Special Orthogonal Groups and Rotations

Christopher Triola

Submitted in partial fulfillment of the requirements for Honors inMathematics at the University of Mary Washington

Fredericksburg, Virginia

April 2009

This thesis by Christopher Triola is accepted in its present form as satisfying the thesis require-ment for Honors in Mathematics.

Date Approved

Randall D. Helmstutler, Ph.D.thesis advisor

Manning G. Collier, Ph.D.committee member

J. Larry Lehman, Ph.D.committee member

Contents

1 Introduction to Rotation Groups 11.1 Linear Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Orthogonal Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Group Properties of SO(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Group Properties of SO(n), n ≥ 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Eigenvalue Structure 62.1 First Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 SO(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 SO(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 SO(4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.5 R2, R3, and R4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Other Representations 173.1 C and SO(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 H and SO(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.3 H×H and SO(4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

References 24

Abstract

The rotational geometry of n-dimensional Euclidean space is characterized by the special or-thogonal groups SO(n). Elements of these groups are length-preserving linear transformationswhose matrix representations possess determinant +1. In this paper we investigate some of thegroup properties of SO(n). We also use linear algebra to study algebraic and geometric prop-erties of SO(2), SO(3), and SO(4). Through this work we find quantifiable differences betweenrotations in R2, R3, and R4. Our focus then shifts to alternative representations of these groups,with emphasis on the ring H of quaternions.

1 Introduction to Rotation Groups

When we think of rotations many examples may come to mind, e.g. the wheels on a car or the Earthspinning as it orbits the Sun. These are all examples of rotations in three dimensions. This setcontains all two-dimensional rotations, and as we will later show, every rotation in three dimensionscan be reduced to something like a two-dimensional rotation in the right basis. What of higherdimensions? It turns out that the set of rotations on Rn and the operation of composition on theserotations constitute a group. It is from this perspective that we will explore rotations in two, three,and four dimensions.

Notice that one of the most fundamental properties of rotations is that they do not changedistances. For instance, rotate a cube about a single axis and the distances between any two pointson or within the cube will remain the same as before the rotation. This property makes rotationson Rn a subset of the isometries on Rn. In fact, the set of rotations on Rn is a normal subgroupof the group of linear isometries on Rn. Naturally, we will start our journey with a discussion ofisometries.

1.1 Linear Isometries

Definition 1.1. An isometry of Rn is a function f : Rn → Rn such that, for any two vectorsx,y ∈ Rn, we have |f(x)− f(y)| = |x− y|. That is, f preserves distances between points in Rn.

Lemma 1.2. Suppose f : Rn → Rn is a function that fixes the origin. If f is an isometry then fpreserves lengths of all vectors in Rn. The converse holds if f is a linear transformation.

Proof. Assume f is an isometry. Then it preserves distances between vectors, that is, |f(x)−f(y)| =|x− y|. Consider the distance between a vector f(x) and the origin. Since f(0) = 0, we have

|f(x)| = |f(x)− f(0)| = |x− 0| = |x|.

Hence f preserves lengths.Suppose that f preserves lengths. Given vectors x,y ∈ Rn, if f is linear we have

|f(x)− f(y)| = |f(x− y)|= |x− y|.

Thus f is an isometry.

The origin-fixing isometries on Rn are called the linear isometries on Rn. This lemma will bevery important when we prove that linear isometries on Rn are in fact linear transformations onRn. The next lemma is important for closure purposes.

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Lemma 1.3. The composition of two isometries is an isometry. Furthermore, if they are bothlinear isometries, then so is the composite.

Proof. Let f and g be two isometries on Rn. We want to show that f ◦ g is another isometry onRn. For all x,y in Rn, we have |f(g(x)) − f(g(y))| = |g(x) − g(y)| = |x − y|. Hence f ◦ g is anisometry.

Furthermore, suppose f and g are both linear isometries. Then we know that f(g(0)) = f(0) =0. So if f and g are both linear, then so is f ◦ g.

Lemma 1.4. Suppose the function f : Rn → Rn is an isometry that moves the origin. Then thefunction g : Rn → Rn given by g(x) = f(x)− f(0) is a linear isometry.

Proof. Consider two vectors x, y in Rn. By definition of g,

|g(x)− g(y)| = |f(x)− f(0)− f(y) + f(0)| = |f(x)− f(y)| = |x− y|.

Hence g is an isometry on Rn.Now, note that g(0) = f(0)−f(0) = 0. So g fixes the origin and is hence a linear isometry.

This allows us to construct a linear isometry g : Rn → Rn given any non-linear isometryf : Rn → Rn by the formula g(x) = f(x) − f(0). Now that we only need to consider linearisometries, we will show that all of these isometries are linear transformations. Before we proceedwith the proof, we must first introduce the orthogonal groups O(n).

1.2 Orthogonal Groups

Consider the following subset of n× n matrices with real entries:

O(n) = {A ∈ GLn | A−1 = AT }.

This set is known as the orthogonal group of n× n matrices.

Theorem 1.5. The set O(n) is a group under matrix multiplication.

Proof. We know that O(n) possesses an identity element I. It is clear that since AT = A−1 everyelement of O(n) possesses an inverse. It is also clear that matrix multiplication is by its very natureassociative, hence O(n) is associative under matrix multiplication. To show that O(n) is closed,consider two arbitrary elements A,B ∈ O(n) and note the following:

(AB)(AB)T = ABBT AT

= ABBT AT

= AAT

= I.

Hence (AB)T = (AB)−1, which makes AB another element of O(n). So, O(n) is closed undermatrix multiplication. Thus we have that it is a group.

Note. Since AT A = I, if we consider the columns of A to be vectors, then they must be orthonormalvectors.

An alternative definition of O(n) is as the set of n×n matrices that preserve inner products onRn. Because we did not use this definition we must prove this using our definition of O(n).

2

Lemma 1.6. Let A be an element of O(n). The transformation associated with A preserves dotproducts.

Proof. If we consider vectors in Rn to be column matrices then we can define the dot product of uwith v in Rn as

u · v = uTv.

Consider the dot product of u,v ∈ Rn after the transformation A:

Au ·Av = (Au)T (Av)

= uT AT Av

= uTv

= u · v.

Hence, elements of O(n) preserve dot products.

Lemma 1.7. If A ∈ O(n) then A is a linear isometry.

Proof. Let A be an element of O(n). Since A preserves dot products, this means it must alsopreserve lengths in Rn, since the length of a vector v ∈ Rn may be defined as |v| =

√v · v.

Furthermore, it is clear that the origin is fixed since A0 = 0. Thus, by Lemma 1.2, A is a linearisometry.

So, we have shown that O(n) is at least a subset of the set of linear isometries. Now, we willshow containment in the other direction.

Proposition 1.8. Every linear isometry is a linear transformation whose matrix is in O(n).

Proof. If we can show that for every origin-fixing isometry f : Rn → Rn there exists an n × nmatrix A such that f(x) = Ax for all x ∈ Rn, then f must be a linear transformation. We willnow construct such a matrix.

