What are the eigenvalues of generators in the irreducible representations of the Lie algebra of $SO(N)$ (and its doublecover)?

For me, the naive generators of $SO(N)$'s fundamental representation are the $1/2 N (N-1)$ antisymmetric matrices with a single $i$ and $-i$ for elements. These have eigenvalues $1$ and $-1$ and $N-2$ zeros. Perhaps I should stipulate that generators in non-fundamental representations should transform under the adjoint representation specified through the fundamental generators' transformation properties - this will help keep the generators fixed up to a similarity transformation, which will preserve the eigenvalues.

My background in representation theory comes largely from physics. I'm not so well versed in roots, weights, and Dynkin diagrams as I should be.

To give a sense of how I see things: If I were asked how to find the eigenvalues of the generators of $SU(2)$ in a $d$-dimensional irreducible representation, I would note that irreps come from taking $d-1$ spin-$1/2$ particles, aligning them in the up direction, and acting the lowering operator on them until they are all aligned in the down direction - this would give an irrep of dimension $d$. The generators of a fixed $SU(2)$ irrep are all unitarily similar, which follows from the generators transforming as "vector" operators and that any $3d$ vector of unit length can be rotated into another, so they have the same eigenvalues. In particular, $S^z$ clearly has eigenvalues stretching from $d/2$ to $-d/2$ in unit steps, so all three generators have these for eigenvalues. I'm interested in the corresponding eigenvalues for general irreps of $SO(N)$.


1 Answer 1


So first a few naming points. Generators of a Lie group are nothing more than a basis of the Lie algebra, probably a nicely chosen one. "Generators of a representation" are simply interpreting those basis vectors as acting on those given representation. Notice this isn't really changing which vectors they are so these are the same "generators". Equally talking about "transforming via the adjoint representation" doesn't make sense here. Clearly the group acts on the basis via the adjoint representation but it would act on any subset similarly so this doesn't really help us fix our particular basis

For a complex or split Lie algebra I would choose root vectors and make everything really simple - eigenvalues would be all $0$ on root vectors and given by the weights on elements of the Cartan subalgebra. So we would just need to pick a basis of the CSA and apply the weights to it. Indeed the action of the the root vectors would be moving between all these eigenspaces (weight spaces that is).

For a compact Lie algebra the story is slightly different however. In the complex/split story the roots live in the (dual of) the Cartan subalgebra. In the compact case they live in $i$ times that. Similarly the root spaces are no longer present in the compact Lie algebra (although they still exist in its complexification). Indeed there are no nilpotent elements at all.

Lets set up our notation. Let $\mathfrak{t}$ be a maximal torus in $\mathfrak{so}(n)$. If you want to think of this as a matrix Lie algebra take this to be the set of block diagonal matrices with $\begin{pmatrix}0 & a_i\\-a_i&0\end{pmatrix}$ as blocks. You have multiplied this by $i$ in your example but that is a physics convention and I am going to leave them as honest elements of $\mathfrak{so}(n)$ to avoid confusing myself. It is a fact that every element in a compact Lie algebra is conjugate to one in a given maximal torus and conjugacy does not change the eigenvalues (in any representation either - although this is perhaps less obvious). The upshot is it is enough to know the eigenvalues on the nice basis of $\mathfrak{t}$ that we have. In fact it should be clear that the elements represented by matrices with a $1$ in one slot and a $-1$ in the corresponding slot across the diagonal are all conjugate to each other. So we only really need to understand the eigenvalues of: $$ X = \begin{pmatrix}0 & 1 & 0 &\cdots\\-1&0 & 0 &\cdots \\ 0 & 0 & 0\\ \vdots & \vdots & & \ddots \end{pmatrix}$$

To do this in general we need to compute the weight lattice for a given representation and apply each weight to this element to find its eigenvalues (just like we would for the split case I mentioned except now this applies to all our basis elements not just the ones in $\mathfrak{t}$ - of course the correponding eigenvectors will change for different elements). It would be a bit much to rehearse all that here so I recommend checking out Fulton and Harris's Representation Theory (or Hall's Lie groups, Lie Algebras and Representations) for more information on how to calculate these.

