# Irreducible tensors concept

This might be a little naive question, but I am having difficulty grasping the concept of irreducible tensors. Particularly, why do we decompose tensors into symmetric and anti-symmetric parts? I have not found a justification for this in my readings and would be happy to gain some intuition here.

You can decompose a rank two tensor $X_{ab}$ into three parts:

$$X_{ab} = X_{[ab]} + (1/n)\delta_{ab}\delta^{cd}X_{cd} + (X_{(ab)}-1/n \delta_{ab}\delta^{cd}X_{cd})$$

The first term is the antisymmetric part (the square brackets denote antisymmetrization). The second term is the trace, and the last term is the trace free symmetric part (the round brackets denote symmetrization). n is the dimension of the vector space.

Now under, say, a rotation $X_{ab}$ is mapped to $\hat{X}_{ab}=R_{a}^{c}R_{b}^{d}X_{cd}$ where $R$ is the rotation matrix. The important thing is that, acting on a generic $X_{ab}$, this rotation will, for example, take symmetric trace free tensors to symmetric trace free tensors etc. So the rotations aren't "mixing" up the whole space of rank 2 tensors, they're keeping certain subspaces intact.

It is in this sense that rotations acting on rank 2 tensors are reducible. It's almost like separate group actions are taking place, the antisymmetric tensors are moving around between themselves, the traceless symmetrics are doing the same. But none of these guys are getting rotated into members "of the other team".

If, however, you look at what the rotations are doing to just, say the symmetric trace free tensors, they're churning them around amongst themselves, but they're not leaving any subspace of them intact. So in this sense, the action of the rotations on the symmetric traceless rank 2 tensors is "irreducible". Ditto for the other subspaces.

• Can you convince me that the symmetric part alone (without subtracting off the trace) is reducible? Is it as simple as saying "there is a subspace $\delta_{ab} Tr(X)$ of $X_{(ab)}$ which transforms to itself under rotations"? Jan 17 '12 at 21:37
• @cduston Yes, the orthogonality property of the rotation matrix means that their action on $\delta_{ab}$ preserves $\delta_{ab}$ and hence the one dimensional subspace of multiples of $\delta_{ab}$ is preserved. Jan 18 '12 at 9:05
• @twistor59: How would you mathematically prove that the traceless symmetric part is irreducible?
– user7757
Feb 21 '13 at 4:01
• @ramanujan_dirac Just thinking about SO3, probably the easiest way is to re-examine the argument that led to the (antisymm+symm_traceless+trace) decomposition and try to apply it again. We started with a 9 dim space of rank 2 tensors, and found the 1, 4 and 5 dimensional subspaces by using the invariant tensors $\delta_{ab}$ and $\epsilon_{abc}$. Given that these are the only invariant tensors, then if we wanted an inv't subspace of the 5dim space of symm traceless tensors, we'd have to get it by applying these tensors to the symm. traceless tensors, which can only give trivial answers. Feb 22 '13 at 13:24
• Is there a reason why one divides by $n$ in the trace part? Sep 28 '14 at 20:13

Physicists are always interested in what properties of a physical system are invariant under symmetries. If it's tricky to see the symmetry then they'll rearrange the system to make the symmetry more obvious.

For example, consider a covariant rank two tensor like $T^{ab}$. In general the components of this tensor will change if the tensor is rotated in 3D. It's hard to see what might be invariant under rotation.

Now consider a symmetric tensor $S^{ab}$. Again, the components of this tensor will change when it is rotated. However, the property of being symmetric is preserved. So from a physicist's point of view this is interesting. Similarly, an antisymmetric tensor $A^{ab}$ remains antisymmetric when rotated.

So we have nice properties that are preserved for these special classes of tensor, but not for $T^{ab}$. But as you probably know, any tensor $T^{ab}$ can be written as a sum of symmetric and antisymmetric parts, $T^{ab}=S^{ab}+A^{ab}$. So now we know that $T^{ab}$ can be written as a sum of two parts, each of which behaves more simply when rotated. This simplifies the analysis of what happens to $T^{ab}$ when it is rotated.

Once we've done that once the obvious question is "can we do this again"? It'd be nice if we could break $T^{ab}$ into more pieces that behave as simply as possible under rotations.

There's another class of tensor that behaves nicely under rotation: the diagonal tensors of the form $\alpha\delta_{ab}$. Under rotations, they simply map to themselves. That's as simple as it gets. There's also a kind of converse class: the symmetric tensors of trace zero. These keep their trace of zero when they are rotated. But here's the nice bit: any symmetric tensor can be written as the sum of a diagonal tensor and a trace-zero symmetric tensor. So now we've broken down $T^{ab}$ into three pieces, each of which has a nice invariance property with respect to rotations.

Can we keep going? Well it turns out for covariant rank two tensors in 3D this is as far as we can go. If we try to break up the antisymmetric matrices, say, as the sum of two pieces from a pair of complementary classes, we'll always find that some rotation will move an element of one class into the other. So three classes is as far as we can go. The elements of these classes are the irreducible tensors.

