# Decomposition of a rank 2 tensor

References: Anthony Zee, Einstein Gravity in a Nutshell, (Princeton University Press, 2013) – Chapter I.4, Problem 2.

In Zee’s book, he defines a tensor as “something that transforms like a tensor”. For a tensor with ${N}$ indices, under a rotation specified by the matrix ${R}$, the transformation of the tensor is given by multiplying the original tensor by one copy of ${R}$ for each index. For a 2-index tensor, for example

$\displaystyle T^{\prime ij}=R^{ik}R^{j\ell}T^{k\ell} \ \ \ \ \ (1)$

Another way of looking at this transformation is to think of each component ${T^{ij}}$ of the tensor as a separate object in its own right. We can then arrange these objects in a column matrix (I’m avoiding calling this column matrix a ‘vector’ since, as Zee points out, vectors have a specific transformation property that this column matrix doesn’t have, namely that it must transform under a rotation by a multiplication by a single instance of a rotation matrix ${R}$). For 3-d, for example, we have the 9-component matrix

$\displaystyle \mathcal{T}=\left[\begin{array}{c} T^{11}\\ T^{12}\\ \vdots\\ T^{33} \end{array}\right] \ \ \ \ \ (2)$

Under a rotation, we see from 1 that the transformed tensor component ${T^{\prime ij}}$ is a linear combination of the original components ${T^{k\ell}}$, where the coefficients of this linear transformation are found from the elements of the rotation matrix ${R}$. This means that we could define a matrix ${\mathcal{D}}$ which, in 3-d, is of size ${9\times9}$ and whose elements are composed of combinations of the elements of ${R}$. That is

$\displaystyle \mathcal{T}^{\prime}=\mathcal{D}\mathcal{T} \ \ \ \ \ (3)$

For example

$\displaystyle \left(\mathcal{T}^{\prime}\right)^{11}=T^{\prime11}=\mathcal{D}^{ij}\mathcal{T}^{j} \ \ \ \ \ (4)$

where the index ${j}$ is summed from ${j=1}$ to ${j=9}$. We can read off the first row of ${\mathcal{D}}$ from 1, as this is the row of ${\mathcal{D}}$ which provides the coefficients for producing the transformed component ${T^{\prime11}}$.

$\displaystyle \mathcal{D}^{1j}=\left[\begin{array}{cccccc} R^{11}R^{11} & R^{11}R^{12} & R^{11}R^{13} & R^{12}R^{11} & \ldots & R^{13}R^{13}\end{array}\right] \ \ \ \ \ (5)$

For a general rank 2 tensor (a tensor having 2 indices), there aren’t any pre-defined symmetries, so all the elements are independent of each other. As such, a transformed component ${T^{\prime ij}}$ could have a contribution from all 9 of the original components ${T^{k\ell}}$. However, it’s possible to create linear combinations of the original ${T^{ij}}$s such that a subset of these linear combinations transform into each other.

One such subset contains the antisymmetric combinations

$\displaystyle A^{ij}\equiv T^{ij}-T^{ji} \ \ \ \ \ (6)$

Zee shows that an antisymmetric component transforms as

$\displaystyle A^{\prime ij}=R^{ik}R^{j\ell}A^{k\ell} \ \ \ \ \ (7)$

That is, antisymmetric components transform as linear combinations of only other antisymmetric components. In 3-d the index ${i}$ in ${A^{ij}}$ can have 3 values, while ${j}$ can have only 2 (since ${A^{ii}}$ is always zero by definition, we don’t count it). Also, since we’re after only components that are linearly independent of each other, we don’t count ${A^{ji}}$ once we’ve counted ${A^{ij}}$, so there are a total of ${\frac{3\times2}{2}=3}$ independent ${A^{ij}}$. In ${D}$ dimensions, there are ${\frac{1}{2}D\left(D-1\right)}$ independent antisymmetric combinations. These components transform entirely within their own private subset.

We can also define a set ${S^{ij}}$ of symmetric components as

$\displaystyle S^{ij}\equiv T^{ij}+T^{ji} \ \ \ \ \ (8)$

These components transform as follows:

 $\displaystyle S^{\prime ij}$ $\displaystyle =$ $\displaystyle T^{\prime ij}+T^{\prime ji}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R^{ik}R^{j\ell}T^{k\ell}+R^{jk}R^{i\ell}T^{k\ell}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R^{ik}R^{j\ell}T^{k\ell}+R^{j\ell}R^{ik}T^{\ell k}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R^{ik}R^{j\ell}\left(T^{k\ell}+T^{\ell k}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R^{ik}R^{j\ell}S^{k\ell} \ \ \ \ \ (13)$

