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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".

Solidification
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2 Answers2

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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 $\mathrm{SU}(2)$.

ACuriousMind
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  • I still have a doubt. Physical operations such as rotations are abstract operations that form a group. On the other hand, SO(3) is a group of $3\times 3$ real orthogonal matrices with determinant $+1$. So I think rather than saying that the rotation group is synonymous with SO(3), what we should really say is that SO(3) is isomorphic to the group of rotations in 3D i.e. to each rotation in 3D we can associate a $3\times 3$ real orthogonal matrix of determinant $+1$. Isn't this correct? Please clarify this. – Solidification Jun 27 '23 at 18:47
  • @Solidification I do not see any difference between saying that a group is "synonymous" with something and that group being isomorphic to that group. By "synonymous", are you asking which we take as a definition? Definitions are arbitrary: When two concrete realizations of a group are isomorphic, you can pick the first as the definition and derive the isomorphism to the second, or you can pick the second as the definition and derive the isomorphism to the first, the difference is merely aesthetic. – ACuriousMind Jun 27 '23 at 18:57
  • @Solidification Then how would you define a rotation (in 3D)? – Tobias Fünke Jun 27 '23 at 21:21
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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.

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 $\mathrm{SO}(3)$.

J. Murray
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  • I still have a doubt. Physical operations such as rotations are abstract operations that form a group. On the other hand, SO(3) is a group of $3\times 3$ real orthogonal matrices with determinant $+1$. So I think rather than saying that the rotation group is synonymous with SO(3), what we should really say is that SO(3) is isomorphic to the group of rotations in 3D i.e. to each rotation in 3D we can associate a $3\times 3$ real orthogonal matrix of determinant $+1$. Isn't this correct? Please clarify this. – Solidification Jun 27 '23 at 18:49
  • @Solidification You're making a distinction without a difference. Saying the rotation group is $SO(3)$ and saying it's isomorphic to $SO(3)$ are the same thing. – d_b Jun 27 '23 at 19:06
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    @Solidification I think you are abstracting yourself beyond the point of usefulness. We generally define the rotation group to be $\mathrm{SO}(3)$, and then when we wish to implement a rotation on some space $V$, we define a representation (or more generally, a group action) $\rho:\mathrm{SO}(3)\rightarrow \mathrm{GL}(V)$. Any generality you think you gain by making a distinction between $\mathrm{SO}(3)$ and the 3D rotation group can be absorbed into the representation/group action anyway, so you lose nothing by giving the rotation group a concrete definition in the first place. – J. Murray Jun 27 '23 at 19:27