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.”

    Since your spam filter catches links, please google “Brad Carlile Gravity Nutshell errata” to see the following link to the Gravity Nutshell errata page:

    “In my discussions with Professor A. Zee … he asked if I could host webpages for the errata and clarifications for two of his books:

    “Einstein Gravity in a Nutshell.” ISBN 978-0691145587 — ERRATA PAGE LINK
    “Quantum Field Theory in a Nutshell (2nd Edition).” ISBN 978-0-691-01019-9 — ERRATA PAGE LINK”

    At that “Errata Page Link” listed above you can find a few, but not many, corrections submitted. I offered a correction and was copied on an email between the author and Brad regarding its inclusion in the errata, so that is probably the official errata page. There may not be many corrections submitted because the author does not appear to be very receptive them: in the email he states “the typos he pointed out are quite trivial and would be picked up by almost any attentive readers.” Over three years after publication no one to date had pointed out the “trivial” error, so either no one is “attentively” reading the “Dynamic Universe” chapter of the book or no one knows or cares where to offer corrections. i hope this helps.

    The actual error/correction is as “trivial” as the difference between A + B versus A – B, where A and B are the two terms of a tensor differential equation.

    Reply

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