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Generators, Relations and Symmetries in Pairs of 3x3 Unimodular Matrices


arXiv:math/0601132v4 [math.AG] 23 Jan 2007

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES
SEAN LAWTON

Abstract. Denote the free group on two letters by F2 and the SL(3, )-representation variety of F2 by R = Hom(F2 , SL(3, )). There is a SL(3, )-action on the coordinate ring of R, and the geometric points of the subring of invariants is an a?ne variety X. We determine explicit minimal generators and de?ning relations for the subring of invariants and show X is a degree 6 hyper-surface in 9 mapping onto 8 . Our choice of generators exhibit Out(F2 ) symmetries which allow for a succinct expression of the de?ning relations.

1. Introduction The purpose of this paper is to describe a minimal generating set and de?ning relations for the ring of invariants [SL(3, ) × SL(3, )]SL(3, ) . This generating set exhibits symmetries which allow for an explicit and succinct expression of the invariant ring as a quotient. Explicit minimal generators have been found by [22] and graphically by [19]; in an unpublished calculation [13] independently describe the de?ning relations. Our treatment provides the most succinct and transparent description by uncovering symmetries which provide a framework for generalization. A related algebra, however di?erent, is the ring of invariants of pairs of 3 × 3 matrices M3 ( ) × M3 ( ) under simultaneous conjugation. The algebra of invariants [M3 ( ) × M3 ( )]GL(3, ) comes to bear on the algebra of invariants [SL(3, ) × SL(3, )]SL(3, ) by restriction. On the other hand, the former ring of invariants may be described, in part, by [sl(3) × sl(3)]SL(3, ) ; the in?nitesimal invariants in the latter ring. In this more general context, similar questions about generators and relations have been addressed. In particular, explicit minimal generators were ?rst found by [5] in 1935, and later by [20, 10, 21]. The much more general results of [1] additionally provide minimal generators. However, [11], and later [2] were the ?rst to explicitly describe the de?ning relations. For the state-of-the-art, see [4].
Date : January 22, 2007. Key words and phrases. character variety, free group.
1

2

SEAN LAWTON

We now describe the main results of this paper. Let X be the variety whose coordinate ring is [X] = [SL(3, ) × SL(3, )]SL(3, ) . Theorem 8 asserts that X is isomorphic to a degree 6 a?ne hyper-surface in 9 which generically maps 2-to-1 onto 8 . Next, Theorem 9 explicitly describes the singular locus of X, and examples of non-singular representations in the branching locus are constructed. Lastly, Theorem 13 describes an 8-fold symmetry on [X] which at once characterizes the algebraically independent generators and allows for a surprisingly simple description of the de?ning relations. We hope that this paper will be of interest to algebraic-geometers, ring theorists, and geometers alike. In particular, results in this paper have recently been used in work concerning the hyperbolic geometry of sphereical CR manifolds (see [17]). With this in mind, some of the exposition, for instance, may be “well-known” to a ring theorist but perhaps not to an algebraic-geometer or a geometer. The reader is encouraged to skip such exposition, as appropriate. Acknowledgements The author thanks William Goldman for overall guidance, substantial suggestions, and edits that always improved the quality of earlier drafts of this paper. He thanks Richard Schwartz, John Millson, and the University of Maryland’s VIGRE-NSF program for generously supporting this work. In particular, he thanks John Millson for suggesting an analysis of the singular locus, which has greatly improved the quality of this paper. The author has bene?ted from fruitful conversations with Ben Howard, Elisha Peterson, Adam Sikora, and Joseph Previte, and thanks them for their time and insight. He especially thanks Joseph Previte and Eugene Xia for generously sharing their calculations with him. Lastly, the author thanks the reviewer, whose suggestions and corrections have helped make this paper more readable. 2. SL(3, ) Invariants 2.1. Algebraic Structure of SL(3, ). The group SL(3, ) has the structure of an algebraic set since it is the zero set of the polynomial ? ? x11 x12 x13 D = det ? x21 x22 x23 ? ? 1 x31 x32 x33 on 9 . Here xij ∈ [x11 , x12 , x13 , x21 , x22 , x23 , x31 , x32 , x33 ], the polynomial ring over in 9 indeterminates. As such denote SL(3, ) by G. The coordinate ring of G is given by [G] = [xij | 1 ≤ i, j ≤ 3]/(D ).

