Solving Linear Tensor Equations

Abstract

We develop a systematic way to solve linear equations involving tensors of arbitrary rank. We start off with the case of a rank 3 tensor, which appears in many applications, and after finding the condition for a unique solution we derive this solution. Subsequently, we generalize our result to tensors of arbitrary rank. Finally, we consider a generalized version of the former case of rank 3 tensors and extend the result when the tensor traces are also included.

1. Introduction

In many applications a tensorial equation of the form

$$a_1{N}_{\alpha \mu \nu }+a_2{N}_{\nu \alpha \mu }+a_3{N}_{\mu \nu \alpha }+a_4{N}_{\alpha \nu \mu }+a_5{N}_{\nu \mu \alpha }+a_6{N}_{\mu \alpha \nu }=B_{\alpha \mu \nu }$$

(1)

appears, where $B_{\alpha \mu \nu }$ is some given (i.e., known) tensor field, $a_i^{\prime }s$, $i=1,2,\cdots ,6$ are some given scalar fields and $N_{\alpha \mu \nu }$ is the unknown tensor field one wishes to solve for. For instance, this tensorial equation is encountered when one varies the quadratic Metric–Affine Gravity action [1,2,3] with respect to the affine connection1. There, $B_{\alpha \mu \nu }$ represents the (known) hypermomentum source and $N_{\alpha \mu \nu }$ is the distortion tensor [8] in which spacetime torsion and non-metricity are encoded. Then, having solved for $N_{\alpha \mu \nu }$ entirely in terms of the sources, that is, combinations of $B_{\alpha \mu \nu }$, one can easily obtain the forms of torsion and non-metricity, namely the non-Riemannian [9] parts of the geometry. In order to solve this equation, one could go about and split $N_{\alpha \mu \nu }$ into its irreducible decomposition and then take contractions, symmetrizations and so forth, in order to find the various pieces in terms of $B_{\alpha \mu \nu }$ and its contractions. Even though this may work in some cases, it will be a difficult task in general. Moreover, this procedure will fall short quickly if one wishes to generalize the above considerations and ask for the general solution (N in terms of B) of the rank-n tensorial equation:

$$\underset{n!-terms}{\underbrace{a_1{N}_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}+a_2{N}_{{\mu }_n{\mu }_1{\mu }_2\cdots {\mu }_{n-1}}+\cdots +a_n{N}_{{\mu }_2,{\mu }_3,\cdots {\mu }_n{\mu }_1}+\cdots ·}}=B_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}.$$

(2)

Evidently, one easily realizes that it would be impossible to solve the latter by resorting to some decomposition scheme for N2. It is then natural to ask, is there a systematic and practical way to solve equations of the form (1) or more generally (2). It is the purpose of this letter to answer this question. As we show, under a fairly general non-degeneracy condition, it is always possible to find the unique solution of (1), or more generally of (2), by following a certain procedure that we develop below along with some extensions/generalizations.

2. The Theorems

In what follows, we present 3 Theorems. In the first one, the systematic way to solve Equation (1) for N is proved. We then extend this result to tensors N of arbitrary rank (i.e., not necessary 3) and solve equations of the form (2). Finally we derive the solution of a generalized version of (1), where the traces of N are also included. We have the following.

Theorem 1.

Consider the tensor equation:

$$a_1{N}_{\alpha \mu \nu }+a_2{N}_{\nu \alpha \mu }+a_3{N}_{\mu \nu \alpha }+a_4{N}_{\alpha \nu \mu }+a_5{N}_{\nu \mu \alpha }+a_6{N}_{\mu \alpha \nu }=B_{\alpha \mu \nu },$$

(3)

where $a_i$, $i=1,2,\cdots ,6$ are scalars, $B_{\alpha \mu \nu }$ is a given (known) tensor and $N_{\alpha \mu \nu }$ are the components of the unknown tensor3 N. Define the matrix:

$$A:=\left(\begin{array}{cccccc}{a}_1& a_2& a_3& a_4& a_5& a_6\\ a_2& a_3& a_1& a_5& a_6& a_4\\ a_3& a_1& a_2& a_6& a_4& a_5\\ a_6& a_5& a_4& a_3& a_2& a_1\\ a_5& a_4& a_6& a_2& a_1& a_3\\ a_4& a_6& a_5& a_1& a_3& a_2\end{array}\right)$$

