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An automorphism of a root system $\Phi \subset E$ is an orthogonal transformation of $E$ mapping $\Phi$ onto itself. The Weyl group is a subgroup of this automorphism group; additional automorphisms can act nontrivially on a Dynkin diagram while preserving a root basis.

The orthogonal transformation ${\psi }_{\pm }=({\varphi }_1\pm {\varphi }_2)/\sqrt{2}$ takes $g({\varphi }_1^4+{\varphi }_2^4+6{\varphi }_1^2{\varphi }_2^2)$ to $2g({\psi }_{+}^4+{\psi }_{-}^4)$. These are the same two decoupled Ising models expressed in rotated fields.

Every Euclidean isometry of ${\mathbb{R}}^n$ is a product of at most $n+1$ operations of reflection in a hyperplane. An isometry fixing zero is orthogonal: the polarization identity preserves inner products, and its values on an orthonormal basis determine its linear action. An orthogonal transformation is a product of at most $n$ linear reflections, by reflecting the image of the first basis vector back to that vector and inducting on its orthogonal complement. For an arbitrary isometry, first reflect its image of zero back to zero, if necessary, and apply the orthogonal result. Affine reflection hyperplanes, rather than only hyperplanes through zero, are essential here.

For unit $u$, put $A=I-2u{u}^T$. Then $A^T=A$ and $A^2=I$, so $A$ is an orthogonal matrix. The affine reflection in a hyperplane is $R(x)=Ax+2cu$. Hence $R(x)-R(y)=A(x-y)$ preserves the Euclidean norm and distance, and $R(x)=x$ whenever $u\cdot x=c$. Thus it is an Euclidean isometry fixing the hyperplane pointwise.

For $p\ne q$, any reflection exchanging them must have normal parallel to $p-q$, and its fixed hyperplane must contain their midpoint. These requirements determine the perpendicular bisector uniquely:Changing both signs of $u,c$ leaves the same reflection. It belongs to the orthogonal group $O(n)$ exactly when its translation term vanishes, that is, $c=0$. Since $c=(\Vert p{\Vert }^2-\Vert q{\Vert }^2)/(2\Vert p-q\Vert )$,To prove the finite reflection decomposition of a Euclidean isometry, first note that a distance-preserving map $g$ fixing zero preserves inner products by the polarization identity. Its values on an orthonormal basis $e_i$ form an orthonormal basis, and $g(x)\cdot g(e_i)=x_i$ shows $g(x)={\sum }_i{x}_{i}g(e_i)$. Thus $g$ is orthogonal.

Every orthogonal transformation is a product of at most $n$ linear reflections: if $g(e_1)\ne e_1$, reflect $g(e_1)$ to $e_1$ using the hyperplane normal $g(e_1)-e_1$, which passes through zero. The resulting orthogonal transformation fixes $e_1$, and its restriction to $e_1^{\perp }$ is handled inductively in dimension $n-1$. If $g(e_1)=e_1$, no initial reflection is necessary. Reflections on the complement extend by fixing $e_1$; the induction starts in dimension zero.

For an arbitrary isometry $f$, if $f(0)\ne 0$, first reflect $f(0)$ to zero across its perpendicular bisector with zero. Composing this one affine reflection with $f$ gives an orthogonal transformation. If $f(0)=0$, skip the initial reflection. ThereforeThe glide reflection $(x,y)\mapsto (x+1,-y)$ cannot use fewer than three: its determinant sign is negative, excluding zero or two reflections, while absence of fixed points excludes a single reflection. Three suffice by reflecting successively in $x=0$, $x=1/2$, and $y=0$.

The linear least-squares problem is to minimize $\Vert Ax-b{\Vert }_2^2$ over $x\in {\mathbb{R}}^n$, allowing an overdetermined system to have a nonzero residual. For a full QR decomposition, $Q$ is $m\times m$ orthogonal and $R$ is $m\times n$ with an upper-triangular top block. Orthogonal transformations preserve the Euclidean norm, soIf $A$ has full column rank, the top triangular block is nonsingular; setting its residual to zero gives the unique minimizer, while the remaining residual is independent of $x$. Without full column rank, minimizers still exist but need not be unique.

For the specified matrix, one Householder reflection already produces an upper-triangular matrix. Take $v=(-1,1,1,1{)}^T$, the difference between the first column and $2e_1$, and setThe formula gives $H^T=H$, $H^2=I$. Direct multiplication yields the Householder QR decompositionAll subdiagonal entries in the remaining columns are already zero, so no further reflection is needed. For the supplied right-hand side, $Q^{T}b=(2,0,2,-2{)}^T$. Back substitution gives $x_3=1$, $x_2=-1$, $x_1=2$, henceThe residual $A{x}^{\ast }-b=(1,-1,-1,1{)}^T$ is orthogonal to every column of $A$, independently confirming the normal equations $A^T(A{x}^{\ast }-b)=0$.

