Matrix Pencils and a Generalized Clifford Algebra Christopher J

Linear Algebra and its Applications 313 (2000) 1–20 www.elsevier.com/locate/laa

Matrix pencils and a generalized Clifford algebra Christopher J. Pappacena1 Department of Mathematics, University of Southern California, Los Angeles, CA 90089-1113, USA Received 16 March 1999; accepted 27 December 1999 Submitted by T.J. Laffey

Abstract

Let F be an algebraically closed field, and x1,...,xm be commuting indeterminates over F. For a given monic polynomial φ(z) ∈ F [x1,...,xm][z] which is homogeneous (viewed as an element of F [x1,...,xm,z]), we construct an associative algebra Cφ which we call the generalized Clifford algebra of the polynomial φ. This construction generalizes that of Roby’s Clifford algebra, and is a universal algebra for the problem of finding matrices A1,...,Am ∈ Mn(F) for some n such that φ(z) is the minimal polynomial of the matrix pencil x1A1 + ···+xmAm. Our main result is that, if φ(z) is quadratic, then Cφ is either a matrix algebra of dimension a power of 2 or a direct sum of two such matrix algebras, and we conclude that the problem of finding m matrices in Mn(F) whose pencil has a prescribed quadratic minimal polynomial can be solved, if and only if n is an appropriate power of 2. We apply this result to the problem of bounding the lengths of generating sets for matrix algebras and discuss some of the difficulties encountered when the degree of φ(z) is >3. © 2000 Elsevier Science Inc. All rights reserved. AMS classification: Primary: 15A22; 15A66; 16S10; Secondary: 17B50; 17C10

Keywords: Matrix pencil; Generalized Clifford algebra; Finite-dimensional representation

1. Matrix pencils

Let F be an algebraically closed ﬁeld. Later we shall place restrictions on the characteristic of F, but for now it can be arbitrary. Denote by Mn(F ) the ring of n × n matrices over F [5,15]. Let A1,...,Am ∈ Mn(F ). Then we deﬁne the pencil of the

E-mail address: [email protected] (C.J. Pappacena). 1 Present address: Department of Mathematics, Baylor University, Waco, TX 76798-7328, USA.

0024-3795/00/$ - see front matter ( 2000 Elsevier Science Inc. All rights reserved. PII:S0024-3795(00)00025-2 2 C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 matrices Ai to be the matrix x1A1 +···+xmAm ∈ Mn(F [x1,...,xm]),wherethe xi are commuting indeterminates over F. We shall denote the pencil of the Ai as P(Ai) or by P if it will not cause confusion. By the degree of the pencil P(Ai ) we shall mean the degree of the minimal polynomial for P over F(x1,...,xm). If A1,...,Am have a degree d pencil, then by specializing the xi to elements in F, we see that every matrix in sp{Ai} has degree at most d over F,wheresp{Ai} denotes the F-linear span of the Ai. Moreover, one can regard the condition that the Ai have a degree d pencil as saying that the elements in sp{Ai} are “generically” degree d over F. Hence, one can always find a basis for sp{Ai } consisting of elements of degree d, and in the sequel when we say that a set of matrices Ai have a degree d pencil P,we shall assume tacitly that each Ai is itself degree d over F. Given these definitions, it is natural to ask whether there exist sets of m matrices in Mn(F ) with a degree d pencil for given integers m and d (with d 6 n). More specifically, one may ask whether or not it is possible to find a set of matrices, whose pencil has not only a specified degree d, but also a specified minimal polynomial φP (z) for P. Our first task shall be to determine as much as possible the form of the minimal polynomial φP of P. For the rest of the paper, we shall use vector notation when convenient. Hence, the notation f(x)E shall denote a function of the m indeterminates x1,...,xm. Since P(Ai ) is in Mn(F (x))E , its minimal polynomial φP (z) will a priori be a monic element of F(x)E [z]. That is, we may write d d−1 φP (z) = z + f1(x)zE +···+fd−1(x)zE + fd (x),E (1) where each fj (x)E ∈ F(x)E . However, as the following proposition shows, we can say a little more about the coefficients fj .

Proposition 1.1. Let φP (z) be as above. Then each of the coefﬁcients fj (x)E is a homogeneous polynomial of degree j in x1,...,xm.

Proof. Since P is an element of Mn(F [Ex]), its characteristic polynomial has coef- ﬁcients in F [Ex]. Hence, the eigenvalues of P, which are the roots of its characteristic polynomial, are integral over F [Ex]. Since the coefﬁcients of the minimal polynomial φP (z) are symmetric functions in the eigenvalues of P, it follows that they are integral as well. Hence, each fj (x)E is both integral over F [Ex] and in F(x)E , and it follows that each fj (x)E ∈ F [Ex]. To prove that each fj is homogeneous of degree j, write fj (x)E = gj (x)E + hj (x)E , where gj is the component of fj that is homogeneous of degree j,andhj is by definition fj − gj . We can then substitute this into (1) to obtain d d−1 φP (z) = z + (g1(x)E + h1(x)E ) z +···+(gd (x)E + hd (x)E ) . (2) Regrouping terms and substituting in P gives the following equation: h i h i d d−1 d−1 P + g1(x)PE +···+gd (x)E + h1(x)PE +···+hd (x)E = 0. (3) C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 3

Now, the entries of P are homogeneous polynomials of degree 1 in the xi,andsothe j entries of P are homogeneous of degree j in the xi. If we look at the entries of each expression in (3), we see that the entries in the ﬁrst term are all homogeneous of degree d in the xi, and that the entries in the second term have no degree d components. It follows that each term must be equal to zero separately. That is, d d−1 P + g1(x)PE +···+gd (x)E = 0. d d−1 So ψ(P ) = 0, where ψ(z) = z + g1(x)zE +···+gd (x)E . By uniqueness of the minimal polynomial it follows that φP = ψ, i.e., fj (x)E = gj (x)E for all j. Hence, the fj are homogeneous of degree j, as claimed.

Note that there are two ways of regarding the minimal polynomial φP (z) of a pencil P. On the one hand, it can be viewed as a polynomial in the single variable z, with coefﬁcients in F [Ex]. On the other hand, it is a homogeneous polynomial of degree d in F [Ex,z]. We shall have occasion to think of it in both contexts. When we wish to emphasize its dependence on the m variables xE, we shall write φP (x,z)E . Proposition 1.1 tells us which types of polynomials φ(x,z)E can be the minimal polynomial of a matrix pencil, and so we are led to ask the following question.