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES 3

Since D is irreducible, (D ) is a prime ideal. So the algebraic set G is in fact an a?ne variety. 2.2. Representation and Character Varieties of a Free Group. Let Fr be the free group of rank r generated by {x1 , ..., xr }. The map Hom(Fr , G) ?→ G×r de?ned by sending ρ → (ρ(x1 ), ρ(x2 ), ..., ρ(xr )) is a bijection. Since G×r is the r -fold product of irreducible algebraic sets, G×r ? = Hom(Fr , G) is an a?ne variety. As such Hom(Fr , G) is denoted by R and referred to as the SL(3, )-representation variety of Fr . Let [R] be the coordinate ring of R. Our preceding remarks imply [R] ? = [G]?r . For 1 ≤ k ≤ r, de?ne a generic matrix of the complex polynomial ring in 9r indeterminates by ? k ? k x11 xk 12 x13 k k ? xk = ? xk . 21 x22 x23 k k k x31 x32 x33 Let ? be the ideal (det(xk ) ? 1 | 1 ≤ k ≤ r ) in [R]. Then [R] = [xk ij | 1 ≤ i, j ≤ 3, 1 ≤ k ≤ r ]/?. Let (x1 , x2 , ..., xr ) be an r -tuple of generic matrices. An element f ∈ [R] is a function de?ned in terms of such r -tuples. There is a polynomial G-action on [R] given by diagonal conjugation; that is, for g ∈ G g · f (x1 , x2 , ..., xr ) = f (g ?1x1 g, ..., g ?1xr g ). The subring of invariants of this action [R]G is a ?nitely generated [3, 14, 15]). Consequently, the character variety X = Specmax ( [R]G) is the irreducible algebraic set whose coordinate ring is the ring of invariants. For r > 1, the Krull dimension of X is 8r ? 8 since generic elements have zero dimensional isotropy (see [3], page 98). More generally, the dimension of the ring of invariants [Mn ( )×r ]GL(n, ) is n2 (r ? 1) + 1 (see [4]). Consequently, the dimension of [sl(n)×r ]SL(n, ) , which equals that of [X], is (n2 ? 1)(r ? 1). π There is a regular map R → X which factors through R/G: let m be a maximal ideal corresponding to a point in R, then the composite isomorphism → [R] → [R]/m implies that the composite map → [R]G → [R]G/(m ∩ [R]G) is an isomorphism as -algebra (see

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SEAN LAWTON

well. Hence the contraction m ∩ [R]G is maximal, and since for any g ∈ G, (g mg ?1) ∩ [R]G = m ∩ [R]G, π factors through R/G (see [6], page 38). Although R/G is not generally an algebraic set, X is the categorical quotient R/ /G, and since G is a (geometrically) reductive algebraic group π is surjective, maps closed G-invariant sets to closed sets, and separates distinct closed orbits (see [3]). 3. Trace Identities for Matrices
+ Let F+ r be the free monoid generated by {x1 , ..., xr }, and let Mr be the monoid generated by {x1 , x2 , ..., xr }, as de?ned in section 2.2, under matrix multiplication and with + identity ? the 3 × 3 identity matrix. There is a surjection F+ r → Mr , de?ned by mapping + xi → xi . Let w ∈ M+ r be the image of w ∈ Fr under this map. Further, let | · | be the function that takes a reduced word in Fr to its word length. Then by [14, 15], we know [X] is not only ?nitely generated, but in fact generated by

(1)

{tr(w) | w ∈ F+ r , |w| ≤ 7}.

More generally, the length of the generators is bounded by the class of nilpotency of nil algebras of class n. With respect to matrix algebras, n is the size of the matrices under consideration. The best known upper bound is that of [15] and is n2 ; the lower bound is n(n + 1)/2 and is conjectured to be equality. For n = 2, 3, 4 this conjecture, known as Kuzmin’s conjecture, has been veri?ed (see [4]). In the proof of the Nagata-Higman theorem (see [4]), the bound is computed to be 2n ? 1, which is how |w| ≤ 7 in (1) arises. th Let x? k be the transpose of the matrix of cofactors of xk . In other words, the (i, j ) ? entry of xk is (?1)i+j Cof ji(xk ); that is, the determinant obtained by removing the j th row and ith column of xk . Let M? r ? ? be the monoid generated by {x1 , x2 , ..., xr } and {x? , x , ..., x } . Observe that 1 2 r (xy)? = y? x? for all x, y ∈ M+ r , and xx? = det(x)?. {det(xk )? | 1 ≤ k ≤ r },
? ?1 and subsequently de?ne Mr = M? r /Nr . Notice in Mr , x = x , and thus Mr is a group. We will need the structure of an algebra, and to that end let Mr be the group algebra de?ned over with respect to matrix addition and scalar multiplication in Mr . Likewise, ? let M? r be the semi-group algebra of the monoid Mr .

