Printed in Great Britain PII: S0020-7225(97)00036-0 0020-7225/97 $17.00 + 0.00
S Y M M E T R Y C L A S S E S A N D H A R M O N I C D E C O M P O S I T I O N F O R
P H O T O E L A S T I C I T Y T E N S O R S
S A N D R A F O R T E and M A U R I Z I O V I A N E L L O *
Dipartimento di Matematica, Politecnico di Milano, Piazza Leonardo da Vinci 32, 20133 Milano, Italy
(Communicated by E. S. SUHUBi)
Abstract--Two different definitions of symmetries for photoelasticity tensors are compared. A count of such symmetries based on an equivalence relation induced on the set of subgroups of SO(3) was presented by Huo and Del Piero, who proved the existence of exactly 12 classes. Here, another viewpoint is chosen, and photoelasticity tensors themselves are divided into symmetry classes, according to a different definition. By use of group theoretical techniques, such as harmonic and Cartan decomposition, it is shown that this approach again leads to 12 classes. ~) 1997 Elsevier Science Ltd.
1. I N T R O D U C T I O N
Photoelasticity tensors map the infinitesimal strain E into the variation of the dielectric tensor D of an ordinary dielectric body. Depending on the number of different eigenvalues of D, dielectric crystals are classified into three symmetry types (biaxial, uniaxial and cubic), which are expected to be altered by the application of a strain field. This phenomenon has been known since the last century: in 1850 Moigno and Soleil noticed that uniaxial crystals may become biaxial under pressure in a suitable direction. The photoelastic effect has relevant applications in the study of the distribution of strains in a loaded transparent body. For an exhaustive treatise on this topic we refer the interested reader to the book by Coker and Filon [1].
In this research we are essentially concerned with the mathematical properties of C, the photoelasticity tensor, viewed as a linear map of the space of symmetric tensors: D - Do = C[E], where D - Do is the infinitesimal increment in the value of the (symmetric) dielectric tensor. More precisely, our interest rests with anisotropic bodies, for which C is supposed to be only invariant under the action of a p r o p e r subgroup of SO(3), the rotation group (for a discussion of the physical implications of anisotropy on dielectric crystals see, for example, Refs [2, 3]).
The space of photoelasticity tensors, which we shall refer to as Gel, can be identified with the space of elasticity tensors when no assumptions about the existence of a stored energy function are made (see, for example, Ref. [4]). However, since this assumption is rarely missing, we prefer to refer our algebraic discussion to a context where it seems to be more acceptable. As expected, there is a relevant amount of literature on symmetry properties of these tensors: some papers by Fieschi and Fumi [5-8], sections in books by Thurston [9], Jagodinski [10] and Schouten [11]. Perhaps surprisingly, the question of determining the number and types of symmetries was explicitly first posed and rigorously answered only in a fairly recent paper by Huo and Del Piero [12]. The problem, which, as we shall see, appears to have a simple and straightforward formulation, and lends itself to a discussion which is subtler than expected. Indeed, as it is not obvious, there are two different and reasonable ways to define symmetries for linear anisotropic constitutive equations. The interesting and perhaps surprising fact is that even if these different definitions share a common physical goal they may lead to different answers. Indeed, Huo and Del Piero proved that there are exactly 10 "symmetr ies" for e las t ic
bodies, but, according to a different definition, symmetry classes are only eight (a complete
* Author to whom all correspondence should be addressed.
1317
1318 S. FORTE and M. VIANELLO
discussion and some historical considerations are contained in a recent paper by Forte and Vianello [13]). We present in detail these different approaches in the next section, after introducing the appropriate notation and terminology.
Photoelasticity tensors were similarly discussed in Ref. [12], and the number of symmetries was found to be 12. Here, our goal is to investigate whether a dichotomy in this number would also emerge for Gel, in view of the different approach taken. We anticipate a negative answer, which should be contrasted with the mentioned situation for linear (hyper) elasticity.
The techniques we use for our proof are based on "harmonic" and "Car tan" decomposition of Gel. These are mathematical tools which, in our opinion, give a deeper understanding and a geometric feeling for the action of SO(3) on Gel, making most of the deductions quite straightforward. As we shall explain in more detail later on, these techniques are mostly used in other contexts, such as group representation theory (see, for example, the book by Golubitski et al. [14]). However, there is a line of research, going back to pioneering papers by Schouten, which is interested in applications to continuum mechanics, as shown by Backus [15], Baerheim [16], Cowin [17], and, most important for our research, by Spencer [18]. The essential idea can be described as the reduction of Gel into a direct sum of subspaces irreducible under the action of SO(3). This, together with a nice geometric view of the action of 0(2) , named after Caftan, yields a useful perspective on the whole problem.