Question 1.2. Given a homogeneous polynomial φ(x,z)E of degree d in m + 1vari- ables, monic with respect to z (i.e., φ(x,z)E = zd + other terms), do there exist A1,...,Am in Mn(F ) for some n so that φ(x,z)E = φP (x,z)E ,whereP = P(Ai ) is the pencil of the Ai?

In order to answer this question we shall construct in Section 2 an associative algebra which is “universal” with respect to pencils with a given minimal polynomial. Before doing this, we briefly give some motivations for studying Question 1.2. Our initial motivation for considering this problem came from a question about the generation of matrix algebras. Specifically, for a given linear subspace V of k Mn(F ) which generates Mn(F ) as an algebra, let V be the F-linear span of all products of at most k elements in V. Then there exists a smallest integer k0 for k which V 0 = Mn(F ), and the problem is to find bounds for k0. One can trivially 2 take k0 6 n − dimF V , and the problem is to find better bounds. The idea is to look at the special case, where V = sp{Ai} for matrices Ai with a degree d pencil, and see whether this additional information enables one to improve this bound. Although we first learned this question from [16], the problem has a long history, and we discuss some of this history in Remark 4.6. The second motivation comes from the theory of partial differential equations and will be mentioned here only briefly (see also [9,10]). Suppose we are given the linear partial differential equation X α Lu = aαD u = cu, (4) |α|6d α where α = (α1,...,αm) is a multi-index, D is the differential operator 4 C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20

Nα1 Nαm Dα = ··· , N α1 N αm x1 xm and the aα and c are scalars with a0 = 0. In analogy with the Dirac equation, we wish to consider when solutions to an appropriate first-order system will also yield solutions to (4). P =− d ∈ C 6 6 E = First, write c b for some b ,andfor1 j d,letfj (x) |α|=j −(d−j) α b aαxE . (Note that fj (x)E is a homogeneous polynomial of degree j.) Let φ(z) = d d−1 z + f1(x)zE +···+fd (x)E . Then, if we can find matrices A1,...,Am in Mn(C) for some n such that φ(P(Ai )) = 0, then we can solve Eq. (4) by solving the first- order system Xm Nu Ai = bu. (5) Nxi i=1 Indeed, since φ(P(Ai )) = 0, we have the formal identity

! ! − Xm N d Xm N d 1 −(d−1) Ai + b L1 Ai Nxi Nxi i=1 i=!1 Xm N −1 +···+b Ld−1 Ai + Ld = 0, (6) Nxi i=1 P = α where Lj is the homogeneous component of L of degree j, i.e., Lj |α|=j aαD . One can then compute directly, using (6), that any solution to (5) is automatical- ly a solution to (4). Note that here we actually require the weaker condition that φ(P(Ai )) = 0, not the stronger condition that φ(z) is actually the minimal poyno- mial of P(Ai ).

2. The generalized Clifford algebra of a polynomial

In this section, we construct an associative algebra whose representation theory is closely linked to the existence of matrix pencils with a given minimal polynomial. This algebra, called the generalized Clifford algebra, is a generalization of a construction due to Roby [1,3,14]. Let φ(x,z)E be a homogeneous polynomial of degree d in m + 1 variables, monic E with respect to z.LetF hX1,...,Xmi=F hXi be the free algebra in m noncom- m E muting indeterminates over F. For each µE ∈ F ,letµE · X = µ1X1 +···+µmXm. E Let Iφ be the two-sided ideal of F hXi generated by all expressions of the form φ(µ,E µE · X)E for every µE ∈ F m. Then the generalized Clifford algebra of the poly- E nomial φ(x,z)E is the quotient algebra F hXi/Iφ . By abuse of notation, we shall also E use Xi to denote the image of Xi ∈ F hXi in Cφ. C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 5

Remark 2.1. If φ(x,z)E = zd − f(x)E for a homogeneous polynomial f of degree d, then Cφ = Cf , the Roby Clifford algebra of the form f.

Remark 2.2. For d > 3, Cφ is inﬁnite-dimensional over F (see Section 5).

The connection between generalized Clifford algebras and matrix pencils is given in the following proposition.

Proposition 2.3. Let φ(x,z)E be a homogeneous polynomial of degree d in m + 1 variables, monic with respect to z, and let Cφ be the generalized Clifford algebra of φ. (a) If φ = φP , where P = P(Ai ) is the pencil of A1,...,Am ∈ Mn(F ), then the map Xi → Ai defines a finite-dimensional representation U : Cφ → Mn(F ). (b) If U : Cφ → Mn(F ) is a finite-dimensional representation, and U(Xi ) = Ai, then φP divides φ in F [Ex][z], where P = P(Ai ) is the pencil of the Ai. In particular if φ is irreducible in F [Ex][z], then φ = φP . E Proof. (a) The map Xi → Ai gives a well-defined homomorphism from F hXi to Mn(F ), so it suffices to check that the kernel of this map contains Iφ .However, m this follows from the definitions: For any µE ∈ F ,wehaveφ(µ,E µ1A1 +···+ µmAm) = 0sinceφ = φP . Hence, the map descends to a well-defined representation U : Cφ → Mn(F ). m (b) By hypothesis, φ(µ,E µ1A1 +···+µmAm) = 0foreveryµE ∈ F .Itthere- fore follows that φ(x,P)E = 0. Indeed, if φ(x,P)E were not identically zero, then there would be some specialization xE →Eµ so that φ(µ,E µ1A1 +···+µmAm)/= 0, contrary to hypothesis. Since the minimal polynomial of a matrix divides any polynomial that the matrix is a root of, we have that φP divides φ. Finally, if φ is irreducible we must have φ = φP .