Now let Nr be the normal sub-monoid generated by

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES 5

The following commutative diagram relates these objects: F+ r ? ? ? ? ? → Fr Fr ? ?

M+ ? ? ? → M? ? ? ? → Mr r ? ?r ? ? ? ? M+ ? ? → r ? M? ? ? → r ?

Mr ? ? ? →

tr

[X].

3.1. Relations. The Cayley-Hamilton theorem applies to this context and so for any x ∈ Mr , (2) x3 ? tr(x)x2 + tr(x? )x ? det(x)? = 0.

By direct calculation, or by Newton’s trace formulas

(3) Together (2) and (3) imply (4)

tr(x? ) =

1 tr(x)2 ? tr(x2 ) . 2

1 1 1 det(x) = tr(x3 ) + tr(x)3 ? tr(x)tr(x2 ). 3 6 2

Remark 1. In general the coe?cients of the characteristic polynomial of an n × n matrix are the elementary symmetric polynomials in the eigenvalues of the matrix. By Newton’s formulas these are trace expressions in powers of the matrix. So one may use this and the general method of polarization, which we demonstrate below, to develop trace identities for larger size matrices. Computations similar to those that follow may be found in [10, 20]; the process is standard and is generally known as (partial) polarization, or multilinearization. For any x, y ∈ Mr and any λ ∈ , equation (2) implies (5) (x + λy)3 ? tr(x + λy)(x + λy)2 + tr((x + λy)? )(x + λy) ? det(x + λy)? = 0.

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SEAN LAWTON

Using equations (2), (3), and (4), equation (5) simpli?es to 1 1 0 =λ2 xy2 + y2 x + yxy ? tr(x)y2 ? tr(y)xy ? tr(y)yx + tr(y)2 x ? tr(y2 )x+ 2 2 1 1 tr(x)tr(y)y ? tr(xy)y ? tr(xy2 )? ? tr(x)tr(y)2 ? + tr(x)tr(y2 )? + tr(y)tr(xy)? + 2 2 1 1 λ yx2 + x2 y + xyx ? tr(y)x2 ? tr(x)yx ? tr(x)xy + tr(x)2 y ? tr(x2 )y+ 2 2 1 1 tr(x)tr(y)x ? tr(xy)x ? tr(yx2 )? ? tr(y)tr(x)2 ? + tr(y)tr(x2 )? + tr(x)tr(xy)? . 2 2 Thus, by Vandermonde arguments (see [16]) we have the partial polarization of (2) yx2 + x2 y + xyx = tr(y)x2 + tr(x)yx + tr(x)xy ? tr(x)tr(y)x + tr(xy)x+ 1 tr(x)2 y ? tr(x2 )y ? tr(y)tr(x)2 ? + tr(y)tr(x2 )? . tr(yx2 )? ? tr(x)tr(xy)? ? 2

(6)

De?ne pol(x, y) to be the right hand side of equation (6); that is, (7) pol(x, y) = yx2 + x2 y + xyx.

Then substituting x by the sum x + z in equation (7), yields the full polarization of (2) (8) xzy + zxy + yxz + yzx + xyz + zyx = pol(x + z, y) ? pol(x, y) ? pol(z, y).

If x, y ∈ Mr then multiplying equation (2) on the right by x?1 y yields, (9) x2 y ? tr(x)xy + tr(x?1 )y ? x?1 y = 0.

Suppose x, y ∈ Mr . Multiplying equation (6) on the left by y?1 x?1 and on the right by x?1 , followed by taking the trace, and using equation (9) appropriately, provides the commutator trace relation tr(xyx?1 y?1 ) = ? tr(yxy?1 x?1 ) + tr(x)tr(x?1 )tr(y)tr(y?1 ) + tr(x)tr(x?1 ) + tr(y)tr(y?1 ) + tr(xy)tr(x?1 y?1 ) (10) + tr(xy?1 )tr(x?1 y) ? tr(x?1 )tr(y)tr(xy?1 ) ? tr(x)tr(y?1 )tr(x?1 y) ? tr(x)tr(y)tr(x?1 y?1 ) ? tr(xy)tr(x?1 )tr(y?1 ) ? 3.

