Solving Linear Tensor Equations

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.


I. INTRODUCTION
In many applications a tensorial equation of the form appears, where B αµν is some given (i.e.known) tensor field, a ′ i s, i = 1, 2, ..., 6 are some given scalar fields and N αµν 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 [1-3] with respect to the affine connection 1 .There, B αµν represents the (known) hypermomentum source and N αµν is the distortion tensor [5] in which spacetime torsion and non-metricity are encoded.Then having solved for N αµν entirely in terms of the sources, that is combinations of B αµν one can easily obtain the forms of torsion and non-metricity, namely the non-Riemannian [6] parts of the geometry.In order to solve this equation one could go about and split N αµν into its irreducible decomposition and then take contractions, symmetrizations etc. in order to find the various pieces in terms of B αµν 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 a 1 N µ1µ2µ3...µn + a 2 N µnµ1µ2...µn−1 + ... + a n N µ2,µ3,...µnµ1 + ....
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.

II. 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 where a i , i = 1, 2, ..., 6 are scalars, B αµν is a given (known) tensor and N αµν are the components of the unknown tensor 3 N .Define the matrix holds true, then the general and unique solution of (3) reads where the ã1i ′ 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 including also (3)  we end up with the system Then, defining the matrix along with the columns and we may express the above system in matrix form as In the above 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(A) = 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 3 Of course the result holds true even when Nαµν are the components of a tensor density instead or even of a connection given that Bαµν are also of the same kind. 4A necessary condition for this to happen is that 6 i=1 a i = 0.However, this condition alone is not sufficient since the latter quantity can be non-vanishing but it may be so that the full determinant still vanishes.See Appendix for more details on this feature.
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 where the ã1i ′ s are the elements of the first row of the inverse matrix A −1 which, of course, depend on a 1 , a 2 , ..., 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 αµν = 0 and the matrix A is non-singular we have that N αµν = 0 as a unique solution.It should be emphasized that the demand that det(A) = 0 is all essential in order for the full N αµν 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 αµν will be vanishing.
where a i , i = 1, 2, ..., n are scalars, B µ1µ2...µn are the components of a given (known) tensor and N µ1µ2µ3...µn are the components of the unknown tensor N of rank n.Define the square n! × n! matrix Given that the system is non-degenerate, that is det A = 0, then the general and unique solution of ( 14) is given by N µ1µ2µ3...µn = ã11 B µ1µ2µ3...µn + ... + ã1n B µ2µ1µ3...µn (16) where the ã′ 1i 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! equations 6 We then define the square n! × n! matrix 5 This is easily realized as follows.Without loss of generality let us suppose that N is symmetric in its first two indices, i.e.Nαµν = Nµαν .Then with this relation and circle permutations of it is is trivial to see that only three combinations of N appear in (3) and as a result the system reduces to a 3 × 3. Of course same goes also when N is antisymmetric in any pair of its indices. 6The first one is eq.( 14) itself.
we may express the above system in the matrix form As in Theorem 1 we then formally multiply the above equation by A −1 from the left to get and by equating the first row element of the left and right hand sides of the above we arrive at the stated result where ã11 , ã11 , ..., ã1n are the elements of the first row of the inverse matrix A −1 .
Remark.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 αµν also appear in (3).Defining the three traces N (1) µ := N αβµ g αβ , N (2) µ := N αµβ g αβ , N the generalized version of (3), still linear in N , including the above traces reads 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 αµν in terms of a modified version of B αµν the includes its traces.We have the following result.where In these cases the determinant criterion would be of great use in determining whether the given tensor equation can give the components of the full N tensor or not.Example 2. Let us now apply the result of our first Theorem in a trivial example one encounters in introductory courses of tensor calculus.There, the metric compatibility condition implies Γ νµα + Γ µνα = ∂ α g µν , where Γ νµα := g λν Γ λ µν (C2) which along with the torsionlessness of the connection Γ λ µν = Γ λ νµ give us the usual Levi-Civita form of the connection.Recall that the trick to solve for Γ λ µν there was to consider two subsequent cyclic permutations of the (C2) and subtract them from the latter.Let us reproduce this result here by applying the Theorem 1.In this case, as we have already mentioned the fact that Γ νµα is symmetric in its last two indices, reduces A to a 3 × 3 matrix and we might as well set a 4 = a 5 = a 6 = 0. Now from (C2) we read off the coefficients a 1 = 0, a 2 = a 3 = 1 and as a result From which we see that det(A) = −2 = 0 and we straightforwardly calculate the first row elements of the inverse matrix to be ã11 = −1/2, ã12 = ã13 = 1/2.Then substituting these into (13) and with the identifications 8 N αµν = Γ αµν and B αµν we arrive at the well known result for the Levi-Civita connection.Of course in this case the same result can be obtained trivially by the classical method we mentioned above.Our intention with these examples here is to illustrate how our general method works.Probably the most useful application of our Theorem to physical systems is the analysis of the connection field equations in Metric-Affine Gravity.There our method will be proven to be all essential in solving for torsion and non-metricity in terms of their sources.