2. SYMMETRY G R O U P S AND SYMMETRY CLASSES
We denote vectors and second-order tensors in a three dimensional Euclidean space V by small and capital boldface letters, while we use Greek letters for scalars. Third-order tensors are written with a sanserif boldface font, such as $ or H, and a blackboard bold font is used for fourth-order tensors: C or H.
Lin is the space of second-order tensors, Sym is the subspace of all symmetric A e Lin. We denote by 0 (3) the group of orthogonal tensors Q, for which Qa.Qb = a.b for any pair of vectors a and b. The unit element is 1, the identity map in Lin. The subgroup of rotations is the set of orthogonal tensors with determinant equal to 1, and is denoted by SO(3).
We fix once and for all an orthonormal basis ei (i = 1,2,3), which, for simplicity, can sometimes be written as i,j,k. Components of tensors with respect to this basis are denoted by subscripts.
Finally, for a ~ V and 0 • [0,27r), we denote by Q(a,0) the right-handed rotation about a of an angle 0 and write 0 for Q(i +j + k,27r/3).
Let Gel be the space of all fourth-order tensors C such that Ci/kt= C/ikt= Cgj~k. We focus our attention on the set of all orthogonal tensors which are symmetry transformations for C • Gel.
More precisely, we define the symmetry group of C to be
g(C): = {Q ~ SO(3 ) IC[QTEQ] = Q T C [ E ] Q VE • Sym};
(the group properties are easily proved). Properly speaking, this definition should be given using the full orthogonal group 0(3) rather than the subgroup of rotations. However, it is easy to verify that, in any case, g(C) is uniquely determined by its intersection with SO(3).
For C E Gel and Q • SO(3) let Q*C be the tensor such that:
(Q*C)[E] : = Q C [ Q T [ E ] Q ] Q T VE ~ Sym; (Q*C)gjkt: = CpqrsQipQ&QkrQts.
The function which maps each Q E SO(3) into the automorphism of Gel given by C---> Q*C is an action of SO(3) on the linear space Gel (see, for example, Ref. [14]). In view of this, the symmetry group of C is the set of all Q such that Q*C = C:
g (C) = {O e SO(3)IQ*C = C}.
An important consequence of this definition is that, by continuity, symmetry groups are closed subgroups of SO(3). Moreover, g(Q*C) is conjugate through Q to g(C): g(Q*C) = Qg(C)Q T.
Symmetry classes and harmonic decomposition 1319
For G a subgroup of SO(3), we say that C has G symmetry if g(C) D G. In order to verify this property it is sufficient to show that Q*C = C for all Q in a set of generators of G.
Here we are interested in the problem of classifying tensors in Gel according to the symmetry group. Following physical intuition, whenever two material bodies can be rigidly rotated so that their symmetry groups become identical, we say they share the same symmetry class. Thus, we define C1 and C2 to be equivalent if there is a rotation Q such that g(Cl)=g(Q*C2). In view of what noticed before, C1 and C2 are equivalent when their symmetry groups are conjugate:
C 1 - C2¢=> {3Q ~ SO(3) ]g (C , ) = Qg(C2)QT].
It is easy to verify that this is an equivalence relation, as it is the conjugacy relation within the set of subgroups of SO(3). Thus, we simply write
C, - C2¢z> g(C, ) - g(C2).
The resulting equivalence classes in Gel are called symmetry classes and we write {C} for the symmetry class of C. Analogously, for G a subgroup of SO(3) we write {G} for the collection of all its conjugates in the set of subgroups of SO(3). The problem we are interested in can simply be phrased as follows: how many and which symmetry classes are there?
It is very important to compare this point of view with a different approach used by Huo and Del Piero [12]. Their idea is to introduce an equivalence relation among subgroups of SO(3). Let G1 and Gz be equivalent when the set of elements of Gel with symmetry group containing GI can be obtained from the set of tensors with symmetry group containing G2 through the action of a rotation. More precisely, for G a subgroup of SO(3), define Gelas(G) as follows:
Gelas(G): = {C ~ Gel[g(C) ~ G}.