We begin by letting the ith column of the matrix be given by the vector f(ei), where ei is theith standard basis vector for Rn. Since f preserves dot products, the columns of A are orthonormaland thus A ∈ O(n). Now we will show that f = A by showing that g = A−1 ◦ f is the identity.

First, it is clear that g is an isometry and that g(0) = 0 so that g preserves length and dotproducts. Also, we can see that g(ei) = ei for all i ∈ {1, 2, . . . , n}. Hence for a vector x ∈ Rn wehave the following (as from [4]):

g(x) =n∑

i=1

[g(x) · ei]ei =n∑

i=1

[g(x) · g(ei)]ei =n∑

i=1

[x · ei]ei = x.

Thus f = A and f is a linear transformation in O(n).

Recall that, in general, det(A) = det(AT ) and det(AB) = det(A)det(B). So, for A ∈ O(n) wefind that the square of its determinant is

det(A)2 = det(A)det(AT )

= det(AAT )= det(I)= 1.

Hence all orthogonal matrices must have a determinant of ±1.

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Note. The set of elements in O(n) with determinant +1 is the set of all proper rotations on Rn.As we will now prove, this set is a subgroup of O(n) and it is called the special orthogonal group,denoted SO(n).

Theorem 1.9. The subset SO(n) = {A ∈ O(n) | det(A) = 1} is a subgroup of O(n).

Proof. It is clear that the identity is in SO(n). Also, since det(A) = det(AT ), every element ofSO(n) inherits its inverse from O(n). We need only check closure. Consider the two elementsA,B ∈ SO(n). Since detA = detB = 1 we know that det(AB) = 1. Hence we have shown thatSO(n) is closed and thus is a subgroup of O(n).

Theorem 1.10. The group SO(n) is a normal subgroup of O(n). Furthermore, O(n)/SO(n) ∼= Z2.

Proof. Let {±1} denote the group of these elements under multiplication. Define the mappingf : O(n) → {±1} by f(A) = det(A). Clearly f is surjective. Also, it is obvious from the propertiesof determinants that this is a homomorphism. Furthermore, the kernel of this homomorphism isSO(n) since these are the only elements of O(n) with determinant +1. Therefore, by the FirstIsomorphism Theorem we have that O(n)/SO(n) ∼= Z2.

1.3 Group Properties of SO(2)

In this section we will discuss several properties of the group SO(2). We know that this is thegroup of rotations in the plane. We want a representation for general elements of this group togain a better understanding of its properties.

Let us start by choosing an orthonormal basis for R2, β = {(1, 0), (0, 1)}. Consider the rotationby an arbitrary angle θ, denoted Rθ. We can see that the vector (1, 0) will transform in the followingmanner:

Rθ(1, 0) = (cos θ, sin θ).

This is basic trigonometry. Now we will rotate the other basis vector (0, 1) to find that

Rθ(0, 1) = (− sin θ, cos θ).

Now we can construct a matrix representation of Rθ with respect to β by using the transformedvectors as columns in the matrix

A =(

cos θ − sin θsin θ cos θ

).

Note that detA = cos2 θ + sin2 θ = +1 and that

AAT =(

cos θ − sin θsin θ cos θ

) (cos θ sin θ− sin θ cos θ

)=

(cos2 θ + sin2 θ cos θ sin θ − cos θ sin θ

sin θ cos θ − cos θ sin θ sin2 θ + cos2 θ

)=

(1 00 1

).

Hence, ordinary rotations in the plane are indeed in SO(2), as we expected.Now, for completeness, we will show the converse. The columns of any arbitrary element of

SO(2) must constitute an orthonormal basis. Therefore, we can think of the columns of an element

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of SO(2) as two orthogonal vectors on the unit circle centered at the origin in the xy-plane. Itis not hard to see that the vectors u = (cos θ, sin θ) parameterize the unit circle centered at theorigin. There are only two vectors on the unit circle that are orthogonal to this vector and theyare v1 = (− sin θ, cos θ) and v2 = (sin θ,− cos θ). We construct a matrix using the vectors u andv1 as the columns: (

cos θ − sin θsin θ cos θ

).

Notice that this is the same matrix that we constructed before so we already know it lies in SO(2).Now, we will examine the matrix constructed using u and v2:(

cos θ sin θsin θ − cos θ

).

See that the determinant of this matrix is − cos2 θ − sin2 θ = −1. Hence, this does not belong toSO(2) and we have shown that all elements of SO(2) are of the form(

cos θ − sin θsin θ cos θ

).

Is SO(2) abelian? We will explore this question using the representation we have just derived.Consider two elements of SO(2):

A =(

cos θ − sin θsin θ cos θ

), B =

(cos φ − sinφsinφ cos φ

).

Now we check commutativity of A and B:

AB =(

cos θ − sin θsin θ cos θ

) (cos φ − sinφsinφ cos φ

)=

(cos θ cos φ− sin θ sinφ − cos θ sinφ− sin θ cos φsin θ cos φ + cos θ sinφ − sin θ sinφ + cos θ cos φ

)=

(cos (θ + φ) − sin (θ + φ)sin (θ + φ) cos (θ + φ)

).

If we swap these elements, we get

BA =(

cos φ − sinφsinφ cos φ

) (cos θ − sin θsin θ cos θ

)=

(cos (θ + φ) − sin (θ + φ)sin (θ + φ) cos (θ + φ)

).

We have just shown that for any A,B ∈ SO(2), AB = BA, hence SO(2) is abelian. This shouldnot be too surprising to anyone who has experience with merry-go-rounds, wheels, or arc length.Now we will consider the question of commutativity of rotations in higher dimensions.

1.4 Group Properties of SO(n), n ≥ 3

We were able to construct simple representations for elements of SO(2). We would like simplerepresentations of SO(3) and SO(4), but as we will discover, the determinants get to be rathercomplicated which makes it harder to check that a matrix is in SO(n) for larger n. However, we

5

may use our intuition about rotations and matrices to make our work easier. Since we know whatelements of SO(2) look like, let us take one and try to insert it in a 3 × 3 matrix. Consider thematrices

A =

cos θ − sin θ 0sin θ cos θ 0

0 0 1

, B =

1 0 00 cos θ − sin θ0 sin θ cos θ

.

Notice that AT A = BT B = I and that detA = detB = 1, hence these matrices exist in SO(3).Furthermore, notice that they possess block elements of SO(2). This is our trick for insertingmatrices from SO(2) into higher dimensions and this will be used frequently when we must constructelements of SO(n) just to give examples of properties.

With these two elements of SO(3) let us consider the question: is SO(3) abelian? Consider theproduct AB:

AB =

cos θ − sin θ cos θ sin2 θsin θ cos2 θ − sin θ cos θ

0 sin θ cos θ

.

Now consider the product BA:

BA =

cos θ − sin θ 0sin θ cos θ cos2 θ − sin θ

sin2 θ sin θ cos θ cos θ

.

Note that in general AB 6= BA and hence SO(3) is not abelian. Furthermore, since elements ofSO(3) may be embedded in SO(n) for n > 3, we know that SO(2) is the only special orthogonalgroup that is abelian. However, it is worth noting that there do exist commutative subgroups ofSO(n) for all n. An example of a commutative subgroup in SO(3) is the set of matrices of the form cos θ − sin θ 0

sin θ cos θ 00 0 1

.