(4)

If the system is non-degenerate, that is, if 4

$$det\left(A\right)\ne 0$$

(5)

holds true, then the general and unique solution of (3) reads:

$$N_{\alpha \mu \nu }={\tilde{a}}_{11}{B}_{\alpha \mu \nu }+{\tilde{a}}_{12}{B}_{\nu \alpha \mu }+{\tilde{a}}_{13}{B}_{\mu \nu \alpha }+{\tilde{a}}_{14}{B}_{\alpha \nu \mu }+{\tilde{a}}_{15}{B}_{\nu \mu \alpha }+{\tilde{a}}_{16}{B}_{\mu \alpha \nu },$$

(6)

where the ${\tilde{a}}_{1i}$${}^{\prime }s$ are the first row elements of the inverse matrix $A^{-1}$.

Proof. 

Starting from (3), we perform the 5 independent possible permutations on the indices, and also including (3), we end up with the system:

$$\begin{array}{c}{a}_1{N}_{\alpha \mu \nu }+a_2{N}_{\nu \alpha \mu }+a_3{N}_{\mu \nu \alpha }+a_4{N}_{\alpha \nu \mu }+a_5{N}_{\nu \mu \alpha }+a_6{N}_{\mu \alpha \nu }=B_{\alpha \mu \nu }\\ a_1{N}_{\nu \alpha \mu }+a_2{N}_{\mu \nu \alpha }+a_3{N}_{\alpha \mu \nu }+a_4{N}_{\mu \alpha \nu }+a_5{N}_{\alpha \nu \mu }+a_6{N}_{\nu \mu \alpha }=B_{\nu \alpha \mu }\\ a_1{N}_{\mu \nu \alpha }+a_2{N}_{\alpha \mu \nu }+a_3{N}_{\nu \alpha \mu }+a_4{N}_{\nu \mu \alpha }+a_5{N}_{\mu \alpha \nu }+a_6{N}_{\alpha \nu \mu }=B_{\mu \nu \alpha }\\ a_1{N}_{\alpha \nu \mu }+a_2{N}_{\mu \alpha \nu }+a_3{N}_{\nu \mu \alpha }+a_4{N}_{\alpha \mu \nu }+a_5{N}_{\mu \nu \alpha }+a_6{N}_{\nu \alpha \mu }=B_{\alpha \nu \mu }\\ a_1{N}_{\nu \mu \alpha }+a_2{N}_{\alpha \nu \mu }+a_3{N}_{\mu \alpha \nu }+a_4{N}_{\nu \alpha \mu }+a_5{N}_{\alpha \mu \nu }+a_6{N}_{\mu \nu \alpha }=B_{\nu \mu \alpha }\\ a_1{N}_{\mu \alpha \nu }+a_2{N}_{\nu \mu \alpha }+a_3{N}_{\alpha \nu \mu }+a_4{N}_{\mu \nu \alpha }+a_5{N}_{\nu \alpha \mu }+a_6{N}_{\alpha \mu \nu }=B_{\mu \alpha \nu }.\end{array}$$

(7)

Then, defining the matrix:

$$A:=\left(\begin{array}{cccccc}{a}_1& a_2& a_3& a_4& a_5& a_6\\ a_2& a_3& a_1& a_5& a_6& a_4\\ a_3& a_1& a_2& a_6& a_4& a_5\\ a_6& a_5& a_4& a_3& a_2& a_1\\ a_5& a_4& a_6& a_2& a_1& a_3\\ a_4& a_6& a_5& a_1& a_3& a_2,\end{array}\right)$$

(8)

along with the columns

$$\mathcal{N}:={(N_{\alpha \mu \nu },N_{\nu \alpha \mu },N_{\mu \nu \alpha },N_{\alpha \nu \mu },N_{\nu \mu \alpha },N_{\mu \alpha \nu })}^T$$

