# A question in Weinberg's QFT

## Main Question or Discussion Point

In &5.6 writes:
"An (A,A) field (A is spin) contains terms with only integer spins 2A,2A-1,...,0, and corresponds to a traceless symmetric tensor of rank 2A.(Note that the number of independent components of a symmetric tensor of rank 2A in four(space-time) dimensions is:
{4.5...(4+2A-1)}/(2A)!=(3+2A)!/6(2A)!
and the tracelessness condition reduces this to:
{(3+2A)!/6(2A)!}-{(1+2A)!/6(2A-2)!}=(2A+1)^2
as expected for an (A,A) field"

I can not derive the number of components of a tensor of rank 2A in four dimensions is
4.5...(4+2A-1) and the number of tracelessnes conditions is (1+2A)!/6(2A-2)!

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And I do not understand why a (1,0) or (0,1) field corresponds to antisymmetric tensor
F$^{\mu\nu}$ and satisfies the condition:
F$^{\mu\nu}$=$\pm$$\epsilon$$^{\mu\nu\lambda\rho}$F$_{\lambda\rho}$
for (1,0) and (0,1) fields,respectively.

Bill_K
And I do not understand why a (1,0) or (0,1) field corresponds to antisymmetric tensor
In a footnote on the previous page, Weinberg mentions an "alternative formalism", in which the representations are written as symmetric SU(2) tensors with two-valued indices of two types - dotted and undotted. Thus, the (1,0) representation is the symmetric tensor φAB, while the (0, 1) representation is φA'B'. In addition we have available the antisymmetric symbol εAB which is used to raise and lower. From these quantities, a Lorentz vector index can be constructed by means of the Pauli matrices σμAB'

So, given a (1, 0) quantity FAB, we construct Fμν = σμAB' σνCD' FAC εB'D' which is the antisymmetric tensor that corresponds to it.

In four dimensions a tensor of rank 2A means there are 2A indexes and each index runs on 0,1,2,3.Then there are all 4^2A components.But in Weinberg's there are 4.5...(4+2A-1) components in this tensor.Where is the wrong in my argument?

Bill_K
In four dimensions a tensor of rank 2A means there are 2A indexes and each index runs on 0,1,2,3.Then there are all 4^2A components.But in Weinberg's there are 4.5...(4+2A-1) components in this tensor.Where is the wrong in my argument?
It's a symmetric tensor. In n dimensions a symmetric tensor of rank r has nr components, but only some of them are independent. The number of independent components is given by the binomial coefficient, (n + r - 1)!/r!(n - 1)!

And what is the difference (in tensor forms) between (1,0) field and (0,1) field?

Bill_K
And what is the difference (in tensor forms) between (1,0) field and (0,1) field?
A (1, 0) field is represented by a rank-two quantity FAB, which has three independent complex components. So does the Fμν that you can construct from it. See this earlier thread.

In &5.6 writes:
"An (A,A) field (A is spin) contains terms with only integer spins 2A,2A-1,...,0, and corresponds to a traceless symmetric tensor of rank 2A.(Note that the number of independent components of a symmetric tensor of rank 2A in four(space-time) dimensions is:
{4.5...(4+2A-1)}/(2A)!=(3+2A)!/6(2A)!
and the tracelessness condition reduces this to:
{(3+2A)!/6(2A)!}-{(1+2A)!/6(2A-2)!}=(2A+1)^2
as expected for an (A,A) field"

I can not derive the number of components of a tensor of rank 2A in four dimensions is
4.5...(4+2A-1) and the number of tracelessnes conditions is (1+2A)!/6(2A-2)!
This is something you will not expect from Weinberg to explain because it is a simple combinatorics.Anyway,when you have a symmetric tensor of order say r into N dimensional space then you can simply specify it by how many indices are zero or 1 and so on.There order does not matter(if you want you can take an ascending order with repetition allowed to find it).So the problem is basically to partition r objects into N sets ignoring the relative positioning.It requires N-1 boundaries among r objects.This boundary and objects together can be arranged in N+r-1! ways and the object itself can be arranged in r! ways.Besides it the boundary can be permuted in N-1! ways.Hence the number of independent components are N+r-1!/(N-1!)(r!).(A simple intution is obtained from second rank tensor in which all the elements above or below the diagonal+diagonal elements are the numbers which is simply 6+4=10).
Also a traceless tensor will have the condition that if you chose any two indices for contraction,then the condition is simply
Tr(aabcd...)=0.so here the elements are now only (r-2) because first two indices are summed over for trace.So just put r-2 in place of r in the above formula and you have N+r-3!/(N-1!)(r-2!).Now this gives your result for N=4.

Bill_K