REDUCED SYMMETRY ELEMENTS IN LINEAR ELASTICITY

In continuum mechanics problems, we have to work in most cases with symmetric tensors, symmetry expressing the conservation of angular momentum. Discretization of symmetric tensors is however difficult and a classical solution is to employ some form of reduced symmetry. We present two ways of introducing elements with reduced symmetry. The first one is based on Stokes problems, and in the two-dimensional case allows to recover practically all interesting elements on the market. This however is (definitely) not true in three dimensions. On the other hand the second approach (based on a very nice property of several interpolation operators) works for three-dimensional problems as well, and allows, in particular, to prove the convergence of the Arnold-Falk-Winther element with simple and standard arguments, without the use of the Berstein-Gelfand-Gelfand resolution.


Introduction
Mixed methods are an appealing technique for the numerical solution of elasticity problems. They ensure the equilibrium condition, a basic property in solid mechanics, and they make the constitutive law more explicit. The stress tensor becomes the main variable but the symmetry of this tensor, however, makes the construction of suitable elements much more complicated than what can be done in the thermal problems where families of elements such as the RT k and BDM k are now classical.
The idea of using stress tensors having only a reduced symmetry goes back to Fraeijs de Veubeke [10], but the introduction and the analysis of specific elements having symmetry only in average was done first in [1], while an even weaker form of symmetry (namely, orthogonality to piecewise linear continuous functions) was proposed and studied in [2]. Since then their use underwent alternate periods of popularity and oblivion. See e.g. [7], [13], [15,16], [12] and many others. See also [8], [5] and the references therein.
Recently, a general construction of elements with reduced symmetry was presented in [4] and [3]. Their construction relies on a very elegant but quite abstract procedure, requiring rather sophisticated instruments. We present here a new proof of their result and related ones, using much more elementary and classical techniques. It is clear to us that the construction in [4] still has the merit of having inspired the choice of these elements (and having provided the first proof of their convergence). Nevertheless we believe that our much simpler construction might be interesting for many readers.
The plan of the paper is as follows. In Section 2 we recall the mixed formulation of linear elasticity problems and the general setting for their approximation. In Section 3 we recall first the non-symmetric formulation, in which the variational problem is posed in a space of non-symmetric tensors, and the symmetry is then imposed by means of a suitable Lagrange multiplier. The well-posedness of such formulation is well known, but it is proved here with a different approach, based on the solution of suitable auxiliary Stokes problems. The discrete counterpart of this approach is then used to construct families of reduced-symmetry elements, first in two dimensions and then in three dimensions, that include in particular most known two-dimensional elements in the literature. An alternative approach for proving stability and error bounds for elements with reduced symmetry is introduced Section 4. This approach is not based on Stokes auxiliary problems, but rather on a new interesting property that is verified by several common interpolation operators: the idea is that when you construct the interpolant of a symmetric tensor in a finite element space made of (a priori) non-symmetric tensors, the interpolant tensor might (and often does) inherit part of the symmetry of the interpolated tensor. This could produce an interpolant having much better symmetry properties than the ones explicitly required during the interpolation process. This new and nice feature is then used to prove the convergence of the Arnold-Falk-Winther family in the three-dimensional case.
2. Linear Elasticity problems; stress methods 2.1. Continuous formulation of stress methods. We consider in this paper a mixed approach to linear elasticity problems, that is we use as main variable a symmetric stress tensor, chosen in a suitable space. We therefore define, We recall the definition of the trace of a tensor and of the deviatoric where I is the identity tensor. Note that tr(I) so that in (2.5) we have tr(τ D ) = 0. Note as well that (2.5) can equally be written as which, applied to the case of a tensor τ = grad v (for some v), gives At this point we can set a(σ, τ ) := and we can write our simple linear elasticity problem as: find (σ, u) ∈ Σ S × U such that Remark. The first equation represents the constitutive law and the second one the equilibrium condition. It must be clear that although we consider a linear model, the results can be transposed to more realistic non linear models.
We thus have to consider the standard conditions for existence and uniqueness of the solution to this problem. It is very easy to check that and that We thus have an inf-sup condition and coercivity so that our problem is well posed. However, trouble arises when we have to deal with a very large λ (nearly incompressible materials).
In fact it is clear that the coercivity constant which appears in (2.12) goes to zero like 1/λ when λ → +∞ so that the stability properties of problem (2.10) seem to deteriorate for large values of λ. Actually, the situation is not as bad as it seems, because, as is well known, we do not need coercivity to hold for every τ ∈ Σ (or Σ h ) but only for τ ∈ KerB (respectively, KerB h for discrete problems). In particular, the continuous formulation (2.10) does not break down when λ → ∞, because of the following proposition, whose proof can be found in [2] or in [8] [5] Proposition 1. There exists a constant C > 0 such that, for every τ ∈ Σ satisfying If we work in the subspace we know that the set is precisely made of tensors satisfying (2.13) and (2.17) divτ = 0 Hence, from Proposition 1 we have The stability constant of our problem is therefore independent of λ.
Remark. It must be noted that condition (2.13) refers to the fact that with Dirichlet boundary conditions, in incompressible problems, pressure is defined only up to an additive constant. The condition can then be applied a posteriori. It disappears whenever Neumann boundary conditions are imposed on a part of the boundary. From the mathematical point of view we can also remark that taking τ = I in the first equation of (2.10) we immediately have that the solution σ belongs toΣ.
To avoid unnecessary complications, we shall often use, in what follows, the spaces Σ and Σ S instead ofΣ andΣ S .