Notice that Gelas(G) is a linear subspace of Gel and that, for different G~ and G2, the corresponding Gelas(G1) and Gelas(G2) are not necessarily disjoint. We define an equivalence relation ~ among the subgroups of SO(3), as follows:
A natural question was asked by Huo and Del Piero in Ref. [12]: how many equivalence classes (elastic symmetries) of rotation subgroups are there? It is by no means obvious that the answer should be the same to our problem, and, indeed, when formulated for classical elasticity tensors, these two different perspectives yield different conclusions. However, as we shall see, this is not the case for Gel, the space of photoelasticity tensors. There are 12 symmetry classes, as there are 12 equivalence classes of rotation subgroups. Notice again that different objects are counted: subsets of Gel, in our case, and collections of subgroups of SO(3), from the point of view of Huo and Del Piero [12]. Thus, it is not surprising that a different mathematical approach is needed. We still feel that it is important to make clear that there are indeed two different notions for symmetries of linear constitutive equations, since it is not uncommon to find quite recent literature where some confusion seems to be present.
In the following part of this section we prepare the groundwork for a precise formulation of our results, which we shall prove later on. It is well known (see, for example, Ref. [14], Chapter XIII, Theorem 6.1) that any closed subgroup of SO(3) is conjugate to one of the groups in the following collection, to which we shall later refer as E:
!
z~
On 5r ¢7
The trivial subgroup, formed by the unit element. (for n -> 2). The set of rotations about k of multiples of 2rr/n. This is a cyclic group with n elements generated by Q(k,2rr/n). (for n -> 2). The dihedral group, generated by Zn and Q(i,Tr), which has 2n elements. The tetrahedral group, generated by D2 and I~, with 12 elements. The octahedral group, with 24 elements, generated by D 4 and t).
1320 S. FORTE and M. VIANELLO
#
SO(2) 0 (2) so(3)
The subgroup of the 60 rotations which take a given dodecahedron centered at the origin onto itself; it can be shown that there is a rotation Q such that # ~ QDsQ~ The subgroup of all rotations such that Qk -- k. The group of all rotations which map k into ± k. The full rotation group.
Since all the symmetry groups are closed subgroups of SO(3), the following result holds.
PROPOSITION 1. For each C • Gel there is exactly one G e ~ such that G is conjugate to g(C).
For G E ~ let Gel(G) be the collection of tensors with symmetry group conjugate to G:
Gel(G): = {C E Gel lg(C) E {G}},G e ~.
Thus, Gel can be viewed as the disjoint union of the equivalence classes Gel(G), as G varies in the collection ~.
The problem of discussing symmetry classes is now reduced to an investigation about which classes Gel(G) are empty and which are not, for each G • ~. Hereafter, we anticipate and make formal our conclusion.
THEOREM 2. For k -> 5, Gel(Zk), Gel(D,) and Gel(N) are all empty.
THEOREM 3. Gel(1), Gel(Z2), Gel(D2), Gel(Z3), Gel(D3), Gel(Z4), Gel(D4), Gel(gY), Gel(~), Gel(SO(2)), Gel(O(2)) and Gel(SO(3)) are not empty.
3. H A R M O N I C D E C O M P O S I T I O N
A tensor of any order is totally symmetric if the components are not changed under permutations of the indexes. A tensor is harmonic if it is totally symmetric and traceless, which means that the trace with respect to every pair of different indexes vanishes. Notice that, in view of total symmetry, it suffices to check this property only for one pair of indexes. This terminology is justified in view of the correspondence between symmetric tensors and polynomials: harmonic tensors are mapped onto harmonic polynomials (see, for example, Refs [15, 16]). We write Dev for the space of second-order harmonic tensors, and denote by Hrm and Hrm, respectively, the spaces of third and second-order harmonic tensors.
THEOREM 4. There is an SO(3)-invariant isomorphism between Gel and the direct sum H r m @ H r m @ D e v @ D e v @ D e v @ V @ R (~) R.
Recall that SO(3)-invariance means that the action of SO(3) commutes with ~0, the isomorphism just introduced, when an appropriate similar action is also defined on vectors and harmonic tensors.
For the action on vectors the obvious choice is (Q*v)i: = Q,v~. For Sym and gym, the spaces of second and third-order harmonic tensors, let
The symmetry group of an element of these spaces is defined as the collection of all rotations which leave this element fixed under the corresponding action.
In view of the isomorphism ¢ we associate to each C e Gel an octuple (H, H, H, I71, I21, v, a, /3) e H r m x H r m x D e v X D e v X D e v x V x R X R , which is called the harmonic decom- position of C. With obvious meaning, we write
^
C = ( ~ , H , H , H , H , v , a , / 3 ) .
An important consequence of this discussion is that the symmetry group of each C e Gel is the intersection of the symmetry groups of all the elements in the decomposition.