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES 7

3.2. Generators. From (1), we need only consider words in F+ r of length 7 or less. In [20] it is shown that this length may be taken to be 6. We give a similar argument here since the development of the result provides many useful relations, and a constructive algorithm for word reduction that is of computational signi?cance. The length of a reduced word is de?ned to be the number of letters, counting multiplicity, in the word. We now de?ne the weighted length, denoted by | · |w , to be the number of letters of a reduced word having positive exponent plus twice the number of letters having negative exponent, again counting multiplicity. ?2 For example, in F2 , we have |x1 x2 | = |x1 x2 |w = 2 but |x3 1 x2 | = 3 + 2 = 5 while ?2 |x3 1 x2 |w = 3 + 2 · 2 = 7. For a polynomial expression e in generic matrices with coe?cients in [X], we de?ne the degree of e, denoted by ||e||, to be the largest weighted length of monomial words in the expression of e that is minimal among all such expressions for e. Additionally, we de?ne the trace degree of e, denoted by ||e||tr , to be the maximal degree over all monomial words within a trace coe?cient of e. For example, when x, y ∈ Mr , ||pol(x, y)|| ≤ max{2||x||, ||x||+||y||}, while ||pol(x, y)||tr ≤ 2||x|| + ||y||. We remark that given two such expressions e1 and e2 , ||e1 e2 || ≤ ||e1 || + ||e2 || and ||e1 e2 ||tr ≤ max{||e1 ||tr , ||e2 ||tr }. We are now prepared to characterize the generators of [X]. Lemma 2. [R]G is generated by tr(w) such that w ∈ Fr is cyclicly reduced, |w|w ≤ 6, and all exponents of letters in w are ±1. Proof. For n ≥ 2, equations (2) and (9) determine equation (11) tr(uxn v) =tr(x)tr(uxn?1 v) ? tr(x?1 )tr(uxn?2 v) + tr(uxn?3 v),

which recursively reduces tr(w) to a polynomial in traces of words having no letter with exponent other than ±1. If however n ≤ ?2 then we ?rst apply equation (9) and then use (11). Hence it follows that w may be taken to be cyclically reduced, having all letters with exponent ±1. It remains to show that the word length may be taken to be less than or equal to 6. Substituting x → y and y → xz in equation (7), and multiplying the resulting expression on the left by x gives (12) x2 zy2 = ?(xy2 x)z ? (xyx)zy + xpol(y, x).

Replacing y → y2 in equation (7) produces y2 x2 + x2 y2 + xy2 x = pol(x, y2 ),

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SEAN LAWTON

which substituted into equation (12) yields equation (13) x2 zy2 = (y2 x2 + x2 y2 ? pol(x, y2 ))z + (yx2 + x2 y ? pol(x, y))zy + xpol(y, xz). Now substituting pol(y, x2z) = x2 zy2 + y2 x2 z + yx2 zy, and x2 pol(y, z) = x2 zy2 + x2 y2 z + x2 yzy into equation (13) results in (14) 3x2 zy2 = pol(y, x2 z) + xpol(y, xz) ? pol(x, y2 )z ? pol(x, y)zy + x2 pol(y, z). ||x2 zy2 || < 2||x|| + ||z|| + 2||y|| and ||x2 zy2 ||tr ≤ 2||x|| + ||z|| + 2||y||. Remark 3. In the proof of the Nagata-Higman theorem, the two-sided ideal of polynomial trace relations, for n = 3, is shown to be generated as a vector space by pol(u, v), u3 , and equation (8) evaluated at monomial words u, v, and w. Equation (14) shows x2 zy2 is in this ideal, and consequently its degree is less than its word length. However, one can conclude x2 zy2 is in this ideal from more general considerations and avoid the above calculation (see [4], page 76). For the remainder of the argument assume x, y, z, u, v, w are of length 1. Replacing y → u + v in equation (14) we deduce ||x2 z(u2 + uv + vu + v2 )|| ≤ 4. This in turn implies ||x2 z(uv + vu)|| ≤ 4 and so ||x2 zw(uv + vu)|| ≤ 5. In a like manner, we have that both ||x2 z(wuv + vwu)|| ≤ 5 and ||x2 z(wv + vw)u|| ≤ 5. Hence we conclude that ||2x2 zwuv|| = ||x2 zw(uv + vu) + x2 z(wuv + vwu) ? x2 z(wv + vw)u|| ≤ 5, and ||2x2 zwuv||tr ≤ 6. Replacing x → x+y in x2 zwuv we come to the conclusion that ||xyzwuv+yxzwuv|| ≤ 5. In other words, permuting x and y introduces a factor of ?1 and a polynomial term of lesser degree. Slight variation in our analysis concludes the same result for any transposition of two adjacent letters in the word xyzwuv. Therefore, if σ is a permutation of the letters x, y, z, u, v, w then ||xyzuvw + sgn(σ )σ (xyzuvw)|| ≤ 5 while ||xyzuvw + sgn(σ )σ (xyzuvw)||tr ≤ 6. Lastly, making the substitutions x → xy, y → zu, and z → vw in the fundamental expression (8), we derive xyvwzu+vwxyzu + zuxyvw + zuvwxy + xyzuvw + vwzuxy = (15) pol(xy + vw, zu) ? pol(xy, zu) ? pol(vw, zu). Thus,

