I'm reading a famous book, "Group Theory and Quantum Mechanics" written by Michael Tinkham. In Chapter 5 of this book, he introduces the relationship between $SU(2)$ and $SO(3)$ and I already understand the reason that $SU(2)$ is covering Lie group of $SO(3)$, which can be regarded as Lie group homomorphism. Then the book says, (p. 106)

Because of the homomorphism, these same matrices(= all irreducible, unitary matrices representing $SU(2)$) will serve as the representation matrices for the three-dimensional rotation group.

My question is "why?". Well, the book firstly derive all possible "irreducible, unitary" representation matrices of $SU(2)$, $U^{(j)}$, having $2j+1$ dimension. With $u\, \in $matrix Lie group $SU(2)$ and $P(u) \in $ Lie group $SU(2)$, we can write

$$P(u)f^{(j)}_m\,=\, \sum_{m'}f_{m'}^{(j)}U^{(j)}(u)_{m'm}$$

where $f^{(j)}_m$ is appropriate basis functions. Accordingly we can possibly find $2j+1$ dimensional matirces $D^{(j)}(R)$ satisfying

$$P(R)f^{(j)}_m\,=\, \sum_{m'}f_{m'}^{(j)}D^{(j)}(R)_{m'm}$$

where $R\in$matrix Lie group $SO(3)$ and $P(R)\in$Lie group $SO(3)$. Then the book says


and this seems to be the part corresponding to boldfaced statement above.

As mentioned, here is a Lie group homomorphism $\Phi$,

$$\Phi\,:\,P(u)\,\rightarrow\,P(R)$$ or $$u\,\rightarrow\,R$$

However I cannot have any idea that relates the existence of $\Phi$ and the fact that $SU(2)$ and $SO(3)$ share same representing matrices, $D^{(j)}(R)=U^{(j)}(u^{-1})$.


This sounds a bit like a misunderstanding. What is true is that any representation of $SO(3)$ gives rise to a representation of $SU(2)$, but not vice versa. This is easy to see if one interprets a representation as a homomorphism to a unitary group. Given a representation $SO(3)\to U(N)$ (for some $N$), you just compose with the homomorphism $SU(2)\to SO(3)$ to obtain a representation $SU(2)\to U(N)$.

For a converse, a bit more information is needed, namely that the homomorphisms $SU(2)\to SO(3)$ acutally induces an isomorphism $SU(2)/\{\pm\mathbb I\}\to SO(3)$. (This basically expresses the fact that $SU(2)$ is a two-fold covering of $SO(3)$). Now if you have a representation of $SU(2)$, it may happen that $-\mathbb I$ is represented by the identity matrix. If this is the case, then the representation defines a representation of $SO(3)$, since it descends to the quotient $SU(2)/\{\pm\mathbb I\}$. In this way, representations of $SO(3)$ are "the same" as representations of $SU(2)$ in which $-\mathbb I$ is represented by the identity. (Roughly speaking this is "half" of the irreducible representations of $SU(2)$. From your question it sounds like the book would only consider these, otherwise the condition on odd dimension of the representing matrices should not be there.) This obvious representation of $SU(2)$ on $\mathbb C^2$ does not give rise to a representation of $SO(3)$ and it is the simplest example of such a representation. This is the basis for spinors.

  • $\begingroup$ I really appreciate your help, especially the part $SU(2)/\{\pm\mathbb I\}\cong SO(3)$. To arrange myself I posted my own conclusions from information you give. $\endgroup$ – Discovery Oct 25 '15 at 2:02

Let me arrage some points. To share same irreducible representation matrices, image sets of two corresponding homomorphisms, $im(\Phi_{SO(3)})$ and $im(\Phi_{SU(2)})$ should be isomorphic. From the first isomorphism theorem, they are isomorphic to $SO(3)/ker(\Phi_{SO(3)})$ and $SU(2)/ker(\Phi_{SU(2)})$ respectively. Therefore I concluded that

Two groups share same irreducible representation matrices if their corresponding quotient groups are isomorphic, i.e. $SO(3)/ker(\Phi_{SO(3)})\cong SU(2)/ker(\Phi_{SU(2)})$.

From Andreas's post, two-to-one homomorphism $SU(2) \rightarrow SO(3)$ induces an isomorphism $SU(2)/\{\pm\mathbb I\}\cong SO(3)$. If $ker(\Phi_{SU(2)})={ \{\mathbb I\}}$ (homomorphism has trivial kernel), $\Phi_{SU(2)}$ is injective and

$$SU(2)/ker(\Phi_{SU(2)})\cong im(\Phi_{SU(2)})\cong SU(2)$$

where first isomorphism arises from the first isomorphism theorem and second one from the injectiveness of $\Phi_{SU(2)}:SU(2)\rightarrow G$ (Cayley's theorem).

Cf. Usage of Cayley's theorem : From the property of image sets, $im(\Phi_{SU(2)})$ is subgroup of $G$. If $\Phi_{SU(2)}$ is injective, $SU(2)\cong im(\Phi_{SU(2)})=$a subgroup of $G$. Cayley's theorem states that

Given two groups $H$ and $G$, and there exist an injective homomorphism $f:H\rightarrow G$, then $H$ is isomorphic to a subgroup of $G$

which is equivalent to $SU(2) \cong$ a subgroup of $G$.

However we cannot find $\Phi_{SO(3)}$ satisfying $$SO(3)/ker(\Phi_{SO(3)})\cong SU(2)/ker(\Phi_{SU(2)})=SU(2)/\{\mathbb I\}\cong SU(2)$$

On the other hand, if $ker(\Phi_{SU(2)})= \{ \pm \mathbb I\}$, $$SO(3)/ker(\Phi_{SO(3)})\cong SU(2)/ker(\Phi_{SU(2)})=SU(2)/\{ \pm \mathbb I\}\cong SO(3)$$ can be satisfied if $ker(\Phi_{SO(3)})=\{\mathbb I \}$ that representation of $SO(3)$ is trivial.

I find that from the formula deriving $U^{(j)}$, representation matrices of $SU(2)$, $ker(\Phi_{SU(2)})= \{ \pm \mathbb I\} $ only if $2j+1$ is odd. Thus I finally conclude that only when the representation matrices of $SU(2)$ have odd dimensions, $SO(3)$ and $SU(2)$ share same representation matrices and corresponding $\Phi_{SO(3)}$ should be trivial, injective homomorphism.

  • $\begingroup$ This looks good. One thing to notice is that since $SU(2)$ and $SO(3)$ both are simple groups, basic Lie thory shows that the kernel of a non-zero homomorphism defined on either of these groups has to be a discrete normal subgroup of the center of the group in question. This implies that the kernel is automatically trivial for $SO(3)$ and either trivial or $\{\pm\mathbb I\}$ for $SU(2)$. $\endgroup$ – Andreas Cap Oct 25 '15 at 10:36

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