I'm sorry for this naive question. I feel that since I was first introduced to the idea of group representation, I did not correctly grasp the idea. Unfortunately, therefore, several confusions keep bothering me. The definition of representation that I know (from Maggiore's QFT book, and Howard Georgi's Lie Algebras in Particle Physics) is highlighted below.

A representation of a group $\mathcal{G}$ on a vector space $\mathbb{V}$ is a mapping $\mathcal{D}: g\rightarrow \mathcal{D}(g)$ (sometimes also written as $\mathcal{D}: \mathcal{G}\to GL(\mathbb{V})$) onto a set of invertible linear operators $GL(\mathbb{V})$ such that $$\mathcal{D}(e)=\mathbb{1}\tag{1}$$ where $\mathbb{1}$ is the identity operator in the space $GL(\mathbb{V})$. For any two elements $g_1,g_2\in\mathcal{G}$,$$\mathcal{D}(g_1)\mathcal{D}(g_2)=\mathcal{D}(g_1*g_2).\tag{2}$$

Now I have a few questions regarding this.

Question 1 The definition of representation that I stated above (i) doesn't require the associativity $$\Big(\mathcal{D}(g_1)\mathcal{D}(g_2)\Big)\mathcal{D}(g_3)=\mathcal{D}(g_1)\Big(\mathcal{D}(g_2)\mathcal{D}(g_3)\Big)$$ has to be satisfied, and (ii) doesn't require that $$\mathcal{D}(g^{-1})=\mathcal{D}^{-1}(g)$$ have to be satisfied. Does it mean that $GL(\mathbb{V})$ does not form a group? That cannot be true. In that case, what am I missing?

Question 2 For concreteness, suppose I want to understand the representation of the group $\mathcal{G}={\rm SO(3)}$. What is $\mathbb{V}$ in this case? And what is $GL(\mathbb{V})$ in this case? To be even more concrete, if $\mathbb{V}=\mathbb{R}^3$, $GL(\mathbb{V})$ has to be to a set of $3\times 3$ matrices. Should these representation matrices need to be orthogonal and unimodular? Does it follow from the definition of representation? Let us consider a different representation. Say a 5-dimensional representation of SO(3). What is $\mathbb{V}$ and $GL(\mathbb{V})$ in this case? Should these representation matrices again need to be orthogonal and unimodular?

  • 1
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Apr 14 '18 at 14:09
  • $\begingroup$ At to you last Question for v2: $\mathbb{V}$ can be as you say but there is - for instance - an irrep of SO3 of dimension $5$: the states of angular momentum $L=2$ are a basis for this. The SO3 matrices would then be orthogonal and unimodular. Note that in many cases we can do a little more by using the complex form $e^{i\theta}$ rather than the real $\cos\theta$ and $\sin\theta$, and complex rather than strictly real basis vectors. For SO3 representations (integer angular momentum) the ${\cal D}$ matrices are just those from en.wikipedia.org/wiki/Wigner_D-matrix $\endgroup$ – ZeroTheHero Apr 14 '18 at 14:34
  1. You are wrong about associativity and compatibility of inverses not being implied by eq. (1) and (2). We have: \begin{align} \left(D(g_1)D(g_2)\right)D(g_3) & \overset{(1)}{=} D(g_1g_2)D(g_3) \overset{(1)}{=} D\left( (g_1g_2)g_3\right) = D\left( g_1 (g_2 g_3) \right) \overset{(1)}{=} D(g_1) D(g_2g_3) \\ & \overset{(1)}{=} D(g_1)\left( D(g_2) D(g_3)\right),\end{align} where the only equality that does not hold because of eq. (1) is just associativity in the original group. Likewise, eq. (2) implies that $D(g^{-1}) = D(g)^{-1}$, try to figure that one out yourself.

  2. There is no "the" representation of $\mathrm{SO}(3)$, there are countably many different ones up to isomorphism, which we physicists usually label by their half-integral spin $s\in\frac{1}{2}\mathbb{Z}$. For each of these representations, the representation space is $\mathbb{V} = \mathbb{C}^{2s + 1}$. I don't understand what you mean by "What is $\mathrm{GL}(\mathbb{V})$?" - this is always the group of invertible matrices on $\mathbb{V}$, i.e. all matrices with non-zero determinant, and the image of $G$ under the representation map $D$ is some subgroup of it. (It is a subgroup precisely because of point 1.)

    Since you seem to actually want to ask whether e.g. a three-dimensional representation of $\mathrm{SO}(3)$ always has $D(\mathrm{SO}(3)$ as being unimodular orthogonal matrices, the answer is no. For instance, you could very well choose the trivial representation $$ D: \mathrm{SO}(3) \to \mathrm{GL}(3), A\mapsto \mathbf{1}_3,$$ where $\mathbf{1}_3$ is the three-dimensional identity.

    If you require the representation to be three-dimensional and irreducible, then $D(\mathrm{SO}(3)$ is always isomorphic as a group to the group of unimodular orthogonal matrices, but it need not be equal to that group as a subgroup of $\mathrm{GL}(3)$. For instance, any conjugate subgroup to the subgroup of unimodular orthogonal matrices would also be an irreducible representation.

  • $\begingroup$ @ACuriousMind I think what the OP might be getting as is: are all representations of SO3 by orthogonal unimodular matrices or just the defining $3\times 3$ rep. $\endgroup$ – ZeroTheHero Apr 14 '18 at 14:53
  • $\begingroup$ @SRS I already said that $D(\mathrm{SO}(3))$ is only a subgroup of $\mathrm{GL}(V)$, since $D$ was never required to be surjective. What the image of $G$ under the representation map - the matrices $G$ is represented by - looks like is a completely different question from what $\mathrm{GL}(V)$ is. $\endgroup$ – ACuriousMind Apr 14 '18 at 14:57
  • $\begingroup$ I already said that D(SO(3)) is only a subgroup of GL(V). Okay. I think I was confusing between D(SO(3)) and GL(V); I was equating the two. My question is about D(SO(3)). Does it have to be orthogonal and unimodular for vector, tensor etc representations? @ACuriousMind $\endgroup$ – SRS Apr 14 '18 at 15:05
  • $\begingroup$ @ACuriousMind Probably it's better to ask why is the subgroup D, of any dimension, have to be unimodular and orthogonal. $\endgroup$ – SRS Apr 14 '18 at 15:11
  • 1
    $\begingroup$ @SRS By pure dimension counting ($\mathrm{SO}(3)$ has dimension 3, but $\mathrm{SO}(n)$ (which is isomorphic to unimodular, orthogonal n-by-n matrices) has dimension $n(n-1)/2$), the image of $\mathrm{SO}(3)$ in higher-dimensional representations cannot be the group of unimodular orthogonal matrices in these dimensions. $\endgroup$ – ACuriousMind Apr 14 '18 at 17:01

Associativity of composition is a general property of maps between sets, it is independent of any group representation. ${D(g)}^{-1}=D(g^{-1})$ is a simple consequence of your properties (1), (2).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.