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES 9

However, each word on the left hand side of equation (15) is an even permutation of the ?rst, so ||6xyvwzu|| ≤ 5 and ||6xyvwzu||tr ≤ 6. Hence, if w is a word of length 7 or more, then ||tr(w)||tr ≤ 6. Moreover, this process gives an iterative algorithm for reducing such an expression. As an immediate consequence we have the following description of su?cient generators of [X]. Corollary 4. [X] is generated by traces of the form

1 ?1 ?1 ?1 ?1 tr(xi ), tr(x? i ), tr(xi xj ), tr(xi xj xk ), tr(xi xj ), tr(xi xj ), tr(xi xj xk ), 1 ?1 ?1 ?1 tr(xi xj xk xl ), tr(xi xj xk xl xm ), tr(xi xj xk x? l ), tr(xi xj xk xj ), tr(xi xj xk ), 1 ?1 ?1 ?1 ?1 ?1 ?1 ?1 ?1 tr(x? i xj xk ), tr(xi xj xk xl ), tr(xi xj xk xj ), tr(xi xj xi xj ), 1 ?1 ?1 tr(xi xj xk xl x? m ), tr(xi xj xk xl xk ), tr(xi xj xk xl xj ), tr(xi xj xk xl xm xn ),

where 1 ≤ i = j = k = l = m = n ≤ r . Proof. First, consider generators of type tr(u?1 wx?1 z). It can be shown that tr(uvwxyz)+ tr(uvwyxz) + tr(vuwxyz) + tr(vuwyxz) has trace degree 5. Setting u = v and x = y and subsequently interchanging words with squares to those with inverses, we ?nd generators of the form tr(u?1 wx?1 z) can be freely eliminated; that is, inverses can be assumed to be adjacent. It remains to show that letters may be taken to be distinct. Equation (7) implies that for any letter x and any monomial words w1 , w2 , w3 , tr(w1 x±1 w2 x±1 w3 ) = ?tr(w1 x±2 w2 w3 ) ? tr(w1 w2 x±2 w3 ) + tr(w1 pol(x±1 , w2 )w3 ). However, by subsequently reducing the words having letters with exponent not ±1, we conclude that expressions of the form tr(w1 x±1 w2 x±1 w3 ) are unnecessary. This result can be re?ned using the work of [1], where explicit minimal generators are formulated in a more general context. In an upcoming paper, we will address the issue of minimality for our generators, as well as provide a maximal subset that is algebraically independent. This subset will allow for a generalization of the symmetry described in section 5. 4. Structure of [G × G]G

4.1. Minimal Generators. As a consequence of Corollary 4, we have

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SEAN LAWTON

Lemma 5.

[G × G]G is generated by
1 ?1 tr(x1 ), tr(x2 ), tr(x1 x2 ), tr(x1 x? 2 ), tr(x1 ), 1 ?1 ?1 ?1 ?1 ?1 tr(x? 2 ), tr(x1 x2 ), tr(x1 x2 ), tr(x1 x2 x1 x2 ).

Proof. The words of weighted length 1, 2, 3, 4 with exponents ±1 are unambiguously cyclically equivalent to one of
1 ?1 ?1 ?1 2 x1 , x2 , x1 x2 , x1 x? 2 , x2 x1 , x1 x2 , (x1 x2 ) .

But equation (9) reduces the latter most of these in terms of the others. All words in two letters of length 5 are cyclically equivalent to a word with an exponent whose magnitude 1 ?1 is greater than 1, except x1 x? 2 x1 x2 , and x2 x1 x2 x1 . Both are cyclically equivalent to 2 (xi xj )2 x? j which in turn, by equation (11) reduces to expressions in the other variables. 1 ?1 The only words of weighted length 6 and with exponents only ±1 are x1 x2 x? 1 x2 , its inverse, and (x1 x2 )3 . But the latter most of these is reduced by equation (2). Lastly, letting x = x1 and y = x2 in equation (10), we have
1 ?1 ?1 ?1 ?1 ?1 tr(x2 x1 x? 2 x1 ) = ? tr(x1 x2 x1 x2 ) + tr(x1 )tr(x1 )tr(x2 )tr(x2 ) 1 ?1 ?1 ?1 + tr(x1 )tr(x? 1 ) + tr(x2 )tr(x2 ) + tr(x1 x2 )tr(x1 x2 )