At the Gaussian fixed point, $[{\varphi }_i]=(d-2)/2$, so all three quartic couplings have engineering dimension $d-4[{\varphi }_i]=4-d$. ThusTake $A>0$ and $ϵ>0$, as in the perturbative epsilon expansion. Define $x=A{g}_1/ϵ$, $y=A{g}_2/ϵ$, $z=A\lambda /ϵ$. The renormalization-group fixed point equations becomeIf $z=0$, each of $x,y$ is independently $0$ or $1/36$, giving four renormalization-group fixed points. If $z\ne 0$, subtracting the first two equations givesThe branch $x+y=1/36$ forces $z=1/12$; inserting this in the first equation yields a repeated root $x=y=1/72$. Hence every nonzero-$z$ solution has $x=y$. Then $z=1/8-3x$, andhas roots $x=1/40$ and $x=1/72$. The complete list of six coupled Ising fixed points near four dimensions, in coordinates $(g_1,g_2,\lambda )$, isThe internal symmetries of these renormalization-group fixed points are:

The last point displays the field-rotation equivalence of decoupled Ising theories. With the orthogonal transformation ${\psi }_{\pm }=({\varphi }_1\pm {\varphi }_2)/\sqrt{2}$,It is the same pair of decoupled Ising models written in fields rotated by $45^{\circ }$, but remains a distinct coordinate solution of the stated renormalization-group beta functions. Here $D_4$ denotes the dihedral group of order eight; an alternative convention calls it $D_8$. We have described linear internal transformations preserving the gradient energy; free massless sectors also have constant scalar-field shift symmetries. At $ϵ=0$, the six coordinates coalesce at the Gaussian fixed point.

A smooth surface in ${\mathbb{R}}^3$ is a subset locally parametrized by a smooth map of two variables whose derivative has rank two and which is a homeomorphism onto its image.

Writing $\rho =\sqrt{x^2+y^2}$, the first surface equation becomesIt is the torus of major radius $R=2\sqrt{2}$ and minor radius $r=1$, with smooth parametrizationSince $R>r$, this is an embedding. The Gaussian curvature of a torus is

For the second surface, use the orthogonal transformationIts equation becomes the same torus equation $(\sqrt{X^2+Y^2}-2\sqrt{2}{)}^2+Z^2=1$. Orthogonal transformations are Euclidean isometries and preserve Gaussian curvature. Curvature vanishes where $\cos v=0$, equivalently $Z=\pm 1$. In the original coordinates the zero-curvature points are therefore exactly

For an edge $ij$, put $\rho =⟨v_i,v_j⟩$ and $\theta =\arccos \rho$. Since $\rho \le -1/2$, we have $\theta \ge 2\pi /3$.

The random normal $a$ has the isotropic Gaussian distribution $N(0,I_p)$. Its distribution is invariant under orthogonal transformations; if $v_i,v_j$ are linearly independent, its projection onto their plane has a uniformly distributed direction. The signs of its inner products with $v_i,v_j$ differ in two sectors of total angle $2\theta$, out of $2\pi$. Thus random hyperplane rounding givesIf $v_j=-v_i$, the signs differ with probability one and the same formula holds with $\theta =\pi$. Zero inner products have probability zero, since each $v_i$ is a unit vector.

If the graph has an edge, its corresponding principal $2\times 2$ block of $U$ forces $t\ge 2$, so the division by $t-1$ used to obtain the Gram matrix is valid. An edgeless graph can instead be colored with one color directly.

The primitive exterior square of a symplectic vector space is the kernel of its symplectic contraction of an exterior square. In dimension four, the invariant inverse-form bivector $\Omega$ has nonzero wedge square. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$; the symmetric bilinear form $\xi \wedge \eta =B(\xi ,\eta )\mathrm{vol}$ is nondegenerate on ${\Lambda }^{2}V$. The primitive subspace is ${\Omega }^{\perp }$, and therefore inherits a nondegenerate form of dimension five. The symplectic Lie algebra acts on it by infinitesimal orthogonal transformations.

For unit vectors $u,v$ and a random normal $a$ with independent standard normal distribution coordinates,The Gaussian distribution is invariant under orthogonal transformations. Projecting onto the plane spanned by $u,v$ therefore gives a uniformly distributed direction. If the angle between $u,v$ is $\theta$, the sign-disagreement directions form two sectors of total angle $2\theta$ out of $2\pi$. The endpoint cases $u=v$ and $u=-v$ give probabilities zero and one directly.