Matrices of this form are commutative since these are rotations in the xy-plane and we alreadyshowed that SO(2) is commutative.

We have established some basic properties of the special orthogonal groups and can use aparameterization of SO(2) to manufacture some elements of SO(n). We will now proceed to delveinto the properties of SO(n) as operations on Rn by investigating their eigenvalue structures.

2 Eigenvalue Structure

2.1 First Results

The objective of this section is to understand the action of elements of SO(n) on Rn. This will giveus a more mathematical intuition for what a rotation actually is. But before we can investigate theeigenvalues of elements of SO(n) we must first understand the tools we will use in our investigation.In this section, we will show that the following properties hold for each A ∈ SO(n) and the rootsof its characteristic polynomial:

• detA = 1 = λ1λ2λ3 · · ·λn, where each λi is a root, counted with multiplicity

• if λ is a root, so is λ

6

• if λi is real then |λi| = 1.

Recall that we define eigenvalues for an n×n matrix A to be all λ ∈ R for which there exists anon-zero vector r such that Ar = λr. The vectors r for which this is true are called the eigenvectorsof A. We call the set of all such eigenvectors for a given eigenvalue λ, along with the zero vector, aneigenspace of A. Before we can begin our investigation of how eigenvalues relate to our discussionof rotations we must establish the properties listed above.

Theorem 2.1. If A is an n×n matrix, the determinant of A is equal to the product of the n roots(counted with multiplicity) of the characteristic polynomial of A.

Proof. Let A be an n× n matrix with characteristic polynomial

p(λ) = det[λI −A] = λn + an−1λn−1 + · · ·+ a1λ + a0.

By the fundamental theorem of algebra, we have

p(λ) = (λ− λ1)(λ− λ2) · · · (λ− λn)

where λi is the ith root of the polynomial. Since p(λ) = det[λI−A], p(0) = det[−A] = (−1)ndetA =a0. Hence a0 = detA for even n and a0 = −detA for odd n. Since p(0) = a0 we have that

a0 = (0− λ1)(0− λ2) · · · (0− λn)= (−λ1)(−λ2) · · · (−λn)= (−1)nλ1λ2 · · ·λn.

Hence, regardless of the value of n, detA = λ1λ2 · · · λn.

We now turn our attention to the second property, that if λ is a root of the characteristicpolynomial of A, so is its conjugate. Note that the characteristic polynomial for a given elementA ∈ SO(n) is a polynomial with real coefficients and hence complex roots of the characteristicpolynomial of A come in conjugate pairs. Thus, if λ is a root of the characteristic polynomial of Athen λ is also a root, though not distinct in the case of real roots.

We now move on to the third property. Recall that any transformation A ∈ SO(n) is a length-preserving isometry, so for all r ∈ Rn we have |Ar| = |r|. It follows that if λ is a real eigenvaluewith eigenvector r then

|Ar| = |λr| = |r|

which implies that |λ| = 1. Since all real roots of the characteristic polynomial must have absolutevalue of 1 and all roots must multiply to 1, the complex roots must all multiply to 1.

2.2 SO(2)

In this section we will find and discuss all the possible roots of the characteristic polynomials ofelements of SO(2). Because we are in SO(2) the characteristic polynomial of each element willhave two roots.

Consider an arbitrary element A ∈ SO(2). We know that detA = 1 = λ1λ2. From the previoussection we know that |λi| = 1 when λi is real. It should be clear that the only possible roots arethose appearing in the following table.

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Roots Rotation Angle1, 1 0

−1,−1 πω, ω θ 6= 0, π

We are fortunate enough to have a simple representation for this group so we will utilize it here.Let us find the roots of the characteristic polynomial of

A =(

cos θ − sin θsin θ cos θ

)∈ SO(2).

Setting the characteristic polynomial equal to 0 gives λ2 − 2λ cos θ + 1 = 0. By the quadraticformula, we have

λ =2 cos θ ±

√4 cos2 θ − 42

=2 cos θ ± 2

√cos2 θ − 1

2= cos θ ±

√− sin2 θ.

Hence the roots of the characteristic polynomial for an element of SO(2) all come in the form

λ = cos θ ± i sin θ.

We can see that the only real value for λ occurs when sin θ = 0, which will happen only whenθ = 0, π. Thus, we have shown that all of the aforementioned cases do in fact arise from thefollowing scenarios:

• if θ = 0 then the roots are 1, 1

• if θ = π then the roots are −1,−1

• if θ 6= 0, π then the roots are complex.

Note that the formula cos θ±i sin θ gives the roots for the characteristic polynomial associated witha rotation by an angle θ. We could just as easily use this process in reverse and calculate the angleof rotation of a transformation based on the roots of its characteristic polynomial.

Now we will calculate the eigenspaces for the real eigenvalues that we found and explain themgeometrically. We will accomplish this by noting that when θ = 0 the rotation matrix is

I =(

1 00 1

).

Hence the eigenvalue is 1 with multiplicity 2. Since all vectors r ∈ R2 will be solutions to Ir = r,all of R2 comprises the eigenspace of this matrix. The dimension of this eigenspace is thus 2.

For θ = π, we note that the matrix representation would be

−I =(−1 00 −1

).

Note that the eigenvalue is −1 with multiplicity 2, and that this is a rotation by π. Therefore, onceagain, all vectors r ∈ R2 are solutions to −Ir = −r. Hence the eigenspace for this case is again allof R2.

For complex roots we do not consider the eigenspaces since this would take us outside of thereal numbers. Hence, we have now described the eigenvalue structure for all elements of SO(2). Itwas what our intuition would tell us: only the identity preserves the whole plane, the negative ofthe identity swaps all vectors about the origin, and except for these two cases, no rotation of theplane maps a vector to a scalar multiple of itself. But we noticed the interesting fact that the rootsof the characteristic polynomial come in the form cos θ ± i sin θ where θ is the angle of rotation,giving a geometric link even to complex roots. Now let us explore higher dimensions.

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2.3 SO(3)

Consider a matrix A ∈ SO(3). By our previously established facts and theorems, the followingtable displays all possible roots for the characteristic polynomial of A.

Roots Rotation Angle1, 1, 1 0

1,−1,−1 π1, ω, ω θ 6= 0, π

Note. When discussing rotations in R2 an angle was enough to describe a rotation. Now that weare dealing with R3, we must specify an angle and a plane. Because we are restricted to R3, thisplane may be specified by a vector orthogonal to the plane. This vector may be thought of as whatwe traditionally call the axis of rotation.

Case 1. The only example of the first case is the identity. This example clearly preserves all of3-space and hence its eigenspace has dimension 3.

Case 2. For this case we shall consider the matrix

A =

1 0 00 −1 00 0 −1

.

We can see that once again its transpose is itself and that AAT = I. We can also see that thedeterminant is +1, hence A ∈ SO(3). Because this is diagonal, we can see that the eigenvalues are1,−1,−1. Now we will compute the eigenspaces for A using the following augmented matrix forλ = 1 1− 1 0 0 0

0 1 + 1 0 00 0 1 + 1 0

=

0 0 0 00 2 0 00 0 2 0

and the following for λ = −1: −1− 1 0 0 0

0 −1 + 1 0 00 0 −1 + 1 0

=

−2 0 0 00 0 0 00 0 0 0

.

