Donn G. Shankland shows how to define tensor fields and their spin content (irreducibility). It is a higher-spin field theory alternative to the Fierz - Pauli, Bargmann - Wigner, Gelfand, Joos - Weinberg, and other formalisms.
Defining a Tensor
There are multiple related inequivalent definitions:
A vector in the tensor product of vector spaces, see e.g. Ruíz-Tolosa and Castillo (2005), Itskov (2025).
Anything that transforms like a tensor, see e.g. Kay (1988), Gelfand et al. (1963).
A composition of various primitives such as \(k_{i}\), \(g_{ij}\), \(\epsilon_{ijkl}\), \(\sigma_{i}\), \(\gamma_{i}\), \(p_{\alpha}\), \(\partial_{i}\), \(A_{i}\), \(\Gamma_{ijk}\)… G. B. Gurevich (1964) and P. Cvitanovic (2008) may shed a dim light here.
Here is yet another approach, with one nontrivial consequence.
Define a tensor as a weighted sum of the Kronecker product of vectors:
\[ \begin{align} X{^i}{^j}{_k}\equiv \sum_{ijk}\,c_{ijk}\,e^{i}\otimes e^{j}\otimes e_{k}\,, \end{align} \]
where \(c_{ijk}\) are some constants. Assume that \(e^{i}\) is a row-vector, and \(e_{k}\) is a column-vector. They need not be unit vectors, may not form a basis.
Note the property of the Kronecker product:
\[ \begin{align} a^{i}\otimes b^{j} &\neq b^{j}\otimes a^{i}\,,\\ a^{i}\otimes b_{j} &= b_{j}\otimes a^{i}\,. \end{align} \]
It holds for any row-vector \(a^{i}\) and column-vector \(b_{j}\).
One can now see that this allows to shift sub and superscript indices while maintaining their type and the order within the type group. In other words, it is possible to have the index positioning freedom:
\[ X{^i}{^j}{_k} = X{^i}{_k}{^j} = X{_k}{^i}{^j}\,. \]
What kind of symmetry is that? It is neither tied to coordinate transformations, nor index symmetrization.
It makes sense to focus on the third way. Gather what you know about the specific primitives with their properties first, then attempt to compose, proceed bottom-up. Remarkably, this also leads to linear algebra.
Shankland’s Tensor Algebras
A problem that ChatGPT-alikes still cannot solve in 2025: Given the four-vector \(k_{\mu}\) and the metric tensor \(g_{\mu\nu}\), write down the most general dimensionless tensor \({T_{\mu\nu}}^{\rho \sigma}\) symmetric under the permutations of the covariant indices \((\mu, \nu)\), and also symmetric w.r.t. the permutations of contravariant indices \((\rho, \sigma)\). It should be a sum of linearly independent terms, each with a manifest symmetry, and at most fourth order in \(k_{\mu}\).
Shankland (1970) jumps into the answer, which is a linear combination of
\[ \begin{align} X_{1} & = \frac{1}{4} {g_{(\mu}}^{(\rho} {g_{\nu)}}^{\sigma)}\,,\\ X_{2} & = g_{\mu \nu} g^{\rho \sigma}\,, \\ X_{3} & = \frac{1}{k^2} g_{\mu \nu} k^{\rho} k^{\sigma}\,, \\ X_{4} & = \frac{1}{k^2} k_{\mu} k_{\nu} g^{\rho \sigma}\,, \\ X_{5} & = \frac{1}{k^4} k_{\mu} k_{\nu} k^{\rho} k^{\sigma}\,, \\ X_{6} & = \frac{1}{k^2} k_{(\mu} {g_{\nu)}}^{(\rho} k^{\sigma)}\,. \end{align} \]
Here the division by 4 of the first basis element is not important, but it turns \(X_{1}\) into a proper index symmetrization operator \(()\) if one acts with it on any two-index tensor. Note that in Shankland (1970) symmetrization and antisymmetrization is not scaled.
Magically, these tensor basis expressions form an algebra under the product defined as
\[ A\, B \equiv {A_{\mu \nu}}^{\alpha \beta} {B_{\alpha \beta}}^{\rho \sigma}\,, \]
with the implied summation over repeated indices.
