Explicit Spinor Representations in $SO(3)$ and $SO(4)$

The beautiful physical interpretation of the decomposition of the elasticity stress tensor into it's irreducible representations (mentioned in context below) has inspired the following question on representation theory.

If you interpret a $3x3$ matrix $A$ as a direct product of vectors, $A = A_jB_k$, then breaking a matrix up into a symmetric and anti-symmetric matrix $A = A_s + A_a$ is a decomposition of a 9 component tensor into a sum of a smaller 6 component and 3 component tensor to get $\bf{3\otimes 3 = 6 \oplus 3}$. Apparently I just showed the matrix representation of orthogonal group $O(3)$ is reducible.

This is as far as you can go in $O(3)$, the group of rotations in 3-space, but if you have a unit determinant you can go further and use the trace to remove some redundancy. When we can do this we call it $SO(3)$, and can further reduce the symmetric matrix into a traceless 5-component symmetric matrix and a scalar multiple of the identity (by the trace) to get a $\bf{3\otimes3 = 5\oplus 3 \oplus 1}$, i.e. $A = A_{ts} + A_t + A_a$.

I apparently just described the tensor decomposition of the $SO(3)$ orthogonal group representation into it's irreducible tensor representations just there, but any orthogonal group also has a spinor representation too.

This can be given a geometric interpretation: we decomposed a rank 2 tensor into the sum of a shearing, (a symmetric rank two tensor representing a unit ellipsoid), a rank 1 tensor rotation (a vector representing a rotation) and a rank $0$ tensor (a scaling factor). (pdf page 10)

1) Is there a similar easy decomposition of a 3x3 matrix into it's spinor representations?*

I mean, starting with an explicit matrix can we do some easy process to it, like taking symmetric and anti-symmetric combinations as above, and magically end up with spinor representations and shows how obvious their existence is? Just so that it's not so abstract.

2) How do we give a nice geometric interpretation to the decomposition of $SO(4)$?

In other words, how do we decompose matrices in $SO(4)$ in such a way that gives a geometric interpretation to every term, illustrates why the $SO(4) \cong [SU(2) \times SU(2)]/\mathbb{Z}_2$ should be obvious, and hopefully makes the symmetries of the Riemann curvature tensor look completely obvious?

(I'm really just hoping to derive and compare $SO(3)$ spinors with $SO(4)$ spinors, since they look so different in QM vs QFT, hence two parts to this question.)

• Comment to the question (v2): The 2nd subquestion is a duplicate of math.stackexchange.com/q/3646/11127 – Qmechanic Feb 9 '15 at 20:15
• "If you interpret a 3x3 matrix $A$ as a direct product of vectors, $A=A_j B_k$" does not make sense to me. $A$ would have 9 independent components but i count only six in $A_j B_k$. – Hector Feb 9 '15 at 20:29
• @Qmechanics it actually isn't, that post does not start with an "explicit" matrix and explicitly work out the decomposition, the way you could decompose a matrix with $1, 2, ..., 9$ in the entries, form a symmetric and anti-symmetric version of this matrix, then remove the trace computationally. That answer is far too abstract for me sadly, I hope for some nice physicist intuition on this one :) – bolbteppa Feb 9 '15 at 20:33
• @Hector if you take a direct product of two vectors, each with $3$ components, you get 9 possible combinations, this is the definition of a direct product, furthermore Kaku in his QFT appendix on representations bases the whole discussion on this idea :) – bolbteppa Feb 9 '15 at 20:35
• Comment to the question (v2): What do you mean by the 1st subquestion decomposition of a 3x3 matrix into it's spinor representations? I would have assumed you were taking about the $SU(2)$ representation $\bf{3\otimes3 \cong 5\oplus 3 \oplus 1}$, but it has no spinor representation. – Qmechanic Feb 9 '15 at 20:41

When we say that $3\otimes 3 \approx 5 \oplus 3 \oplus 1$, this means that under an $SO(3)$ rotation, the first five components mix together alone "traceless symmetric tensor", so do the next 3, while the last component "the trace" is invariant.

If you prove that each of these three parts of the sum is indeed "irreducible", this means that this decomposition is UNIQUE! because in any other decomposition you are going to mix some of the "5" components with some of the "3", which would not work because these parts that you took out will definitely mix with the ones you left out (I'm trying to be as colloquial as possible).

We can also give physical arguments why a composite particle of two sub particles each having spin one (boson) cannot possibly be a half integer particle (fermion), which is what would happen if any of the representations being summed on the right hand side was even.