(16)

?1 ?1 ?1 1 + tr(x1 x? 2 )tr(x1 x2 ) ? tr(x1 )tr(x2 )tr(x1 x2 ) 1 ?1 ?1 ?1 ? tr(x1 )tr(x? 2 )tr(x1 x2 ) ? tr(x1 )tr(x2 )tr(x1 x2 ) 1 ?1 ? tr(x1 x2 )tr(x? 1 )tr(x2 ) ? 3,

which expresses the trace of the inverse of the commutator in terms of the other expressions.
2 3 ? 4.2. × 3 -Grading. The center of G is ζ (G) = {ω ? | ω = 1} = ×2 of ζ (G) on [X] given by 3.

There is an action tr(w(x1 , x2 )).

(ω1 ?, ω2 ?) · tr(w(x1 , x2 )) = tr(w(ω1 x1 , ω2 x2 )) = ω1

|w(x1 ,?)|w

ω2

|w(?,x2 )|w

Applying this action to the generators and recording the orbit by a 9-tuple, all generators are distinguished. Consequently, we have Proposition 6. [X] =
(ω1 ,ω2 )∈
3× 3

[X](ω1 ,ω2 )

is a 3 × 3 -graded ring. The summand [X](ω1 ,ω2 ) is the linear span over of all monomials whose orbit under 3 × 3 equals one of the orbits of the nine orbit types corresponding to the minimal generators.

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES11
r In fact the situation is general. For a rank r free group, × 3 acts on the generators of [X] and gives a ?ltration. However, since the relations are polarizations of the Cayley-Hamilton polynomial, which itself has a zero grading, no relation can compromise summands. So the ?ltration is a grading.

4.3. Hypersurface in

9

. Let

R = [t(1) , t(?1) , t(2) , t(?2) , t(3) , t(?3) , t(4) , t(?4) , t(5) , t(?5) ] be the complex polynomial ring freely generated by {t(±i) , 1 ≤ i ≤ 5}, and let R = [t(1) , t(?1) , t(2) , t(?2) , t(3) , t(?3) , t(4) , t(?4) ] be its subring generated by {t(±i) , 1 ≤ i ≤ 4}, so R = R[t(5) , t(?5) ]. De?ne the following ring homomorphism, R[t(5) , t(?5) ] ?→ [G × G]G by t(1) → tr(x1 ) t(2) → tr(x2 ) t(3) → tr(x1 x2 ) 1 t(4) → tr(x1 x? 2 ) 1 ?1 t(5) → tr(x1 x2 x? 1 x2 ) It follows from Lemma 5 that t(?1) t(?2) t(?3) t(?4) t(?5)
1 → tr(x? 1 ) 1 → tr(x? 2 ) 1 ?1 → tr(x? 1 x2 ) ?1 → tr(x1 x2 ) 1 ?1 → tr(x2 x1 x? 2 x1 ). Π

[X] ? = R[t(5) , t(?5) ]/ ker(Π). In other words, Π is a surjective algebra morphism. We de?ne P = t(1) t(?1) t(2) t(?2) ? t(1) t(2) t(?3) ? t(?1) t(?2) t(3) ? t(1) t(?2) t(?4) ? t(?1) t(2) t(4) +t(1) t(?1) + t(2) t(?2) + t(3) t(?3) + t(4) t(?4) ? 3, and so P ∈ R. Moreover, by equation (16), P ? (t(5) + t(?5) ) ∈ ker(Π). Hence it follows that the composite map R[t(5) ] ?→ R[t(5) , t(?5) ] ? R[t(5) , t(?5) ]/ ker(Π), is an epimorphism. Let I be the kernel of this composite map, and suppose there exists Q ∈ R so Q ? t(5) t(?5) ∈ ker(Π) as well. Then under this assumption, we prove

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SEAN LAWTON

Lemma 7. I is principally generated by the polynomial (17) t2 (5) ? P t(5) + Q.