(9)

and

$$\mathcal{B}:={(B_{\alpha \mu \nu },B_{\nu \alpha \mu },B_{\mu \nu \alpha },B_{\alpha \nu \mu },B_{\nu \mu \alpha },B_{\mu \alpha \nu })}^T$$

(10)

we may express the above system in matrix form as:

$$A\mathcal{N}=\mathcal{B}.$$

(11)

In the above, $\mathcal{N}$ is the column consisting of the unknown elements we wish to find. Since, by hypothesis, we have a non-degenerate system, it follows that $det\left(A\right)\ne 0$ and as a result the inverse $A^{-1}$ exists. We then formally multiply the above equation by $A^{-1}$ from the left to get:

$$\mathcal{N}=A^{-1}\mathcal{B}.$$

(12)

The above is a column equation and of course each element on the left column must be equal to each element on the right. Equating the first element we arrive at the stated result:

$$N_{\alpha \mu \nu }={\tilde{a}}_{11}{B}_{\alpha \mu \nu }+{\tilde{a}}_{12}{B}_{\nu \alpha \mu }+{\tilde{a}}_{13}{B}_{\mu \nu \alpha }+{\tilde{a}}_{14}{B}_{\alpha \nu \mu }+{\tilde{a}}_{15}{B}_{\nu \mu \alpha }+{\tilde{a}}_{16}{B}_{\mu \alpha \nu },$$

(13)

where the ${\tilde{a}}_{1i}$${}^{\prime }s$ are the elements of the first row of the inverse matrix $A^{-1}$ which, of course, depend on $a_1,a_2,\cdots ,a_6$. Note that the equations we get for the rest of the column elements will be related to the above one with cyclic permutations and will therefore give nothing new. Concluding, (13) is the general solution of (3). Some comments are now in order.

Comment 1. Note that if $B_{\alpha \mu \nu }=0$ and the matrix A is non-singular we have that $N_{\alpha \mu \nu }=0$ as a unique solution. It should be emphasized that the demand that $det\left(A\right)\ne 0$ is all essential in order for the full $N_{\alpha \mu \nu }$ tensor field to be vanishing. If the last requirement is not fulfilled, the full N tensor may as well not be identically vanishing since in this case, not the full N but certain (anti)-symmetrizations of it appear in (3). In such an occasion, only certain parts of $N_{\alpha \mu \nu }$ will be vanishing.

Comment 2. If the components $N_{\alpha \mu \nu }$ are symmetric or antisymmetric in any pair of indices then the system is greatly simplified and the $6\times 6$ matrix A is reduced to a $3\times 3$ matrix instead.5 □

Theorem 2.

In a $d-$dimensional space, consider the tensor equation (with $n\le d$):

$$\underset{n!-terms}{\underbrace{a_1{N}_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}+a_2{N}_{{\mu }_n{\mu }_1{\mu }_2\cdots {\mu }_{n-1}}+\cdots +a_n{N}_{{\mu }_2,{\mu }_3,\cdots {\mu }_n{\mu }_1}+\cdots ·}}=B_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n},$$

(14)

where $a_i$, $i=1,2,\cdots ,n$ are scalars, $B_{{\mu }_1{\mu }_2\cdots {\mu }_n}$ are the components of a given (known) tensor and $N_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}$ are the components of the unknown tensor N of rank n. Define the square $n!\times n!$ matrix

$$A=\left(\begin{array}{cccc}{a}_1& a_2& \cdots & a_{n!}\\ a_2& a_n& \cdots & a_{n+1}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n+1}& a_{n!}& \cdots & a_2.\end{array}\right)$$

(15)

Given that the system is non-degenerate, that is, $detA\ne 0$, then the general and unique solution of (14) is given by:

$$N_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}={\tilde{a}}_{11}{B}_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}+\cdots +{\tilde{a}}_{1n}{B}_{{\mu }_2{\mu }_1{\mu }_3\cdots {\mu }_n},$$

(16)

where the ${\tilde{a}}_{1i}^{\prime }s$ are the first row elements of the inverse matrix $A^{-1}$.