Symmetry classes and harmonic decomposition 1321
The proof of Theorem 4 is based on a general method introduced by Spencer [18]: (1) we decompose C into symmetric tensors; (2) we decompose each symmetric tensor into harmonic tensors. For the reader 's convenience, before presenting the complete deduction, we briefly sketch an outline of Spencer's technique, and later go on to adapt his argument to our context.
Let A~,~i~...~,, be the components of a tensor of order n, for which we denote by Au,~...~,)~.+,...~" the sum of the r[ components obtained from Ai,~...i , by all permutations of the first r indexes, divided by r!. Thus, Au,~...~,)i.+,...~, is the symmetrization of Ag, i.~...i,, with respect to the first r indices.
where e~,/~, are the components of the third-order alternating tensor.
Let B~,i~,...#_~,+,...~,i : = eip, i A~,~,i2...~,_~)i~,+,...~. Thus, A(6i~...~,_,)~,...i ° is equal to the sum of the tensor A6~3...~" symmetrized with respect to the first r indices and some terms obtained by contraction of B6~6...i._~i,+,...i,~, with the alternating tensor. Applying by recurrence the identity above, we are finally able to express the given tensor A6~i3...~° as a sum of totally symmetric tensors of order equal or less than n, suitably contracted with isotropic tensors.
PRoov ov T~EORE~ 4. For C ~ Gel, applying to C the identity equation (1) and taking into account its symmetries, for r = 2 we have the trivial result: Cuk ~ = C(q~k ~. For r = 3:
4 C ( i p ) t = 4Ctqko + eilsepqsCfpjk)q q- ejlsepqsC(pik)q Jr ektsepqsC(pij)q.
In view of this identity, C,jkt) are the components of a fourth-order totally symmetric tensor, which we call ~, and Bqk: =epqkCwij) q are the components of a third-order tensor, symmetric with respect to the first two indexes, which we call B. Applying to B the identity equation (1) we have
Thus, SUk:=B(uk) are the components of a third-order totally symmetric tensor $ and Dij: = epqjBpi q are the components of a second-order tensor D. Simple calculations show that D is the sum of a second-order traceless symmetric tensor Hij: = l / 3 [Cqpp - Cppq] and a second-order skew-symmetric tensor W~i: = 1/3[Cpiip - C~eip ]. We denote by v the axial vector v~ = ers~Wrs.
Now consider the identity equation (2). Let Guk: =epq~Cpiqi be the components of a third- order tensor G to which we apply identity equation (1), with n = 3. For r = 2:
2Gok = 2Guj)k + eijsepqsGpq k.
Simple calculations show that Sks: -'- epqsGpqk is symmetric with trace a: = Cppqq - C p q p q . For r = 3:
Let TUk:=G(qk) be the components of a third-order totally symmetric tensor T and Pis: = epqsG(pi )q be the components of a second-order tensor P. It can be easily checked that P = 3D/2 and T = 35/2.
In conclusion, we have been able to express C in terms of a fourth-order totally symmetric tensor §, a third-order totally symmetric tensor S, a second-order symmetric tensor S, a second- order harmonic tensor H and a vector v. We now proceed to the reduction of ~, S and S into harmonic tensors. There is a unique harmonic tensor/4u such that
So = f lu + a6u13, (3)
1322 S. FORTE and M. VIANELLO
where 6ij is the Kronecker delta. In view of equation (3) we may associate to S ~ Sym a unique pair (171,a) where t71 ~ Dev and a is a real number.
For each third-order totally symmetric tensor $ there is a unique third-order harmonic tensor Hij~ such that
Sijk = Hij~ + ( aj~sppi + aiksppj + a i j s ~ ) / 5 .
Easy calculations show that Sppi = - v J3. For the fourth-order totally symmetric tensor ~ there is a unique fourth-order harmonic
where/:/~j: = Spp~i - /38ij /3 of a second-order harmonic tensor Ill and/3 is the real number Sppqq.
Therefore, we may associate to ~ ~ ~ym a tensor H ~ Hrm, a tensor 111 E Dev and a real
number/3. In conclusion, each C corresponds to a unique element of Hrm × Hrm × D e v × D e v
× Dev × V x R × R. This correspondence, described by the map ~p, is linear, since the decomposition is based on the repeated application of identity equation (1), involving linear operations. The dimension of the spaces is 36. The kernel of ~0 is {0}, and therefore, we conclude that q~ is an isomorphism. The SO(3)-invariance is simply verified checking that the following equality holds for all rotations Q:
Q*C = ( Q * ~ ,Q*H ,Q*H,Q*IT-I,Q*I~I,Q*v,a,13).