Proof. The following argument is an adaptation of one found in [11]. Certainly, t2 (5) ? P t(5) + Q ∈ I for it maps into R[t(5) , t(?5) ]/ ker(Π) to the coset repre2 sentative t(5) ? (t(5) + t(?5) )t(5) + t(5) t(?5) = 0. On the other hand, observe R[t(5) ]/I ? = R[t(5) , t(?5) ]/ ker(Π) ? = [X], the dimension of X is 8, and R[t(5) ]/I has at most 9 generators. Then it must be the case that I is principally generated since R[t(5) ] is a U.F.D., and thus a co-dimension 1 irreducible subvariety of 9 must be given by one equation (see [18] page 69). Moreover, I is non-zero since otherwise the resulting dimension would necessarily be too large. Seeking a contradiction, suppose there exists a polynomial identity comprised of only elements of R. Then Krull’s dimension theorem (see page 68 in [18]) implies t(5) is free. In other words, given any restriction of the generators of R, t(5) is not determined. Consider ? ? a b 0 SL3 (SL(2, ))×2 ? G×2 ; that is, matrices of the form ? c d 0 ? so ad ? bc = 1. Then 0 0 1 by restricting to pairs of such matrices, we deduce that
1 ?1 ?1 ?1 tr(x1 x2 x? 1 x2 ) = tr(x2 x1 x2 x1 ),

since for all x ∈ SL(2, ), tr(x) = tr(x?1 ). Then equation (16) becomes t(5) = P/2, which is decidedly not free of the generators of R. Thus, the generators of R are algebraically independent in R[t(5) ]/I . Since I is principal and contains a monic quadratic over R, its generator is expression (17), or a factor thereof. We have just showed that there are no degree zero relations, with respect to t(5) . However, if I is generated by a linear polynomial over R then t(5) is determined by the generators of R alone. However this in turn would imply that all representations who agree by evaluation in R also agree by evaluation under t(5) . Consider the representations

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES13

F2 ?→ G F2 ?→ G ? ? ? ? a 0 0 a 0 0 x1 ?→ ? 0 b and x1 ?→ ? 0 b 0 ? 0 ? 0 0 1/ab 0 0 1/ab ? ? ? ? 1 1 ?1 1 ?1 1 x2 ?→ 411/3 ? ?1 ?1 ?1 ? . x2 ?→ 411/3 ? 1 ?1 1 ? ?1 ?1 ?1 1 1 ?1 It is a direct calculation to verify that they agree upon evaluation in R but disagree under t(5) . Lemmas 5 and 7 together imply the following theorem whose result, in part, was given by [22], and later by [19], and may also be inferred by the work of [21]. Theorem 8. G×2 / /G is isomorphic to a degree 6 a?ne hyper-surface in 8 onto .
9

ρ1

ρ2

, which maps

Proof. The degree of Q will be apparent when we explicitly write it down. It remains to show that X → 8 is a surjection. To this end, let (z1 ? ζ1 , ..., z8 ? ζ8 ) be a maximal ideal in the coordinate ring of 8 . Moreover, let ζ9 be de?ned to be a solution to t2 ? P (ζ1, ..., ζ8 )t + Q(ζ1 , ..., ζ8 ) = 0. Then (t(1) ? ζ1 , t(?1) ? ζ2 , ..., t(?4) ? ζ8 , t(5) ? ζ9 ) + I is a maximal ideal in [X], and so all maximal ideals of [ 8 ] are images of such in [X]. 4.4. Singular Locus of X. The surjection X → are exactly two solutions to
8

is generically 2-to-1; that is, there

t2 ? P (ζ1, ..., ζ8 )t + Q(ζ1 , ..., ζ8 ) = 0 for every point in
8

except where P 2 ? 4Q = 0. In this case, 0 = (t(5) + t(?5) )2 ? 4t(5) t(?5) = (t(5) ? t(?5) )2

which implies t(5) = t(?5) = P/2. On the other hand, at points in X if t(5) = P/2, then P 2 ? 4Q = 0. Let L denote the locus of solutions to P 2 ? 4Q = 0 in X, which is a closed subset of X. It is readily observed that the partial derivative with respect to t(5) of t2 (5) ? P t(5) + Q is zero if and only if t(5) = P/2. The singular set in X, denoted by J, is the closed subset cut out by the Jacobian ideal; that is, the ideal generated by the formal partial derivatives of t2 (5) ? P t(5) + Q. Thus J ? L. If H ?→ G is a sub-algebraic group, then we de?ne H×r / /G to be the image of H×r ?→ R → X.

14

SEAN LAWTON

/G ? L. Additionally, since In the proof of Lemma 7, we observed SL3 (SL(2, ))×2/ ? ? a 0 0 matrices of the form ? 0 b 0 ? commute, restricting to pairs of such matrices 0 0 1/ab enforces the relation
1 ?1 ?1 ?1 tr(x1 x2 x? 1 x2 ) = 3 = tr(x2 x1 x2 x1 ).