So the existence of representations of Cφ is closely connected with ﬁnding matrices with pencil P satisfying φP = φ.Whenφ is irreducible over F [Ex] this corre- spondence is in fact one-to-one. We shall therefore study the problem of determining the ﬁnite-dimensional representations of the algebras Cφ. To begin with, we shall look at the effect of a linear change of variables in E the free algebra F hXi on the Clifford algebra Cφ. Specifically, suppose we are given the change of variables Xi → aiXi + bi = Yi for ai,bi ∈ F . Moreover, we presume that this change of variables in nonsingular, i.e., we have an isomorphism E ∼ E F hXi → F hY i. First, let us write the inverse of this isomorphism as Yi → αi Yi + βi = Xi. P P E ∈ m = + = For any µ F ,wehaveP i µiXi P i µi(αi Yi βi). Setting νi αi µi,we + = −1 can write this latter sum as i νiYi i γiνi,whereγi βiαi .Now,wecom- pute 6 C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20

Xd Xd E E j ˜ E j φ µ,E µE · X = fd−j (µ)E µE · X = fd−j (ν)E νE · Y + g(ν)E , (7) j=0 j=0 P ˜ E = E E = where fj (ν) Pfj (µ) and g(ν) i γiνi . If we expand out the ﬁnal sum in (7), we E E · E j E may write it as j gd−j (ν)(ν Y) , whereP each gj (ν) is a homogeneous polynomial j of degree j in the νi .Letψ(y,Z)E = gd−j (y)ZE . Then we have the equality j φ µ,E µE · XE = ψ ν,E νE · YE .

That is, under the isomorphism Xi → Yi,wehaveIφ → Iψ . Hence, we have the following result.

E ∼ E Proposition 2.4. Under the isomorphism F hXi → F hY i above, we have Iφ → Iψ , ∼ and hence we have an induced isomorphism Cφ → Cψ .

We shall make use of this change of variables in Section 3, when we determine the Clifford algebra of a homogeneous quadratic polynomial φ(x,z)E .

3. The quadratic case

In this section, we completely determine the structure of the Clifford algebra Cφ of a homogeneous quadratic polynomial φ(x,z)E in m + 1 variables, monic with respect to z. Whenever we write a polynomial φ in this section it is tacitly assumed to be quadratic. Our analysis divides into two cases, depending on whether char(F )= / 2 or char(F ) = 2. In the ﬁrst case, we shall see that the Clifford algebra Cφ is isomorphic to the (classical) Clifford algebra of a quadratic form q, whose definition we give presently. Recall that a quadratic form on an F-vector space V is a map q : V → F satisfying q(rv) = r2q(v) for all r ∈ F , v ∈ V , such that the map b : V × V → F deﬁned by b(u, v) = q(u+ v) − q(u) − q(v) is a symmetric bilinear form. If T(V) denotes the tensor algebra of V, then the Clifford algebra Cq of q is the quotient of T(V) by the ideal generated by {v ⊗ v − q(v) : v ∈ V }. If we choose a basis X1,...,Xm for V, then the quadratic form q can be realized as a homogeneousP quadratic polynomial in m variables, which we also denote by q.Thus,ifv = µiXi, ∼ i then q(v) = q(µ1,...,µm). Under the isomorphism T(V)= F hX1,...,Xmi,the Clifford algebra of a quadratic form q is the same as the generalized Clifford algebra 2 Cφ of the polynomial φ(z) = z − q(x)E (which as noted above, is the same as the Roby Clifford algebra Cq of the polynomial q). When the characteristic of F is 2, it is no longer always possible to give Cφ as the classical Clifford algebra of a quadratic form. In this case, Cφ is best regarded as the restricted enveloping algebra of a restricted Lie algebra. We shall recall these definitions when we treat this case. C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 7

The following proposition will be useful:

Proposition 3.1. The generalized Clifford algebra Cφ of a quadratic polynomial is ﬁnite-dimensional over F.

Proof. Cφ is generated as an algebra over F by X1 ...,Xm. By the deﬁning re- 2 + ∈ lations for Cφ, each Xi can be written as αiXi βi for αi ,βi F . Similarly, for 2 i

It follows at once that Xj Xi =−XiXj + terms linear in Xi,Xj . Hence, the or- ··· ··· 6 6 dered monomials Xi1 Xik , with i1 <

In particular, by the regular representation we can regard Cφ as a subring of Mt (F ) for some integer t.

3.1. Characteristic of F different from 2

Let φ be as above, and consider the Clifford algebra Cφ of φ. Each of the elements Xi is quadratic over F, and so has two eigenvalues which are the roots of its minimal polynomial. Denote the eigenvalues of Xi as αi and βi, and note that αi may equal βi for some (perhaps all) i. Our ﬁrst task is to ﬁnd a linear change of variables Xi → Yi so that the eigenvalues of Yi are easier to work with.

Lemma 3.2. There exists a linear change of variables Xi → Yi so that either Yi has 0 as a double eigenvalue for all i or Yi has eigenvalues 1 for all i.

Proof. Using Proposition 3.1 and the regular representation, we view Cφ as a subring of Mt (F ) for some t.IfXi has a single eigenvalue αi of multiplicity 2, then set Yi = Xi − αi .IfXi has two eigenvalues αi , βi,thenset 2 αi + βi Yi = Xi − . αi − βi 2

It is routine to check that Yi has 0 as a double eigenvalue in the ﬁrst case, and that Yi has eigenvalues 1 in the second. If all of the Yi have the same eigenvalues, then we are done. 2 = 2 = + If not, then there exist Yi,Yj such that Yi 1, Yj 0. Consider the matrix Yi λYj ∈ Mt (F [λ]) for an indeterminate λ. One can check that Yi + λYj has quadratic minimal polynomial over Mt (F (λ)) and so has two eigenvalues, which cannot be identically equal since they differ on the specialization λ → 0. These eigenvalues are integral over F [λ] since they satisfy the characteristic polynomial for Yi + λYj . In particular they agree for only ﬁnitely many specializations λ → α ∈ F . Choosing 8 C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20

α so that Yi + αYj has different eigenvalues, we may proceed as above, normalizing 2 = them to 1. Continuing in this manner we will eventually have Yi 1foralli.

The advantage of this change of variables is that ψ(y,Z)E is particularly easy to compute. (Recall how ψ is obtained from φ.)

Lemma 3.3. Under the change of variables in Lemma 3.2, we have ψ(y,Z)E = Z2 − q(y),E where q(y)E is a homogeneous polynomial of degree 2.

d Proof. By hypothesis ψ(y,Z)E = Z + f1(y)ZE + f2(y)E ,wherefj (y)E is homoge- Pneous of degree j for j = 1, 2. It sufﬁces to show that f1(y)E = 0. Write f1(y)E = ∈ ∈ m i aiyi for ai F .Letei F be the vector with zeroes everywhere, except for a 1 E E in the ith component. Then we have ei · Y = Yi ,andsoψ(ei ,ei · Y) = ψ(Yi ) (where 2 = 2 = ψ(Yi ) is the minimal polynomial for Yi ). By Lemma 3.2, Yi 0foralli or Yi 1 = 2 = 2 − = for all i, hence ψ(Yi ) Yi or ψ(Yi ) Yi 1. In either case we see that f1(ei) 0 2 for all i.Butf1(ei) = ai. Hence, f1 is identically zero and ψ(y,Z)E = Z − q(y)E , where q(y)E =−f2(y)E .