A complete characterization of the symmetry groups of traceless second-order tensors is a useful result, which is easily proved (see Ref. [13]).
THEOREM 5. Let A,B E Dev. Then:
(i) {g(A)} is either {De}, {O(2)} or {SO(3)}; (ii) {g(A)} fq {g(B)} is either {I}, {Z2}, {D2}, {0(2)} or {SO(3)}.
4. CARTAN D E C O M P O S I T I O N
Let r: =xi +yj + zk and let Pn be the space of homogeneous polynomials of degree n in three variables (1 -< n -< 4). There is a well-known isomorphism ~0 with the space of totally symmetric tensors of order n:
Ti,i2...i,, ~ ~l(Tiliz...i,,): = Tili2...i, rilri2...ri,, E Pn. (4)
The space of harmonic tensors is isomorphic under ~0 to Hn, the space of homogeneous harmonic polynomials of corresponding degree.
Through the isomorphism equation (4) we define an action of SO(3) on P,. For p = ~(Ti,i2...in) ~ P, and Q E SO(3), we define Q*p as
( Q*~O)(Ti, i2...i.) = ~O(Ti,i2...i,,aj, i, Qj2i2...aj, i.).
It is a classical result (see, for example, Ref. [14]) that the space of homogeneous polynomials of degree n is the direct sum of H , and Q,, the space of polynomials which are multiple of p:=xZ+y2+zZ . For each p ~ P, there exists a unique q ~ Q, such that h : = ( p + p q ) ~ H, . We call h the harmonic part of p.
The space of harmonic polynomials H , has an SO(2)-invariant decomposition known as Cartan decomposi t ion (see Ref. [14]). Let w: =x + iy. For 0 - l-< n we denote by st(0 <--l <-n) the
Symmetry classes and harmonic decomposition 1323
harmonic part of z" -Z ,~ (w t) and by tt the harmonic part of z " - ~ ( w ~ ) . For simplicity we write u instead of So and notice that to = 0. Finally we define
K0 = span (u), K / = span (st,tt), 1 <-- 1 <-- n.
CARTAN DECOMPOSITION. The space Hn of harmonic polynomials of degree n has an SO(2) invariant decomposition:
K0 (~ K, @ ... (~ K,. (5)
Moreover, each rotation Q(k,0) acts on Ko as the identity and, for 1 -< 1 ~ n, as a rotation of 10 on K~. Precisely:
Q ( k , O ) * s t = s t c o s ( 1 0 ) + t t s in (10) , Q ( k , 0 ) * t t = - s l s i n ( l O ) + t t cos ( lO) . (6)
Finally, Q(i,Tr) acts as a reflection on KI (0 -< l -< n). Precisely:
Q ( i , T r ) * u = ( - 1)nu, Q ( i , r r ) * s , = ( - 1)~-ts,, Q ( i , ~ ) * t ~ = ( - 1) (n - t)+ ltt. (7)
It is interesting to notice the close link between Cartan decomposition and some aspects of the theory of K r o n e c k e r p o w e rs , as presented by Zheng and Spencer [19].
The straightforward consequence of Cartan decomposition is that to each h e g2, correspond 2n + 1 real numbers A, at,/3~ (l -< l -< n), such that
h = Au + o ~ l + ~ttt, l <- l <- n.
In view of the SO(3)-invariant isomorphism ~b, Cartan decomposition may be transferred from the space Hn into the space of harmonic tensors of order n. This tensor space is the direct sum of n + 1 subspaces ~0-~(K~)(0 ~ l-< n), for which, without danger of confusion we use again the same symbol K~. We discuss more in detail Cartan decomposition for n = 3, since we borrow the relevant results for n = 4 from Ref. [13].
For H ~ Hrm, let ~ ( H ) = h = A u + a ~ + ~ t t (1-<l-<3). Simple computations show (see Ref. [14], Chapter XIII, Section 7, p. 111) that a base for Kt(0 -< l - < 3) is
u = z 3 - 3 p z / 5 , (s~,t~) = (z2x - px /5 , z2y - py /5 ,
(s2,t2) = (z(x 2 - y2), x y z , (s3,t3) = (x 3 - 3xy 2, 3x2y - y3).
