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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 $n=3$?

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    $\begingroup$ Instead of dimensions, you should be looking at weights of the representations. $\endgroup$ Jul 20, 2020 at 4:09

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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?

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  • $\begingroup$ Can you provide some refs on "the root lattice and the weight lattice of these groups"? $\endgroup$ Jul 22, 2020 at 18:53
  • $\begingroup$ 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. $\endgroup$ Jul 22, 2020 at 20:15
  • $\begingroup$ OK - recommend one single best textbook for undergrad or 1st year grad level please! $\endgroup$ Jul 22, 2020 at 20:17
  • $\begingroup$ May I make sure one thing "the representations of 𝑆𝑂(𝑛) correspond exactly to those representations of 𝑆𝑝𝑖𝑛(𝑛) which are trivial on −1." ---- (a) Does it mean if all the 𝑆𝑂(𝑛) representations will be part of 𝑆𝑝𝑖𝑛(𝑛) representations? ------ (b) But some 𝑆𝑝𝑖𝑛(𝑛) representations may not be 𝑆𝑂(𝑛) representations ? $\endgroup$ Jul 23, 2020 at 1:31
  • $\begingroup$ 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. $\endgroup$ Jul 23, 2020 at 4:48

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