Proof. 

In an identical manner to the proof of Theorem 1, we now start from (14) and perform the $(n!-1)$ possible independent permutations to end up with the system of $n!$ equations6

$$\begin{array}{c}{a}_1{N}_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}+a_2{N}_{{\mu }_n{\mu }_1{\mu }_2\cdots {\mu }_{n-1}}+\cdots +a_n{N}_{{\mu }_2,{\mu }_3,\cdots {\mu }_n{\mu }_1}+\cdots ·=B_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}\\ a_1{N}_{{\mu }_2{\mu }_3{\mu }_4\cdots {\mu }_1}+a_2{N}_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}+\cdots +a_n{N}_{{\mu }_3{\mu }_4\cdots {\mu }_1{\mu }_2}+\cdots ·=B_{{\mu }_2{\mu }_3{\mu }_4\cdots {\mu }_1}\\ \cdots \\ \cdots \\ \cdots \\ \cdots \end{array}$$

(17)

$$\begin{array}{c}(all-possible-permutations)\end{array}$$

(18)

We then define the square $n!\times n!$ matrix:

$$A=\left(\begin{array}{cccc}{a}_1& a_2& \cdots & a_{n!}\\ a_2& a_n& \cdots & a_{n+1}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n+1}& a_{n!}& \cdots & a_2\end{array}\right)$$

(19)

and the columns

$$\mathcal{N}:={(N_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n},N_{{\mu }_n{\mu }_1{\mu }_2\cdots {\mu }_{n-1}},\cdots ·)}^T$$

(20)

$$\mathcal{B}:={(B_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n},B_{{\mu }_n{\mu }_1{\mu }_2\cdots {\mu }_{n-1}},\cdots ·)}^T,$$

(21)

we may express the above system in the matrix form:

$$A\mathcal{N}=\mathcal{B}.$$

(22)

As in Theorem 1, we then formally multiply the above equation by $A^{-1}$ from the left to get:

$$\mathcal{N}=A^{-1}\mathcal{B}$$

(23)

and by equating the first row element of the left and right hand sides of the above we arrive at the stated result:

$$N_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}={\tilde{a}}_{11}{B}_{{\mu }_1{\mu }_2{\mu }_3\cdots {\mu }_n}+{\tilde{a}}_{12}{B}_{{\mu }_n{\mu }_1\cdots {\mu }_{n-1}}\cdots +{\tilde{a}}_{1n}{B}_{{\mu }_2{\mu }_1{\mu }_3\cdots {\mu }_n},$$

(24)

where ${\tilde{a}}_{11},{\tilde{a}}_{11},\cdots ,{\tilde{a}}_{1n}$ are the elements of the first row of the inverse matrix $A^{-1}$. □

Remark 1.

Again, if the tensor N has some symmetry property over some pair(s) of its indices, the dimension of the matrix A will be lowered accordingly.

Now, going back to the case of a rank 3 tensor, one may ask how does the situation change when the traces of $N_{\alpha \mu \nu }$ also appear in (3). Defining the three traces,

$$N_{\mu }^{\left(1\right)}:=N_{\alpha \beta \mu }{g}^{\alpha \beta },N_{\mu }^{\left(2\right)}:=N_{\alpha \mu \beta }{g}^{\alpha \beta },N_{\mu }^{\left(3\right)}:=N_{\mu \alpha \beta }{g}^{\alpha \beta },$$

(25)

the generalized version of (3), still linear in N, including the above traces reads

$$a_1{N}_{\alpha \mu \nu }+a_2{N}_{\nu \alpha \mu }+a_3{N}_{\mu \nu \alpha }+a_4{N}_{\alpha \nu \mu }+a_5{N}_{\nu \mu \alpha }+a_6{N}_{\mu \alpha \nu }+\sum _{i=1}^3\left(a_{7i}{N}_{\mu }^{\left(i\right)}{g}_{\alpha \nu }+a_{8i}{N}_{\nu }^{\left(i\right)}{g}_{\alpha \mu }+a_{9i}{N}_{\alpha }^{\left(i\right)}{g}_{\mu \nu }\right)=B_{\alpha \mu \nu }.$$

(26)

As we show below the appearance of the these extra terms does not introduce any serious technical difficulty and one can always solve for $N_{\alpha \mu \nu }$ in terms of a modified version of $B_{\alpha \mu \nu }$ that includes its traces. We have the following result.

