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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?

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  • $\begingroup$ 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

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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 $V^{(2)}$.

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