From this we can tell that for λ = 1 the eigenspace is generated by 100

and for λ = −1 the generators are 0

10

,

001

.

Hence for λ = 1 we have a one-dimensional eigenspace. That is, all eigenvectors lie along an axis.For λ = −1 we see that the eigenvectors make up a plane. Notice however that for λ = 1 thevectors are invariant while for λ = −1 the vectors are flipped about the origin. This tells us thatfor this rotation, an axis stays the same and the yz-plane is rotated by π.

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Case 3. We will now deal with the most general case of λ = 1, ω, ω. Consider the matrix

A =

0 −1 01 0 00 0 1

.

Note that AAT = I so AT = A−1 and that det(A) = 1, so A ∈ SO(3). Now we will solve for itseigenvalues. Setting det(λI −A) = 0 gives λ3 − λ2 + λ− 1 = 0. Since we know from before that 1must be an eigenvalue, we divide this polynomial by λ− 1 to obtain

λ2 + 1 = 0

which has solutionsλ = ±i.

Hence we have a rotation whose characteristic polynomial has roots λ = 1, i,−i.Now we will find the eigenspace for λ = 1. We start with the following augmented matrix 1 1 0 0

−1 1 0 00 0 1− 1 0

=

1 1 0 0−1 1 0 00 0 0 0

which row reduces to 1 0 0 0

0 1 0 00 0 0 0

.

Thus, the eigenspace is generated by 001

.

So the entire z-axis is preserved but the rest of 3-space is not. If we inspect what this does toa vector in the xy-plane, e.g. (1, 1, 0), we find the transformed vector is (−1, 1, 0). Using the dotproduct we find that the cosine of the angle θ between this vector and its transformed counterpartis equal to 0. This tells us that this is a rotation by ±π

2 . This corresponds to sin θ = ±1.Recall that in SO(2) we could find all roots using the formula cos θ± i sin θ, where θ is the angle

of rotation. Recall also that the complex roots for this rotation were ±i. These complex roots areexactly what we would expect if the same formula applied to roots in SO(3). We will need moreevidence to support this claim before we attempt a proof.

Though we have given examples of each possible combination of roots for SO(3), we will nowdemonstrate the eigenvalue properties of SO(3) using a more complex example.

Example. Consider the matrix

A =

−58 −3

√3

8

√3

43√

38 −1

834

−√

34

34

12

.

10

Note that AAT = I and that detA = +1 so that A is in SO(3). Now let us compute its eigenvalues.We will solve the usual equation for determining eigenvalues:

det

λ + 58

3√

38 −

√3

4

−3√

38 λ + 1

8 −34√

34 −3

4 λ− 12

= λ3 +14λ2 − 1

4λ− 1 = 0.

As we have shown, +1 must be an eigenvalue so we will divide this polynomial by λ− 1 to obtain

λ2 +54λ + 1 = 0.

We will now use the quadratic formula to solve for the remaining roots. We find that

λ =−5±

√−39

8=−5± i

√39

8.

Hence the characteristic polynomial of A has two complex roots that are conjugates of each other.Now we will compute the eigenspace for λ = 1 using the augmented matrix 1 + 5

83√

38 −

√3

4 0−3

√3

8 1 + 18 −3

4 0√3

4 −34 1− 1

2 0

=

138

3√

38 −

√3

4 0−3

√3

898 −3

4 0√3

4 −34

12 0

which row reduces to 1 0 0 0

0 0 0 00 −3

2 1 0

.

Hence, the eigenspace is generated by 0132

.

Note that this transformation still leaves a line invariant.

Leonhard Euler also noticed this fact and proved his famous theorem about three-dimensionalrotations that, put in our language, is:

Euler’s Rotation Theorem. If A is an element of SO(3) where A 6= I then A has a one-dimensional eigenspace.

This is not to say that A does not also have a two-dimensional eigenspace but that A musthave a one-dimensional eigenspace. It is this eigenspace that is known as the axis of rotation. Thework we have done in this section should be enough to convince us that this is true. However, aswe will see in SO(4), this way of viewing rotations does not generalize at all. In fact, having anaxis of rotation is a special case of how we may define planes in 3-space that only applies to R3. Itis for this reason that we have used the term “axis” sparingly.

Before we proceed with SO(4), we noted earlier that there might be some geometric significanceto the complex roots but needed another example to motivate a proof. We would like to see whatangle of rotation corresponds to the matrix example above. The generator of the eigenspace,(0, 1, 3

2), can be seen as the axis of rotation, so we will see what this matrix does to an element ofthe orthogonal complement of this eigenspace. We can see from the dot product that v = (0,−3

2 , 1)

11

is orthogonal to the axis of rotation. Also note that after being transformed this vector becomesv′ = (13

√3

16 , 1516 ,−5

8). Now we will determine the cosine of the angle between these two vectors:

v · v′ = 134

cos θ = −6532

.

Solving for the cosine term we see that cos θ = −58 , which is the real part of the complex roots. This

tells us that even a rotation about a more general axis still follows this tendency. In a later sectionwe will prove that this same geometric significance exists for all roots of characteristic polynomialsof matrices in SO(3).

2.4 SO(4)

To begin our discussion of SO(4) we will mimic the procedure we used to discuss SO(3). We beginby noting the possible eigenvalues and other complex roots of the characteristic polynomials. Weproceed by giving examples of each case and discussing their geometric significance.

Note. In R2 there is only one plane, so we only needed to specify an angle. In R3 we needed tospecify a plane and an angle. In R4 we must specify two planes and an angle for each plane. Thereasoning behind this will become apparent in this section.

Roots Rotation Angles1, 1; 1, 1 0; 0

−1,−1;−1,−1 π;π1, 1;−1,−1 0;π1, 1;ω, ω 0; θ

−1,−1;ω, ω π; θω, ω; z, z θ;φ

Notice that there are twice the number of cases for SO(4) roots than for SO(3) or SO(2). Thiscomes from the fact that we may rotate in, say, the x1x2-plane and then rotate in the x3x4-planewithout disturbing our previous rotation. As we will see, R4 is an interesting place.

Case 1. As usual, the identity handles the first case and it is clear that the eigenspace for I ∈ SO(4)is all of R4.

Case 2. It is simple to construct an example by considering −I. Since −Ix = −x for all x ∈ R4,we know that the eigenspace is once again all of R4. In this case, each vector in R4 is swappedabout the origin.

