$SO(3)$ 5×5 irrep: what does it look like explicitly? (Zee IV.1)

Question

BACKGROUND.

So my understanding is this: the $3\times3$ rep of $SO(3)$ are matrices $R$ that rotate a vector $x$, for example:

$$ x \rightarrow Rx = \left(\begin{matrix} 1&0&0 \\ 0&C&S \\ 0&-S&C \end{matrix}\right) \left(\begin{matrix}{x\\y\\z}\end{matrix}\right) ,$$

with $C=\cos\theta$ and $S=\sin\theta$. Instead of acting on vectors $x$, $R$ can act on a 2-rank $3\times 3 $ tensor $t$:

$$ t\rightarrow RtR^{-1},~~ t=\left(\begin{matrix} a&b&c \\ d&e&f \\ g&h&i \end{matrix}\right) .$$

Or we can go from a $3\times3$ rep to a $9\times9$ and $9\times1$ rep:

$$ T \rightarrow (R\otimes R) T$$

with

$$~~(R\otimes R) = \left(\begin{matrix} 1&0&0&0&0&0&0&0&0\\ 0&C&S&0&0&0&0&0&0\\ 0&-S&C&0&0&0&0&0&0\\ 0&0&0&C&0&0&S&0&0\\ 0&0&0&0&C^2&C S&0&C S&S^2\\ 0&0&0&0&-C S&C^2&0&-S^2&C S\\ 0&0&0&-S&0&0&C&0&0\\ 0&0&0&0&-C S&-S^2&0&C^2&C S\\ 0&0&0&0&S^2&-C S&0&-C S&C^2\\ \end{matrix}\right) ,~~T=\left(\begin{matrix} a\\b\\c\\d\\e\\f\\g\\h\\i \end{matrix}\right). $$

Next, this 9 rep decomposes into $1\oplus 3\oplus 5$ so we should be able to block-diagonalize it into 1×1, 3×3 and 5×5 blocks. To do this, Zee looks at the objects being transformed, and groups them into diagonal, asymmetric and traceless symmetric:

$$ \left(\begin{matrix} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{matrix}\right)\\ \left(\begin{matrix} 0& 0& 0\\ 0& 0& 1\\ 0& -1& 0 \end{matrix}\right) \left(\begin{matrix} 0& 0& -1\\ 0& 0& 0\\ 1& 0& 0 \end{matrix}\right) \left(\begin{matrix} 0& 1& 0\\-1& 0& 0\\ 0& 0& 0 \end{matrix}\right)\\ \left(\begin{matrix} 0& 0& 0\\ 0& 0& 1\\ 0& 1& 0 \end{matrix}\right) \left(\begin{matrix} 0& 0& 1\\ 0& 0& 0\\ 1& 0& 0 \end{matrix}\right) \left(\begin{matrix} 0& 1& 0\\ 1& 0& 0\\ 0& 0& 0 \end{matrix}\right) \left(\begin{matrix} 1& 0& 0\\ 0& 1& 0\\ 0& 0& -2 \end{matrix}\right) \left(\begin{matrix} 1& 0& 0\\ 0& -1& 0\\ 0& 0& 0 \end{matrix}\right) $$

The same tensors in the 9×1 form:

$$ \left(\begin{matrix} 1\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0\\ 0\\ 1 \end{matrix}\right) ~~~~~ \left(\begin{matrix} 0\\ 0\\ 0\\ 0\\ 0\\ 1\\ 0\\ -1\\ 0 \end{matrix}\right) \left(\begin{matrix} 0\\ 0\\ -1\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0 \end{matrix}\right) \left(\begin{matrix} 0\\ 1\\ 0\\-1\\ 0\\ 0\\ 0\\ 0\\ 0 \end{matrix}\right) ~~~~~ \left(\begin{matrix} 0\\ 0\\ 0\\ 0\\ 0\\ 1\\ 0\\ 1\\ 0 \end{matrix}\right) \left(\begin{matrix} 0\\ 0\\ 1\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0 \end{matrix}\right) \left(\begin{matrix} 0\\ 1\\ 0\\ 1\\ 0\\ 0\\ 0\\ 0\\ 0 \end{matrix}\right) \left(\begin{matrix} 1\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0\\ 0\\ -2 \end{matrix}\right) \left(\begin{matrix} 1\\ 0\\ 0\\ 0\\ -1\\ 0\\ 0\\ 0\\ 0 \end{matrix}\right) $$

The vectors above are mutually orthogonal. They provide a complete basis. Normalize them, and make them columns in a matrix $S$. Then I expected the following to be block diagonal (1,3,5):

$$ S^{-1}(R\otimes R)S $$

However, I get a 2-block and a 7-block.

