Orthogonal group ($O(n)$)

The real orthogonal group is $O(n)=\{R:R^{T}R=I\}$. Its tangent space at the identity is the vector space of skew-symmetric matrices.

A finite subgroup of $O(2)$ is cyclic or dihedral. Its orientation-preserving part consists of rotations and is cyclic: choose the smallest positive rotation angle, divide any other angle by it with remainder, and use minimality to force zero remainder. The smallest angle is therefore $2\pi /n$ for some $n$. A trivial rotational subgroup is included by $n=1$.

If there is no orientation-reversing element, this is the whole group. Otherwise choose a reflection $s$. Every orientation-reversing element is $r^{j}s$, where $r$ generates the rotations; these form the second coset. Since $srs=r^{-1}$, the group is a dihedral group of order $2n$. Its involutions are all $n$ reflections and, when $n$ is even, the extra half-turn $r^{n/2}$. Consequently it has $n$ order-two cyclic subgroups for odd $n$, and $n+1$ for even $n$.

The smooth Gram map $F(A)=A^{T}A$ takes real matrices to symmetric matrices. At an orthogonal $A$, its derivative sends $AH/2$ to any prescribed symmetric matrix $H$. The regular level set $F^{-1}(I)$ is therefore a smooth manifold of the displayed dimension.

// Target: geometry-and-topology.bigb

Every orthogonal transformation of a finite-dimensional nondegenerate quadratic space over a field of characteristic different from two is a product of at most its dimension many hyperplane reflections. In the real positive-definite case, a reflection normal to $gv-v$ sends $gv$ to $v$ whenever $gv\ne v$. Composing with that reflection fixes $v$ and leaves an orthogonal transformation of $v^{\perp }$, so induction proves the bound. Each reflection has determinant minus one, hence orientation-preserving transformations use an even number of reflections.

For real orthogonal matrices in odd dimension $d$, multiplying by their determinant defines a group homomorphism from $O(d)$ onto $SO(d)$. Scalar factors commute, so $\pi (AB)=\pi (A)\pi (B)$, while $\det \pi (A)=(\det A{)}^{d+1}=1$. The map fixes $SO(d)$ and has kernel $\{I,-I\}$. Since $-I$ is central and has determinant $-1$ in odd dimension, $O(d)\cong SO(d)\times C_2$. In even dimension the same expression does not map every orthogonal matrix into the special orthogonal group.

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.

A real square matrix is orthogonal when $R^{T}R=I$, equivalently when its columns form an orthonormal basis.

An invertible matrix over the complex numbers preserving the symmetric bilinear form $v^{T}w$: $A^{T}A=I$. The transpose here does not include complex conjugation, unlike the defining condition for a unitary matrix. For example, $\left(\begin{array}{cc}\cosh t& i\sinh t\\ -i\sinh t& \cosh t\end{array}\right)$ is complex orthogonal for real $t$, but is not unitary when $t\ne 0$.

Every two-by-two complex orthogonal matrix has form $A=\left(\begin{array}{cc}a& b\\ -\epsilon b& \epsilon a\end{array}\right)$, where $\epsilon =\det A\in \{+1,-1\}$ and $a^2+b^2=1$. This follows by equating $A^{-1}$ to $A^T$ in the two-by-two inverse formula, and direct multiplication proves the converse. A complete parametrization is $a=\cosh t$, $b=i\sinh t$ with complex $t$; restricting $t$ to real values selects only a subfamily.

An orientation-reversing orthogonal transformation. In three dimensions it is a rotation about an axis followed by a reflection perpendicular to that axis. Its eigenvalues are $-1,e^{i\theta },e^{-i\theta }$, giving $\mathrm{tr}R=-1+2\cos \theta$. A plane reflection is the case $\theta =0$, while inversion through the origin is the case $\theta =\pi$.

An orientation-preserving orthogonal transformation. In three dimensions it has an invariant axis and acts as a planar rotation through angle $\theta$ perpendicular to that axis. Its eigenvalues are $1,e^{i\theta },e^{-i\theta }$, so $\mathrm{tr}R=1+2\cos \theta$. In higher dimensions a proper rotation can involve several mutually orthogonal planes and need not fix an axis.