In the third line, we swapped the dummy summed indices ${k}$ and ${\ell}$. Thus the symmetric combinations also transform within their own subset. There are ${\frac{1}{2}D\left(D-1\right)}$ plus the ${D}$ diagonal components ${S^{ii}}$ (no sum) which are, in general, non-zero, for a total of ${\frac{1}{2}D\left(D+1\right)}$ symmetric components. Together the antisymmetric and symmetric components contain all ${\frac{1}{2}D\left(D-1\right)+\frac{1}{2}D\left(D+1\right)=D^{2}}$ independent linear combinations in the original tensor ${T^{ij}}$. This means that any of the original tensor components can be written as a combination of the ${A^{ij}}$ and ${S^{ij}}$ as

$\displaystyle T^{ij}=\frac{1}{2}\left(A^{ij}+S^{ij}\right) \ \ \ \ \ (14)$

This decomposition also works for diagonal elements since ${A^{ii}=0}$ and ${S^{ii}=2T^{ii}}$ (no sums).

If we write the original tensor in terms of the ${A^{ij}}$ and ${S^{ij}}$, then (in 3-d) the matrix ${\mathcal{D}}$ decomposes into a block diagonal matrix with a ${3\times3}$ block for the ${A^{ij}}$ and a ${6\times6}$ block for the ${S^{ij}}$. That is, the transformation equation becomes

 $\displaystyle T^{\prime ij}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(A^{\prime ij}+S^{\prime ij}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}R^{ik}R^{j\ell}\left(A^{k\ell}+S^{k\ell}\right) \ \ \ \ \ (16)$

For example, if we want ${T^{\prime32}}$ we have

 $\displaystyle T^{\prime32}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(A^{\prime32}+S^{\prime32}\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(-A^{\prime23}+S^{\prime23}\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}R^{2k}R^{3\ell}\left(-A^{k\ell}+S^{k\ell}\right) \ \ \ \ \ (19)$

The sums over ${A^{k\ell}}$ and ${S^{k\ell}}$ can now be worked out using the symmetry properties of these elements. For ${A^{k\ell}}$ we have

 $\displaystyle -R^{2k}R^{3\ell}A^{k\ell}$ $\displaystyle =$ $\displaystyle -R^{21}R^{32}A^{12}-R^{21}R^{33}A^{13}-R^{22}R^{31}A^{21}-\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle$ $\displaystyle R^{22}R^{33}A^{23}-R^{23}R^{31}A^{31}-R^{23}R^{32}A^{32}\nonumber$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(R^{22}R^{31}-R^{21}R^{32}\right)A^{12}+\left(R^{23}R^{31}-R^{21}R^{33}\right)A^{13}+\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(R^{23}R^{32}-R^{22}R^{33}\right)A^{23}\nonumber$

Thus the third row of the ${3\times3}$ block in the matrix ${\mathcal{D}}$ (which is used to calculate ${A^{\prime23}}$) is

$\displaystyle \left[\begin{array}{ccc} \left(R^{21}R^{32}-R^{22}R^{31}\right) & \left(R^{21}R^{33}-R^{23}R^{31}\right) & \left(R^{22}R^{33}-R^{23}R^{32}\right)\end{array}\right] \ \ \ \ \ (22)$

We could do a similar calculation for ${S^{ij}}$ except this time we’d get 6 terms in the transformation.

In fact the symmetric part of ${\mathcal{D}}$ can be decomposed further by observing that the trace of the symmetric submatrix is invariant under rotation, as Zee shows in his equation 6 (sum implied over ${i}$):

$\displaystyle S^{\prime ii}=S^{ii} \ \ \ \ \ (23)$

Therefore the ${6\times6}$ matrix breaks into a ${1\times1}$ matrix and a ${5\times5}$ matrix. Zee shows that the components of the ${5\times5}$ block (or ${D-1\times D-1}$ in the ${D}$-dimensional case) are given by

$\displaystyle \tilde{S}^{ij}=S^{ij}-\delta^{ij}\frac{S^{kk}}{D} \ \ \ \ \ (24)$

Zee gives an example in 3-d showing that the components of ${\tilde{S}^{ij}}$ do indeed transform into themselves.

## 2 thoughts on “Decomposition of a rank 2 tensor”

1. Steve

I was unable to find a “Leave a Reply” button on the main “Zee: Einstein Gravity in a Nutshell” page, so I clicked on the link to chapter I.4, problem 2 (4.2) to make this reply.

On the main page you state:
“I’m not aware of any online list of errata for this book … If anyone knows of an official errata site for this book, please leave a comment below.”