# What does “transform among themselves” mean?

I'm reading a script on atomic physics, and there's a chapter on irreducible tensors. I can't understand the meaning of "transform among themselves" in this context:

An arbitrary rotation of the coordinate frame will transform the tensor $T$ into a tensor $T'$, whose components $T_{ij}'$ are, in the most general case, linear combinations of all the components $T_{ij }$. However, it is always possible to find certain subgroups of the components $T_{ij}$ (formed by linear combinations thereof) that transform among themselves under a rotation. These components are called the irreducible tensor components.

Can you please explain that scientifically and linguistically?

• Crossposted from math.stackexchange.com/q/521734/11127 Oct 10 '13 at 19:25
• Which script are you reading? Oct 10 '13 at 19:33
• Oct 10 '13 at 22:05

The author is most likely considering the following scenario:

1. The tensor components $T_{ij}$ are coordinates in a vector space $V$ consisting of vectors $$T~=~\sum_{ij}T_{ij}~ e^i \otimes e^j$$ with basis elements $e^i \otimes e^j\in V$. [Here the word vector is used in the sense of linear algebra (as opposed to e.g. the sense of a (1,0) contravariant tensor.)]

2. A group $G$ acts on the vector space $V$. In other words, there exists a group action $G\times V\to V$; and $V$ is a group representation of $G$.

3. The vector space $V$ has a $G$-invariant subspace $W\subseteq V$. In other words, the group action can be restricted to a group action $G\times W\to W$ on the linear subspace $W$; and $W$ is a group representation of $G$. We would then say that the tensors in the linear subspace $W$ transform among themselves under the action of the group $G$.

The author is being imprecise for at least two reasons:

1. He should for clarity write subspace rather than subgroup. [Of course a linear subspace $(W,+)$ is a subgroup of the group $(V,+)$, but no-one (except the author) states it that way.]

2. There is no reason that the subspace/subrepresentation $W$ should be an irreducible subspace/representation. This requires additional conditions, namely that $W$ itself only has trivial $G$-invariant subspaces.

• Something else that might be worth mentioning. I found this very confusing (but then I am easily confused): many physicists speak of $V$ or $W$ as the representation whereas mathematicians speak of the mapping (group action) $G\times V\to V$ as the representation (to my mind the latter is more natural, but then I can see that "watching what happens on the other end of the arrow" is too). If you don't ken an author and are reading him/her for the first time, it's worth scanning a text to test which of these they mean, or it can do your head in. Oct 10 '13 at 23:46
• Right, your point is basically my second item in e.g. this Phys.SE answer, but it should probably be mentioned here as well for clarity. Oct 11 '13 at 0:02

A simple example: for a rotation about the Z axis, the X and Y components of a vector will change, their new values being linear combinations of their previous, but no change occurs to their Z components.

In higher dimensions you can have transforms affecting some components but not others in more general ways, including components "transforming among themselves" in separate groups. For example in eight dimensions you could write a tensor to make the 1st and 2nd components transform like X and Y did, and also let's say their 5th and 8th components, while all other components of a vector sit still.

• +1 (more if I could). Such simplicity! Puts us all thoroughly to shame. John Stillwell (also see here and here ) would be proud of you. John Stillwell is someone I think of as the Feynman of teaching mathematics, in that he seeks to hammer and hammer at topics until the most wonderful and understandable pieces of technical writing you ever saw are the result. I was lucky enough to take several of his courses. Oct 10 '13 at 23:38

This is an explanation ripped more or less wholesale from Zee's field theory book, from which I first understood what everyone was banging on about with irreducible whatnots. Let's consider a rank two tensor in three dimensions, which you can always think of as the tensor product of two fictitious vectors, or in components: $$T_{ij} = x_i y_j.$$ How does this object transform under rotations? Rotations are matrices that transform vectors into a new coordinate system as $x^{\prime}_i = O_{ij}x_j$, where the Einstein summation notation is implied. The fact that $O$ represents a rotation tells you that it rotates angles but doesn't change lengths: a global rotation must preserve the inner product between two vectors. Therefore $O$ is an orthogonal matrix, satisfying $O^T = O^{-1}$. Orthogonal matrices form the group $O(3)$; $O$ for orthogonal and 3 for 3D. Now it should be clear how $T$ transforms under rotations, you just apply an orthogonal matrix to each of these fictitious vectors:

$$T^{\prime}_{kl} = O_{ki}x_i O_{lj}y_j = O_{ki} T_{ij} O^{-1}_{jl} = (O T O^{-1})_{kl}$$

using the defining property of orthogonal matrices and denoting the components of the matrix inverse of $O$ by $O^{-1}_{jl}$.

There are 3 interesting "pieces" to the tensor $T$, the trace $$T_{ii} ,$$ the antisymmetric part $$A_{ij} = \frac{1}{2}(T_{ij} - T_{ji}) = - A_{ji},$$ and a traceless symmetric part $$S_{ij} = \frac{1}{2}(T_{ij} + T_{ji}) - \frac{1}{3}\delta_{ij}T_{ii} = S_{ji}.$$ Let's see how these each transform. The trace is just a number so obviously $$T^{\prime}_{ii} = O T_{ii} O^{-1} = T_{ii} O O^{-1} = T_{ii},$$ or in other words the trace is invariant, or equivalently a scalar, or equivalently it transforms trivially. You can show just by manipulating indices that after transforming the antisymmetric part $$A_{ij}^{\prime} = O_{ik}A_{kl}O_{jl} = -A^{\prime}_{ji},$$ i.e. it is still antisymmetric. (The orthogonal property of $O$ is essential here.) Likewise, the transformed $S^{\prime}_{ij} = S^{\prime}_{ji}$ is still traceless and symmetric. This shows that these three parts are independent in the sense that orthogonal rotations do not mix them together. These form the irreducible components of the tensor that your author refers to.

Notice that the trace has 1 element, the antisymmetric part has 3 independent elements, and the symmetric traceless part has 5 independent elements. $1 + 3 + 5 = 9$, the total number of elements of a general rank-two tensor $T_{ij}$ in 3D. These 3 bits roughly correspond to the three "interesting" ways of combining two vectors $\mathbf{x}$ & $\mathbf{y}$. You can tensor them together with a kronecker product $T = \mathbf{x}\otimes\mathbf{y}^T$, which has 9 independent components. However, taking the trace of this gives you the dot product $\mathbf{x}\cdot\mathbf{y} = T_{ii}$, a one-component scalar. Another three components form the cross product, giving you a 3-component (pseudo)vector $(\mathbf{x}\times\mathbf{y})_i = \epsilon_{ijk}A_{jk}$ (you might want to look up the Levi-Civita symbol). The remaining 5 independent components form a traceless symmetric rank-two tensor $S_{ij}$.

• +1 "banging on about with irreducible whatnots" made me giggle. It frightens me how long I take to understand things sometimes and a well placed paragraph can sometimes stand in the stead of months of work. Oct 10 '13 at 23:41