While reading Polchinski's Chapter 1, I encountered the following on page 24,

"For example, the $(D-1)$ dimensional vector representation of $SO(D-1)$ breaks up into an invariant and a $(D-2)$-vector under the $SO(D-2)$ acting on the transverse directions, $$ \vec{v} = (v^1, 0, 0, \dots) + (0. v^2, v^2, \dots, v^{D-1}) $$

Thus, if a massive particle is in the vector representation of $SO(D-1)$, we will see a scalar and a vector when we look at the transformation properties under $S0(D-2)$. This idea extends to any representation : one can always reconstruct the full $SO(D-1)$ spin representation from the behaviour under $SO(D-2)$. "

I can show that the second excited states which is given by, $$ \alpha_{-1}^i \alpha_{-1}^j |0\rangle \bigoplus \alpha_{-2}^i |0\rangle $$ where $i,j$ runs from $\{2,D-1\}$ and treating them symmetric, the no. of excited states would be $\dfrac{(D+1)(D-2)}{2}$ which matches with the dimensions of a traceless symmetric irrep of $SO(D-1)$.

My question is how can we be sure by just matching numbers, and what does this Physically mean and is there a mechanism to do this consistently ? What does this business of "reconstructing $SO(D-1)$ representation from $SO(D-2)$" mean ?

I know a little bit about group theory like, Cartan Matrices, Dynkin Diagrams and Young Tableaux method for SU(N) theory, so it is fine if someone could give me a good reference. A precise answer would ofcourse be great :).


You may decompose (irreductible) representation of groups as a sum of (irreductible)representations of subgroups.

Starting from a traceless symmetric irreductible representation of $SO(D-1)$:

$$R_{ij} = \frac{1}{2} (v^iv^j+v^jv^i) - \frac{1}{D-1} \delta_{ij} ( \sum\limits_{i=1}^{D-1} v_k v_k),\, \text{with} \, (i,j) \, \text{in} \,[1, D-1] \tag{1}$$

You consider $x_1$ as a scalar under a $SO(D-2)$ transformation on $x_2,x_3...x_{D-1}$, then you got, decomposing $R_{ij}$ in irreductible representations of $SO(D-2)$:

A) The trivial representation : $R_{11}$

B) The vector representation : $R_{1i}$, with $i$ in $[2, D-1]$

C) The tracelesss symmetric representation : $R_{ij} + \delta_{ij} \frac{1}{D-2}R_{11}$, with $i,j$ in $[2, D-1]$

  • $\begingroup$ Hi, I understood it for the second/third excited states, I wanted to know how it generalizes. $\endgroup$ – Jaswin Aug 4 '14 at 16:46
  • $\begingroup$ Did you mean : How to decompose any irreductible representation of $SO(D-1)$ in irreductible representations of $SO(D-2)$ ? $\endgroup$ – Trimok Aug 5 '14 at 8:21
  • $\begingroup$ No, how to extend it for higher excited states. And how to identify such a mapping uniquely. $\endgroup$ – Jaswin Aug 5 '14 at 16:47
  • $\begingroup$ IMHO, it is the same thing. For instance, $\alpha_{-1}^i$ is in the irreductible fundamental (vectorial) representation of $SO(D-1)$, $(\alpha_{-1}^i \alpha_{-2}^j - \alpha_{-1}^j \alpha_{-2}^i)$ is in the irreductible antisymmetric representation of $SO(D-1)$ and so on. $\endgroup$ – Trimok Aug 6 '14 at 8:14

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