For a real three-dimensional linear map $A$, preservation of every surface area is equivalent to being an orthogonal matrix. For necessity, diagonalize the symmetric matrix $A^{T}A$ in an orthonormal basis, with nonnegative eigenvalues ${\lambda }_i$. The three unit coordinate parallelograms have image areas $\sqrt{{\lambda }_1{\lambda }_2}$, $\sqrt{{\lambda }_1{\lambda }_3}$, and $\sqrt{{\lambda }_2{\lambda }_3}$. Setting all three equal to $1$ forces every ${\lambda }_i=1$. Sufficiency follows from preservation of dot products and $\Vert \mathbf{u}\times \mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2\Vert \mathbf{v}{\Vert }^2-(\mathbf{u}\cdot \mathbf{v}{)}^2$. This specifically uses dimension three; in dimension two, preservation of planar area alone only requires $|\det A|=1$.

Suppose real symmetric matrices $E,F$ satisfy $E^2=F^2=I$ and $EF=-FE$. The involutions are orthogonal matrices, and $J=EF$ satisfies $J^T=-J$, $J^2=-I$. The matrix exponential therefore reduces to $R(t)=e^{tJ}=\cos tI+\sin tJ$. Multiplication gives $R(t{)}^{T}R(t)=I$ and the conjugation identitiesIndeed $JE=-F$, $EJ=F$, $JF=E$, and $FJ=-E$, which give both identities by expansion. The angle in this two-dimensional space of matrices is twice the parameter in the orthogonal matrix. These formulas hold in any finite dimension where the hypotheses can be satisfied.

For a real orthogonal involution, $A^{-1}=A=A^T$, so it is a symmetric matrix. Its eigenvalues are $\pm 1$. In two dimensions the possibilities are the identity, the half-turn $-I$, and reflectionsThe reflection fixes the line making angle $\theta /2$ with the first coordinate axis and negates its perpendicular direction. This follows either from the two corresponding eigenvectors or by rotating $\mathrm{diag}(1,-1)$. A nonidentity element of order two in a faithful representation must be either the half-turn or a reflection; the identity matrix is excluded by faithfulness.

An orthogonal transformation of a real inner-product space is a linear map $Q$ satisfying $Q^{T}Q=I$. It preserves inner products, lengths, angles, and therefore Gaussian curvature when applied to an embedded surface in Euclidean space.

Let $b_1,\dots ,b_n$ be an arbitrary basis of a real inner product space, with Gram matrix $G=(⟨b_i,b_j⟩)$. If a linear map $T$ has matrix $A$ in this basis, thenConsequently $T$ is an orthogonal transformation exactly when $A^{T}GA=G$. The simpler orthogonal matrix condition $A^{T}A=I$ applies directly when the chosen basis is an orthonormal basis, so that $G=I$.

For example, take a two-dimensional vector space with weighted Euclidean inner product having $G=\mathrm{diag}(1,4)$. Thensatisfies $A^{T}GA=G$, but $A^{T}A=\mathrm{diag}(1/4,4)\ne I$. Thus it represents an orthogonal transformation even though it is not an orthogonal matrix.

An improper orthogonal transformation has determinant $-1$. In three dimensions it may be a plane reflection or a rotation composed with a reflection perpendicular to the rotation axis.

A real orthogonal matrix $T$ of size three with $\det T=-1$ has an eigenvector $n$ of eigenvalue $-1$. Its orthogonal complement is invariant, and the restriction there is an orientation-preserving plane rotation. Thus $T$ is a rotation about the axis $\mathbb{R}n$ composed with a reflection across the plane perpendicular to $n$. Writing the plane angle as $\varphi$, $\mathrm{tr}T=-1+2\cos \varphi$. A pure plane reflection is the case $\varphi =0$; if $\det (T-I)\ne 0$, it cannot be a pure plane reflection.

The special orthogonal group isEvery element of $SO(2)$ is a rotation in the plane. Every element of $SO(3)$ fixes an axis and restricts to a planar rotation on its perpendicular plane.

The compact group of real seven-by-seven orthogonal matrices with determinant one. Its Lie algebra has B3 root system. Its simply connected double cover is the Spin group $\mathrm{Spin}(7)$; the eight-dimensional spin representation belongs to that cover and does not descend to SO(7).