Does this rotation take place about an axis? In three dimensions we had a similar case in whicha plane of vectors was rotated by π. However that had an axis because only a plane was involvedand there was an extra degree of freedom which we called the axis. In this case, however, all vectorsare included in the eigenspace so there is no favored axis of rotation here: the entire space is flipped.In three-space we think of rotations as occurring about one-dimensional axes, but what is reallygoing on is that a plane is being rotated and any vector with a non-trivial projection in that planewill also be rotated. However, only the component in the direction of the projection will be rotated:the component orthogonal to the plane of rotation will remain unchanged. If we remember, thisis what happened in two-space as well. In fact, it is by embedding the two-dimensional rotationsinto the higher dimensions that we have obtained most of the examples in this section. So, lookingat it from this perspective, the first case would be rotation by 0 radians. The second would be a

12

rotation by π of, say, the x1x2-plane followed by a rotation of the x3x4-plane by π. We can seethat this acts exactly as we claim it does:

cos π − sinπ 0 0sinπ cos π 0 0

0 0 1 00 0 0 1

1 0 0 00 1 0 00 0 cos π − sinπ0 0 sin π cos π

x1

x2

x3

x4

=

−1 0 0 00 −1 0 00 0 1 00 0 0 1

1 0 0 00 1 0 00 0 −1 00 0 0 −1

x1

x2

x3

x4

=

−1 0 0 00 −1 0 00 0 1 00 0 0 1

x1

x2

−x3

−x4

=

−x1

−x2

−x3

−x4

.

Thinking about rotations in this way may help to alleviate the confusion that is common whentrying to visualize higher dimensions.

Case 3. For this case, we can use the matrix

A =

1 0 0 00 1 0 00 0 −1 00 0 0 −1

.

It should be clear that the eigenvalues for this matrix are 1, 1,−1,−1. Solving for the eigenspaces,we will need to use two augmented matrices: (I − A|0) and (−I − A|0). In the first case we findan eigenspace generated by

1000

,

0100

.

This eigenspace is simply a copy of the x1x2-plane. For the second augmented matrix we obtainan eigenspace generated by

0010

,

0001

.

This eigenspace is a copy of the x3x4-plane. However, if we once again use this matrix to see wherean arbitrary vector is sent we find

Ax =

1 0 0 00 1 0 00 0 −1 00 0 0 −1

x1

x2

x3

x4

=

x1

x2

−x3

−x4

.

Hence we have a rotation by π of the x3x4-plane and a rotation by 0 of the x1x2-plane.

13

Case 4. We will show that the following matrix satisfies this case and then examine the implicationsof such a rotation:

B =

1 0 0 00 1 0 00 0 0 −10 0 1 0

.

It is not hard to see that BBT = I and det(B) = 1 and hence B ∈ SO(4). Now we can move onto its eigenvalues which we will find in the usual way:

det(λI −B) = det

λ− 1 0 0 0

0 λ− 1 0 00 0 λ 10 0 −1 λ

= (λ− 1)(λ− 1)(λ2 + 1) = 0.

We see that λ = 1 appears as a solution with multiplicity 2 and we have λ = ±i as the other twosolutions. We have thus shown that this is an example of Case 4. Now we must find the eigenspacefor λ = 1 using the augmented matrix

1− 1 0 0 0 00 1− 1 0 0 00 0 1 1 00 0 −1 1 0

=

0 0 0 0 00 0 0 0 00 0 1 1 00 0 −1 1 0

which row reduces to

0 0 0 0 00 0 0 0 00 0 1 0 00 0 0 1 0

.

Hence, the eigenspace is generated by 1000

,

0100

which generates a copy of the x1x2-plane. To avoid confusion and to see what happens to the othertwo parameters, let us see where B maps an arbitrary vector:

Bx =

1 0 0 00 1 0 00 0 0 −10 0 1 0

x1

x2

x3

x4

=

x1

x2

−x4

x3

.

This rotation leaves the x1x2-plane invariant (as we said before) and it rotates the x3x4-plane by π2 .

Case 5. For the case where λ = −1,−1, ω, ω, we will use our previous example as inspiration toconstruct the following matrix:

B′ =

−1 0 0 00 −1 0 00 0 0 −10 0 1 0

.

14

Once again, it is easy to show that this is an element of SO(4). When we compute its eigenvalueswe find that they are λ = −1,−1, i,−i, so this is in fact an example of Case 5. Watching what thismatrix does to an arbitrary vector, however, we see that:

B′x =

−1 0 0 00 −1 0 00 0 0 −10 0 1 0

x1

x2

x3

x4

=

−x1

−x2

−x4

x3

.

This still rotates the x3x4-plane by π2 but it also rotates the x1x2-plane by π.

Case 6. This is the case where there are no real eigenvalues. We might think that such an examplewould have to be a very exotic matrix. However, using what we’ve learned about rotations, all weneed is a matrix that doesn’t map any vector in R4 to a scalar multiple of itself. So, if we rotatethe x1x2-plane by π

2 and do the same to the x3x4-plane, that action should satisfy this case. Let’sdo that using the matrix

C =

0 −1 0 01 0 0 00 0 0 −10 0 1 0

.

We now compute the eigenvalues of this matrix:

det(λI − C) = det

λ 1 0 0−1 λ 0 00 0 λ 10 0 −1 λ

= λ4 + 2λ2 + 1 = 0.

This factors into(λ2 + 1)2 = 0.

Hence, the eigenvalues are λ = ±i, each with multiplicity two, which gives us an example of Case 6.For completeness, we will now verify that it transforms vectors the way we constructed it to:

Cx =

0 −1 0 01 0 0 00 0 0 −10 0 1 0

x1

x2

x3

x4

=

−x2

x1

−x4

x3

.

As we can see this is a rotation of the x1x2-plane by π2 and then a rotation of the x3x4-plane by π

2 ,just as we claimed during its construction.

2.5 R2, R3, and R4

The following are important properties to note about the relationships between rotations in twodimensions and those in three and four dimensions. With these properties we will be able to simplifyour work in the final section while remaining mathematically rigorous in the theorems to come.

Theorem 2.2. Any arbitrary element of SO(3) may be written as the composition of rotations inthe planes generated by the three standard orthogonal basis vectors of R3.

15

Proof. It is easy to show that the following matrices represent rotations by θ1, θ2, and θ3 in thexy-, yz- and xz-planes respectively:

Rz =

cos θ1 − sin θ1 0sin θ1 cos θ1 0

0 0 1

, Rx =

1 0 00 cos θ2 − sin θ2

0 sin θ2 cos θ2

, Ry =

cos θ3 0 − sin θ3

0 1 0sin θ3 0 cos θ3

.

As we know, all rotations in R3—aside from the identity—have a one-dimensional eigenspace aboutwhich we rotate by an angle θ. A unit vector n in this eigenspace may be parameterized as

n = (cos α sin β, sinα sin β, cos β).

This is because the one-dimensional eigenspace, or axis of rotation, must intersect the unit spherecentered at the origin and the above can be shown to parameterize the unit sphere.

We may align the axis of rotation with the z-axis as follows:

Ry(β)Rz(α)Tn =

cos β 0 − sinβ0 1 0

sinβ 0 cos β

cos α sinα 0− sinα cos α 0

0 0 1

cos α sinβsin α sinβ

cos β

=

001

.

We can thus perform the rotation by θ in the xy-plane and then transform back. All we need toknow is the angle of rotation θ and the angles α, β which specify the rotation’s one-dimensionaleigenspace to write any element of SO(3) as

Rz(α)Ry(β)T Rz(θ)Ry(β)Rz(α)T .