Combining these two lemmas we have the following main result.

Theorem 3.4. Let φ(x,z)E be a homogeneous polynomial in m + 1 variables, monic with respect to z. Then the generalized Clifford algebra Cφ is isomorphic to the (classical) Clifford algebra Cq of a quadratic form q.

∼ Proof. The change of variables in Lemma 3.2 induces an isomorphism Cφ → Cψ , 2 by Proposition 2.4. By Lemma 3.3, ψ(y,Z)E = Z − q(y)E , and thus Cψ = Cq ,where Cq is the Clifford algebra of the quadratic form q. Note that the vector space V here is just sp{Y1,...,Ym}.

Because of Theorem 3.4, we can use the well-developed structure theory of Clif- ford algebras to determine the structure of Cφ. We begin by listing some classical results on Clifford algebras and quadratic forms. The radical of a quadratic form q on a vector space V is the radical of its associated bilinear form b deﬁned above. That is, rad(q) = rad(b) ={v ∈ V : b(v, V ) = 0}. The image of the radical of the form q in Cq generates rad(Cq ), where “rad” here is the Jacobson radical. A form with zero radical is called nondegenerate.Sincerad(q) is a subspace of V, modding out by the radical gives a nondegenerate quadratic form q¯ on the vector space V¯ = V/rad(q). ∼ ¯ Hence, we have Cq /rad(Cq ) = Cq¯ . We shall denote the dimension of V as m¯ .We ¯ may always choose a basis X1,...,Xm for V so that X1,...,Xm¯ is a basis for V . The structure of the Clifford algebra for a nondegenerate quadratic form q¯ is known: If m¯ = 2k is even, then Cq¯ is a central simple algebra over F of dimension m¯ 2 ,andifm¯ = 2k + 1 is odd, then Cq¯ is a direct sum of two central simple algebras, C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 9

m¯ −1 m¯ each of dimension 2 , so in this case as well the dimension of Cq¯ is 2 [7,18]. Since in our case F is algebraically closed, we have the following corollary.

Corollary 3.5. Let φ and q be as in Theorem 3.4, and let m¯ be the dimension of V¯ . Write m¯ as 2k or 2k + 1 depending on whether it is even or odd. Then we have = ¯ Cφ/rad(Cφ) M2k (F ) if m is even, (8) = ⊕ ¯ Cφ/rad(Cφ) M2k (F ) M2k (F ) if m is odd. (9)

Proof. Since F is algebraically closed, the only central simple algebras over F are matrix algebras, and the dimensions come from the appropriate dimension for Cq¯ .

We close this subsection by giving an alternate description of the Clifford algebra Cφ, which will serve to motivate the discussion when char(F ) = 2.

Proposition 3.6. The vector subspace J = sp{1,X1,...,Xm} of Cφ is a Jordan ◦ = 1 + algebra under the Jordan product u v 2 (uv vu).

Proof. One simply expands the expression φ(ei + ej ,Xi + Xj ) in Cφ.Sinceφ is quadratic, we obtain an equation of the form 2 + + + 2 = + + Xi XiXj Xj Xi Xj γij (Xi Xj ) δij ,

2 2 for γij , δij in F.SinceXi and Xj are linear combinations of Xi (respectively, Xj ) and scalars, it follows that

Xi ◦ Xj ∈ sp{1,Xi ,Xj }⊂J.

Since the Jordan product ◦ extends linearly the result follows.

Note that the linear change of variables described in Lemma 3.2 corresponds to a change of basis for J,takingittosp{1,Y1,...,Ym}. In fact, we can compute the 2 = Jordan product relative to this basis in terms of the quadratic form q.WhenYi 0 for all i, then multiplying out ψ(ei + ej ,Yi + Yj ) gives 2Yi ◦ Yj = q(ei + ej ) = + = = 2 = q(Yi Yj ) b(Yi,Yj ),sinceq(Yi) 0foreveryi. Similarly, if Yi 1foralli then 2Yi ◦ Yj = q(ei + ej ) − 2 = q(Yi + Yj ) − 2 = b(Yi,Yj ). ◦ = 1 J So in either case Yi Yj 2 b(Yi,Yj ), which shows that is the Jordan algebra of the bilinear form b. In this case, the Clifford algebra of q is the special enveloping algebra Us(J) of J, i.e., the associative algebra such that every Jordan homomorphism from J into a special Jordan algebra is induced by an algebra homomorphism from Us(J) into the associative algebra of the special Jordan algebra [6,8]. 10 C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20

3.2. F has characteristic 2

To treat the case when the characteristic of F is 2, we begin with an analogue of Lemma 3.2. However, in this case +1 =−1,andsointhecasewhereXi has two eigenvalues we must normalize them to 0 and 1. With this modiﬁcation, we can prove exactly as in the characteristic different from 2 case the following:

Lemma 3.7. There exists a linear change of variables Xi → Yi so that either Yi has 0 as a double eigenvalue for all i or Yi has eigenvalues 0 and 1 for all i.

Also, we can compute the polynomial ψ(y,Z)E as in Lemma 3.3. However, the ﬁnal computation is different because of the different eigenvalues. In particular, 2 = 2 = we have either Yi 0foralli or Yi Yi for all i. So we have this analogous result.

2 = Lemma 3.8. Consider the change of variables in Lemma 3.7.IfYi 0 for all 2 i, then we have ψ(y,Z)E = Z − q(y),E where q(y)E is aP homogeneous polynomial 2 = E = 2 + + E of degree 2.IfYi Yi for all i, then ψ(y,Z) Z ( i yi)Z f2(y) for f2 a homogeneous polynomial of degree 2 in the yi.

2 Proof. As above, write ψ(y,Z)E = Z + f1(y)ZE + f2(y)E . Then the proof in the 2 = 2 = case Yi 0foralli carries over exactly as in Lemma 3.3. When Yi Yi for all i, then the computationP of f1(y)E follows the same method, with different result. E = = Writing f1(y) i aiyi and specializing to ψ(ei ,Yi ) gives ai 1foralli. Hence, the polynomial ψ(y,Z)E has the stated form.

