# Why does a space of symmetric traceless tensors form an irreducible representation of $SO(3)$?

I have asked a similar question on Physics StackExchange, but did not get an answer. In the chapter IV.1 "Reducible or Irreducible?" of Zee's Group Theory book (p. 188-), the author breaks a 2nd rank tensor $$T^{ij}$$ into invariant subspaces with respect to the action of $$\mathrm{SO(3)}$$ group. The tensor $$T^{ij}$$ breaks into a five-dimensional (symmetric traceless), three-dimensional (antisymmetric) and one-dimensional invariant subspaces. Zee claims this five-dimensional space to be irreducible, i.e. it is implied that it does not have non-trivial invariant subspaces. Unfortunately, there is no proof.

Can someone explain to me why symmetric traceless tensors form an irreducible representation $$\mathrm{SO(3)}$$? References to relevant literature would be helpful as well.

Denote $$S^{Tr}$$ to be the space of symmetric traceless tensors. Let $$D\subset S^{Tr}$$ be a non-trivial subspace of $$S^{Tr}$$ that is irreducible with respect to $$SO(3)$$. We will prove $$D = S^{Tr}$$.

$$D$$ is closed with respect to $$SO(3)$$ transformations. Specifically: $$\forall \; T^{ab} \in D, \; R_{x}^{a} \in SO(3)$$, we have $$R_{x}^{a}T^{xy}R_{y}^{b}\in D$$.

If $$T^{ab}\in D$$ then $$T^{ab}\in S^{Tr}$$. Also, if $$\;T_{1}^{ab}, T_{2}^{ab}\dots T_{n}^{ab} \in D$$ then $$\text{span}\{T_{1}^{ab}, T_{2}^{ab}\dots T_{n}^{ab}\} \subset D$$.

We must show that $$\forall T\in S^{Tr}$$ with $$T\neq 0\;$$, $$\exists R_{1},R_{2} \dots R_{n}\in SO(3)$$ such that $$\text{span}\{R_{1x}^{a}T^{xy}R_{1y}^{b}, R_{2x}^{a}T^{xy}R_{2y}^{b}\dots R_{nx}^{a}T^{xy}R_{ny}^{b}\} = S^{Tr}$$. In short, any non-zero tensor of $$S^{Tr}$$ can generate a basis for $$S^{Tr}$$ under the action of $$SO(3)$$, thus proving the ireductibility .

From now it's a problem of linear algebra. A symmetric traceless tensor $$T^{ab}$$ can be bought in the form $$\left(\begin{smallmatrix} a & 0 & 0\\ 0 & b & 0\\ 0 & 0 & -(a+b) \end{smallmatrix}\right)$$ by a suitable rotation matix (symetric matrices are diagnoalizable). With the rotation matrix $$\left(\begin{smallmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{smallmatrix}\right)$$ we bring it to the form $$\left(\begin{smallmatrix} b & 0 & 0\\ 0 & -(a+b) & 0\\ 0 & 0 & a \end{smallmatrix}\right)$$, with which we form a basis for all traceless diagonal tensors. By linear combinations we obtain the matrix $$\left(\begin{smallmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 0 \end{smallmatrix}\right)$$, which can be transformed via suitable rotation matrices in $$\left(\begin{smallmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{smallmatrix}\right)$$,$$\left(\begin{smallmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 & 0 \end{smallmatrix}\right)$$,$$\left(\begin{smallmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{smallmatrix}\right)$$.

Starting with any traceceless symetric tensor, and applying only $$SO(3)$$ transformations and linear combinations, we can form a basis for $$S^{Tr}$$, thus obtain any possible traceceless symmetric tensor. Thus $$S^{Tr}$$ is irreducible with respect to $$SO(3)$$.

• Wonderful! It is a nice pedestrian way of demonstrating the irreducibility. Also, quite straightforward to expand it to $SO(N)$. Oct 6, 2021 at 16:34

The "theory of highest-weight vectors" definitively answers many questions about finite-dimensional representations of (reductive, e.g....) compact real Lie groups, up to some fooling around with connectedness.

Namely, the complexified Lie algebra of the real Lie algebra of a compact real Lie group is (provably) reductive (often semi-simple), so has a Cartan ("diagonalizable") subalgebra $$\mathfrak h$$, and (with a choice) positive and negative root spaces $$\mathfrak g_\alpha$$ for roots $$\alpha$$.

In the finite-dimensional situation, a representation $$V$$ decomposes into simultaneous eigenspaces ("weight spaces") $$V_\lambda$$ for $$\mathfrak h$$ with $$\lambda$$ a linear functional ("weight") on $$\mathfrak h$$, and the action of the root spaces moves things around in an intelligible fashion. In particular, one proves that for irreducible $$V$$ there is a unique "weight" $$\lambda$$ such that $$V_\lambda$$ (is non-zero and) is annihilated by the root-space $$\mathfrak g_\alpha$$ for all positive roots $$\alpha$$, and this $$V_\lambda$$ is one-dimensional.

And conversely.

So in small situations we can look for highest-weight vectors, and if we find exactly one (up to scalar multiples), then the repn is irreducible.

• I appreciate the answer, but I am affraid it is way above my level at this point. Roots and weights stuff is still more than hundred pages away in the book. I may return to it when I learn more Oct 5, 2021 at 20:55
• @Pavlo.B., ah, well, I do think this is the way to understand the example you mention. Maybe this can be one of several motivations to understand roots-and-weights... which otherwise might be perceived as "just abstract things..." :) Oct 5, 2021 at 21:15
• @Pavlo.B. On a first encounter in quantum mechanics the roots are the ladder operators and the weights are the eigenvalues of $L_z$ (scaled by a multiple of h-bar IIRC). True, when the Lie algebra becomes more complicated (think $SU(5)$ and friends), so does the definition of a weight. But for $SO(3)$ (which is the same thing as $SU(2)$ or $SL_2$ after the Lie algebra is complexified), this suffices. Oct 6, 2021 at 12:28