This is not unique since we could just as easily transform the eigenspace to align with either ofthe other two axes. Hence, as we claimed, we may write any rotation in SO(3) as the product ofrotations in the orthogonal planes of R3.

Recall earlier we noticed a pattern indicative of a possible geometric interpretation of complexeigenvalues. Now we will state and prove this fact in the following theorem.

Theorem 2.3. For any element A ∈ SO(3) with rotation angle θ, the roots of the characteristicpolynomial of A are 1 and cos θ ± i sin θ.

Proof. Let A be an element of SO(3) with axis generated by n = (cos α sinβ, sinα sinβ, cos β). ByTheorem 2.2, we may represent A as the product

A = Rz(α)Ry(β)T Rz(θ)Ry(β)Rz(α)T

= (Rz(α)Ry(β)T )Rz(θ)(Rz(α)Ry(β)T )T

= PRz(θ)P T .

This means that the matrix A is similar to the matrix

Rz(θ) =

cos θ − sin θ 0sin θ cos θ 0

0 0 1

.

16

The characteristic polynomial of this matrix is

(λ− 1)(λ2 − 2λ cos θ + 1).

Dividing by λ − 1 we find that the roots that are not equal to 1 are cos θ ± i sin θ. Since A andRz(θ) are similar they share the same characteristic polynomial. Hence for any rotation by θ in R3

the roots of the characteristic polynomial are 1, cos θ + i sin θ, and cos θ − i sin θ.

Proposition 2.4. Any arbitrary element of SO(4) may be written as the composition of rotationsin the planes generated by the four orthogonal basis vectors of R4.

The proof involves changing the basis of an arbitrary element of SO(4) such that the two planesof rotation are made to coincide with the x1x2-plane and the x3x4-plane. We would change thebasis using products of the matrices

cos θ1 − sin θ1 0 0sin θ1 cos θ1 0 0

0 0 1 00 0 0 1

,

cos θ2 0 − sin θ2 0

0 1 0 0sin θ2 0 cos θ2 0

0 0 0 1

,

cos θ3 0 0 − sin θ3

0 1 0 00 0 1 0

sin θ3 0 0 cos θ3

1 0 0 00 cos θ4 − sin θ4 00 sin θ4 cos θ4 00 0 0 1

,

1 0 0 00 cos θ5 0 − sin θ5

0 0 1 00 sin θ5 0 cos θ5

,

1 0 0 00 1 0 00 0 cos θ6 − sin θ6

0 0 sin θ6 cos θ6

.

We would then perform our rotation using the matricescos φ1 − sinφ1 0 0sinφ1 cos φ1 0 0

0 0 1 00 0 0 1

,

1 0 0 00 1 0 00 0 cos φ2 − sinφ2

0 0 sin φ2 cos φ2

and then change back using the appropriate transposes of the matrices above. Without an axis itis more difficult to see how to change the basis appropriately, so an actual proof will be omitted.However, this is a known fact that will be assumed later.

3 Other Representations

Recall that the reason we have been using matrices was that rotations turned out to be lineartransformations and matrices are natural representations of linear transformations. However, wemust not believe that these are the only possible ways of expressing rotations. After all, we caneasily represent a rotation in R3 by using just an axis and an angle. In this section we will turnour attention to alternative representations of the special orthogonal groups. Specifically, we willconsider those representations given by complex numbers and quaternions.

3.1 C and SO(2)

A complex number is a number with a real and an imaginary component, written z = a + bi. Forall elements z1 = a + bi, z2 = c + di ∈ C we define addition as z1 + z2 = (a + c) + (b + d)i. Itis simple to check that with this addition operation C is a two-dimensional vector space over thereal numbers which makes it isomorphic to R2 as a vector space. If z = x + yi then we define

17

the conjugate to be z = x− yi. Also, we have a multiplication operation defined for every pair ofelements as above by z1z2 = (ac− bd) + (ad + bc)i. With this knowledge we may define the normof z ∈ C as

|z| =√

zz.

Recall that the norm is the length of the vector. Another useful property to note is that the normis multiplicative, that is,

|z1z2| = |z1||z2|.With this knowledge we are able to express the unit circle, or one-sphere S1, as

S1 = {z ∈ C : |z| = 1}.

Lemma 3.1. The one-sphere is an abelian group under complex multiplication.

Proof. Recall that the set of non-zero complex numbers is a group under multiplication. Thismeans we only need to check that S1 is a subgroup of C− {0}. We start by checking closure. Letz1, z2 be elements of S1. Then we have that

|z1z2| = |z1||z2| = 1 · 1 = 1.

Hence S1 is closed under the multiplication operation of C. It is clear that |1| = 1 and it should alsobe clear that this is the identity under multiplication. Hence S1 possesses an identity. Also, everyelement z ∈ S1 has an inverse in C such that zz−1 = 1. Then |zz−1| = |1| = 1 and |z| = |z−1| = 1,hence if z is in S1 so is its inverse. Associativity is inherited from the complex numbers and we haveshown that S1 is a group under complex multiplication. Complex multiplication is commutative,so S1 is abelian.

Recall that the set of unit complex numbers may be written in the form eiθ = cos θ + i sin θ.This new form will aid us in the proof of the main theorem for this section.

Theorem 3.2. The group SO(2) is isomorphic to S1.

Proof. Recall that any element in SO(2) may be represented as a matrix

A(θ) =(

cos θ − sin θsin θ cos θ

)where θ is the angle of rotation, chosen to be in [0, 2π). Consider the mapping f : S1 → SO(2),where f(eiθ) = A(θ). See that

f(eiθ)f(eiφ) =(

cos θ − sin θsin θ cos θ

) (cos φ − sinφsin φ cos φ

)=

(cos θ cos φ− sin θ sinφ −(cos θ sinφ + sin θ cos φ)sin θ cos φ + cos θ sinφ cos θ cos φ− sin θ sinφ

)=

(cos (θ + φ) − sin (θ + φ)sin (θ + φ) cos (θ + φ)

).

Now see that

f(eiθeiφ) = f(ei(θ+φ))

=(

cos (θ + φ) − sin (θ + φ)sin (θ + φ) cos (θ + φ)

)= f(eiθ)f(eiφ).

18

Thus f is a homomorphism. By our representation, every rotation in SO(2) is of the form A(θ)for some angle θ, and in this case f(eiθ) = A(θ). Hence f is surjective. Since θ is chosen to be in[0, 2π) it is uniquely determined, and hence f is injective. Therefore S1 is isomorphic to SO(2).

3.2 H and SO(3)

We have shown that S1 is a group under complex multiplication and that it is isomorphic to SO(2).We would now like to find representations for higher-dimensional rotations. Since the elements ofSO(3) can be seen as invariant points on the unit sphere S2 in R3, we turn our attention to S2. Notethat an element of SO(3) is not just a point on a sphere: it contains another piece of information,the angle of rotation. Therefore, we look to the next dimension up, the three-sphere S3. Butbefore we can hope to find a representation using elements of the three-sphere, we must establishits multiplication rules. To begin we would like to find a set that is a four-dimensional vectorspace and whose non-zero elements form a group under multiplication. Thus we must introduceHamilton’s quaternions.

