What is meant by "rotation group"?

Question

What do physicists mean by the term "rotation group"? Is it synonymous with $SO(3)$? Is it synonymous with $SU(2)$?

I am confused because rotations in real 3D Euclidean space can also be described in terms of $2\times 2$ unitary matrices with determinant +1. Let me explain how. We can assemble the coordinates of a point $(x_1,x_2,x_3)$ in 3D space in a $2\times 2$ matrix as $$ H= \begin{pmatrix} x_3 & x_1-ix_2\\ x_1+ix_2 & -x_3 \end{pmatrix} $$ Now consider the transformation $$ H^{\prime}=UHU^{\dagger} $$ where U is any $2\times 2$ unitary matrix with deteminant $=+1$. If we now take the determinants of both sides, $$ x_1^{\prime 2}+x_2^{\prime 2}+x_3^{\prime 2}=x_1^2+x_2^2+x_3^2. $$ So it looks like $2\times 2$ unitary matrix with deteminant=+1 are capable of describing rotations in real 3D space. So I wonder whether people also mean SU(2) when they refer to the "rotation group".

Answer 1

This is just the restriction of the well-known double covering map $\pi: \mathrm{SL}(2,\mathbb{C})\to \mathrm{SO}_+(1,3)$ to $\mathrm{SU}(2)$, whose image is $\mathrm{SO}(3)$. See the first few paragraphs in this answer by Qmechanic for more details on that.

Note that the map $\pi\vert_{\mathrm{SU}(2)}$ here is not bijective, because for any $U$, $-U$ has the same action on your $H$ since $(-U)H(-U)^{-\dagger} = UHU^\dagger$. This is one way to establish that $\mathrm{SU}(2)$ is a double cover of $\mathrm{SO}(3)$ - for each rotation there are two distinct 2-by-2 unitary matrices corresponding to it.

So, no, the rotation group is not the same as $\mathrm{SU}(2)$, the rotation group is $\mathrm{SO}(3)$ and it is double-covered by $\mathrm{SU}(2)$.

Answer 2

The rotation group in $N$ spatial dimensions is $\mathrm{SO}(N)$ by definition.

[...] rotations in real 3D Euclidean can also be described in terms of 2x2 unitary, matrices with determinant +1.

We should be precise here. What we want are projective representations of $\mathrm{SO}(3)$. We can obtain these by exponentiating a representation of the Lie algebra $\mathfrak{so}(3)$.

However, projective representations are annoying to work with. Because $\mathfrak{so}(3)\simeq \mathfrak{su}(2)$, any representation of $\mathfrak{so}(3)$ can be translated into a representation of $\mathfrak{su}(2)$, and exponentiating the latter yields a genuine (not projective) representation of $\mathrm{SU}(2)$.

In this sense, there is a direct correspondence between projective representations of $\mathrm{SO}(3)$ (which are what we want) and genuine representations of $\mathrm{SU}(2)$. Since they are equivalent and the latter are much nicer to work with, we use them instead - but that doesn't mean that $\mathrm{SU}(2)$ is the rotation group.

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I am confused because rotations in real 3D Euclidean space can also be described in terms of 2×2 unitary matrices with determinant +1.

If we associate $$\mathbf x \equiv \pmatrix{x_1\\x_2\\x_3} \leftrightarrow \pmatrix{x_3 & x_1-ix_2 \\ x_1+ix_2& -x_3} \equiv \tilde{\mathbf x}$$

Then for each $U\in \mathrm{SU}(2)$, $$U \tilde{\mathbf x} U^\dagger \leftrightarrow R \mathbf x$$ for some $R\in \mathrm{SO}(3)$. However, note that $U$ and $-U$ correspond to precisely the same $R$. In other words, the association of elements of $\mathrm{SU}(2)$ to elements of $\mathrm{SO}(3)$ is two-to-one (each element of the latter corresponds to two elements of the former which differ by a minus sign). This is what we mean when we say that $\mathrm{SU}(2)$ is a double-cover of $\mathrm{SO}(3)$.

Answer 3

SU(2) can be represented by Euler parameters (not to be confused with Euler angles), which can be thought of as points on the unit sphere in four dimensions. (Note that that sphere has three dimensions, analogous to a sphere in three dimensions having two dimensions.)

SO(3) can be represented by axis-angles, which can be thought of as points inside a three dimensional ball of radius pi, with polar opposite points on the ball’s surface identified, that is, the pair considered as one point.

For any pair of polar points in SU(2) – the sphere in four dimensions – there is one corresponding point in SO(3) – the ball in three dimensions. The animated movies shown by the computer program Antitwister illustrate the two-to-one correspondence graphically in its “Orientation Space” areas to the left (Axis-Angle) and right (Euler Parameters) side of the program’s window.

Answer 4

What do physicists mean by the term "rotation group"? Is it synonymous with $SO(3)$?

Let $V$ be a finite dimensional inner product space then $O(V) = Aut(V)$ is the orthogonal group and $SO(V)$ is the subgroup whose determinant is +1. This is the rotation group of $V$.

Now $\mathbb{R}^d$ is a finite dimensional inner product space and hence the above applies to this. Then we define $SO(d) := SO(\mathbb{R}^d)$. Now for any finite dimensional inner product space $V$ of dimension $d$ we do not have $SO(V) = SO(d)$ but they are isomorphic, $SO(V) \cong SO(d)$.

Note that the rotation group depends upon the vector space on which it lives - it isn't canonical. However, since $\mathbb{R}^d$ is canonical - there is only one such space for each $d$ then $SO(d)$ is also canonical and we call it the rotation group.

Is it synonymous with $SU(2)$?

Now for each finite dimensional inner product space $V$ of dimension greater than 2 there is a double covering of $SO(V)$ called the spin group and designated as $Spin(V)$. Thus we also have the double covering $Spin(d)$ of $SO(3)$.

It turns out in dimension 3 and in this dimension only that $Spin(3) \cong SU(2)$. This is why often $SU(2)$ is called the spin group. It is not the rotation group.

(It's worth noting that the elements of representations of spin groups are called spinors).