Exercise 1. Show that
\[ X_{6} X_{6} = 8 X_{5} + 2 X_{6}, \]
which is not \(8 X_{4} + 2 X_{6}\) stated by Shankland (1970).
Hint:
The definitions of the product and index symmetrization \((\cdot, \cdot)\) lead to \(X_{6} X_{6}\) expressed as
\[ \begin{align} & \equiv \frac{1}{k^2} k_{(\mu} {g_{\nu)}}^{(\alpha} k^{\beta)} \frac{1}{k^2} k_{(\alpha} {g_{\beta)}}^{(\rho} k^{\sigma)}\\ &= \frac{1}{k^4} \Big( k_{\mu} {g_{\nu}}^{\alpha} k^{\beta} + k_{\mu} {g_{\nu}}^{\beta} k^{\alpha} \\ & \;\;\;\;+ k_{\nu} {g_{\mu}}^{\alpha} k^{\beta} + k_{\nu} {g_{\mu}}^{\beta} k^{\alpha} \Big)\\ & \;\;\;\;\cdot \Big( k_{\alpha} {g_{\beta}}^{\rho} k^{\sigma} + k_{\alpha} {g_{\beta}}^{\sigma} k^{\rho} \\ & \;\;\;\;+ k_{\beta} {g_{\alpha}}^{\rho} k^{\sigma} + k_{\beta} {g_{\alpha}}^{\sigma} k^{\rho} \Big). \end{align} \]
This will produce 16 terms, but they will further simplify with the use of
\[ \begin{align} k_{\alpha} k^{\alpha} & = k^2 \\ {g_{\nu}}^{\alpha}k_{\alpha} &= k_{\nu} \\ {g_{\nu}}^{\beta} {g_{\beta}}^{\sigma} &= {g_{\nu}}^{\sigma}\;, \end{align} \]
e.g.
\[ \begin{align} t_{1} & = k_{\mu} {g_{\nu}}^{\alpha} k^{\beta} \cdot k_{\alpha} {g_{\beta}}^{\rho} k^{\sigma} \\ & = k_{\mu} \big({g_{\nu}}^{\alpha} k_{\alpha}\big) \big({g_{\beta}}^{\rho} k^{\beta}\big) k^{\sigma} \\ &= k_{\mu} k_{\nu} k^{\rho} k^{\sigma}. \end{align} \]
A page later ;) there will be 8 such terms and a doubled \(X_{6}\) left.
The Spectrum of a Tensor Field
The products \(X_{i}X_{j}\) and the traces \(\textit{tr} X_{i}\) determine the spectrum of the algebra. Shankland does not define a tensor field, but it is assumed that \(X_{i}\) will form an operator applied to build a quadratic form for a field, where \(k_{i}\) becomes a four-nabla. The PhD thesis of K.J. Barnes (1963) sheds more light here, but his notation requires getting used to.
The traces are defined as
\[ \begin{align} \textit{tr}\,{X_{\mu \nu}}^{\rho \sigma} \equiv {X_{\mu \nu}}^{\mu \nu}\,. \end{align} \]
They demand the values of the metric tensor \(g_{ij}\), along with the mixed tensor
\[ \begin{align} {g_{i}}^{j}={\delta_{i}}^{j}=\begin{cases} 0 & i \ne j \\ 1 & i = j \end{cases}\;\;. \end{align} \]
There is no need to know these values when getting the product tables \(X_{i}X_{j}\).
Shankland (1970) applies the Faddeev - LeVerrier algorithm, or rather its advanced variant extended to tackle multiple eigenvalues, see e.g. Helmberg and Wagner (1993). Here there are no eigenvectors in a traditional sense, they are weighted sums of the basis \(X_{i}\), not some matrix columns.
K.J. Barnes (1963) seeks the spectrum differently, with the matrix projection operators.
Mysteriously, the eigenvalues will have multiplicities which can be deduced independently from the Lorentz group theory (Lorentz with “t”), without any iterations and polynomial equations. The group theory alone, however, will not get us to the eigenvector equations leading to the Lorenz gauge condition for spin 1 (Lorenz without “t”).