But we can compute some of the straightforward stuff more easily: Let $V$ denote the defining representation of $SO(n)$ (or $\mathfrak{so}(n)$) (what Physicists call the fundamental representation but mathematicians mean something else when they say fundamental representation). Clearly our basis elements all have eigenvalues $i$, $0$ and $-i$ on $V$ although it is important to note that the eigenvectors are all complex. We can define new representations by tensoring together old ones. So lets look at $V \otimes V$. The natural actions on tensor powers look like: $$ g(v\otimes w) = gv\otimes gw$$ $$ X(v\otimes w) = Xv\otimes w + v\otimes Xw$$ For $g \in SO(n)$ and $X \in \mathfrak{so}(n)$. In particular, if $v$ and $w$ are eigenvectors of $X$ with eigenvalues $\lambda$, $\mu$ then $v\otimes w$ is also an eigenvector with eigenvalue $\lambda + \mu$. So the eigenvalues of $X$ on $V \otimes V$ must be $2i,i,0,-i,-2i$. Of course $V \otimes V$ is not irreducible, it breaks up into the alternating part $\bigwedge^2 V$ which turns out to be the adjoint representation in fact, and the symmetric part $S^2V$ which breaks down even further into a 1-dimensional trivial representation (roughly because there is a symmetric bilinear form on $V$ preserved by $SO(n)$) and the "tracefree" symmetric part $S^2_0 V$. We can explore these individually in fact since $X(v\wedge w) = Xv \wedge w + v\wedge Xw$ and so on. We can then see that the eigenvalues on $\bigwedge^2 V$ will be $i,0,-i$. Meanwhile on the trivial rep it is only $0$ naturally and on $S^2_0 V$ we have the full $2i,i,0,-i,-2i$.

For a Lie algebra of type $A$ such as $\mathfrak{su}(n)$ we can find all irreducible representations contained inside tensor powers of the defining representation. But for $\mathfrak{so}(n)$ this is not possible. For example, the spin representations do not arise this way so this will require delving into the theory of weights.

Edit: I've thought about my claim that all (irreducible) representations can be found in tensor powers of the defining and spin representations and realised that it is not too hard to show. Firstly, we note that for irreps $V(\lambda), V(\mu)$ with highest weights $\lambda, \mu$ their tensor product contains the irrep $V(\lambda + \mu)$ (with multiplicity 1 in fact). Secondly, by the theory of highest weight every irrep has highest weight $\sum_{i=1}^r k_i\omega_i$ where $k_i$ are positive integers, $\omega_i$ are the fundamental weights, $r$ is the rank of the Lie algebra/group.

So any irrep is found in some tensor product of the $V(\omega_i)$. Note, a mathematician would call the $V(\omega_i)$ the fundamental representations and they include the defining representation $V(\omega_1)$ and the spin representations $V(\omega_r)$, or $V(\omega_{r-1})$ and $V(\omega_r)$. So now we just need to see that the other fundamental representations can be found in tensor products of these specific ones. In fact, $V(\omega_2) = \bigwedge^{2}V(\omega_1) \leq \bigotimes^{2}V(\omega_1)$ and in general $V(\omega_k) = \bigwedge^{k}V(\omega_1) \leq \bigotimes^{k}V(\omega_1)$ as long as $V(\omega_k)$ is not a spin representation.

  • $\begingroup$ Thank you, this helps set many things straight. The natural action on tensor powers I think captures what I wanted for fixing the basis for the Lie algebra based on the choice for the fundamental, and the choice for the fundamental is also what I had in mind. I hadn't appreciated before (though I probably should have :)) how general it was for the eigenvalues to add on tensor products, which was very nice to see. That really constrains the behavior of the irreps contained within the reducible tensor products representations. $\endgroup$
    – user196574
    Commented Jun 6, 2022 at 23:03
  • $\begingroup$ I'm curious, you noted that tensor products of the defining representation aren't enough to get the irreps. Would one be able to get all the irreps via tensor products of the "smallest" spin representation? $\endgroup$
    – user196574
    Commented Jun 6, 2022 at 23:08
  • $\begingroup$ Yes all representations can be found as sub representations of tensor powers of the defining representation and the spin representation (or the two half-spin representations for $n$ even). This is not an easy fact to show but I believe it is true. More practically you can compute how the tensor powers decompose using the program LiE although you will need to understand weights a bit to follow the notation $\endgroup$
    – Callum
    Commented Jun 7, 2022 at 5:58
  • 1
    $\begingroup$ @user196574 I've added a quick proof that you can find all irreducible representations inside tensor products of the defining and spin representations (I'm pretty sure you need both in general) $\endgroup$
    – Callum
    Commented Jun 10, 2022 at 8:32

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