While studying representations of Simple Lie Algebras the procedure is quite straight forward.

  1. Find the Cartan Sub-Algebra
  2. Label states with the corresponding eigenvalues
  3. Use the remaining generators to construct SU(2) blocks and use them as raising/lowering operators.

This leds to the notion of roots and in general it's straight forward to find the corresponding root structure or decomposition for the classical groups.

Having all this information (root system, Cartan Matrix, and so on) one can in principle use them to construct new representations of the algebra. My question is the following, given all the information regarding the roots of a classical Lie algebra how can I build the weights of representation with fixed symmetry properties. For example say you are in SP(N) and you know all the root system and Cartan Matrix, how do you build the Weights corresponding to the antisymmetric representation?. Is there a systematic way to do so?.

Any bibliography will be appreciated!


It's actually very straightforward. Let's go through a simple example. Suppose $V$ is a $3$-dimensional representation of your Lie algebra, and suppose that $V$ has the basis $$v_1, v_2, v_3, $$ where $v_1$, $v_2$, $v_3$ each lie in a weight space, and the corresponding weights are $w_1$, $w_2$, $w_3$ respectively. (So for any $H$ in the Cartan subalgebra, we have $H(v_i) = w_i(H) v_i$ for $i \in \{ 1, 2, 3 \}$.)

Then the symmetric tensor product $Sym^2 (V)$ is a six-dimensional representation, with basis $$ v_1 \otimes v_1, v_2 \otimes v_2, v_3 \otimes v_3, \tfrac 1 2 (v_1 \otimes v_2 + v_2 \otimes v_1), \tfrac 1 2 (v_2 \otimes v_3 + v_3 \otimes v_2) , \tfrac 1 2 (v_3 \otimes v_1 + v_1 \otimes v_3) $$ These basis vectors each lie in a weight space, and the corresponding weights are $2w_1, 2w_2, 2w_3, w_1 + w_2, w_2 + w_3, w_3 + w_1$ respectively.

[In practice, you may find that some of the weights coincide. If this happens, then it means you have degenerate weight spaces (i.e. a weight space corresponding to a given weight has dimension greater than one) - it's not a problem!]

Returning to our example, the antisymmetric tensor product $\wedge^2 (V)$ is a three-dimensional representation, with basis $$ \tfrac 1 2 (v_1 \otimes v_2 - v_2 \otimes v_1), \tfrac 1 2 (v_2 \otimes v_3 - v_3 \otimes v_2) , \tfrac 1 2 (v_3 \otimes v_1 - v_1 \otimes v_3) $$ and these basis vectors each lie in a weight space, and the corresponding weights are $w_1 + w_2, w_2 + w_3, w_3 + w_1$ respectively.

This method generalises to every example you'll ever encounter, and it is as systematic a method as you can hope for!

  • $\begingroup$ So for SU(2) taking as a basis $J_3$ and $J_{\pm}$ I form the $\Lambda^2(V)$ antisymmetric representation by making the corresponding antisymmetric tensor product?. The cartan subalgebra of this irrep corresponds to $J_3 \otimes J_3$ or there are any new elements?. If instead of taking the (anti)symmetric tensor product I would make a reducible representation that I can decompose with Clebsch-Gordan like coefficients? $\endgroup$ – Jasimud Jun 24 '18 at 15:47
  • $\begingroup$ @Jasimud Are you confusing the basis of the representation space $V$ with the basis of the Lie algebra itself? Let's do the example where $V$ is the fundamental, two-dimensional representation (think of a spin-half particle). The states are $v_{\uparrow}, v_{\downarrow}$, where $J_3 (v_{\uparrow} ) = + \tfrac 1 2 v_{\uparrow}$ and $J_3 (v_{\downarrow}) = - \tfrac 1 2 v_{\downarrow}$. The antisymmetric tensor product $\wedge^2(V)$ is a one-dimensional representation, spanned by $\tfrac 1 {\sqrt{2}} \left( v_{\uparrow} \otimes v_{\downarrow} - v_{\downarrow} \otimes v_{\uparrow} \right)$. $\endgroup$ – Kenny Wong Jun 24 '18 at 16:14
  • $\begingroup$ … and the action of the Cartan generator $J_3$ is $J_3 \left( \tfrac 1 {\sqrt{2}} \left( v_{\uparrow} \otimes v_{\downarrow} - v_{\downarrow} \otimes v_{\uparrow} \right) \right) = 0$. So $\wedge^2 (V)$ is equivalent to a spin-zero representation. [Similarly, you can show that ${\rm Sym}^2 (V)$ is equivalent to a direct sum of a spin-1 representation and a spin-0 representation.] Finally, I'll remark that the Cartan subalgebra is always spanned by $J_3$. This statement does not depend on your choice of representation! $\endgroup$ – Kenny Wong Jun 24 '18 at 16:16
  • $\begingroup$ But to reiterate, if $\mathfrak g = su(2)$ is your Lie algebra, then there are infinitely many possible choices of $V$! I've just done an example where $V$ is two-dimensional. I could just as well have done an example where $V$ is three-dimensional (for example, representing the state space of a spin-1 particle). And so on... $\endgroup$ – Kenny Wong Jun 24 '18 at 16:21
  • $\begingroup$ Yes, I was confused about both basis, but once I started working out the SU(2) example I realized it was wrong. Now, in general for the classical groups one has the information about the root system and the corresponding dynkin diagram but not of the particular states. Is there a way to construct representations in a way similar to this one but with the Dynkin diagrams? $\endgroup$ – Jasimud Jun 24 '18 at 16:53

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