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 indices, under a rotation specified by the matrix , the transformation of the tensor is given by multiplying the original tensor by one copy of for each index. For a 2-index tensor, for example

Another way of looking at this transformation is to think of each component 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 ). For 3-d, for example, we have the 9-component matrix

Under a rotation, we see from 1 that the transformed tensor component is a *linear* combination of the original components , where the coefficients of this linear transformation are found from the elements of the rotation matrix . This means that we could define a matrix which, in 3-d, is of size and whose elements are composed of combinations of the elements of . That is

For example

where the index is summed from to . We can read off the first row of from 1, as this is the row of which provides the coefficients for producing the transformed component .

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 could have a contribution from all 9 of the original components . However, it’s possible to create linear combinations of the original s such that a subset of these linear combinations transform into each other.

One such subset contains the antisymmetric combinations

Zee shows that an antisymmetric component transforms as

That is, antisymmetric components transform as linear combinations of *only* other antisymmetric components. In 3-d the index in can have 3 values, while can have only 2 (since 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 once we’ve counted , so there are a total of independent . In dimensions, there are independent antisymmetric combinations. These components transform entirely within their own private subset.

We can also define a set of symmetric components as

These components transform as follows:

In the third line, we swapped the dummy summed indices and . Thus the symmetric combinations also transform within their own subset. There are plus the diagonal components (no sum) which are, in general, non-zero, for a total of symmetric components. Together the antisymmetric and symmetric components contain all independent linear combinations in the original tensor . This means that any of the original tensor components can be written as a combination of the and as

This decomposition also works for diagonal elements since and (no sums).

If we write the original tensor in terms of the and , then (in 3-d) the matrix decomposes into a block diagonal matrix with a block for the and a block for the . That is, the transformation equation becomes

For example, if we want we have

The sums over and can now be worked out using the symmetry properties of these elements. For we have

Thus the third row of the block in the matrix (which is used to calculate ) is

We could do a similar calculation for except this time we’d get 6 terms in the transformation.

In fact the symmetric part of 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 ):

Therefore the matrix breaks into a matrix and a matrix. Zee shows that the components of the block (or in the -dimensional case) are given by

Zee gives an example in 3-d showing that the components of do indeed transform into themselves.

SteveI 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.

gwrowePost authorSomehow comments got switched off for all my pages. I’ve now fixed this.

I’ve copied your post to the main “Zee: Einstein Gravity in a Nutshell” page, so please go there for my reply.