dimension of the special orthogonal group

Let V be a n-dimensional real inner product spaceMathworldPlanetmath. The group of orthogonalMathworldPlanetmathPlanetmathPlanetmath operators on V with positivePlanetmathPlanetmath determinantMathworldPlanetmath (i.e. the group of “rotationsMathworldPlanetmath” on V) is called the special orthogonal groupMathworldPlanetmath, denoted SO(n).

The theorem on decomposing orthogonal operators as rotations and reflectionsMathworldPlanetmath (http://planetmath.org/DecompositionOfOrthogonalOperatorsAsRotationsAndReflections) suggests that all elements of SO(n) are all fundamentally two-dimensional in some sense. This article is an elementary exploration of this aspect.

First observe that the set of orthogonal operators O(n) is a manifold embedded in the real vector space GL(V)n×n, defined by the condition f(X)=XX*-I=0, where XGL(V), and X* is the transposeMathworldPlanetmath of X (represented as a matrix in orthonormal coordinatesPlanetmathPlanetmath). f(X)=0 is simply shorthand for the condition that the columns of X are orthonormal — we see there are n+n(n-1)/2=n(n+1)/2 scalar equations11f(X)=0 viewed as a matrix equation of course has n2 scalar equations, but many of them are duplicated. For simplicity, we prefer to view f’s codomain to be the space of symmetric matricesMathworldPlanetmath, which is n(n+1)/2-dimensional. in n2 variables. Then if we can show that the derivativeMathworldPlanetmathPlanetmath Df(X) as full rank everywhere, then we will have established that O(n) is a manifold of dimensionPlanetmathPlanetmathPlanetmath n(n-1)/2. This is an easy computation if organized in the right way: for any yGL(V), by the product ruleMathworldPlanetmath we have


Consider first the case that X=I. Then Df(I)Y=0 implies Y*=-Y; in other words, Y is skew-symmetric. But the skew-symmetric matrices have dimension (n2-n)/2, so the kernel of Df(I) has this dimension. By the dimension theorem of linear algebra, the rank of Df(I) is n2-(n2-n)/2=n(n+1)/2, which is the full rank.

For general XO(n), let Y=X*Y. Then Df(X)Y=0 reduces to Y*=-Y, again the skew-symmetric condition. The set of all such Y have dimension (n2-n)/2; but multiplicationPlanetmathPlanetmath by X* is an automorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of GL(V), so the set of all Y also have the same dimension. Thus Df(X) has full rank, as before.

SO(n) is of course just the part of O(n) that satisfies the additional condition detX=1. It has the same dimension as a manifold.

Notice that n(n-1)/2=(n2), the number of ways to choose 2 basis elements out of n. We might naïvely think that an elementMathworldMathworld X of SO(n) can be decomposed into rotations on each of the coordinate 2-dimensional planes in n-dimensional space. This does not quite work, because rotations do not commute, even though we expect that the order of the decomposition should not matter (after all, there is no canonical order on the pairs of basis elements of V).

Nevertheless, on an “infinitesimalMathworldPlanetmathPlanetmath level”, the decomposition does work. Suppose we are given a one-parameter subgroup22Given XSO(n), it can be shown with the orthogonal operator decomposition theorem that there is a one-parameter subgroup containing X. Hint: eA=(cosθ-sinθsinθcosθ),if A=(0-θθ0). Xt of SO(n). From ODE theory, it is known that Xt is given by the matrix exponentialMathworldPlanetmath etA, where A is derivative of Xt at t=0. Also, A must be skew-symmetric, because it is a tangent vector to the curve Xt and so must lie in the tangent space to SO(n) (the kernel of Df(X)).

Since Xt=etA, we may reasonably ask: in what way does the matrix A represent the rotations Xt? The answer is suggested by the simpler case of n=3; by Rodrigues’ rotation formula, A is the operator Av=ω×v where ω is the angular velocity vector of the rotation.

The wedge productMathworldPlanetmath of the exterior algebra on V generalizes the cross productMathworldPlanetmath in higher dimensions (n>3), so we should start looking there, and one cannot but help notice that Λ2(V) also has dimension (n2). The basis elements of this space are of the form x^y^, where x and y are orthonormal basis vectors of V, and x^,y^V* are defined by x^(w)=x,w, y^(w)=y,w. By attempting to generalize the cross product representation for A, we can derive that the natural isomorphism between elements ηΛ2(V) and the skew-symmetric matrices A should be given by:


If η=ωx^y^, a simple calculation shows Xt=etA is a rotation on the plane spanned by x and y with angular velocity ω!

So one can think of the infinitesimal decomposition of Xt, as the “angular velocity” with its (n2) components. The fact that angular velocity in our physical world can be represented by one vector, and the fact that rotations in 3 have a single axis, is a consequence of the fact that (32)=3.

So although rotations do not commute in general, the correspondence between rotations and alternating 2-tensors, shows that rotations do add up and commute at the infinitesimal level33This really just boils down to the fact that a manifold looks like a vector spaceMathworldPlanetmath locally..

Title dimension of the special orthogonal group
Canonical name DimensionOfTheSpecialOrthogonalGroup
Date of creation 2013-03-22 15:24:22
Last modified on 2013-03-22 15:24:22
Owner stevecheng (10074)
Last modified by stevecheng (10074)
Numerical id 8
Author stevecheng (10074)
Entry type Result
Classification msc 15A04
Classification msc 15A75
Classification msc 20G20
Related topic ProofOfRodriguesRotationFormula
Related topic DecompositionOfOrthogonalOperatorsAsRotationsAndReflections
Related topic OrthogonalMatrices
Related topic ExteriorAlgebra
Related topic OrthogonalGroup
Defines special orthogonal group