Every real orthogonal two-by-two matrix of determinant one has the unique formIndeed the first column is a unit vector, the perpendicular second column has one of two signs, and positive determinant selects the displayed one. The map $z\mapsto R(z)$ from the circle group is bijective and satisfies $R(zw)=R(z)R(w)$ by direct multiplication, proving a group isomorphism. Writing $z=e^{i\theta }$ parametrizes these matrices as planar rotations, with the angle defined modulo $2\pi$.

For positive distinct diagonal entries $c_1<\cdots <c_n$, the critical points of this function are diagonal sign matrices $D=\mathrm{diag}({\epsilon }_1,\dots ,{\epsilon }_n)$ with $\prod {\epsilon }_i=1$. Criticality forces $CX$ symmetric; its square is $C^2$, so it commutes with the distinct-spectrum diagonal matrix $C^2$ and is diagonal. In the skew coordinates $a_{ij}$ its Hessian is $-{\sum }_{i<j}(c_i{\epsilon }_i+c_j{\epsilon }_j)a_{ij}^2$. The sign of each parenthesis is that of ${\epsilon }_j$, giving Morse index ${\sum }_{j:{\epsilon }_j=1}(j-1)$.

All the special orthogonal groups are connected, and their second homotopy groups vanish. The rotation circle gives $SO(2)\cong S^1$. Conjugation by unit quaternions gives the twofold universal cover $S^3\to SO(3)$, so ${\pi }_{1}SO(3)=\mathbb{Z}/2$ and ${\pi }_{3}SO(3)=\mathbb{Z}$. The special orthogonal sphere fibration $SO(n-1)\to SO(n)\to S^{n-1}$ propagates the first two homotopy groups for $n\ge 4$. In dimension four its exact sequence has a zero map from the torsion group ${\pi }_4{S}^3=\mathbb{Z}/2$ to ${\pi }_{3}SO(3)=\mathbb{Z}$, giving ${\pi }_{3}SO(4)={\mathbb{Z}}^2$. For $SO(1)$ all positive groups vanish, and ${\pi }_{3}SO(2)=0$.

The last-column map $p(A)=A{e}_{s+1}$ has fiber $SO(s)$, included as $\mathrm{diag}(B,1)$. Local completion of a unit vector by the Gram-Schmidt process gives local fiber bundle charts. Consequently it has the homotopy lifting property and the long exact sequence of homotopy groups of a fibration.

Project a based map $\alpha :S^r\to SO(s+1)$ to $S^s$. Since $r<s$, the resulting homotopy class vanishes. Lift a based null-homotopy by the special orthogonal sphere fibration. Its endpoint lies in the fiber $SO(s)$. If the lift moves its base point, right-multiply at time $t$ by the inverse of its base-point value, which lies in the fiber. This keeps the projected homotopy unchanged and makes the lift based. Thus $\alpha$ is based-homotopic to a map into $SO(s)$, proving the asserted surjectivity.

The SO(4) group consists of real orthogonal four-by-four matrices of determinant one. It preserves orientation and the Euclidean metric. Its Spin(4) double cover identifies it with the quotient of two copies of SU(2) by their simultaneous central sign.

If $[L_a,L_b]={ϵ}_{abc}{L}_c$, $[L_a,K_b]={ϵ}_{abc}{K}_c$ and $[K_a,K_b]={ϵ}_{abc}{L}_c$, then $L_a^{\pm }=(L_a\pm K_a)/2$ generate commuting copies of $\mathfrak{su}(2)$. This Lie-algebra decomposition does not remove the central quotient in the corresponding global group.

Tensor products $V_{j_L}\otimes V_{j_R}$ give the irreducible representations of the Spin(4) double cover. The simultaneous central sign acts by $(-1{)}^{2j_L+2j_R}$, so exactly the integer-sum pairs descend to SO(4). Their dimensions are $(2j_L+1)(2j_R+1)$. Under the diagonal spatial rotation subgroup, the Clebsch-Gordan decomposition for SU2 gives spins from $|j_L-j_R|$ through $j_L+j_R$ in steps of one.