Let IJ=Cb-l(u), St=g, 1(s/), Tt=g,-l(t /) . Therefore, we may write H=H0+H~+H2+H3 where H0=AIJ and, for 1-<1--<3, Ht=cetS~+/3tT~. A straightforward consequence of Cartan decomposition, as described by equations (5)-(7), is the characterization of the symmetry group of H ~ Hrm through the properties of the "components" H~. The reader is encouraged to draw a sketch of the action of 0 (2) on Hrm as described by the above results, and directly check how this point of view makes most of the following proofs trivial. Keeping our geometric approach, we shall refer to Ht as "horizontal" if it is a multiple of S~ and as "vertical" if it is a multiple of 1",.
PROPOS~TXON 6. For I'1 = H0 + H~ + H2 + ]'13 the following implications hold:
(i) g ( H ) = SO(2) O H = H0; (ii) g ( H ) = O(2) ¢:>H = 0;
(iii) g (H) = ZkC:>H = H0, (k -> 4); (iv) g (H) = Dk ¢¢' H = 0, (k -> 4); (v) g ( n ) = Z2Cz>H1 = H 3 = 0 ;
(vi) g ( H ) ~ D2C::~H = H2, with H2 vertical in K2; (vii) g (H) D Z3c~H~ = H2 = 0 ;
(viii) g (H) = D3C::,H = H~ with H3 horizontal in K3; (ix) g (H) ~ 5 ¢ ¢ , H = Hz with H2 vertical in Kz.
PROOF. The only part of this Proposition which needs a real proof is part (ix). Assume that g(H) ~ ft. Since 5- = D~, part (vi) implies that there is a real number r such that H = zT2. Vice
1324 s. FORTE and M. VIANELLO
versa, assume H = r T 2. Then, g ( H ) ~ D 2 follows from part (vi). In order to show that 0 = Q(i +j + k,2~-/3) ~ g(H) consider the harmonic polynomial h corresponding to H through the isomorphism ~O. The polynomial h = ~9(H) can be written as h = rt2. We now show that h satisfies the condition (~*h=h. Notice that for any harmonic polynomial p of degree n: ((~*p)(x,y,z) =p(y,z ,x) , and therefore
(Q*h) (x ,y , z ) = h(y,z ,x) = rxyz = h(x,y,z).
Thus, 0 E g(H). Since 3- is generated by D2 and 0 , we conclude that g(H) = 3-. A similar result holds for Hrm (as shown in Ref. [13], Proposition 2 and Lemma 1).
PROPOSITtON 7. For N = N0 + ~ + 1-12 + 1-13 + 1-~4 the following implications hold:
(i) j ( H ) ~ SO(2)¢¢,H = Hoe::> g ( ~ ) ~ 0(2); (ii) g ( ~ ) ~ Zk¢=>H = ~0, (k-> 5);
(iii) g ( H ) ~ DkC=>N = H0, (k--> 5); (iv) g ( H ) ~ Z2C:>H, = H 3 = 0 ; (v) g ( H) ~ De <=> H ~ = H3 = 0, with both H2 and H4 horizontal in K 2 and K4;
( v i ) g ( l " t ) ~ Z3,~z:~[H] I = [H ]2= 1-14 ~'-~- 0 ;
(vii) g ( H) ~ D3C=>H~ = ~-12 = 1-~ 4 = 0 , with 1-13 vertical in K3; (viii) g ( ~ ) ~ Za<=>H1 = H2 = 1-13=0;
(ix) g ( H) ~ D4¢=> H1 = 1-t 2 ~- ~-13 = 0, with 1-14 horizontal in K4; (x) g ( H ) ~GCz> g (H) ~ 3-¢:> {3A ~ RIH = A(U + 5~4)}.
5. F INAL PROOFS AND C O N S I D E R A T I O N S
The first conclusion we draw from our results is about tensors which are invariant under
8. For C ~ Gel, the following implications hold:
g(C) s Zk ¢=>g(C) ~ SO(2), (k -> 5); g(C) = DkC=>g(C) s O(2), (k -> 5);
Let C=(H,H,H,ft,12I,v,a,/3) be the harmonic decomposition of a given C 6 Gel. Therefore, g (C)=g(H) f3 g(H) n g(rl) n g(fi) n g(H) n g(v).
(i) Suppose g(C) = Zk for k - 5. Part (ii) of Theorem 5, part (iii) of Proposition 6 and part (ii) of Proposition 7 show that g( l t ) f3 g(171) N g(121) = O(2), g(H) = SO(2) and g(H) = SO(2); thus, g(C) = SO(2). The reverse implication is trivial.
(ii) Suppose g(C) = Dk for k - 5. Part (ii) of Theorem 5, part (iv) of Proposition 6 and part (iii) of Proposition 7 show that g( l t ) f'l g(f l) N g(12I) s O(2), H = 0 and g(H) = 0(2) ; thus, g(C) ~ 0(2) . The reverse implication is trivial.