2 = = In the case Yi 0foralli, we can prove exactly as in the case char(F ) / 2that =∼ 2 = Cφ Cq . However, in the latter case when Yi Yi , this is no longer true. To see this, note that in the Clifford algebra of a quadratic form q, the equation v2 = q(v) for every vector v implies that v has a single eigenvalue: Since F is algebraically closed, we can write the number q(v) as α2 for some α.Thenv2 − α2 = (v − α)2 = 0and 2 = so v has a single eigenvalue α with multiplicity 2. Hence, in the case Yi Yi we cannot, by change of variables, take Cψ to the Clifford algebra of a quadratic form. ∼ Remembering how Cφ = Cψ could be interpreted as an enveloping algebra of a Jor- dan algebra when char(F )= / 2, we are led to examine if it is possible to describe Cφ as an enveloping algebra of a Lie algebra when char(F ) = 2. We begin with a proposition that is the analogue of Proposition 3.6. Recall here and throughout this section that since F has characteristic 2, +1 =−1 and we shall consistently use ‘+’ throughout.

Proposition 3.9. The vector subspace L = sp{1,X1,...,Xm} of Cφ is a restricted Lie algebra under the Lie bracket [u, v]=uv + vu and power operation u[2] = u2. C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 11

Proof. The fact that L is a Lie algebra follows in a manner analogous to Propo- sition 3.6. Moreover, one simply notes that since u2 ∈ L for every u ∈ L by the [2] definition of Cφ,thatL is in fact a restricted Lie algebra under the operation u = u2.

Recall that the restricted enveloping algebra u(L) of a restricted Lie algebra L in characteristic p is the quotient of the enveloping algebra U(L) by the ideal I generated by all expressions of the form vp − v[p],wherev ∈ L and [p] denotes the power mapping [7]. Thus, the map v → v[p] is just the regular pth power map in u(L). We have the following basic result.

Theorem 3.10. Let L be the restricted Lie algebra in Proposition 3.9.ThenCφ is the restricted enveloping algebra u(L) of L.

Proof. The universal enveloping algebra U(L) of L is just the free algebra F hXEi. 2 [2] So we must show that the ideal generated by v − v for all v ∈ L is Iφ .This E m [2] follows because if we take v =Eµ · X for µE ∈ F ,thenv = f1(µ)vE + f2(µ)E .So v[2] = v2 is equivalent to φ(µ,E v) = 0.

So, to determine the structure of Cφ we must study the restricted enveloping algebra of L. One tool which will prove useful is the Poincaré–Birkhoff–Witt theorem, ··· ··· which says that the ordered monomials of the form Xi1 Xik , with i1 < Weyl algebra over F. Ap(F ) has 2p generators x1,...,xp; y1,...,yp subject to the commutator relations [xi,xj ]= [yi,yj ]=0forallN i, j and [xi,yj ]=δij ,whereδij is the Kronecker delta. We note A = p A A that p(F ) i=1 1(F ), and also that p(F ) has ﬁnite-dimensional representations when the characteristic of F is positive.

Theorem 3.11. Let φ(x,z)E be a homogeneous polynomial in m + 1 variables, mon- = ic with respect to z, over a ﬁeld F of characteristic 2.ThenCφ/rad(Cφ) M2k (F ) = ⊕ 6 6 b c or Cφ/rad(Cφ) M2k (F ) M2k (F ) for some 1 k m/2 . 12 C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 2 = Proof. The proof divides naturally into two cases, depending on whether Yi 0 2 = for all i or Yi Yi for all i. 2 = L [ ]= ∈ Case I. Yi 0: In this case the bracket for is Yi ,Yj αij for some αij F . We shall change basis as follows. First, group the Yi into pairs Y2j−1,Y2j for j 6 bm/2c.Ifm is odd, then Ym will be unpaired. If [Y1,Y2] =/ 0, then multiply Y1 by an appropriate scalar so that [Y1,Y2]=1. Then replace Y3 with Y3 + c1Y1 + c2Y2 for scalars c1,c2 so that [Y1,Y3]=[Y2,Y3]=0. We proceed in a similar fashion with Y4,...,Ym, so that when we are ﬁnished Y1 and Y2 commute with all Yi for i/= 1, 2. We make the analogous changes for the pair Y3,Y4, noting that any subsequent changes will not alter the conditions [Y1,Y3]=[Y2,Y3]=[Y1,Y4]=[Y2,Y4]=0. Continuing for each pair Y2j−1,Y2j (and for the unpaired Ym if m is odd), we have a much simpler bracket. Note that, if m is odd, then Ym commutes with all of the Yi.Also,wemayhave[Y2j−1,Y2j ]=0 for one or more values of j. Reindexing the Yi if necessary, we may assume that for some k 6 m,wehave[Y2j−1,Y2j ] =/ 0for j 6 k. Under this new basis the bracket for L has the following simple form:

[Y2j−1,Y2j ]=1forj 6 k, [Yi,Yj ]=0 for all other i, j.

2 One may verify that Yi is scalar in this new basis, and since F is algebraically 2 = 2 2 = 2 6 closed write Y2j−1 αj and Y2j βj for j k.SinceY2k+l commutes with each + 2 = 2 Yi, we may replace Y2k+l with Y2k+l γ2k+l,whereY2k+l γ2k+l. This will have 2 = no effect on the above bracket (as can be checked) and allow us to write Y2k+l 0 for l = 1,...,m− 2k. We may now explicitly write down the restricted enveloping algebra u(L) as

u(L) = F [t1,...,tm−2k]hx1,...,xk; y1,...,yki, where xj corresponds to Y2j−1, yj corresponds to Y2j ,andtl corresponds to Y2k+l, and the following relations are satisfied: (a) [xi,xj ]=[yi,yj ]=0, [xi,yj ]=δij for all i, j, 2 = 2 2 = 2 2 = (b) xi αi ,yi βi ,tj 0. Let T be the two-sided ideal of u(L) generated by the tj . Clearly T ⊆ rad(u(L)) 2 = since tj 0forallj and each tj is central. Suppose first that k = 0, i.e., that u(L) is commutative. Then, since u(L)/T = F [t1,...,tm]/(t1,...,tm) = F , we see that T = rad(u(L)) and the theorem follows in this case. When k > 1 the relations above show that u(L)/T = Ak(F )/I,whereI is the 2 + 2 2 + 2 A = A 2 + 2 2 + 2 ideal generated by xi αi and yi βi .Ifweset 1(F )/(x α ,y β ), A =⊗k A then the isomorphism k(F ) i=1 1(F ) descends to an isomorphism L =⊗k A = u( )/T i=1 . So it suffices to treat the case k 1. C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 13