The set of quaternions H is the set of generalized complex numbers q = a0 + a1i + a2j + a3k,where i, j, k are imaginary numbers satisfying the properties:

• i2 = j2 = k2 = ijk = −1

• ij = k

• jk = i

• ki = j

• ji = −k

• kj = −i

• ik = −j.

We define the addition of quaternions similarly to the way we defined complex addition. Explicitly,for every pair of elements q1 = a0 + a1i + a2j + a3k, q2 = b0 + b1i + b2j + b3k ∈ H we define theiraddition component-wise:

q1 + q2 = (a0 + b0) + (a1 + b1)i + (a2 + b2)j + (a3 + b3)k

where all coefficients are real numbers. Thus we can see that addition is commutative in H. Infact, it is not hard to see that H makes up a four-dimensional vector space and hence is isomorphicto R4 as a vector space.

Just like in the complex numbers we have a conjugate and it is defined in a similar fashion. Ifq = a0 + a1i + a2j + a3k is a quaternion we say that the conjugate of q, denoted q, is given byq = a0 − a1i− a2j − a3k. Another similarity between the quaternions and the complex numbers isthat we may define a multiplication operation on H. This quaternionic multiplication works usingthe regular distribution laws combined with the identities above, so that for every pair of elementsq1 = a0 + a1i + a2j + a3k, q2 = b0 + b1i + b2j + b3k ∈ H we define q1q2 by

q1q2 = (a0b0 − a1b1 − a2b2 − a3b3) + (a0b1 + a1b0 + a2b3 − a3b2)i + (a0b2 + a2b0 + a3b1 − a1b3)j+ (a0b3 + a3b0 + a1b2 − a2b1)k.

19

Notice that this does not always commute. Now let us see what happens when q2 = q1:

q1q1 = (a20 + a2

1 + a22 + a2

3) + (0)i + (0)j + (0)k = a20 + a2

1 + a22 + a2

3.

This expression happens to be the square of the norm for a vector in R4, hence we define the normof a quaternion q to be

|q| =√

qq.

It is not hard to show that the following property holds for all quaternions q1, q2:

|q1q2| = |q1||q2|.

Now we are prepared to define the three-sphere using quaternions. Since the three-sphere S3 isthe set of all unit length vectors in R4 (or H) we may define S3 as follows:

S3 = {q ∈ H : |q| = 1}.

We can now prove the following result.

Lemma 3.3. The three-sphere is a non-abelian group under quaternionic multiplication.

Proof. Consider the norm of the multiplication of two elements q1, q2 ∈ S3:

|q1q2| = |q1||q2| = 1.

Hence S3 is closed under quaternionic multiplication.Next, we can see that the set contains an identity, namely the number 1. We know this acts

as the identity because of how real numbers distribute over the quaternions. Also note that for allq ∈ S3 we have |q|2 = qq = 1, hence q is the inverse of q. It is clear that q−1 ∈ S3.

Finally, S3 inherits associativity from H. Thus S3 is a group under quaternionic multiplication.Furthermore, this group is not abelian, since quaternions rarely commute.

Now that we know that this is a group we are in a position to prove the following theorem.

Theorem 3.4. There exists a two-to-one homomorphism from S3 onto SO(3).

Proof. To show this we must demonstrate that every quaternion of unit length may be mapped toa rotation in R3 and that every rotation in R3 has two quaternion representations.

Let n = a1i + a2j + a3k be such that |n| = 1. Consider a unit quaternion written in the form

q = cos φ + n sinφ.

It is not hard to show that by varying both the direction of n and the angle φ we may representany element of S3 in this form.

We will show that we can use quaternionic multiplication to construct rotations in R3. Letr = (x, y, z) be a vector in R3. We could just as well write this vector as r = (0, x, y, z) in R4.Since R4 is isomorphic to H as a vector space we could also write this vector as r = xi + yj + zk.If we multiply by a unit quaternion q then we see that |qr| = |r|, so that multiplication by q islength-preserving. However, this multiplication will in general yield a non-zero real component inqr, which will prevent us from mapping the result back to R3. So we need to find a multiplicationwhose result does not contain a real part. We will try the following:

qrq = (cos φ + n sinφ)r(cos φ− n sinφ).

20

When we expand this we see that it does not contain a real part:

qrq = [cos2 φx + 2 cos φ sinφ(za2 − ya3)− sin2 φ(a1(−xa1 − ya2 − za3) + a2(xa2 − ya1)− a3(za1 − xa3))]i

+ [cos2 φy + 2 cos φ sinφ(xa3 − za1)− sin2 φ(a2(−xa1 − ya2 − za3) + a3(ya3 − za2)− a1(xa2 − ya1))]j

+ [cos2 φz + 2 cos φ sinφ(ya1 − xa2)− sin2 φ(a3(−xa1 − ya2 − za3) + a1(za1 − xa3)− a2(ya3 − za2))]k.

This is a step in the right direction because this vector has only three components.We will now construct the matrix representation of this linear transformation on R3. By evalu-

ating qrq as r ranges over the standard basis vectors, we see that the matrix representation of thisoperation is

A =

cos2 φ + sin2 φ(a21 − a2

2 − a23) −2 cos φ sinφa3 + 2 sin2 φa1a2 2 cos φ sinφa2 + 2 sin2 φa1a3

2 cos φ sinφa3 + 2 sin2 φa1a2 cos2 φ + sin2 φ(a22 − a2

1 − a23) −2 cos φ sinφa1 + 2 sin2 φa2a3

−2 cos φ sinφa2 + 2 sin2 φa1a3 2 cos φ sinφa1 + 2 sin2 φa2a3 cos2 φ + sin2 φ(a23 − a2

2 − a21)

.

Furthermore, it can be shown that detA = 1 and that AT A = I. Given a unit length quaternion q,we define a transformation fq on R3 by fq(r) = qrq. The argument above proves that fq ∈ SO(3).

Define a map f : S3 → SO(3) by f(q) = fq. We claim that f is a two-to-one surjectivehomomorphism. Consider two elements q1, q2 ∈ S3. Note the following:

f(q1q2)(r) = q1q2rq1q2

= q1q2r(q1q2)−1

= q1q2rq−12 q−1

1

= (f(q1) ◦ f(q2))(r).

Thus, f is a homomorphism. Now we will show that this map is onto.Consider the matrix associated with the quaternion q = cos φ + n sinφ when n = (1, 0, 0). We

find this by setting a1 = 1 and a2 = a3 = 0 in the matrix A above. We find that this matrix is 1 0 00 cos2 φ− sin2 φ −2 cos φ sinφ0 2 cos φ sinφ cos2 φ− sin2 φ

=

1 0 00 cos 2φ − sin 2φ0 sin 2φ cos 2φ

.

Note that it is not a rotation by φ but instead a rotation by 2φ in the yz-plane. Notice that if n hadcoincided with the y- or z-axis we would have obtained a similar matrix, except it would represent arotation of 2φ about the chosen axis. Since we can generate rotations about the standard orthogonalbasis vectors using quaternions, we may invoke Theorem 2.2 to conclude that this homomorphismis onto.