QUESTION:

1) Is something wrong in my reasoning? (I'm newish to group theory.)

2) Is there a standard form for the 5×5 rep? Even if I correctly got a 5-block above, it might be some scrambled, non-standard version.

Answer 1

OK, I'll transcribe the homework problem 4 I refer to in the comments, adapting it to your specific problem, as a hint, since there is no space in serial comments... I have never laid eyes on Zee's text, however, and my discussion below does not follow Tony's symmetry driven shortcut.

Working in Lie algebra generator space is easier, in my mind, and SO(3) is so neat, in that you may ignore $j_-$, as its effects follow by hermitean conjugation of $j_+$. Only $[j_0,j_+]= j_+$ and $[j_+, j_-]=j_0$ need checking. For the singlet, triplet, and quintet representations we know from Wikipedia that, respectively, $$ \mathbf{1}: \qquad j_0=0=j_+ , \\ \mathbf{3}: \qquad j_0=\begin{pmatrix} 1&0&0 \\ 0&0&0 \\ 0&0&-1 \end{pmatrix} \qquad j_+=\begin{pmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{pmatrix} , \\ \mathbf{5}: \qquad j_0= \begin{pmatrix} 2&0&0&0&0\\ 0&1&0&0&0\\ 0&0&0&0&0\\0&0&0&-1&0\\0&0&0&0&-2\\ \end{pmatrix} \qquad j_+= \begin{pmatrix} 0&\sqrt{2}&0&0&0\\ 0&0&\sqrt{3}&0&0\\ 0&0&0&\sqrt{3}&0\\0&0&0&0&\sqrt{2}\\0&0&0&0&0\\ \end{pmatrix} . $$ The reducible coproduct of generators on $\mathbf{3}\otimes \mathbf{3}$ (necessarily obeying the same Lie algebra!) is $$ \mathbf{9}:\qquad \Delta(j_0) =1\!\!1 \otimes j_0+ j_0\otimes 1\!\! 1= \operatorname {diag} (2,1,0,1,0,-1,0,-1,-2), \qquad \\ \Delta(j_+)=1\!\!1 \otimes j_+ + j_+ \otimes 1\!\! 1 = \begin{pmatrix} 0&1&0&1&0&0&0&0&0\\ 0&0&1&0&1&0&0&0&0\\ 0&0&0&0&0&1&0&0&0\\ 0&0&0&0&1&0&1&0&0\\ 0&0&0&0&0&1&0&1&0\\ 0&0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&0&0&0\\ \end{pmatrix} . $$

It Clebsch-reduces to $\mathbf{1}\oplus \mathbf{3}\oplus \mathbf{5}$ blocks, by a similarity transformation like your S, the orthogonal Clebsch matrix. It is very sparse, and its components ($1/\sqrt 2, 1/\sqrt 6, \sqrt{2/3},...$) should remind you of the C-G coefficients in standard texts.

You know it is sparse, since it basically operates in the space mixing all three 0 components, and both 1 components, and -1 components of $\Delta (j_0)$. That very same matrix will perforce block reduce your rotation matrix, like any rotation matrix, since these should be but exponentials of arbitrary linear combinations of $j_0,j_+,j_-$, all reduced here. In the end, you might do a relabeling transformation of rows and columns for elegance, i.e. have the blocks visibly isolated.

Where is the singlet hiding? Note the null vector of $\Delta (j_0)$ and $\Delta (j_+)$ is the transpose of (0,0,1,0,-1,0,1,0,0), so you may project it out to reduce the dimension of the space to 8.

But I will not work out C for you...

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Here is a hint of one of its factors--the one that moves the singlet on the central position, 5, of the 9-vector, and as result the two coproducts looked at have zeros in their 5-th rows and columns and can thereby by ignored: you simply chuck out 5th components of everything and you reduced your problem to an 8×8 one, which you must now reduce to 3×3 and 5×5 blocks. The partial C matrix that achieves that is the orthogonal one $$ c_0= \begin{pmatrix} 1/\sqrt 2 &1/\sqrt 3&1/\sqrt 6 \\ 0&-1/\sqrt 3&\sqrt{2/3} \\ -1/\sqrt 2&1/\sqrt 3&1/\sqrt 6 \end{pmatrix} $$ in the 3,5,7 subspace and with zeros implied in the 1,2,4,6,8,9 entries. This moves the null vector found above for the singlet to the central position 5, so $c_0^T \Delta (j_+) c_0$ has vanishing 5th row and column, and, of course, so does $c_0^T \Delta (j_0) c_0$, which was actually left invariant. You may now delete anything connected to the 5th entry, and handle the 8×8 matrices.