When k = 1, we have (by the Poincaré–Birkhoff–Witt theorem) that {1,x,y,xy} is a basis for A, and hence A is a four-dimensional F-algebra. One may verify that the map A → M (F ) given by 2 α 1 β 0 x → ,y→ , 0 α 1 β is a surjective homomorphism. Comparing dimensions we have A = M2(F ).Itfol- L = L ⊆ L lows that u( )/T M2k (F ).Sinceu( )/T is semisimple and T rad(u( )), we have that T = rad(u(L)). The theorem now follows in Case I since by Theorem 3.10, Cφ = u(L). 2 = L [ ]= + + Case II. Yi Yi: For this case, the Lie bracket of is Yi,Yj Yi Yj αij . 0 = + 0 = + 0 We make the following change of variables: Y1 Y1 Y2, Y2 Y2 Y3,...,Ym−1 = + 0 = [ 0 0 ]= Ym−1 Ym, Ym Ym. Then for i, j < m we have Yi ,Yj βij for some scalar 0 [ 0 0 ] βij . Now we may proceed as in the ﬁrst case, normalizing the Yi so that Y2j−1,Y2j = 0 0 0 or 1, and all other brackets among Yi ,Yj for i, j < m are zero. Note that even [ 0 0 ]= 0 + after these normalizations we have Yi ,Ym Yi βim for i

[ 0 0 ]= 6 Y2j−1,Y2j 1forj k, [ 0 ]= 0 Yi ,Ym Yi for i

02 0 2 = 0 = − − Additionally, each Y is scalar for i

u(L) = F hX1,...,Xq ; Y1,...,Yq ; T1,...,Tr ; Zi/I, where I is the ideal corresponding to the relations summarized in the following: (a) [tj ,xi]=[tj ,yi]=0foralli, j,and[tj ,tk]=0forallj,k. (b) The xi and yi satisfy the relations for the Weyl algebra Aq (F ). 2 2 2 2 = 2 2 = 2 2 = 2 2 = (c) Each xi ,yi ,tj is scalar, say xi αi ,yi βi ,tj γj ,andz z. (d) [tj ,z]=tj , [xi,z]=xi,and[yi,z]=yi for all i, j. 2 By relation (c), each tj is scalar. Suppose that one of them, say without loss of generality t1, has nonzero square. Then by replacing tj with tj + λj t1 for an 2 = 2 = appropriate scalar λj we can assume that tj 0forj>1. So, either tj 0for 2 all j or only t1 is nonzero. As is the ﬁrst case, the following lemma describes the radical of u(L) in terms of the tj .

Lemma 3.12. Let T be the two-sided ideal of u(L) generated by the elements tj 2 = = L such that tj 0.ThenT rad(u( )). 14 C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20

Lemma 3.12 is proven below. Thus, the algebra u(L)/rad(u(L)) has two forms, 2 = 2 = 2 = depending on whether t1 0ort1 / 0 (for simplicity write t for t1). If t 0, ∼ 2 then u(L)/rad(u(L)) = F hx1,...,xq ; y1,...,yq ; zi,andift =/ 0, then u(L)/rad ∼ (u(L)) = F hx1,...,xq ; y1,...,yq ; t; zi. We again consider ﬁrst the case q = 0. If t2 = 0, then u(L)/rad(u(L)) =∼ F [z] =∼ F ⊕ F , and in this case the irreducible F-representations of u(L) are one-dimensional. If t2 = γ 2 =/ 0, we have the algebra u(L)/rad(u(L)) =∼ F ht,zi.There- lations above show that {1,t,z,tz} spans u(L)/rad(u(L)), so it is at most four- dimensional. The representation given by γ 1 10 t → ,z→ 0 γ γ 0 is an irreducible two-dimensional representation, and is easily seen to generatee ∼ M2(F ). A dimension count then shows u(L)/rad(u(L)) = M2(F ), and the theorem is true when q = 0. 2 ∼ Now suppose that q > 1. When t = 0, u(L)/rad(u(L)) = F hxi,yi; z : i = 1, ...,qi, which has as a subalgebra F hxi ,yi : i = 1,...,qi.Soanyu(L)-module, and in particular any irreducible u(L)-module, must also be an F h{xi,yi}i-module. ∼ ⊗q But F h{xi,yi}i = M2q (F ), since it is nothing more than the algebra A studied in the ﬁrst case. Hence, an irreducible representation of u(L) must have dimension at least 2q. On the other hand, relations (a)–(d) above show that the ordered monomials in the 2q + 1 variables xi, yi, z with exponents 0 or 1 span F h{xi ,yi}; zi. Hence, the total dimension of u(L)/rad(u(L)) is at most 22q+1, and so an irreducible representation of u(L) must have dimension at most 2q . So in this case the structure of u(L) is ∼ ∼ either u(L)/rad(u(L)) = M2q (F ) or u(L)/rad(u(L)) = M2q (F ) ⊕ M2q (F ). 2 2 If t = γ =/ 0, then u(L)/rad(u(L)) = F h{xi,yi}; t; zi.LetM be an irreducible u(L)-module, and set N ={m ∈ M : tm = γm}.Sincet commutes with each xi and yi by relation (a), we see that N is an F h{xi,yi }i-module. One can readily verify that N + zN is a nonzero submodule of M,sothatM = N + zN by irreducibility. We claim that this sum is direct, and that dim N = dim zN. For the former, let zn ∈ N. Then on the one hand, t(zn) = γ(zn), by the definition of N. But on the other hand t(zn) = (tz)n = (zt + t)n = ztn + tn = z(γ n) + γn= γ(zn)+ γn.Soγn= 0, hence n = 0, and so M = N ⊕ zN.Now,letn1,..., np be a basis for N. We claim that zn1,...,znp is a basis for zN. Clearly, the zni span, and Xp X X λi zni = 0 ⇒ t · λi zni = 0 ⇒ λi tzni = 0. i=1 By relation (d) we may write tz = zt + t and so X X X λi (ztni + tni )=0 ⇒ γ · λizni + γ · λini X =0 ⇒ γ · λini = 0. C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 15