To see that f is two-to-one, we will show that the kernel of f contains only two elements. Itis clear that f1 and f−1 are both the identity rotation. We claim that ±1 are the only elementsin the kernel. Suppose that q ∈ ker(f), so that fq is the identity rotation. Then qrq = r for allr = xi+yj +zk. This would imply that qr = rq for all such r. The only quaternions that commutewith all pure quaternions are the reals, so q must be real. Since q must be unit length, we concludethat q = ±1. Hence f is two-to-one.

While this is not an isomorphism because it is two-to-one, by the First Isomorphism Theoremwe see that S3/Z2

∼= SO(3). This result is important because it allows us to prove a famousresult from topology using only linear and abstract algebra. First, note that in S3 every point(x1, x2, x3, x4) has an antipode (−x1,−x2,−x3,−x4). This relation leads to a normal subgroup oforder 2 which is isomorphic to Z2. If we take S3 modulo this subgroup we obtain real projectivespace RP 3. There is a famous result from topology that states RP 3 ∼= SO(3). However, we havearrived at this result using mostly linear algebra.

21

3.3 H×H and SO(4)

We showed that S1 is isomorphic to SO(2). We then showed that there is a two-to-one homo-morphism from S3 onto SO(3). We will now attempt to find a similar representation of SO(4).Intuition might suggest we attempt to find another sphere to represent SO(4). However, from ourprevious work with SO(4), we know there are fundamental differences in the behavior of rotationsin R4. We understand from earlier that SO(3) is like SO(2) except we may specify the plane inwhich to rotate by giving the orthogonal complement of that plane, namely the axis. This had theeffect of taking away commutativity for SO(3). However, elements of SO(4) are not just singleelements of SO(2) oriented in odd directions, they are more like pairs of noncommutative elementsof SO(2). Elements of SO(3) can be seen as noncommutative elements of SO(2) so perhaps weshould investigate S3 × S3. As we will soon see, this choice is, in fact, correct.

Theorem 3.5. There exists a two-to-one, surjective homomorphism f : S3 × S3 → SO(4).

Proof. Let x = (x1, x2, x3, x4) be a vector in R4. Since H is a vector space isomorphic to R4 we mayrepresent this vector as the quaternion x = x1 +x2i+x3j +x4k. Given two unit length quaternionsq1, q2, we define a transformation fq1,q2 on R4 by

fq1,q2(x) = q1xq2.

We claim that fq1,q2 ∈ SO(4). We know that |q1xq2| = |x| so the transformation fq1,q2 preserveslength. Also note that if we let x = 0 then q1xq2 = 0. Thus such a transformation is also origin-preserving and hence linear. Since fq1,q2 is a length-preserving linear transformation on R4, weknow that it is an element of O(4).

We will now construct the matrix representation of a general transformation fq1,q2 . Let q1 =a0 + a1i + a2j + a3k and q2 = b0 − b1i− b2j − b3k be two unit quaternions. Then fq1,q2(x) = q1xq2

where x is an arbitrary vector in R4. Notice that |xq2| = |x| and that 0q2 = 0, so that f1,q2 alone isa length-preserving linear transformation on R4. Hence it is in O(4) and may be represented witha matrix, specifically

b0 −b1 −b2 −b3

b1 b0 b3 −b2

b2 −b3 b0 b1

b3 b2 −b1 b0

.

Solving for the determinant we find that it is 1, so the linear transformation f1,q2 is an element ofSO(4). Along a similar argument we can see that fq1,1 is a linear transformation that preserveslength. Hence we may find that its matrix representation is

a0 −a1 −a2 −a3

a1 a0 −a3 a2

a2 a3 a0 −a1

a3 −a2 a1 a0

.

The determinant of this matrix is also 1, so fq1,1 is in SO(4). Now a general transformation fq1,q2

is simply the composition of these two transformations. Hence its matrix representation is theproduct of the matrix representations of f1,q2 and fq1,1. When we multiply these two matricestogether we find that the matrix representation for a general transformation fq1,q2 is0BB@

a0b0 − a1b1 − a2b2 − a3b3 −a0b1 − a1b0 + a2b3 − a3b2 −a0b2 − a1b3 − a2b0 + a3b1 −a0b3 + a1b2 − a2b1 − a3b0

a1b0 + a0b1 − a3b2 + a2b3 −a1b1 + a0b0 + a3b3 + a2b2 −a1b2 + a0b3 − a3b0 − a2b1 −a1b3 − a0b2 − a3b1 + a2b0

a2b0 + a3b1 + a0b2 − a1b3 −a2b1 + a3b0 − a0b3 − a1b2 −a2b2 + a3b3 + a0b0 + a1b1 −a2b3 − a3b2 + a0b1 − a1b0

a3b0 − a2b1 + a1b2 + a0b3 −a3b1 − a2b0 − a1b3 + a0b2 −a3b2 − a2b3 + a1b0 − a0b1 −a3b3 + a2b2 + a1b1 + a0b0

1CCA .

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Since this matrix is the product of two elements of SO(4) it must also be an element of SO(4).We define f : S3×S3 → SO(4) by f(q1, q2) = fq1,q2 . We claim that f is a surjective, two-to-one

homomorphism. Now consider quaternions q1, q2, q3, q4 ∈ S3. We check that f is a homomorphism:

f(q1q3, q2q4)(x) = q1q3xq2q4

= q1q3x(q2q4)−1

= q1q3xq−14 q−1

2

= (f(q1, q2) ◦ f(q3, q4))(x).

Hence f is a homomorphism, as claimed.Now, letting a0 = b0 = cos θ and a3 = b3 = sin θ in the matrix above we obtain

cos 2θ 0 0 − sin 2θ0 1 0 00 0 1 0

sin 2θ 0 0 cos 2θ

.

Notice that this is a rotation by 2θ in the x1x4-plane. We can similarly select rotations to occurin any of the six coordinate planes of R4. Recall that Proposition 2.4 stated that every rotation inSO(4) may be expressed as a product of the resulting six matrices. Thus, by Proposition 2.4 wemay conclude that f is onto.

Furthermore, notice that f(1, 1) is the identity and so is f(−1,−1). Assume that f(q1, q2) isthe identity. Then it must be true that q1xq2 = x for all x ∈ R4. Let x = 1 to see that q1 mustequal q2. Now we have q1xq1 = x and so q1 commutes with every quaternion x. Hence q1 is real.Since q1 is unit length, it must be ±1. Thus the kernel of f contains only two elements, so f istwo-to-one.

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References

[1] Simon Altmann, Rotations, Quaternions, and Double Groups, Dover Publications, Mineola,NY, 1986.

[2] Andrew Hanson, Visualizing Quaternions, Morgan Kaufmann, San Francisco, CA, 2006.

[3] Ron Larson, Bruce Edwards, and David Falvo, Elementary Linear Algebra, 5 ed., HoughtonMifflin, Boston, MA, 2004.

[4] Kristopher Tapp, Matrix Groups for Undergraduates, Student Mathematical Library, vol. 29,American Mathematical Society, Providence, RI, 2005.

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