Exercise 2. Verify Shankland’s spectral results, esp. the case with one vector and one spinor index: “… we find, together with their antiparticles, the following groups of particles: a quadruplet, and two doublets.”
Hint:
According to group theory, combining indices means taking “tensor products \((m,n)\otimes (k,l)\) of the Lorentz irreps”, clf. Weinberg’s QFT, Vol. 1, pages 229-233. What is relevant here is that each such an irrep constitutes a set of subspaces with multiplicities \(2s+1\) for \(s=|m-n|, |m-n+1|,\,\ldots, m+n\).
\((0,0)\): A scalar. Shankland’s singlet: A single subspace with eigenvalue multiplicity \(2\cdot 0+1=1\).
\((\frac{1}{2},\frac{1}{2})\): A single four-vector index. Shankland’s singlet and a triplet: two subspaces \(0, 1\) with multiplicities 1 and 3.
\((\frac{1}{2},0)\oplus (0,\frac{1}{2})\): A full single spinor index. Shankland’s doublet and its antidoublet: \(\frac{1}{2}, \frac{1}{2}\) subspaces with multiplicites 2 and 2.
\((1,1)\): Two symmetric tensor indices. A mismatch with Shankland’s pentuplet, triplet, and two singlets: Subspaces \(0, 1, 2\) with multiplicities 1, 3, and 5. Where is the missing singlet? In a symmetric two-index tensor case, to remove a singlet also means to make the tensor traceless, so the group theory still matches Shankland with this assumption.
\((1,0)\oplus (0,1)\): Two asymmetric tensor indices. Shankland’s two particle triplets: Subspaces \(1\) and \(1\) with multiplicities 3 and 3.
\((\frac{1}{2},\frac{1}{2}) \otimes \Big((\frac{1}{2},0)\oplus (0,\frac{1}{2})\Big)\), i.e. combining a vector and a spinor index?
The last case, spin \(\frac{3}{2}\), splits into a spinor and \((1,\frac{1}{2}) \oplus (\frac{1}{2},1)\), clf. Weinberg’s QFT, Vol. 1, page 232. The latter brings subspaces \(\frac{1}{2}\) and \(\frac{3}{2}\) with multiplicities 2 and 4, along with their “antisubspaces”. All of this combined perfectly matches the result of Shankland.
Note that the construction of algebras is skipped, but it is not trivial. For spin \(\frac{3}{2}\), Shankland had to spot that the combination \(\gamma_{\mu}p^{\mu}\) acted independently of \(p\), \(\gamma\), and \(g\). This has effectively doubled the basis dimension of the vector-spinor algebra from 5 to 10.
Other Relevant Algebras
One can find some other mildly successful uses/hints of tensor algebras in Phys. Rev. 106, 1345 (1957); Nuovo Cimento, 43, 475 (1966); Nuovo Cimento 47, 145 (1967); Phys. Rev. 153, 1652 (1967); Phys. Rev. 161, 1631 (1967); Phys. Rev. D 8, 2650 (1973); Nuovo Cimento 28, 409 (1975); Phys. Lett. B 301 4 339 (1993); Phys. Rev. C 64, 015203 (2001); Phys. Rev. D 64, 125013 (2001); Hadronic J. 26, 351 (2003); Phys. Rev. D 67, 085021 (2003); Phys. Rev. D 67, 125011 (2003); Nucl. Phys. B724, 453 (2005); Phys. Rev. D 74, 084036 (2006); P. Cvitanović (2008); V. Monchiet and G. Bonnet (2010); Phys. Rev. D 97, 115043 (2018); SUGRA and CDC…
It is tough to read this literature, and the results may not always justify the complexity.
Why Shankland?
To sum up, we are given a field with tensor/spinor indices and their permutation symmetries. A Lorentz-invariant operator is then constructed. It may serve as a quadratic form operator for the field. The spin content of the field is discovered as the eigenvalue multiplicities of that operator. One test of this formalism confirms that removing spin 0 from a vector field leads to “apesanteur” \(A\) aka vector potential.
Considering the vast literature on group theory, irreducible representations, angular momentum, higher-spin field theories, spin projection operators, tensors, spinors, Weyl, Wigner, Weinberg… Shankland’s system is the most open and inviting.