2.2.
Numerical approximations of stress formulations. If we now choose some finitedimensional subspaces Σ Sh of Σ S and U h of U, we must be careful to have the discrete analogues of (2.11) and (2.18) verified. However we have to face a delicate point. In order to prove an inequality of type (2.18) we needed, in Proposition 1, to have divτ = 0. Hence our life would be a lot easier if we had the "inclusion of the kernels property": KerB h ⊂ KerB. In other words, we would like our spaces Σ Sh and U h to satisfy the following property: At the same time, the inf-sup condition (2.11) is related to the existence of an operator There are in the literature many examples of discrete spaces satisfying (2.19), (2.20), and (2.21) for the approximation of spaces of type H(div; Ω) and L 2 (Ω) when dealing with mixed formulations. The most popular are surely the Raviart Thomas elements [14] and the Brezzi-Douglas-Marini [6], but many others are available: see e.g. [8], [5]. It seems, at the first sight, that we could just use a pair of vectors in H(div; Ω) to approximate Σ, but we should not forget the symmetry of the tensors in Σ S . The problem of finding subspaces of Σ S and U satisfying (2.19), (2.20), and (2.21) is actually very difficult. One of the (nowadays) classical remedies is to give up the symmetry of τ and enforce it back in a weaker form by some Lagrange multiplier. This is what we are going to do in the next section.

Relaxed symmetry
3.1. Continuous formulation of the relaxed symmetry approach. The idea of relaxing symmetry was, to our knowledge, first used by [11] and his school; it was then used by [1] and then by [2]. Other recent results can be found in [7], [13] and [15,16].
Remark. It is worth recalling that the symmetry of the stress tensor is in fact a simplified way of expressing a conservation law, namely the conservation of angular momentum. This should make it easier to understand why symmetry is difficult to enforce. Conservation laws are not easily imposed exactly.
In fact the point of using spaces like H(div; Ω) and its discrete counterparts is to get a strong form for conservation of momentum. For the moment, let us try to define a variational formulation suitable for our purpose. Given a tensor τ ∈ Σ, we define its skew-symmetric part as, We now define a space of skew-symmetric tensors, and we introduce a new bilinear form: We can see that, in general, an approximation of (2.10) with relaxed symmetry requirements corresponds to a conforming approximation of the following continuous problem: find We now have to prove an existence result for this problem. With respect to the general theory, what we need is an inf-sup condition of the form, where, here and in all the sequel, C denotes a generic constant that is independent of h. The above condition is indeed satisfied, as we can see in the following proposition.