For the second part of your question then the unique decomposition of a tensor in $4D$ is

$4\otimes 4 \approx 9 \oplus \bar{3} \oplus 3 \oplus 1$

which are the same as before, but now the antisymmetric component splits into a dual and anti-self dual irreducible representations. (which from the point of view of the $SO(3)$ subgroup behave like a pseduo-vector and a polar vector)

REPLY TO YOUR COMMENT: Here is where it might get confusing.. If the group itself is a product, then its representation will also be $\otimes$ but you can't convert them into sums because the final answer should always be a product. This is why you see physicists denote a representation of $SU(2)\otimes SU(2)$ by $(n,m)$.. because it ensures that whatever manipulations you do, you will end up with something of the form $(n,m)$ even though this is just $n\otimes m$ in a sense.. Now to give you some examples of how representations of $SO(4)$ are related to those of $SU(2)\otimes SU(2)/Z2$: $4 \longrightarrow (2,2)$ for a four vector which can be thought of as a product representation of two spinors, each one in different part of the $SU(2)$. An antisymmetric 4-tensor: $(4\otimes 4)_A \longrightarrow (3,1)\oplus (1,3)$ and so on

• Thanks a lot, wow maybe the difference is this - in taking tensor product representations we are using sums $\oplus$, but the spinor representation satisfies $SO(4) \cong [SU(2) \times SU(2)]/\mathbb{Z}_2$ so maybe I should be using multiplication to decompose the matrix instead? Should I be factoring the matrix? Maybe we get something like $A = (r_1e^{i\theta})(r_2e^{i0_2}) = (r_1e^{-i\theta})(r_2e^{-i0_2})$ or $A = (r_1e^{i\theta})(r_2e^{-i0_2}) = (r_1e^{-i\theta})(r_2e^{i0_2})$ and this explains why we have to quotient? Or perhaps because the order of the factors shouldn't matter? – bolbteppa Feb 9 '15 at 23:41
• I edited my answer to reply because it was too long – Ali Moh Feb 10 '15 at 0:04
• Hmm I seen something in Morse and Feshbach a long time ago that is finally starting to make sense now, thanks so much! What you said makes total sense, I am still just unable to intuitively see how we naturally arive at $SO(4) \cong [SU(2) \times SU(2)]/\mathbb{Z}_2$ by starting with an explicit matrix $A$? Surely I should be able to factor it into $A = BC$ and try to say $SO(4) \cong [SU(2) \times SU(2)]$, except $A = CB$ forces me to say $SO(4) \cong [SU(2) \times SU(2)]/\mathbb{Z}_2$ or something? There is a notion of projection operators, might explain it, i.e. $B$ & $C$ are projections? – bolbteppa Feb 10 '15 at 0:19
• Okay I got it, finally! :D – bolbteppa Feb 10 '15 at 1:52

1) No.

When we do a decomposition $3\otimes 3\cong 5 \oplus 3 \oplus 1$, we are moving from a composite system of two vector particles $3\otimes 3$ into interpreting it as a whole system whose angular momentum can be in a superposition of states of spin 2, 1, and 0. Note that these values are separated by 1. This is a consequence of the commutators of angular momentum. The eigenvalues of $J^2$, the angular momentum of the total system, are separated by 1.

If the whole system was to have some spinor components, that would mean some eigenvalues of $J^2$ should be half-integer. There are some possible combinations of spaces that when added would give a total of 9 dimensions ($3\otimes 3$). For example, $4 \oplus 3 \oplus 2$. But this would correspond to eigenvalues of $J^2$ of 3/2, 1, and 1/2. They are not separated by 1. Hence, this system would not be equivalent to the $3\otimes 3$ one.

Consider for example the transformation It takes a matrix that transforms as a spin 1 object and returns a $1\oplus 1\oplus 1$ object. Say that the initial matrix belongs to space $P$ and the second matrix belongs to space $Q$. The transformation is linear and in each space the matrices can be considered to be irreducibly decomposed. However, the two spaces cannot be used to describe the same Physical system.

2) See link posted in a comment.

You also wrote "hopefully make[s] the symmetries of the Riemann curvature tensor look completely obvious". The symmetries of the Riemann tensor are not related to SO(4). The Riemann tensor can be decomposed into irreducible representations of the orthogonal group though. But that decomposition, which is valid for metrics of any signature, has nothing to do with the usage of $SO(4)$ as a symmetry of the spacetime metric.