October 8th, 2026

Recall the standard Euclidean structure of $\bR^n$. For any $x, y \in \bR^n$, our standard inner product (or the “dot product”) is given by

where $x = (x_1, \dots, x_n)$ and $y = (y_1, \dots, y_n)$ are coordinates with respect to the standard basis of $\bR^n$. Note that if $x$ and $y$ are written as column vectors, then the standard dot product is given by $x^T y$, which can be quickly seen by performing the matrix multiplication by hand.

Recall that $\norm{x}^2 = x^T x$, and so $\norm{x} = \sqrt{x^T x}$. Note that $x, y \in \bR^n$ are orthogonal if $x^T y = 0$. For nonzero vectors in $\bR^2$, we also have that $x^T y = \norm{x} \norm{y} \cos(\theta)$, where $\theta$ is the angle between the vectors.

By induction, this generalizes to any finite sum of orthogonal vectors.

We have the tools to define an orthonormal basis.

Since each vector is mutually orthogonal, they are all linearly independent. Thus $n$ of them are enough to form a basis for $\bR^n$, and normalizing them gives us an orthonormal basis. An equivalent characterization of an orthonormal basis of $\bR^n$ is a list of vectors $v_1, \dots, v_n$ satisfying

It is quick to show that orthogonal matrices satisfy a variety of properties, such as the fact that it preserves inner products and norms, its determinant is $\pm 1$, and that its rows and columns form an orthonormal basis of $\bR^n$. These properties are easy to verify by matrix multiplication.

Note that $[\Orth_n: \SO_n] = 2$, so normality comes for free (recall index $2$ subgroups are normal).

We have characterized orthogonal operators algebraically, but we have not yet discussed what these operators actually do geometrically. Recall that orthogonal operators preserve inner products, and consequently preserve lengths, angles, and distances. Thus they should correspond to rigid transformations of Euclidean space that fix the origin.

We now seek to classify these operators explicitly in low dimensions. More precisely, we want to understand what the elements of $\Orth_2$ and $\Orth_3$ look like, and how their determinants distinguish the possible geometric transformations. We begin with $\Orth_2$, where the classification is particularly simple.

Note that a reflection in $\bR62$ is completely determined by its fixed line $L$. Indeed, letting $V = L \oplus L^\perp$, note that $L = E(1, T)$ and $L^\perp = E(-1, T)$, both of which are one-dimensional $T$-invariant subspaces of $V$. Letting $\{e_1, e_2\}$ be the orthogonal eigenbasis of $V$, the matrix is given by $\diag\{1, -1\}$. Rotations in $\bR^2$ are also uniquely determined by their angle $\theta$, modulo $2\pi$.

Thus every orthogonal operator on $\bR^2$ is either a rotation or a reflection. The determinant distinguishes these two possibilities: determinant $1$ corresponds to rotations, while determinant $-1$ corresponds to reflections. Note that we can also think of the determinant $-1$ case as a reflection over the $x$-axis, followed by a usual rotation.

In particular, $\SO_2$ consists precisely of the rotations about the origin. Since $R_\alpha R_\beta = R_{\alpha + \beta}$, we have that $\SO_2$ is abelian. This is geometrically unsurprising, since composing two rotations of the plane about the same point simply adds their angles.

Notice that reflections do not form a subgroup of $\Orth_2$, since the product of two reflections has determinant $1$ and is therefore a rotation. Indeed, every element of $\Orth_2$ can be written either as $R_\theta$ or as $R_\theta S_0$. Thus $\Orth_2$ is generated by rotations together with a single reflection.

We now turn to $\bR^3$. Unlike rotations of $\bR^2$, which are determined entirely by an angle, rotations of $\bR^3$ require specifying both an angle and an axis. We first define what it means for a linear operator on $\bR^3$ to be a rotation.

The geometric interpretation is straightforward. Since $T(u) = u$ and $T$ is linear, we immediately have that $T(cu) = cT(u) = cu$ for every $c \in \bR$. Thus $T$ fixes the entire line $\Span\{u\}$ pointwise, not merely the vector $u$. Meanwhile, it rotates every vector in the orthogonal plane $u^\perp$ through some fixed angle. This is precisely the familiar notion of rotation about an axis in three-dimensional space.

The requirement that $u$ be a unit vector is simply a normalization convention. Indeed, if $T$ fixes any nonzero vector $v$, then it also fixes the unit vector $u = v/\norm{v}$ by linearity. We use unit vectors because they give us a convenient way to specify a direction along the axis.

Consider the vector $u \in \bR^3$ and its orthogonal plane $u^\perp$.

Suppose, for example, that $u = e_1$. Then $u^\perp = \Span\{e_2, e_3\}$, and a rotation through angle $\theta$ about this axis is represented by

The first coordinate is fixed, while the remaining two coordinates undergo the ordinary two-dimensional rotation we classified earlier. More generally, given any unit vector $u$, we can extend it to an orthonormal basis $(u, v_2, v_3)$ of $\bR^3$, and the matrix of the rotation with respect to a suitably oriented such basis takes this exact same form.