Proposition 2.
There exists a constant C such that for any v ∈ U and γ ∈ X, there exists τ ∈ Σ such that Proof. We first give the proof for the two-dimensional case and afterwards for the threedimensional one. We give it in detail because the technique will be relevant for the construction of the discrete approximations. The construction of τ will be done in two steps. The first one is to build a tensor τ 1 ∈ Σ such that, . This is easily done, even with a symmetric τ 1 . One could, for instance, solve a classical elasticity problem and take the associated stress field. The second step is to correct this tensor by a divergence-free tensor τ 2 such that as(τ 2 ) = γ − as(τ 1 ). In the two-dimensional case this divergence free tensor is obtained by taking the i − th row (i = 1, 2) made by the curl of the i − th component of a vector Ψ ≡ (ψ 1 , ψ 2 ), that is, One sees immediately that the condition as(τ 2 ) = γ − as(τ 1 ) is equivalent to where for every scalar q the skew-symmetric tensor S 2 (q) is defined by To satisfy equation (3.10) with the required continuity condition, it is then sufficient to solve a Stokes problem for Ψ. In the three-dimensional case, the situation is slightly more complex. The first step (3.8) holds unchanged, but the divergence-free tensor τ 2 will now be the curl of another tensor of the form This means that we will look for a τ 2 of the form: whose skew-symmetric part as(τ 2 ) is individuated by the vector (3.14) The construction of τ 2 such that as(τ 2 ) = γ − as(τ 1 ) is thus equivalent to solving three Stokes problems, (2) and φ (3) have been found, we can construct Ψ such that 3 . (3.17) 3 , 3 , Denoting by Φ the tensor having φ (1) , φ (2) , and φ (3) as columns, it is immediate to see that We may write the algebraic relation (3.20) as It is also immediate to verify that the above construction, which relies on the solution of well posed Stokes problems, satisfies the required continuity conditions as well.
Remark. The above construction is not the most general in the three-dimensional case. It is not necessary to get Ψ with all the regularity implied by our procedure. This will have as a consequence that some approximations cannot be generated with the discrete equivalent procedure.

3.2.
Numerical approximation of relaxed-symmetry formulations. We can now start considering the approximation of the variational formulation (3.4). We want to choose subspaces Σ h , U h , X h of Σ, U, X and to solve the problem: Here again, we shall rely on the general theory. We shall try to build approximations satisfying the "inclusion of kernels property"(2.19). To do so, we can use some of the finite element discretizations of H(div, Ω) available in the literature. Assuming that we made a choice that takes care of that, we still must check the discrete inf-sup condition: It is well known (see e.g. [9] or [8]) that this can be done by building an interpolation operator Π h : Σ → Σ h satisfying, To do this we shall try to proceed in the same way that we used to prove the continuous inf-sup condition: we shall first build τ 1 h so that its divergence satisfies the first requirement h Σ ≤ C τ Σ , and then we correct this tensor by a divergence-free tensor τ 2 h to obtain the required asymmetry, Referring to the continuous case, we can try to build τ 2 h by solving (discrete) Stokes problems (one in two dimensions and three in three dimensions) returning a tensor Ψ h and then by taking τ 2 h = curl(Ψ h ). It follows that one possible key to our constructions will be stable elements for the Stokes problems, together with the inclusion of the kernels property, that we now have to require in Σ h (rather than Σ Sh as in (2.19)): But the approach through Stokes problems is less effective for the three-dimensional case, and we shall present an alternative one in Section 4. Before considering specific constructions, let us state the error estimate that one can expect for the discrete problem (3.22).
The proof is an easy consequence of the general theory on mixed formulation (see e.g. [5]).
One can see from (3.28) that it is important to balance the quality of the approximation for the three components of the solution. In particular, symmetry must be imposed at least to the same precision as the approximation properties for the other variables Remark. It is clear that the inclusion of kernels (3.27) is not necessary, but it makes the theory easier.
Remark. It is easy to see that if the space Σ h contains (as we implicitly assume) the constant identity tensor I, then solving the problem inΣ h × U h × X h or in Σ h × U h × X h gives exactly the same result.
Remark. In a few cases, we shall be able to build explicitly a basis for the space Σ Sh of discrete symmetric tensors. We shall then be able to consider the problem A first family of relaxed symmetry elements in two dimensions. We start by considering the two-dimensional case. Let us suppose that we can choose a pair Σ h × U h such that (3.27) is satisfied, and moreover the "first part"of the inf-sup condition holds that is This is equivalent to be able to build, for any τ , a tensor τ 1 h such that, To get the full inf-sup condition, we must control symmetry. Let us suppose that we have a stable approximation V h × Q h for the Stokes problem such that Considering (3.30) and (3.34), it is then natural to take X h := S 2 (Q h ) where S 2 (q) is always defined in (3.11) hence we easily obtain the required inf-sup condition (3.23) for the triplet (Σ h , U h , X h ) . With respect to our elasticity problem, we work then with a reduced symmetry property, "symmetry weighted by q h ". We summarize the result in the following proposition Proposition 3. For n = 2, for any couple of spaces Σ h × U h satisfying conditions (3.27) and (2.11), and any couple V h × Q h stable for the Stokes problem and satisfying (3.32), the triplet satisfies the conditions of Theorem 1.
This analysis through a Stokes problem was first introduced in [12]. There are several examples in the literature of approximations that could be inserted in the above theory. These include the PEERS element of Arnold-Brezzi-Douglas [2], or its variant by Brezzi-Douglas-Marini [7]. These include as well the Amara-Thomas element [1] and other possible elements that could be developed along these lines.

