The rotation group in two Euclidean dimensions is known as $\SO(2)$. How many representations of this group can you think of?

The first representation of $\SO(2)$ we consider is in terms of $2\times2$ matrices, of the form \begin{equation} M(\phi) = \begin{pmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{pmatrix} . \end{equation} The “O” in $\SO(2)$ stands for orthogonal. Orthogonal matrices satisfy \begin{equation} M^T M = 1 \end{equation} where $T$ denotes matrix transpose, and where we write simply $1$ for the identity matrix (rather than $I$, for which we will have another use). Such matrices preserve the (squared) magnitude \begin{equation} |v|^2 = v^T v \end{equation} of a vector $v\in\RR^2$, since \begin{equation} (Mv)^T (Mv) = (v^TM^T) (Mv) = v^T v . \end{equation} The “S” in $\SO(2)$ stands for special, and refers to the additional condition that \begin{equation} |M| = \det(M) = 1 . \end{equation} Orthogonal matrices $M$ with $|M|=1$ are rotations; if $|M|=-1$, the only other possibility, they are reflections.

Our second representation of $\SO(2)$ is in terms of the complex numbers, of the form \begin{equation} w(\phi) = e^{i\phi} . \end{equation} Such complex numbers have norm 1, that is \begin{equation} |w(\phi)|^2 = w\,\bar{w} = 1 \end{equation} and preserve the magnitude $|z|$ of any complex number $z\in\CC$, since \begin{equation} |wz| = |w|\,|z| = |z| . \end{equation} Sound familiar?

Our third representation of $\SO(2)$ is purely geometric. Rotations are rigid transformations of $\RR^2$, obtained by, well, rotating the plane through a given angle $\phi$. In other words, the rotations in $\SO(2)$ are in one-to-one correspondence with the angles in the (unit) circle, that is, with the circle itself. Thus, $\SO(2)$ can be thought of as the circle $\SS^1$.

Take a moment to compare and contrast these various representations of $\SO(2)$. What are their properties?

The geometric representation makes clear that $\SO(2)$ is a group; the composition of two rotations is another rotation. In matrix language, we have \begin{equation} M(\alpha+\beta) = M(\alpha) M(\beta) \end{equation} and similarly \begin{equation} e^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta} \end{equation} for complex numbers. Setting $\alpha=0$ corresponds to the identity element, and setting $\beta=-\alpha$ leads immediately to inverse elements.

Our two algebraic representations are clearly closely related. The identification of $M(\phi)$ with $e^{i\phi}$ suggests the further identification \[ x+iy \longleftrightarrow \begin{pmatrix}x & -y \\ y & x\end{pmatrix} . \] This identification seems even more reasonable after writing \begin{equation} \begin{pmatrix}x & -y \\ y & x\end{pmatrix} = x 1 + y \Omega \end{equation} where again $1$ denotes the identity matrix, and \begin{equation} \Omega = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} . \end{equation} Notice that $\Omega^2=-1$!

We note for future reference both that \begin{equation} \Omega = M'(0) \end{equation} and that \begin{equation} M(\phi) = e^{\Omega\phi} \end{equation} where matrix exponentiation is formally defined in terms of a power series (which always converges).

Returning to our geometric representation, since $\SO(2)$ can be thought of as $\SS^1$, it is a smooth manifold, that is, it is a smooth 1-surface (i.e. a curve), on which one can introduce coordinates (e.g. $\phi$). Thus, $\SO(2)$ is both a group and a manifold; it is our first example of a Lie group.

In the language of vector calculus, we can introduce a vector field that is tangent to $\SS^1$. One possible choice would be the unit vector tangent to the circle, often written $\hat\phi$. In the language of differential geometry, however, vector fields are interpreted as directional derivative operators, so that \begin{equation} \vv(f) = \vv\cdot\grad f . \end{equation} Since \begin{equation} \hat\phi\cdot\grad f = \frac{1}{r}\frac{\partial f}{\partial \phi} \end{equation} we choose instead the tangent vector \begin{equation} r\hat\phi = x\,\hat{y} - y\,\hat{x} \end{equation} Equivalently, as differential operators we choose \begin{equation} \partial_\phi = x\,\partial_y - y\,\partial_x \end{equation} where we have introduced the notation $\partial_q$ for $\frac{\partial}{\partial q}$.

What do tangent vectors look like in the complex representation of $\SO(2)$? Take the derivative! We have \begin{equation} \frac{dw}{d\phi} = iw = ie^{i\phi} \end{equation} What does this result mean geometrically?

Evaluate this derivative first at the identity element, where $\phi=0$. At the point $z=1$, this derivative is $i$. But the $i$-direction is vertical; this direction is tangent to the circle at $z=1$. A similar argument works at any point on the circle; $iw$ always represents the direction rotated $\frac{\pi}{2}$ counterclockwise from $w$—precisely the direction tangent to the circle.

Finally, consider the matrix representation of $\SO(2)$. Again, take the derivative, yielding \begin{equation} A = M'(0) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} . \end{equation} At other points on the circle, we have \begin{equation} M'(\alpha) = \begin{pmatrix} -\sin\phi & -\cos\phi \\ \cos\phi & -\sin\phi \end{pmatrix} = M(\alpha) A . \end{equation} This relationship between the derivative of a path in the group at any point and its derivative at the identity element is a hallmark of the study of Lie groups, and allows us to study such groups by studying their derivatives at the identity element, a much simpler process.