# Irreducible tensors concept

+ 8 like - 0 dislike
1202 views

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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user yayu

retagged Mar 25, 2014

Ireeducible with respect to wich group representation?

@dilaton Where's the stack exchange links?

@physicsnewbie In the question.

+ 9 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user twistor59
answered Dec 13, 2011 by (2,500 points)
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"?

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user levitopher
@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.

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user twistor59
@twistor59: How would you mathematically prove that the traceless symmetric part is irreducible?

This post imported from StackExchange Physics at 2014-03-22 17:17 (UCT), posted by SE-user ramanujan_dirac
@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.

This post imported from StackExchange Physics at 2014-03-22 17:18 (UCT), posted by SE-user twistor59
+ 6 like - 0 dislike

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 diagonal tensors (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.

This post imported from StackExchange Physics at 2014-03-22 17:18 (UCT), posted by SE-user Dan Piponi
answered Dec 13, 2011 by (100 points)
+ 3 like - 0 dislike

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

This post imported from StackExchange Physics at 2014-03-22 17:18 (UCT), posted by SE-user Misha
answered Dec 13, 2011 by (40 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.