Since the ni form a basis for N, λi = 0foralli and the claim is proved. As noted above, N is an F h{xi ,yi}i-module. If N = N1 ⊕ N2 (as F h{xi,yi}i- modules), then each of N1 ⊕ zN1, N2 ⊕ zN2 is a submodule of M.SinceM is irreducible we have N1 = 0orN2 = 0andsoN is an irreducible F h{xi,yi}i-module, which by Case I has dimension 2q .SodimM = 2q + 2q = 2q+1. Relations (a)–(d) above show that the ordered monomials in xi, yi, z,andt with exponents 0or1spanu(L)/rad(u(L)), and so the total dimension of u(L)/rad(u(L)) is at most 22q+2. So as above there are two possibilities for the structure of u(L): ∼ ∼ u(L)/rad(u(L)) = M q+1 (F ) or u(L)/rad(u(L)) = M q+1 (F ) ⊕ M q+1 (F ). 2 ∼ 2 2 The theorem follows since Cφ = u(L).

Proof of Lemma 3.12. Using the relations (a)–(d) from Case II of Theorem 3.11, = 2 = = 2 = = + = we observe that tj xitj tj xi 0, tj yitj tj yi 0andtj ztj tj (tj z tj ) 2 + = 2 = = ∈ tj (z 1) 0whentj 0. Hence, (u1tj v1)(u2tj v2) 0foranyu1, u2, v1, v2 u(L). This shows that T ⊆ rad(u(L)). Now let I be a two-sided ideal of u(L)/T . In the proof of Theorem 3.11, we saw ∼ that u(L)/T contained the subalgebra F h{xi,yi}i = M2q (F ).SinceI ∩ F h{xi,yi}i is an ideal of F h{xi,yi}i,wehaveI ∩ F h{xi,yi}i=0orI ∩ F h{xi,yi}i=F h{xi ,yi}i. In the latter case this implies that I = u(L)/T since u(L)/T and F h{xi,yi }i have the same identity. ∩ h{ }i = 2 = If I F xi,yi 0andalltj 0, then I is the two-sided ideal generated by z, which is not nil since z` = z/= 0forall`.Sointhiscaseu(L)/T is semisimple = L 2 = and T rad(u( )).Whent1 / 0 then (writing t for t1), I must be generated by elements of the form a + bt + cz + dtz for scalars a, b, c, d. Note that if I contains t,thenI = u(L)/T ,sincet2 = γ 2 =/ 0, and if I contains z or tz then it is not nil. (If z ∈ I, then as above z` =/ 0, and if tz ∈ I,thentz(tz − 1) = γ 2z ∈ I,soz ∈ I.) So we may assume when we write a + bt + cz + dtz ∈ I that at least two of a, b, c, d are nonzero. If a/= 0, then multiply by a−1 to obtain 1 + a−1bt + a−1cz + a−1dtz ∈ I.Ifa = 0butb/= 0 multiply on the left-hand side by (bγ 2)−1t to get 1 + (bγ 2)−1ctz + b−1dz ∈ I.Ifa = b = 0, then c and d are both nonzero. In this casewehave(cz + dtz)t = czt + dtzt = c(tz + t) + d(t2z + t2) ∈ I. Simplifying, we have dγ2 + ct + dγ2z + ctz ∈ I,anddγ2 =/ 0, so we are again in the ﬁrst case (with dγ2 = a). So we have reduced to the case where I is a two-sided ideal of u(L)/T containing an element of the form 1 + at + bz + ctz. We claim that this element is not nilpotent. Indeed, it is easy to see from relations (a)–(d) in the proof of Theorem 3.11 that (1 + at + bz + ctz)` is of the form 1 + pt + qz + rtz for scalars p, q, r for any `. We will show that pt + qz + rtz is never a scalar unless p = q = r = 0. Suppose pt + qz + rtz is a scalar. Then in particular it commutes with z,sowe have (pt + qz + rtz)z = z(pt + qz + rtz). Expanding the two sides gives

ptz + qz + rtz=pzt + qz + rztz 16 C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20

=ptz + pt + qz + r(tz + z)z =ptz + pt + qz + rtz + rtz. Simplifying the above (remember we are in characteristic 2) yields rtz = pt. Multiplying both sides by t on the left-hand side gives rγ2z = pγ 2, and the only multiple of z which is a scalar is 0. So rγ2 = 0, hence r = 0. This also yields pγ 2 = 0, whence p = 0. Now, this says that qz is a scalar, whence q = 0. So the element 1 + at + bz + ctz is not nilpotent. This shows that u(L)/T is semisimple, and so rad(u(L)) = T as claimed.

We close this section by recording a common corollary to Corollary 3.5 and The- orem 3.11.

Corollary 3.13. Let m be a ﬁxed positive integer. Then there exist matrices A1,..., k Am in Mn(F ) with a quadratic pencil if and only if n = 2 for some k 6 bm/2c. Moreover, the Ai generate Mn(F ) in this case.

Remark 3.14. The results of this section are related to the proof of Hurwitz’s theorem on sums of squares given in [4, pp. 141–144].

4. An application to the generation of matrix algebras

In this section, we describe an application of the results of Section 3. The prob- lems considered here were our original motivation for studying the problem of ﬁnd- ing a set of matrices with a degree d pencil. We ﬁrst learned of this question in [16] and the following version of it is from [2].

Question 4.1. Given matrices A1,...,Am that generate Mn(F ) as an algebra (i.e., such that F {A1,...,Am}=Mn(F )), can we ﬁnd an upper bound for the length of words in the Ai needed to span Mn(F ) that depends only on n?

As mentioned above, one can trivially take the bound n2 − m. The real issue is whether or not a bound can be chosen that is asymptotically better than n2 or, more optimistically, one that is linear in n. For notational convenience, let S = {A1,...,Am}. Then we shall write l(S) 6 f(n)if words in the Ai of length 6f(n) span Mn(F ). The following are Theorem 3.6 of [2] and Theorem 4.1 of [11], respectively.

Theorem 4.2. If some matrix in sp{Ai} has distinct eigenvalues, then l(S) 6 2n − 2.