We are finally in a position to complete our discussion with the proofs of the main results: Theorems 2 and 3.
PROOF OF THEOREM 2. TO prove that a symmetry class is empty we assume the existence of a tensor in that class and show that the assumption leads to a contradiction.
(i) Assume that C ~ G e l ( Z k ) , for k>-5. For some rotation Q, Q g ( C ) Q r = Z , , or equivalently g(Q*C)= Zk. Part (i) of the Lemma implies that g ( Q * C ) ~ SO(2), which means g ( C ) ~ QTSO(2)Q. But it is not possible for a conjugate of Zk to contain a conjugate of SO(2).
(ii) Assume that C ~ Gel(Dk), for k -> 5. As in part (i), with help of part (ii) of the Lemma, we reach again a contradiction.
SO(2) (transversely hemitropic) and under 0 (2) (transversely isotropic).
LEMMA
(i) (ii)
PROOF.
Symmetry classes and harmonic decomposition 1325
(iii) Suppose C ~ Gel(#). For some rotation Q,Qg(C)Qr=~, or equivalently g(Q*C)=#. Since there is a rotation () such that the symmetry group 5~ of a dodecahedron contains (~Ds(~ T, we deduce that, for (): = ()TQ,g(Q*C) = Ds. Part (ii) of the Lemma implies that g(0*C) = O(2), which means g(C) ~ QTo(2)(). However, it is not possible for a conjugate of # to contain a conjugate of 0(2).
At this stage, we have shown that most symmetry classes are empty, except for 12 of them. The final result consists in proving that each one of these remaining classes contains at least one element.
PROOF OF THEOREM 3. We prove the Theorem showing how to construct, with the help of harmonic and Cartan decomposition, tensors with exactly the required symmetry. At each step of the proof ce and/3 denote arbitrary real numbers.
(i) Let C: = (H, H, H, 17-I, l~I, v, a, fl), with H and 171 such that g(H) (3 g(17-1) = I. For H, H, I~I and v arbitrarily chosen, g(C) = I.
(ii) Choose H, 171, l?l such that g(H) fqg(H)fqg(H)=Z2, H with HI=H3=0, and H such ^
that Ht = 1-13 =0. For v parallel to k, let C: = (H, H, H, H, H, v, a,/3). Then, g(C) =Z> (iii) Let H be an element of Dev with symmetry group 02. Choose H such that fl~ 1 = H3 = 0,
with both H2 and H4 horizontal in K2 and K4, and H such that Ho = H1 = H3 = 0 with H2 vertical in K2. Then, the symmetry group of C: = (H, H, H, H, H, 0, or, fl) is D2.
(iv) For the rest of the proof, we let H be such that g(H)= 0(2). Choose H with only Ho and H3 different from zero, and H with only Ho ~ 0. For v parallel to k, let C = (H, H, H, H, H, v, a, /3). Then g(C) = Z3.
(v) Choose H with Ht = H2= H4=0, and H3 vertical in K3. Moreover, let H be such that Ho = Hi = H2 = 0 with H3 horizontal in K3. Then, the symmetry group of C: = (H, H, H, H, H, 0, a,/3) is D 3.
Let Ho and H4 be the only non-zero components of H and let H be such that only Ho # 0. For v parallel of k, C: = (H, H, H, H, H, v, a , /3) has symmetry group Z4. For k0 with Ht = H2= ~3---0, and ~4 horizontal in K4, let C:=(H, 0, H, H, H, 0, c~,/3). Then g(C) = D4. Choose H: = a(U + 5~4) for some a ~ 0 and H: = r'r2, for some r ~ 0. Let C: -- (8, H, 0, 0, 0, 0, a, /3). Then g(C) =3 .
(ix) As above, let H: = a(U + 5S4) for a # 0. Then g(C) =6, for C: -- (H, 0, 0, 0, 0, 0, a,/3). (x) Choose H with only Ho # 0, and H such that only Ho ~ 0. Let v be parallel to k and
C: -- (H, H, H, H, H, v, ce,/3). Then g(C) = SO(2). (xi) Let C: = (H, 0, H, H, H, 0, a, /3), where H0 is the only non-zero component of H. Then
g(C) = 0(2). (xii) C: = (0, O, O, O, O, O, a , /3) is obviously isotropic.
