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The Lagrangian Grassmannian is the space of all Lagrangian subspaces of a symplectic vector space. The unitary group acts transitively on $\mathrm{LGr}({\mathbb{R}}^{2n})$, and the stabilizer of ${\mathbb{R}}^n$ is the orthogonal group $O(n)$, giving

In odd dimension, the map $A\mapsto (\det A)A$ is a surjective group homomorphism from the orthogonal group to the special orthogonal group, with kernel of a group homomorphism $\{I,-I\}$. Scalar matrices commute, and $\det [(\det A)A]=(\det A{)}^{2m+2}=1$. Together with the determinant sign, it identifies $O(2m+1)$ with the direct product of groups $SO(2m+1)\times C_2$. In even dimension, scalar multiplication by $\det A$ does not repair the determinant, so this construction does not give the same splitting.

For particles in the vector representation of the orthogonal group, the two-body S-matrix is a linear combination of the identity $I$, permutation $P$ and trace contraction $K$: $S=S_{2}I+S_{3}P+S_{1}K$. Here $K(e_i\otimes e_j)={\delta }_{ij}{\sum }_a{e}_a\otimes e_a$, $P^2=I$, $PK=KP=K$ and $K^2=NK$. The orthogonal-invariant scattering channels diagonalize these three operators simultaneously.

The $O(N)$ model has an $N$-component real field and an orthogonal group symmetry. Its standard Landau-Ginzburg interaction is $g(\mathbf{\varphi }\cdot \mathbf{\varphi }{)}^2$; below the upper critical dimension its continuous transition is governed by an $O(N)$ Wilson-Fisher fixed point.

For a continuous real representation $\rho :G\to GL(V)$ of a compact group, average any positive-definite real inner product against normalized Haar measure:The result is positive definite and $G$-invariant. An orthonormal basis for it conjugates the representation into an orthogonal group.

True. The determinant gives a group homomorphismFor an orthogonal matrix, $A^{T}A=I$ implies $(\det A{)}^2=1$, and a reflection has determinant $-1$, so the target has order two and the map is onto. Its kernel of a group homomorphism is exactlyThis is the defining special orthogonal group inside the orthogonal group.

Interpret the paired indices $(i,k)$ as incoming and $(j,l)$ as outgoing. For self-conjugate particles in the real vector representation of the orthogonal group, crossing the second incoming particle with the second outgoing one givesThe continuation $\theta \mapsto i\pi -\theta$ exchanges the trace and permutation tensor structures and leaves the identity structure fixed. Therefore crossing symmetry imposesThese are analytic relations; they do not determine the three functions completely.

For unitarity, act on the two-species tensor product with the O(N)-invariant S-matrix $S=S_{2}I+S_{3}P+S_{1}K$. The invariant operators obey $P^2=I$, $PK=KP=K$ and $K^2=NK$. Writing $S_a^{-}=S_a(-\theta )$ and multiplying $S(\theta )S(-\theta )=I$ givesThe three invariant operators are independent for $N\ge 2$. Equivalently, the orthogonal-invariant scattering channels have amplitudesand each satisfies $s_a(\theta )s_a(-\theta )=1$. The trace singlet, symmetric traceless and antisymmetric channels have dimensions $1$, $N(N+1)/2-1$ and $N(N-1)/2$. For $N=2$ these are $1,2,1$. With Hermitian analyticity of a two-particle S-matrix, $S_a(-\theta )=S_a(\theta {)}^{\ast }$ on the real axis, so each channel has unit modulus. This expresses conservation of scattering probability.

The Faddeev-Zamolodchikov algebra orders particle operators by rapidity. Its associativity requires that a product of three operators be independent of parentheses and, in particular, that both sequences of adjacent exchanges give the same final ordered species word with the same coefficient. This is the Faddeev-Zamolodchikov associativity constraint, or spectral Yang-Baxter equation. The exchanges do not use the ordinary creation/annihilation normal ordering convention.

Take ${\theta }_1>{\theta }_2>{\theta }_3$, with $\theta ={\theta }_1-{\theta }_2$, ${\theta }^{\prime }={\theta }_2-{\theta }_3$, and abbreviateFor $N=2$, start with $A_1({\theta }_1)A_1({\theta }_2)A_2({\theta }_3)$ and compare the coefficient of $A_1({\theta }_3)A_2({\theta }_2)A_1({\theta }_1)$. First exchange positions $1,2$, then $2,3$, then $1,2$. The initial equal-index exchange givesFrom the first term, the specified final word is reached through $b_2{c}_3$; from the second term, the intermediate equal-index exchange uses $b_1$ to produce species $1$, followed by $c_2$. Thus this exchange route has coefficientFor the other route, first exchange positions $2,3$, then $1,2$, then $2,3$. The first exchange is between different indices and gives $c_2{A}_1({\theta }_1)A_2({\theta }_3)A_1({\theta }_2)+c_3{A}_1({\theta }_1)A_1({\theta }_3)A_2({\theta }_2)$. The first term reaches the target through $b_3{a}_3$, and the second reaches it through $(b_1+b_2+b_3)a_2$. ThereforeEquate these two coefficients, cancel the common term $a_2{b}_2{c}_3$, and rearrange:Restoring the three arguments gives the required scalar Yang-Baxter equation. The derivation starts in the ordered physical region $\theta ,{\theta }^{\prime }>0$; the identity extends to other values by the same scattering analytic continuation, wherever its factors are defined. Associativity must hold coefficient by coefficient for every species word; this displayed relation is one necessary component.

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 $y=0$.

Write $r_i={\mu }_i^2$ and assume $g>0$, as required for quartic stability. In the mean-field approximation, the two-component radial quartic Landau potential isSet $x={\varphi }_1^2\ge 0$, $y={\varphi }_2^2\ge 0$. At fixed $\rho =x+y$, the quartic term is fixed and the quadratic term is minimized by assigning all of $\rho$ to the component with smaller $r_i$. Minimizing over $\rho$ then givesThe resulting phase diagram has these regions:

On $r_1=r_2=r<0$, the symmetry is enhanced to the orthogonal group $O(2)$, and the minima form the circleMixed-component minima occur on this line, rather than in an open coexistence region. Crossing the negative diagonal swaps the selected component discontinuously: this is a first-order phase transition, also called a spin-flop transition. For example, holding $r_2=r<0$ fixed and moving $r_1$ across $r$, the derivative $\partial f_{\min }/\partial r_1$ jumps from $-r/(8g)$ to zero. The angular degeneracy exactly on the line does not remove this discontinuity.

The positive half-axis $r_1=0$, $r_2>0$ is a continuous phase transition, as is $r_2=0$, $r_1>0$. Their endpoint at $(0,0)$ is a bicritical point, where the two continuous boundaries meet the first-order negative diagonal. Approaching the origin radially makes the ordered amplitude vanish continuously. The negative half-axes lie inside ordered phases and are not additional transition lines. There are two continuous half-axis boundaries and one first-order negative diagonal, meeting at the origin.

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.