Theorem 3.

Consider the 15 parameter linear tensor Equation (26), where $N_{\alpha \mu \nu }$ are the components of the unknown tensor field. Define the matrices7

$$\Gamma :=\left(\begin{array}{ccc}{\gamma }_{11}& {\gamma }_{12}& {\gamma }_{13}\\ {\gamma }_{21}& {\gamma }_{22}& {\gamma }_{23}\\ {\gamma }_{31}& {\gamma }_{32}& {\gamma }_{33}\end{array}\right)$$

(27)

and

$$A:=\left(\begin{array}{cccccc}{a}_1& a_2& a_3& a_4& a_5& a_6\\ a_3& a_1& a_2& a_5& a_6& a_4\\ a_2& a_3& a_1& a_6& a_4& a_5\\ a_4& a_6& a_5& a_1& a_3& a_2\\ a_5& a_4& a_6& a_2& a_1& a_3\\ a_6& a_5& a_4& a_3& a_2& a_1.\end{array}\right)$$

(28)

Then, given that both of the above matrices are non-singular, the unique solution to (26) reads:

$$N_{\alpha \mu \nu }={\tilde{a}}_{11}{B}_{\alpha \mu \nu }+{\tilde{a}}_{12}{\widehat{B}}_{\nu \alpha \mu }+{\tilde{a}}_{13}{\widehat{B}}_{\mu \nu \alpha }+{\tilde{a}}_{14}{\widehat{B}}_{\alpha \nu \mu }+{\tilde{a}}_{15}{\widehat{B}}_{\nu \mu \alpha }+{\tilde{a}}_{16}{\widehat{B}}_{\mu \alpha \nu },$$

(29)

where

$${\widehat{B}}_{\alpha \mu \nu }=B_{\alpha \mu \nu }-\sum _{i=1}^3\sum _{j=1}^3\left(a_{7i}{\tilde{\gamma }}_{ij}{B}_{\mu }^{\left(j\right)}{g}_{\alpha \nu }+a_{8i}{\tilde{\gamma }}_{ij}{B}_{\nu }^{\left(j\right)}{g}_{\alpha \mu }+a_{9i}{\tilde{\gamma }}_{ij}{B}_{\alpha }^{\left(j\right)}{g}_{\mu \nu }\right).$$

(30)

Proof. 

We begin by tracing Equation (26) three times independently with $g^{\alpha \mu }$, $g^{\alpha \nu }$ and $g^{\mu \nu }$ to get:

$$\begin{array}{c}\sum _{i=1}^3{\gamma }_{1i}{N}_{\mu }^{\left(i\right)}=B_{\mu }^{\left(1\right)},\sum _{i=1}^3{\gamma }_{2i}{N}_{\mu }^{\left(i\right)}=B_{\mu }^{\left(2\right)},\sum _{i=1}^3{\gamma }_{3i}{N}_{\mu }^{\left(i\right)}=B_{\mu }^{\left(3\right)},\end{array}$$

(31)

where we have renamed all free indices to $\mu$ and the ${\gamma }_{ij}^{\prime }s$ are some shorthand notations for certain linear combinations of the $a_i^{\prime }s$, whose relations are given in the Appendix A. Now, defining the matrix $\Gamma$ with coefficients the ${\gamma }_{ij}^{\prime }s$ along with the columns $\eta ={(N_{\mu }^{\left(1\right)},N_{\mu }^{\left(2\right)},N_{\mu }^{\left(3\right)})}^T$ and $b={(B_{\mu }^{\left(1\right)},B_{\mu }^{\left(2\right)},B_{\mu }^{\left(3\right)})}^T$ we may express the above system in matrix form as:

$$\Gamma \eta =b.$$

(32)