There is one subtlety we have not yet addressed. In $\bR^2$, the sign of the rotation angle is determined by the standard orientation of the plane. In $\bR^3$, however, the sign of a rotation angle depends on which direction we choose along the axis. This motivates the following definition.

Note that the same axis admits two choices of unit pole, namely $u$ and $-u$. Reversing the direction of the pole reverses the orientation of the orthogonal plane, and consequently reverses the sign of the rotation angle. Thus

In other words, the two choices of pole describe the same rotation, provided we reverse the sign of the angle accordingly. As usual, rotation angles are understood modulo $2\pi$.

Having defined rotations on $\bR^3$, we now return to our original classification problem. We want to determine which matrices in $\Orth_3$ represent these rotations. Recall that in dimension $2$, every matrix in $\SO_2$ represents a rotation. It turns out that the analogous statement is true in dimension $3$, although the proof is less immediate.

We first need a useful lemma.

Note that this result is special to odd dimensions. The argument relied on the fact that $\det(-A) = -\det(A)$ for a $3 \times 3$ matrix, which would not hold in even dimensions. More generally, the same reasoning shows that every matrix in $\SO_{2n+1}$ has eigenvalue $1$.

We now have everything needed to classify $\SO_3$.

Thus every matrix in $\SO_3$ represents a rotation about some axis, and every rotation about the origin is represented by an element of $\SO_3$. In particular, the determinant condition $\det M = 1$, together with orthogonality, is sufficient to guarantee the existence of a fixed axis.

This is a somewhat surprising result. In $\bR^2$, the composition of rotations is easy to understand because all rotations share the same center and their angles simply add. In $\bR^3$, two rotations may have completely different axes, yet their composition is still a rotation about some third axis. Note, however, that $\SO_3$ is generally not abelian, since rotating about different axes in different orders need not produce the same transformation.

We now discuss two useful properties of three-dimensional rotations, namely how their angles relate to their traces and how their axes behave under conjugation.

Thus we can recover the cosine of the rotation angle directly from the trace of its matrix, since $\cos(\theta) = (\operatorname{tr}(M)-1)/2$. Note that this determines the angle only up to sign and multiples of $2\pi$, since $\cos(\theta) = \cos(-\theta)$. To recover an oriented angle, we must also specify the direction of the pole.

Geometrically, conjugation by $B$ amounts to rotating our entire coordinate system by $B$, performing the original rotation, and then translating the result back into the original coordinates. The axis is transported from $\Span\{u\}$ to $\Span\{Bu\}$, while the oriented rotation angle remains unchanged. This is a concrete example of how conjugation changes the geometric representation of an operator without altering its underlying structure.

This theorem also shows that the trace uniquely identifies a class of similar matrices that each represent the same rotation of some plane isomorphic to $\bR^2$, at least up to orientation (since $1 + 2\cos(\theta)$ is unique for $0 \le \theta \le \pi$).

Now we come to an important application of groups. Up to this point, we have considered groups largely as standalone objects—as sets of values equipped with an operation that possesses certain regularity properties (like closure and associativity). We have focused primarily on their internal structures, subgroups, and the maps between them. We now shift our perspective to how groups act on other mathematical objects. This viewpoint serves two main purposes: describing the symmetries of certain structures and understanding group representations. By treating groups as transformations rather than static sets, we will see that they naturally induce actions on various mathematical spaces. When these transformations operate on a geometric or combinatorial space, they encode the intrinsic symmetries of that object. Furthermore, when a group acts on a vector space by invertible linear transformations, it yields a group representation, allowing us to translate abstract group elements into linear operators and leverage the machinery of linear algebra.

As we explore these actions, we will again consider the tradeoff between the generality and the strength of a result. We will soon see, for instance, that every group can be realized as a group of permutations, and thus as a group of symmetries of some arbitrary set. However, this broad notion of symmetry is rather weak, since arbitrary permutations need not preserve distances, angles, or any other underlying geometry. By requiring our transformations to preserve additional structure, we restrict the kinds of actions we may consider, but in return, we obtain a much richer description of mathematical objects. Throughout our study of group actions, we will repeatedly navigate this tradeoff.

For now, we will focus on group symmetries, beginning with those of a geometric nature. Recall that an object is said to be symmetric under an operation if it remains invariant under that operation. In the plane, the fundamental symmetries are bilateral symmetry (reflection about an axis), rotational symmetry (rotation through some angle), translational symmetry (shifting an object in space), and glide symmetry (a specific combination of translation and reflection). More formally, any rigid motion of the plane is called an isometry, and an isometry that preserves a particular subset of the plane is called a symmetry of that subset. For example, while a $60^\circ$ rotation is always an isometry of the plane, it is specifically a symmetry of a regular hexagon.

Note that isometries need not be linear operators, they are simply maps. (We will soon show, however, that all isometries of $\bR^n$ are affine transformations.)

Note that any isometry $f: \bR^n \to \bR^n$ can also be written as $f(x) = T(x) + b$, where $T$ is the orthogonal operator and $b$ is the translation.