The subgroup of three-dimensional proper rotations preserving a chosen unoriented axis consists of all rotations about it and all half-turns about perpendicular axes. It is isomorphic to $O(2)$; the continuous molecular point-group notation is $D_{\infty }$. This uncountable group is different from the abstract countable infinite dihedral group $\mathbb{Z}⋊C_2$.

The SO(3) group consists of real three-by-three orthogonal matrices with determinant one. It acts by rotation in three dimensions. Its group manifold is ${\mathbb{RP}}^3$, since the Adjoint double cover from SU(2) to SO(3) identifies antipodal points of the three-sphere.

The Adjoint double cover from SU(2) to SO(3) has kernel $\{I,-I\}$. Under SU(2) as the three-sphere, multiplication by $-I$ is the antipodal map, so the quotient is Real projective space ${\mathbb{RP}}^3$. Equivalently, an axis-angle ball of radius $\pi$ has opposite boundary points identified. The fundamental group is ${\mathbb{Z}}_2$.

Every real orthogonal transformation of odd dimension with determinant one has eigenvalue $+1$. Nonreal eigenvalues come in conjugate pairs with product one, while real eigenvalues are $+1$ or $-1$. There are an odd number of real eigenvalues and an even number of negative ones, leaving at least one positive eigenvalue. The corresponding eigenvector is fixed.

The map $A\mapsto A^{T}A$ from real matrices to symmetric matrices has derivative $H\mapsto R^{T}H+H^{T}R$ at $R\in O(n)$, which is surjective. Thus $SO(n)$ is a submanifold of dimension $n(n-1)/2$, and

For $v\in S^n$, the stabilizer $S_v=\{R\in SO(n+1):Rv=v\}$ is conjugate to the block subgroup $SO(n)$ acting on $v^{\perp }$. Hence it is a submanifold of dimension $n(n-1)/2$, with

For distinct $v,w\in S^n$, the submanifolds $S_v,S_w\subseteq SO(n+1)$ are not transverse. They both contain the identity. If $v,w$ are independent, the bivector $v\wedge w$ is a nonzero normal direction common to both tangent spaces there; if $w=-v$, the two stabilizers coincide.

A rotation in three dimensions is an element of $SO(3)$. It fixes an axis and acts as a two-dimensional rotation on the plane perpendicular to that axis.

In the three-dimensional vector representation, the Hermitian matrices $(J_k{)}_{ij}=-i{ϵ}_{kij}$ generate active rotations by the matrix exponential $R(\theta ,\mathbf{n})=\exp (-i\theta n^k{J}_k)$. They satisfy $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$. The real antisymmetric matrices $A_k=-i{J}_k$ are the corresponding Lie algebra generators of $SO(3)$. Their commutator gives the leading change produced by interchanging two infinitesimal rotations.

A matrix is skew-symmetric when $A^T=-A$.

Proof. If $(I+A)x=0$, the quadratic-form criterion for a skew-symmetric matrix gives $0=x^{\top }(I+A)x=\Vert x{\Vert }^2$, so $x=0$. The rank-nullity theorem makes the square matrix $I+A$ invertible. More generally $tI+A$ is invertible for every nonzero real $t$, since its null space would imply $t\Vert x{\Vert }^2=0$. Equivalently, the only possible real eigenvalue of a real skew-symmetric matrix is zero.

An odd integer number of rows and columns forces a real skew-symmetric matrix to be a singular matrix. The determinant satisfies $\det A=\det A^{\top }=\det (-A)=(-1{)}^n\det A=-\det A$, so it is zero. The argument also works over any field of characteristic different from two.

Proof. For a skew-symmetric matrix, the scalar $x^{\top }Ax$ equals its transpose $x^{\top }{A}^{\top }x=-x^{\top }Ax$, so it vanishes. Conversely, substituting standard basis vectors $e_i$ into the vanishing quadratic form gives $a_{ii}=0$. Substituting $e_i+e_j$ gives $a_{ij}+a_{ji}=0$. Thus $A^{\top }=-A$. In particular, a real quadratic form depends only on a matrix's symmetric part.