3.4.
Relaxed symmetry, the three-dimensional case. Building stable elements with relaxed symmetry is somewhat more tricky for the three-dimensional case. The basic idea is again to start from elements satisfying the first part (3.30) of the inf-sup condition and then to correct for symmetry. We can still use the same trick as in the two-dimensional case and rely, following the continuous case, on solving three Stokes problems. In more details, we start from a pair Σ h × U h (in three dimensions this time) such that (3.27) is satisfied, and moreover the "first part"of the inf-sup condition holds: that is We recall once more that this is equivalent to be able to build, for any τ , a tensor τ 1 h such that, To get the full inf-sup condition, we have again to work on the symmetry condition. Let us suppose that we have a stable approximation V h × Q h for the Stokes problem in three dimensions. For any triplet of vectors φ (1) , φ (2) , and φ (3) , we denote by Φ(φ (1) , φ (2) , φ (3) ) the tensor having them as columns, and we denote by [V h |V h |V h ] the tensor space containing all possible tensors built in his way. Our requirement (in place of the two-dimensional (3.32)) is now that where A is still defined as in (3.21). If this is satisfied, we can then proceed as we did for the continuous case. We do not repeat the procedure here. We just point out that, instead of S 2 (Q h ), in three dimensions we define, for every vector q, the tensor S 3 (q) given by We summarize the result for the three-dimensional case.
satisfies the conditions of Theorem 1.
Remark. We already noted that using a Stokes problem was not the most general way of obtaining the continuous inf-sup condition. In the same way, the above result will enable us to obtain some useful constructions of relaxed symmetry tensors. However, it does not yield all constructions.

Example 1.
As examples of applications of the above strategy we could consider the threedimensional version of the PEERS element. We define Σ h as the space of tensors (in 3 dimensions) where each line is an element of the lowest order Raviart-Thomas space on tetrahedra. Using for U h a space of piecewise constant vectors, it is immediate that we have (3.27) and (3.36). For the Stokes problem we use the three-dimensional MINI element: where B 4 is the space generated by the elementwise quartic bubbles b 4 , obtained as the product of the equations of the four faces. We can then augment the space Σ h by adding, in each element, curl(A([V h |V h |V h ]). This will leave (3.27) and (3.36) still holding true, and (3.38) will now hold as well (we hammered it into the method). All the assumptions of Proposition 4 will then be satisfied. Example 2. Consider, for each face f of each tetrahedron K, the cubic function b f 3 that vanishes identically on the three other faces and has value 1 at the barycenter of f . Then consider the vector valued function b f 3 n f where n f is the unit normal vector to f . Let finally B f aces be the space generated by all these vector valued functions all over the domain, and consider a Stokes element where V h = (L 1 1 ) 3 B f aces and Q h is made by piecewise constant functions. We consider now Σ h as the tensor space obtained using a lowest-order Raviart-Thomas element per line (as above), augmented with curl (A(B f aces )), and U h made by piecewise constant vectors. Then all the assumptions of Proposition 4 will be satisfied. The space of tensors thus obained is unfortunately not easy to make explicit.
Example 3. The three-dimensional equivalent of the Amara-Thomas element. As in the two-dimensional case, it is easy to obtain a suitable construction based on the Crouzeix-Raviart element for the Stokes problem. This construction yields a very rich space of tensors which is not likely to be of practical use and that we shall not try to make explicit here.

Example 4. A simple second order element
We define Σ h as the space of tensors (in 3 dimensions) where each line is an element of the Raviart-Thomas RT 1 space on tetrahedra. We can then use U h = L 0 1 , the space of discontinuous piecewise linear vectors; it is immediate that we have (3.27) and (3.36). For the Stokes problem we consider the Taylor-Hood element in which velocity is approximated by a space of quadratic elements (L 1 2 ) 3 and where pressure is continuous piecewise linear (L 1 1 ). This immediately yields a second order elasticity element in which we have . Symmetry is enforced as in the PEERS element but we now have second order accuracy. This construction is obviously a member of an infinite family, using higher order Raviart-Thomas elements and corresponding generalized Taylor-Hood elements.