The space of covariant rank two tensors has dimension 9. It is the sum of three spaces: the multiples of the identity (a space of dimension 1), the antisymmetric tensors (dimension 3) and the symmetric trace-zero tensors (dimension 5). 1+3+5=9.

For tensors of different rank, and in different dimensions, you get different irreducible tensors.

To get the intuition, it is better to become familiar with Wigner–Eckart theorem and spherical tensors which generalize these ideas.

For intuition, consider the Inertia Tensor, which is symmetric and hence has six components. The scalar $|0,0\rangle$ part (~trace) is just the mass of the object: clearly rotationally invariant, and hence, physically significant. That leaves us with the five $|2, m\rangle$ parts. If you rotate to the principle axes, the $|2,\pm 1\rangle$ are zero, and you are left with $m=0$ and $|m| = 2$. The latter is non-zero if the object is not cylindrically symmetric, and the former tells you if the object is oblate/prolate-- and zero for a sphere. So: each irreducible component is directly interpretable as a useful geometric property (unlike, say, the dyadic $\bf \hat x \hat y$).

Conversely, consider an antisymmetric rank-2 tensor like $\vec r\vec p - \vec p \vec r$, which has 3 non zero components. That looks like a vector (under rotations) and that vector is the angular momentum---hence you have isolated a significant geometric quantity by considering the irreducible part of the 9-component tensor.

Note that higher order tensors have more complicated symmetries--all related to rotationally invariant subspace. A rank 3 tensor has the following:

10 Dimensional Symmetric:

$$S = \frac 1 6 [T_{ijk} +T_{ikj} +T_{jik} +T_{jki} +T_{kij} +T_{kji}]$$

This can be reduced further by taking the (vector) trace, which transforms like a vector:

$$\nu^{(0)} = S_{iij} = S_{iji}=S_{jig}$$

and subtracting that from $S$ to form a 7 dimensional natural (e.g. trace-free) symmetric rank-3 tensor that transforms like the seven $Y_3^m(\theta, \phi)$:

$$T^{(3)} = S_{ijk} -\frac 3 5 \delta_{ij}\nu^{(0)}_k$$

A 1 Dimensional Antisymmetric part (proportional to $\epsilon_{ijk}$):

$$A = \frac 1 6 [T_{ijk} -T_{ikj} -T_{jik} +T_{jki} +T_{kij} -T_{kji}]$$

The remains 16 degree-of-freedom are split between two 8 dimensional subspaces with mixed symmetry:

$$M^1= \frac 1 3 [ T_{ijk} +T_{jik} -T_{kji} -T_{kij}]$$

$$M^2= \frac 1 3 [T_{ijk} +T_{kji} -T_{jik} -T_{jki} ]$$

Each of these can also have a vector trace subtracted (each of which transforms like a vector), making two 3 dimensional subspaces:

$$\nu_i^{(\lambda)} = M^{\lambda}_{ijj}$$

which can be subtracted from the $M$ to give two 5 dimensional natural form (trace-free) rank-2 tensors that transform like the five $Y_3^m(\theta, \phi)$

$$M^{\lambda}_{ijk}-\frac 3 2 \delta_{ij}\nu^{(\lambda)}$$

Rank 4 tensors are more complicated. In summary, the 81 degrees of freedom are divided into invariant subspaces as follows:

$${\bf 3}\otimes{\bf 3}\otimes{\bf 3}\otimes{\bf 3} = {\bf 15} \oplus 3\cdot{\bf 15}\oplus 2\cdot{\bf 6}+3\cdot{\bf 3}$$

where the bold number refer to the dimension of the subspace and the regular font is their multiplicity. A suitable subtraction of traces leads to a further break up into subspaces that transform as natural form (trace free) rank $N$ tensors with multiplicity $2N+1$:

$${\bf 3}\otimes{\bf 3}\otimes{\bf 3}\otimes{\bf 3} = {\bf 9+5+1} \oplus 3\cdot{\bf 7+5+3}\oplus 2\cdot{\bf 5+1}+3\cdot{\bf 3}$$

Note that there is no antisymmetric (scalar) combination. For example, the symmetric combinations is:

$$S=\frac 1 {24}[ T_{ijkl}+T_{ijlk}+T_{ikjl}+T_{iklj}+T_{iljk}+T_{ilkj}+T_{jikl}+T_{jilk}+T_{jkil}+T_{jkli}+T_{jlik}+T_{jlki}+T_{kijl}+T_{kilj}+T_{kjil}+T_{kjli}+T_{klij}+T_{klji}+T_{lijk}+T_{likj}+T_{ljik}+T_{ljki}+T_{lkij}+T_{lkji} ]$$

The scalar subspace is:

$$\omega = S_{iijj}$$

The trace free rank 2 tensor subspace is:

$$S^{(2)} = S_{ijkk} - \frac 1 3 \omega\delta{ij}$$

Subtracting this gives the natural form rank-4 tensor:

$$\Phi_{ijkl} = S_{ijkl}-\frac 6 7 \delta_{ij}S^{(2)}_{kl}-\frac 1 5 \omega \delta_{ij}\delta_{kl}$$.

After that, it just gets tedious.