For real $n\times n$ skew-symmetric matrices, each upper-triangular entry can be chosen freely, the corresponding lower-triangular entry is its negative, and diagonal entries vanish. Thus the displayed matrix unit differences span the space and are linearly independent: inspecting entry $(i,j)$ with $i<j$ recovers its coefficient. There are $n(n-1)/2$ such pairs, giving the dimension.

Every real square matrix decomposes as $A=S+K$, where $S=(A+A^T)/2$ is its symmetric part and $K=(A-A^T)/2$ is a skew-symmetric matrix. For a real column vector $x$, $x^{T}Kx$ equals its transpose $x^T{K}^{T}x=-x^{T}Kx$, hence vanishes. Thus a real quadratic form $x^{T}Ax$ depends only on $S$.

Every real $2\times 2$ antisymmetric matrix has the form $A=sJ$, where $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Thus $A^2=-s^{2}I$, $\mathrm{tr}A=0$, and $\det A=s^2$. In particular,A convergent scalar power series applied to $A$ can therefore be reduced into an even scalar multiple of $I$ plus an odd scalar multiple of $A$. For real coefficients and $s\ne 0$, this is $f(A)=\mathrm{Re}f(is)I+[\mathrm{Im}f(is)/s]A$, provided the series converges at $\pm is$. Absolute convergence justifies separating even and odd terms. This converts many small matrix series problems into ordinary scalar series, without computing eigenvectors.

A real skew-symmetric matrix is an orthogonal matrix exactly when its square is minus the identity matrix. Indeed, $B^{\mathsf{T}}B=-B^2$. Such a matrix can exist only in even dimension: taking determinants of $B^2=-I$ would otherwise give $(\det B{)}^2=-1$. In two dimensions every example isThese are rotation matrices through opposite right angles. Their orthogonal similarity transforms preserve eigenvalues, but do not in general produce a diagonal matrix.

For a real skew-symmetric matrix $M$, $M^2$ is symmetric and $x^T{M}^{2}x=-\Vert Mx{\Vert }^2$. On an eigenspace with eigenvalue $-{\mu }^2<0$, every unit vector $u$ has unit partner $v=Mu/\mu$, orthogonal to $u$, with $Mu=\mu v$ and $Mv=-\mu u$. Its plane and its orthogonal complement are invariant under $M$. Repeating gives real orthogonal two-dimensional blocks $\left(\begin{array}{cc}0& -\mu \\ \mu & 0\end{array}\right)$ and a zero block. Thus every nonzero eigenvalue of $M^2$ has even multiplicity, the zero multiplicity has the parity of $n$, and the number of distinct eigenvalues is at most $n/2$ for even $n$ and $(n+1)/2$ for odd $n$. Distinct positive block parameters attain the bound, with one zero direction added when $n$ is odd.

A complex matrix $Z=-Z^T$ can be put by unitary congruence into two-by-two skew blocks and zero blocks; the magnitudes $|z_r|$ are its singular values, each repeated twice. One proof uses the antilinear map $A(v)=Z\bar{v}$. Because $Z^{†}=-\bar{Z}$, $A^2=-Z{Z}^{†}$. On a nonzero eigenspace of $Z{Z}^{†}$ with eigenvalue $s^2$, any unit vector $v$ has orthogonal partner $A(v)/s$, and $A(A(v)/s)=-sv$. These orthonormal pairs give the blocks; the orthogonal complement is invariant, so the procedure repeats. Phases of the paired vectors set the block phases.

For $a=(a_1,a_2,a_3{)}^T$, the map $b\mapsto a\times b$ has matrixIt has trace zero and annihilates $a$, so its determinant is zero.

The cross-product matrix $A\mathbf{x}=\mathbf{n}\times \mathbf{x}$ obeys $A\mathbf{n}=0$ and $A^2=\mathbf{n}{\mathbf{n}}^T-|\mathbf{n}{|}^{2}I$, by the vector triple product. On the orthogonal complement of a nonzero $\mathbf{n}$, this is $-|\mathbf{n}{|}^{2}I$; hence the two remaining complex eigenvalues are $\pm i|\mathbf{n}|$. In an orthonormal basis adapted to $\mathbf{n}$, $A$ is a planar quarter-turn multiplied by $|\mathbf{n}|$, with zero action along its axis.