(vi)
(vii)
(viii)
As a concluding remark, we stress again that it is quite possible to conceive a different technique to prove our theorems. One should first reduce to a finite number of cases the collection of possible non-empty symmetry classes, with direct computations, and then show that no further reductions are possible. Very likely, this program can be implemented along the line proposed in the literature on elasticity tensors, as discussed in Ref. [13], (Section 6, Historical remarks). Here, we do not wish to enter the details of this alternative approach, but, still, would like to draw the reader's attention to the conceptual advantages gained through the use of harmonic and Cartan decompositions. Indeed, as remarked in Section 1, this geometric approach sheds considerable light on the action of the group of rotations, making quite obvious the mathematical reasons which justify our conclusion. More interestingly, we feel that this approach might be of great help in suggesting a useful technique for discussing and obtaining invariants of tensors under the action of different subgroups of SO(3), a problem which recently
1326 S. FORTE and M. VIANELLO
raised the interest of some researchers (see the contributions by Ting [20], Boehler et al. [21], Zheng et al. [22, 23], Betten and Helisch [24-26], Blinowski et al. [27]).
Acknowledgements--Research supported by GNFM of CNR (Italy).
REFERENCES
1. Coker, E. G. and Filon, U N. G., A Treatise on Photoelasticity, revised by H. T. Hessop. Cambridge University Press, Cambridge, MA, 1957.
2. Eringen, A. C. and Maugin, G. A., Electrodynamics of Continua I. Springer, New York, 1990. 3. Landau, L. D. and Lifshiz, E. M., Electrodynamics of Continuous Media, volume 8 of Course of Theoretical Physics.
Pergamon, Oxford, 1960. 4. Gurtin, M. E., in Mechanics of Solids IL volume Via/2 of Handbiich der Physik, ed. C. Truesdell. Springer, Berlin,
1972. p. 1. 5, Fieschi, R. Physica 1957, 24, 972 6, Fieschi, R. and Fumi, E G. Nuovo Cimento 1952, 9, 9 739 7. Fieschi, R. and Fumi, E G. Nuovo Cimento 1953, 10, 7 865 8. Fumi, E G. Nuovo Cimento 1952, 9, 9 184 9. Thurston, R. N., in Mechanics of Solids IV, volume Via/4 of Handbiich der Physick, ed. C. Truesdell. Springer,
Berlin, 1974, p. 109. 10. Jagodinski, H., in Kristallphysik 1, volume VII/1 of Handbiich der Physik, ed. S. Fl0gge. Springer, Berlin, 1995, p. 1.
11. Schouten, J. A., Tensor Analysis for Physicists. Oxford University Press, London, 1951. 12. Huo, Y.-Z. and Del Piero, G. Journal of Elasticity 1991, 25, 203 13. Forte, S. and Vianello, M. Journal of Elasticity 1996, 43, 2 81 14. Golubitsky, M., Stewart, I. and Schaeffer, D. G., Singularities and Groups in Bifurcation Theory, Vol. 2. Springer,
New York, 1985. 15. Backus, G. Reviews of Geophysics and Space Physics 1970, 8, 3 633 16. Baerheim, R. Quarterly Journal of Mechanics and Applied Mathematics 1993, 46, 3 391 17. Cowin, S. C. Quarterly Journal of Mechanics and Applied Mathematics 1989, 42, 2 249 18. Spencer, A. J. M. International Journal of Engineering Science 1970, 8, 475 19. Zheng, Q.-S. and Spencer, A. J. M. International Journal of Engineering Science 1993, 31, 4 617 20. Ting, T. C. T. Quarterly Journal of Mechanics and Applied Mathematics 1987, 40, 3 431 21. Boehler, J. P., Kirillov, A. A. and Onat, E. T. Journal of Elasticity 1994, 34, 97 22. Zheng, Q.-S. ZeitschriftAngewandt Mathematik und Mechaniks (ZAMM) 1994, 74, 8 357 23. Zheng, Q.-S. AMR Applied Mechanics Reviews 1994, 47, 11 545 24. Betten, J. and Helisch, W. ZeitschriftAngewandt Mathematik und Mechaniks (ZAMM) 1992, 72, 1 45 25. Betten, J. and Helisch, W. ZeitschriftAngewandt Mathematik und Mechaniks (ZAMM) 1995, 75, 10 753 26. Betten, J. and Helisch, W. ZeitschriftAngewandt Mathematik und Mechaniks (ZAMM) 1996, 76, 2 87 27. Blinowski. A., Ostrowska-Maciejewska, J. and Rychlewski, J. Archives of Mechanics 1996, 48, 2 325
(Received 27 September 1996; accepted 13 December 1996)