An alternative approach
In order to explain the following variant of the above strategy, it is convenient to recall the definition of the permutation tensor (or pseudo-tensor) in two and three dimensions: for n = 2 the double tensor P is given by Note that we could have summarized (4.1) and (4.2) in one formula (as in most textbooks on tensors), using the concept of even and odd permutations. We chose the above presentation for the sake of clarity, considering that the cases n = 2 and n = 3 will be the only ones of interest for us.
Then we denote (4.5) x ∧ y the external product of two vectors. In two dimensions, this is a scalar given by (4.6) x ∧ y := (P · y) · x = (P ij y j )x i and in three dimensions it is a vector given by where in (4.6) and (4.7) (and in all the rest of the paper) the Einstein convention of summation of repeated indices is employed.
In a similar way we can define the wedge product of a double tensor τ with a vector v as in two and three dimensions, respectively. We recall now a useful property of tensor calculus. We denote by x ≡ (x 1 , x 2 ) T or x ≡ (x 1 , x 2 , x 3 ) T (respectively in two or three dimensions) the vector containing the independent variables. Let τ be given in H(div; Ω). In two dimensions we have (4.9) div(τ ∧ x) = (divτ ) ∧ x + P : τ as it can be easily seen by (4.10) (P ij x j τ ir ) /r = P ij δ jr τ ir + P ij x j (τ ir ) /r = P : τ + (divτ ) ∧ x.
In particular we have (4.13) div(τ ∧ x) = τ : P whenever div τ = 0 with its obvious analogue in dimension n = 2. Multiplying (4.13) times a vector p ∈ (H 1 (K)) n , we get (4.14) K div(τ ∧ x) · p dx = K τ : P · p dx whenever div τ = 0 that integrated by parts reads where n K is the outward unit normal vector to ∂K. In two dimensions, instead, (4.15) obviously becomes (4.16) Remark. Using the following identities we can write (4.15) in the form which will be more suitable to define degrees of freedom for a tensor τ .
The idea, now, is to use (4.19) or (4.16) with τ = σ − Π h σ. Indeed, this is what we shall do in the sequel. We will deal first with the three-dimensional case, and then we will briefly present the corresponding two-dimensional results.
Theorem 2. Assume that n = 3, and assume, for simplicity, that Ω is a convex polyhedron.
We now introduce a space W h of vector functions so that X h = S 3 (W h ) defines the reduced symmetry. We now suppose that there exists a mapping Π h from (H 1 (Ω)) n×n S into Σ h satisfying: Π h τ H(div;Ω) ≤ C τ (H 1 (Ω)) n×n S with the following properties: Then the triplet satisfies the conditions of Theorem 1.
Remark. It is clear that the convexity assumption is unnecessarily strong. Indeed, as is well known, a little bit of regularity more than H(div; Ω) will be sufficient, in applications, to build Π h . See e.g. [8] or [5].
Remark. The choice of the space W h is arbitrary even though we shall use below W h = U h . We could also take for W h a space of continuous functions. In this case condition (4.23) reduces to so that reduced symmetry then relies only on internal degrees of freedom.
Theorem 2 opens a new way of getting reduced symmetry. In particular we can recover, with a simpler proof, the elements with reduced symmetry of Arnold-Falk-Winther [4].
Example 5. The Arnold-Falk-Winther family. Let us consider for k ≥ 1 the following approximation (4.29) It is well known that for BDM k vectors we can use, as d.o.f. for each tetrahedron K, • f τ · n f K p k ds for every face f and every polynomial p k of degree k • For k ≥ 2, K τ · φ k−2 for every element φ k−2 of the first Nédéléc family N k−2 , defined as It is easy to check that, taking (Σ h , U h , X h ) as in (4.29), the above properties imply that the assumptions of Theorem 2 are verified.
Remark. The first member of the the Arnold-Falk-Winther family is To specify the tensor τ h we need 9 d.o.f. on each face. This is obviously not optimal in regard of (4.22) and (4.23) where only 6 conditions are generated by v h and w h in U h = (L 0 0 ) 3 . It is indeed stated in [4] that is possible to reduce the number of degrees of freedom on each face from nine to 6. This is not as good as in the two-dimensional case where the tangential part can be taken constant. The proper choice of space, using the right number of degrees of freedom will be a consequence of the following general result.
Let f be a face of some tetrahedron of a mesh. Let P k (f ) andP k (f ) denote respectively the set of polynomials and homogeneous polynomials of degree k on f . Let n be the normal to the face and denote τ nn and τ nT = τ n − τ nn n the normal and tangential part of the vector τ n . We also denote x ⊥ T := x ∧ n. With an abuse of language, we shall consider τ nT and x ⊥ T as two-dimensional vectors, and more generally we will identify, whenever convenient, all vectors tangential to the face f with their two-dimensional projection on f .  τ nn ∈ P k (f ) and for the tangential part It is easy to see that in order to generate this space, it is sufficient to consider functions of the form (4.34) p + x ∧q where p ∈ P k−1 is a general vector valued polynomial of degree k − 1 butq ∈P k−1 is a vector valued homogeneous polynomial of degree k − 1. Now we want to evaluate on a face f of some tetrahedron To do so, we use on the face a set of orthogonal co-ordinates defined by the normal n and two tangential vectors, s and t. We then write, (4.36) x = x n n + x s s + x t t p = p n n + p s s + p t t q =q n n +q s s +q t t τ · n = τ nn n + τ ns s + τ nt t An elementary computation then yields (4.37) (p + (x ∧q)) · τ · n = (p n − x tqs + x sqt )τ nn + (p s − x nqt + x tqn )τ ns + (p t + x nqs − x sqn )τ nt Now we see that τ nn is multiplied by a full polynomial of degree k and we need it to be of the same order. For the tangential terms, we recall that x n is constant on the face so that the terms x nqt and x nqs can be absorbed by p s and p t respectively. That leaves us with something of the form (4.38) τ ns τ nt · p s + x tqn p t − x sqn which shows that the tangential part of τ n needs only to be in N k−1 (f ) which is a space smaller than P k (f ).
Remark. A simple count shows that the number of degrees of freedom on each face is then (k + 1)(k + 2)/2 + k(k + 1) + k while for the whole (BDM k ) 3 , we would have 3(k + 1)(k + 2)/2 For k = 1 we have 6 instead of 9 and for k = 2, 14 instead of 18.