Theorem 4.3. If the Ai have a degree n pencil, then l(S) 6 3n − 3. C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 17

The condition that the Ai have a degree n pencil comes into play because, as noted in Section 1, it implies that “most” elements of sp{Ai} have minimal polynomial of degree n (i.e., a Zariski-open subset of sp{Ai }). Theorem 4.3 suggests that, in order to ﬁnd examples, where l(S) is large, one should look at the opposite extreme, namely, the case where P(Ai ) has a small degree minimal polynomial. However, the results of Section 3 show that in this case one actually obtains an even sharper bound.

Theorem 4.4. Suppose that A1,...,Am have a quadratic pencil, and that F {A1, }= 6 + ...,Am Mn(F ).Thenl(S) 2log2 n 1.

Proof. Let φP (z) be the minimal polynomial of P(Ai). By the results of the pre- → vious section there is a homomorphism CφP Mn(F ) sending Xi to Ai.Wemay assume without loss of generality that, for some m¯ 6 m,wehavethatXm¯ +1,...,Xm { ··· generate rad(CφP ).SinceCφP /rad(CφP ) is spanned as an F-vector space by Xi1 : ··· 6 ¯ } =¯ = k = Xit i1

This result reveals the dichotomous nature of this problem: In the two extreme cases, where the Ai have a pencil of high degree and where the Ai have a pencil of low degree, the bound obtained for l(S) is (sub)linear in n. By playing off this dichotomy it is possible to prove a bound of approximately n2/3 for arbitrary sets { } A1,...,Am that is slightly weaker than a bound due to Paz [12]. √ Using other methods, the author proved in [11] that l(S) ∈ O(n n) for any set {A1,...,Am}. Of course, the above evidence suggests that l(S) is always at worst linear in n. An example of [2] shows that one can do no better than l(S) 6 2n − 2in general, and so we have the following conjecture:

Conjecture 4.5. Let l(n) be the maximum over all l(S) such that F {Ai}=Mn(F ). Then l(n) = 2n − 2.

Remark 4.6. As we mentioned in the introduction, the problem of finding a good bound for l(n) (or, more generally, a similar bound for subalgebras A of Mn(F ))has been around a long time, and has applications to many areas. We shall recall here only some of these applications. The motivation in [16] was to find a shorter proof of the fact that any affine, semiprime F-algebra of Gelfand–Kirillov dimension 1 satisfies a polynomial identity. This requires a bound for l(n) that is asymptotically better than n2. See [11] for a further discussion of this application. On the other hand, this problem also has applications to the theory of unitary matrices, where a sharper bound for l(n) makes certain theorems more effective. See [2,13]. 18 C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20

5. Cubic and beyond

The study of Cφ when φ(z) is no longer a quadratic polynomial is substantially more difﬁcult, and there is very little to say. One obstacle is the fact that, for φ(z) of degree at least 3, Cφ is no longer ﬁnite-dimensional over F. We shall give a sketch of the proof here. The proof is an unabashed adaptation of the corresponding result for Roby Clifford algebras proved in [17], and we make no claim of originality.

Theorem 5.1. Let φ(z) be a polynomial of degree at least 3.ThendimF Cφ =∞.

Proof. Suppose first that φ = φ(x1,x2,z)is a polynomial of degree d in three vari- d d−1 ables. Write φ(x1,x2,z)= z + f1(x1,x2)z +···+fd (x1,x2). Then the defin- ing relations for the Clifford algebra Cφ are given by d d−1 (µ1X1 + µ2X2) + f1(µ1,µ2)(µ1X1 + µ2X2) +···+fd (µ1,µ2) = 0. Expanding this polynomial and grouping the terms by powers of µ1 and µ2 gives the following set of relations for 0 6 i 6 d: i d−i = µ1µ2 gi (X1,X2) 0, where gi (X1,X2) is a polynomial of degree d. Moreover, all of the terms of degree d in g have degree i in X1 and degree d − i in X2. d−i i 6 6 In particular, one can rewrite the monomials of the form X2 X1,for0 i d, in terms of other monomials of degree d with degree i in X1 and degree d − i in X2, plus lower-order terms. Hence, every element of Cφ can be written as an F-linear n1 a1 b1 ··· at bt n2 6 6 combination of terms of the form X1 (X2 X1 ) (X2 X1 )X2 , with 0 n1,n2 d − 1, and, if t>0, ai + bi 6 d − 1, 0 6 ai,bi 6 d − 2fori 6 t. ai bi It follows that Cφ contains the free algebra on the symbols X2 X1 for ai,bi as above, and so Cφ is infinite-dimensional. For a general polynomial φ(x,z)E in m + 1 variables, one simply notes that specializing µ3,...,µm = 0showsthatCψ is a subalgebra of Cφ,whereψ is now a form in three variables.

The fact that Cφ is infinite-dimensional for d > 3 makes the problem of finding finite representations of Cφ, and hence that of finding matrices with pencil P such that φ = φP , difficult. In the case of the Roby Clifford algebra Cf of the polynomial d z − f(x)E , it is known that Cf has finite representations and moreover, if the form f is nondegenerate, then the degree of f divides the degree of the representation. See [17] for proofs of these facts. Unfortunately, the proofs of these results do not carry over to the more general setting of Clifford algebras of polynomials, and so we still have the following basic question:

Question 5.2. Does Cφ have a ﬁnite-dimensional representation for any polynomial φ?IfCφ has ﬁnite-dimensional representations, how do the degrees of the representations relate to the degree of φ? C.J. Pappacena / Linear Algebra and its Applications 313 (2000) 1–20 19

There is a result due to Van den Bergh in the case of a cubic polynomial φ(x1,x2,z) which is nonsingular over a ﬁeld F of characteristic zero. Although Van den Bergh only states and proves his result for Roby Clifford algebras, his proof goes through in this more general setting as well.

Theorem 5.3 (Van den Bergh). Let F be a ﬁeld of characteristic zero, and let φ(x1,x2, = P3 z) be a cubic polynomial. Suppose that the subvariety φ(x1,x2,z) 0 of F is nonsingular (i.e., is an elliptic curve). Then the Clifford algebra Cφ has ﬁnite- dimensional representations. Moreover, all irreducible representations of Cφ are degree 3.

Proof. See Section 2 of [19].

Acknowledgements

The results of this paper were a part of my Ph.D. thesis, written under the direction of Professor Robert M. Guralnick. I would like to thank Professor Guralnick for all of his help and guidance. Also, I would like to thank the referee for many helpful comments. Finally, I was supported in part by an NSF grant while working on this project, and I gratefully acknowledge this support.

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