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Earlier I asked this Comparison between SO(n) and Spin(n) representation theory which is closed. I think the question is certainly valid and a good one. But my comments are too many and too long, so someone did not like that or got bored. So let me focus on one thing ONLY this time.

We know that $Spin(n)/\mathbb{Z}_2=SO(n)$. The $SO(n)$ and $Spin(n)$ have the same Lie algebra. When it comes to the representation of $SO(n)$ and $Spin(n)$, does it make any difference?

Since Spin group is a double cover of SO group, how does this global structure being reflected in the case of representation? (if their representations are the same? or differed also by a double cover? perhaps the parameters of Lie group are "doubled" in some way?) Am I correct to say that SO group has integer spin representations, while Spin group has both integer and half-integer spin representations? For example, the SO(3) group has a trivial representation, and other odd-rank dimensional matrix representation: $$ 1,3,5,7,\dots. $$ In contrast, the Spin(3) group has a trivial representation, and other odd and even-rank dimensional matrix representation: $$ 1,2,3,4,5,6,7,\dots. $$ The odd and even-rank dimensional matrix representation is related to what physicists call the integer and half-integer spin representations.

How about the more general cases for $SO(n)$ and $Spin(n)$, other than $n=3$?

Taylor Rendon
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annie marie cœur
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1 Answers1

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If $G$ is a (Lie) group and $H$ a (closed) normal subgroup, then for any given (say, complex) vector space $V$ there is a bijection

$$\text{(cont.) rep's of }G \text{ on } V \text{ such that } \rho_{\vert H}=id \leftrightarrow \text{ (cont.) rep's of } G/H \text { on }V$$

given by mapping a representation $\rho :G \rightarrow Aut(V)$ on the left to the induced $\tilde \rho: G/H \rightarrow Aut(V)$ on the right, and in the other direction, given a representation $\rho: G/H \rightarrow Aut(V)$, pulling it back to $\hat\rho:G \twoheadrightarrow G/H \stackrel{\rho}\rightarrow Aut(V)$.

This bijection respects irreducibility.

Apply this to $G= Spin(n)$ and $H=\{\pm1\}$ where you know that $SO(n) \simeq G/H$.

You'll find that the representations of $SO(n)$ correspond exactly to those representations of $Spin(n)$ which are trivial on $-1$.

And now you need to know a little more theory to tell which representations of $Spin(n)$ that are. What do you know about the root lattice and the weight lattice of these groups?

Torsten Schoeneberg
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  • Can you provide some refs on "the root lattice and the weight lattice of these groups"? – annie marie cœur Jul 22 '20 at 18:53
  • Whatever source on representation theory and those groups you are using, if it does not talk about weights and roots and e.g. these things, then you should look for a different source and learn that theory from the ground up. Books by Knapp, Hall, Fulton&Harris, Helgason come to mind. – Torsten Schoeneberg Jul 22 '20 at 20:15
  • OK - recommend one single best textbook for undergrad or 1st year grad level please! – annie marie cœur Jul 22 '20 at 20:17
  • May I make sure one thing "the representations of () correspond exactly to those representations of () which are trivial on −1."

    ---- (a) Does it mean if all the () representations will be part of () representations?

    ------ (b) But some () representations may not be () representations ?

     – annie marie cœur Jul 23 '20 at 01:31
  • Yes. All representations of $SO(n)$ are quotients of $Spin(n)$-representations; but some $Spin(n)$-representations do not factor through $SO(n)$, hence do not give $SO(n)$-representations. – Torsten Schoeneberg Jul 23 '20 at 04:48