4.2.
The two-dimensional case. For the sake of completeness, we now show the twodimensional results that can be obtained by the corresponding version of Theorem 2. The only (minor) difficulty in going from the three-dimensional to the two-dimensional case is to make explicit the special form of some formulas. To start with, we denote from now on by x ⊥ := (x 2 , −x 1 ) the orthogonal of x. The starting point is now (4.16). A simple computation shows that instead of (4.17) we now have: where we used the notation (4.40) x ⊥ ⊗ ∇p := x 2 ∂p/∂x 1 x 2 ∂p/∂x 2 −x 1 ∂p/∂x 1 −x 1 ∂p/∂x 2 .
Theorem 4. Assume that n = 2, and assume, for simplicity, that Ω is a convex polygon. Assume that Σ h × U h are such that We now choose a space of scalar functions W h such that X h = S 2 (W h ) defines a reduced symmetry. If the following properties are satisfied: then the triplet satisfies the conditions of Theorem 1.
The proof is the same as for Theorem 4.
Example 6. We can take (4.47) Σ h = (BDM k ) 2 U h = (L 0 k−1 ) 2 Q h = L 0 k−1 . As in the three-dimensional case, we can introduce on an edge e a local basis with normal and tangential vectors n and t. On every edge e we denote as previouslyP k−1 (e) the homogeneous polynomials of degree k − 1 on e. Taking into account that x n is constant on e and forgetting redundant conditions, we get from (4.45): (4.48) e (τ − Π h τ ) nn x tq ds = 0 ∀q ∈P k−1 (e) and (4.49) e (τ − Π h τ ) · n · q ds = 0 ∀q ∈ (P k−1 (e)) 2 .
We thus see that the normal component must be one degree higher than the tangential component.
Remark. In the two-dimensional case, the case k=1 of Example 5 was introduced in [12]. The proof is based on the stability of the P 2 − P 0 element for the Stokes problem. In the three-dimensional case, the P 2 − P 0 element is not a stable element for the Stokes problem which justifies our alternative construction. The reduced case was also considered and it was shown that it was sufficient to have, on each edge e, τ nn ∈ P 1 (e) for the normal component but only τ nt ∈ P 0 (e) for the tangential component.

Conclusion
We have been able to obtain by elementary methods many constructions of tensors with relaxed symmetry for mixed elasticity problems. Some of them have a large number of degrees of freedom which makes them difficult to use. However it seems that the price of symmetry is high, especially if inclusion of kernels, which implies strong conservation of momentum, is required.