By hypothesis, the matrix $\Gamma$ is non-singular (i.e., $det(\Gamma )\ne 0$) and therefore the inverse ${\Gamma }^{-1}$ exists and we may formally solve for $\eta$ as

$$\eta ={\Gamma }^{-1}b,$$

(33)

which in component notation translates to

$$N_{\mu }^{\left(i\right)}=\sum _{j=1}^3{\tilde{\gamma }}_{ij}{B}_{\mu }^{\left(i\right)}$$

(34)

with the ${\tilde{\gamma }}_{ij}^{\prime }s$ being the elements of ${\Gamma }^{-1}$. Then, substituting these last relations back in (26), we fully eliminate the N traces in favour of the traces of B, ending up with

$$a_1{N}_{\alpha \mu \nu }+a_2{N}_{\nu \alpha \mu }+a_3{N}_{\mu \nu \alpha }+a_4{N}_{\alpha \nu \mu }+a_5{N}_{\nu \mu \alpha }+a_6{N}_{\mu \alpha \nu }={\widehat{B}}_{\alpha \mu \nu },$$

(35)

where

$${\widehat{B}}_{\alpha \mu \nu }=B_{\alpha \mu \nu }-\sum _{i=1}^3\sum _{j=1}^3\left(a_{7i}{\tilde{\gamma }}_{ij}{B}_{\mu }^{\left(j\right)}{g}_{\alpha \nu }+a_{8i}{\tilde{\gamma }}_{ij}{B}_{\nu }^{\left(j\right)}{g}_{\alpha \mu }+a_{9i}{\tilde{\gamma }}_{ij}{B}_{\alpha }^{\left(j\right)}{g}_{\mu \nu }\right).$$

(36)

We are pretty much done now since we can apply the result of Theorem 1 to the modified tensor field ${\widehat{B}}_{\alpha \mu \nu }$ in place of $B_{\alpha \mu \nu }$, completing therefore the proof

$$N_{\alpha \mu \nu }={\tilde{a}}_{11}{\widehat{B}}_{\alpha \mu \nu }+{\tilde{a}}_{12}{\widehat{B}}_{\nu \alpha \mu }+{\tilde{a}}_{13}{\widehat{B}}_{\mu \nu \alpha }+{\tilde{a}}_{14}{\widehat{B}}_{\alpha \nu \mu }+{\tilde{a}}_{15}{\widehat{B}}_{\nu \mu \alpha }+{\tilde{a}}_{16}{\widehat{B}}_{\mu \alpha \nu },$$

(37)

where the modified components ${\widehat{B}}_{\alpha \mu \nu }$ are given by (36). □

3. Conclusions

We have formulated an analytical method that allows one to solve tensorial equations of the form (1), for the unknown tensor components $N_{\alpha \mu \nu }$ of the rank-3 tensor N. In particular, we proved that under a fairly general non-degeneracy condition (i.e., $det\left(A\right)\ne 0$) one can always solve equations of the form ((1) by simply finding the inverse of the matrix A, which is built from the coefficients $a_i$ appearing in the same equation. Subsequently, we generalized our result for arbitrary rank tensors and similarly obtained the solution of (14). Finally, we extended the result we obtained for the first case (Equation (1)) to the 15 parameter linear tensor Equation (26), including also the traces of $N_{\alpha \mu \nu }$ and obtained the unique solution for this case as well.

As we already mentioned in the introduction, these results find a natural application in geometric extensions of General Relativity that take into account the non-Riemannian structure of spacetime (torsion and non-metricity). In this context of Metric-Affine Theories of Gravitation, equations of the form (1), or more generally (26), relate the distortion tensor to the hypermomentum (source). Therefore, the technique we developed here will be proven to be essential for finding how the matter sources produce spacetime torsion and non-metricity. This last point is under consideration now.