. But it is also true that an ...


6

The automorphism group of this configuration $C'$ is the Mathieu group $M_{11}$. Firstly, we construct a larger configuration $C$ consisting of a 12-dimensional orthoplex inscribed in a 12-dimensional hypercube. In particular, the orthoplex in $C$ consists of the vectors $\{ \pm v : v \textrm{ is a column of } H \}$, where $H$ is a 12-by-12 Hadamard matrix ...


6

The OP doesn't say what is meant by a 'geometric object', so it's hard to give a definitive answer. However, if one assumes that the geometric object is a smooth manifold $M^7$ and that the action is smooth, then there are a few things one can say: First, the $\mathrm{G}_2$-stabilizer of any point $p\in M^7$ has to be a closed subgroup $H_p\subset \mathrm{G}...


5

Sorry to join so late to the party, but I couldn't help noticing there is a missing class of rich matrix norms (which are not operator norms). These are the p-Schatten norms on $R^{n^2}$, which see only the singular values of a matrix. Its immediate to see that these norms are invariant under left and right multiplication by orthogonal matrices. Hence in ...


4

This is Lemma 6.2 in Gonzalez, Fulton B.; Kakehi, Tomoyuki, Dual Radon transforms on affine Grassmann manifolds, Trans. Am. Math. Soc. 356, No. 10, 4161-4180 (2004). ZBL1049.44001. (actually, the lemma is for arbitrary dimension, so the special functionology might be simplified further for $\mathbb{S}^2$) For convenience here is the Lemma (complete with ...


4

There are mainly two ways to get symmetry of minimizers, at least when the functional is associated to an elliptic PDE. The first one is the rearrangement argument (Schwarz symmetrization, Steiner symmetrization etc.), which shows in some cases symmetric objects have lower energy. The other is the moving plane method, which involves some maximum principle ...


4

The geometric reason for the ubiquity of $\sqrt 5$ in problems involving a pentagon is that it is the diagonal of a $1\times 2$ rectangle (a "half-square"). This links $\sqrt 5$ to the construction of a pentagon from its side, which may be at the origin of the first three geometric observations in the OP. The fourth observation on Hurwitz theorem is not ...


4

First, a simple remark: If a polytope with congruent facets is inscribed in a sphere, then it is circumscribed about a sphere as well, and the two spheres are concentric. Next, there is a series of examples described and pictured in my old question Can the sphere be partitioned into small congruent cells? . Each of these examples is what you want in $R^3$. ...


4

Complements don't even have to be edge transitive. Perhaps the simplest example is the wreath graph $W_5$ (which is obtained by applying the construction below to $C_5$). Call two vertices "twins", if they have the same neighbourhood in $G$. Since the twin-relation is preserved under automorphisms, it suffices to construct an arc-transitive graph $G$ of ...


4

With most such maps you should get some sort of uniform distribution result. For example, J. Beck and I did this for the map x^{-1} mod n. See "On the uniform distribution of inverses modulo n", Periodica Math. Hungarica, Vol 44 (2) 2002, 147-155. (Forgive the self-serving PR, but as one gets older one loses all sense of propriety or modesty! Also ...


4

Here's the example I mean: I'm pretty sure the transposition symmetry here is not a power of another automorphism. Also, it should be vertex/edge transitive.


4

This answer builds upon Sam Hopkins' suggestion of an oriented toroidal grid, but it's more than a comment. Bottom line: need the grid to be $2^s$ by $2^s$ where $s\ge2$ to get the requested involution property. So here's a sketch of a grid: WLOG we can label some vertex as $(0,0)$, and then its two out-neighbours as $(1,0)$ & $(0,1)$. But how do we ...


3

Here is a counterexample (from p.76 of my thesis): $$ A = A^T = \begin{bmatrix} 0 & 3 & 2 \\ 3 & 0 & 2 \\ 2 & 2 & 3 \end{bmatrix}. $$ Labeling the strategies in order as $a$, $b$, and $c$, there are asymmetric Nash equilibria $(a,b)$ and $(b,a)$ with support $\{a,b\}$, but there is no symmetric Nash equilibrium supported on this set....


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