Rank two Cartesian tensors can be decomposed into $L=0,1,2$ spin like things 3x3=1+3+5.

But the second equation below does not transform like a "tensor", it looks more like a vector transform in five dimensional Cartesian coordinate.

$$\mathfrak{D}^{+}(R)V_{i}\mathfrak{D}(R) = \sum_{j=1,2,3}R_{ij}V_{j}\tag{1} $$ (three dimensional vector.)

$$\mathfrak{D}(R)T_{i}^{(2)}\mathfrak{D}^{+}(R) = \sum_{j=-2,-1,0,1,2}\mathfrak{D}_{ij}^{(2)}T_{j}^{(2)}\tag{2}$$ (rank two spherical harmonic tensor.)

$$ \mathfrak{B}^{+}(F)V_{i}\mathfrak{B}(F) = \sum_{j=1,2,3,4,5}F_{ij}V_{j}\tag{3}$$ (a vector transform in 5 dimensional Cartesian space.)

$(2)$ and $(3)$ above looks the same.

My question is can 3d spherical tensors be regarded as high dimensional vectors? Are there some examples?

  • $\begingroup$ Vectors in $\mathbb R^5$ transform under $SO({\bf 5})$, instead all the representations you are considering (for every value of $l=0,1,2,\ldots$) are irreducible representations of $SO({\bf 3})$, how can you think that there is any relation between them? $\endgroup$ Jan 16, 2014 at 9:32
  • $\begingroup$ The $l=2$ representation $D^{(2)}_{mm'}$, up to a transformation of the form $VD^{(2)}V^{-1}$, for a certain fixed matrix $V$, coincides to the representation of $SO(3)$ in the irreducible space of real symmetric traceless $3\times 3$ matrices that, in fact, has dimension $5$. $\endgroup$ Jan 16, 2014 at 9:49

1 Answer 1


Vectors in $\mathbb R^5$ transform under the action of $SO(5)$, instead all the representations you are considering (for every value of $l=0,1,2,…$) are irreducible representations of $SO(3)$, so what you suppose is not possible. However, barring the standard spherical harmonics intepretation, the $l=2$ representation has a precise geometric meaning as the action of $SO(3)$ on an irreducible space of 2nd order real tensors in $\mathbb R^3 \otimes \mathbb R^3$ with dimension $5$. I go to illustrate it.

Since you are considering integer values of the spin, the arising $SU(2)$ representation are also representations of $SO(3)$. More precisely the matrices in the right-hand side of (1) and (2) corresponds to unitary irreducible representations of $SO(3)$: $$U(R) T^{(l)}_m U(R)^\dagger = \sum_{|m'| \leq l} D^{(l)}_{mm'}(R)T_{m'}$$ These matrices work in invariant and irreducible subspace of complex tensor representations of $SO(3)$. The maps $SO(3) \ni R \mapsto D^{(l)}(R)$, varying $l=0,1,2,\ldots$ label all complex unitary irreducible representations of $SO(3)$ (as a byproduct one sees that these representations are necessarily finite dimensional all that is a consequence of Peter-Weyl theorem).

Actually one can say more: all complex irreducible representations of $SO(3)$ are obtained form these unitary irreducible representations via a transformation of the form: $$D^{(l)}(R) \to V D^{(l)}(R) V^{-1}\qquad (1)$$ for suitable non singular $(2l+1) \times (2l+1)$ matrices $V$ independent form $R$.

Let us consider the action of $SO(3)$ on $\mathbb R^3\otimes \mathbb R^3$. If $SO(3) \ni R \to R : \mathbb R \to \mathbb R$ is the fundamental representation, the tensor representation on $\mathbb R^3\otimes \mathbb R^3$ is $SO(3) \ni R \mapsto R \otimes R$. However $\mathbb R^3\otimes \mathbb R^3$ is not irreducible, but it consists of a direct sum of $3$ real irreducible subspaces: $$\mathbb R^3\otimes \mathbb R^3 = \mathbb S_1 \oplus \mathbb S_2 \oplus \mathbb S_3$$ $\mathbb S_1$ is the subspace of scalars: the $3\times 3$ real matrices of the form $sI$, $\mathbb S_2$ is the space of antisymmetric $3\times 3$ real matrices, $\mathbb S_3$ is the space of traceless symmetric $3\times 3$ real matrices. If $M \in \mathbb R^3\otimes \mathbb R^3$, its decomposition in the three spaces above is:

$$M = \frac{1}{3}(tr M) I + \frac{M-M^t}{2} + \left(\frac{M+M^t}{2}- \frac{1}{3}(tr M)I\right)$$

These three spaces are separately irreducible and invariant under $SO(3)$ and transform with corresponding real irreducible representations of $SO(3)$ that I indicate with $SO(3) \ni R \mapsto L^{(i)}(R)$.

NOTE: If $M_{ij}= -M_{ji}$, $t_k := \sum_{i,j=1}^3\epsilon_{kij}M_{ij}$ transform as a real $3$-vector under $SO(3)$ (the situation would be different if $SO(3)$ were replaced by $O(3)$). So actually $\mathbb S_2$ is a vector representation of $SO(3)$. In components of elements of $\mathbb S_2$, the representation $L^{(2)}$ is nothing but the fundamental representation $SO(3) \ni R \mapsto R$.

If we complexificate these spaces, $\mathbb S_i +i \mathbb S_i$, every $SO(3) \ni R \mapsto L^{(i)}(R)$ becomes a complex irreducible representation of $SO(3)$. So:

up to a transformation like in (1) for some fixed matrix $V^{(i)}$, each representation $L^{(i)}$ must coincide with one of the $D^{(l)}$ ones.

The first irreducible space $\mathbb S_1$ has real dimension $1$ so that $\mathbb S_1 + i \mathbb S_1$ has complex dimension $1$. It means that $L^{(1)}= D^{(0)}$. It is evident that $V^{(0)}=1$.

The second irreducible space $\mathbb S_2$ has real dimension $3$ so that $\mathbb S_2 + i \mathbb S_2$ has complex dimension $3$. It means that $L^{(2)}$ coincides to $D^{(1)}$ up to a similitude transformation as in (1). The form of $V^{(1)}$ can be found in the literature, is the matrix used to pass from spherical to Cartesian components.

The third irreducible space $\mathbb S_3$ has real dimension $5$ so that $\mathbb S_3 + i \mathbb S_3$ has complex dimension $5$. It means that $L^{(3)}$ coincides to $D^{(2)}$ up to a similitude transformation as in (1). I do not know the explicit expression of $V^{(2)}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.