Let SL3 ( ? × ? ) denote the subset of such matrices in G. Consequently, SL3 ( ? × ? ×2 ) / /G ? L as well. We claim both sets satisfy all the generators of the Jacobian ideal, ?Q ?P and so are singular. The Jacobian ideal is generated by the polynomials ?t(5) ?t + ?t for (i) (i) 1 ≤ |i| ≤ 4, and 2t(5) ? P . Using the formulas for P and Q (see section 4.5), we explicitly write out the generators of the Jacobian ideal (see [8] for details). Then evaluating these polynomials at pairs of generic matrices in either SL3 (SL(2, )) or SL3 ( ? × ? ) we verify that all partials vanish using Mathematica ([23]). It turns out these examples are prototypical. ? ? a b 0 0 ?. Let SL3 (GL(2, )) be the subset of G consisting of elements of the form ? c d 1 0 0 ad? bc /G and SL3 (SL(2, ))×2/ /G are contained in SL3 (GL(2, ))×2 / /G. Notice that SL3 ( ? × ? )×2 / Again, using Mathematica we evaluate all generating polynomials of the Jacobian ideal on pairs of generic matrices in SL3 (GL(2, )). Since all partials vanish, SL3 (GL(2, ))×2/ /G is singular in X as well. In general, if [ρ] ∈ G×r / /G is singular, then its orbit has positive-dimensional isotropy. Any completely reducible representation (these parameterize G×r / /G as an orbit space) that is not irreducible is conjugate to an element in SL3 (GL(2, ))×r . This follows since there must be a shared eigenvector with respect to its generic matrices, if the representation reduces at all. Irreducible representations are known to be non-singular, and their isotropy is zero dimensional. Consequently, it follows that in general the singular set of G×r / /G is contained in SL3 (GL(2, ))×r / /G. In the case of a free group of rank 1, there are no singular points in the quotient and so the identity, which has maximal isotropy, remains non-singular. Hence the converse inclusion does not generally hold. In the case of a free group of rank 2, the situation is much better. In fact, we have already established Theorem 9. A completely reducible representation in G×2 / /G is singular if and only if /G. its orbit has positive-dimensional isotropy; that is, J = SL3 (GL(2, ))×2 / As a ?nal note, we give an example of a non-singular representation in the branching locus (actually we give a 2-dimensional family in L ? J):

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES15

F2 ?→ G ? ? ? a 0 0 1 1 ?1 c1/3 ? x1 ?→ ? 0 a 0 ? x2 ?→ 4 1 ?1 1 ?, 1/3 2 0 0 1/a ?1/c ?1/c ?1/c so long as a3 = 1 and c = 0. Calculating the Jacobian relations we determine that all partial derivatives are 0 except for ?t(5) ?Q (?1 + a3 )3 ?Q (?1 + a3 )3 ?P ?P + =? + = and ? t , (5) ?t(1) ?t(1) 4a4 ?t(?1) ?t(?1) 4a5 ?

ρ

which are clearly not always 0. 4.5. Determining Q. For the proofs of Lemma 7 and subsequently Theorem 8 to be complete, it only remains to establish that there exists Q ∈ R so Q ? t(5) t(?5) ∈ ker(Π). Before doing so, we state and prove the following technical fact, which may be found in [11]. Fact 10. De?ne a bilinear form on the vector space of n × n matrices over (A, B ) = ntr(AB ) ? tr(A)tr(B ). Then given vectors A1 , ..., An2 , B1 , ..., Bn2 , the n2×n2 matrix ? = Proof. Consider the co-vector ? ? (A1 , ) (A2 , ) ? ? ?. . . ? . (An2 , ) (Ai , Bj ) is singular. by

? ? v( ) = ? ?

If B1 , ..., Bn2 are linearly dependent then so are v (B1 ), v (B2 ), ..., v (Bn2 ), which implies the columns of ? are linearly dependent. Otherwise there exists coe?cients, not all zero, so c1 B1 + c2 B2 + · · · + cn2 Bn2 = ?, which implies c1 v (B1 ) + c2 v (B2 ) + · · · + cn2 v (Bn2 ) = 0 since the identity ? is in the kernel of dependent. Either way, ? is singular. (A, ). So again the columns of

? are linearly

16

SEAN LAWTON

Lemma 11. There exists a polynomial Q ∈ R so Q ? t(5) t(?5) ∈ ker(Π), and in particular
3 3 Q =9 ? 6t(1) t(?1) ? 6t(2) t(?2) ? 6t(3) t(?3) ? 6t(4) t(?4) + t3 (1) + t(2) + t(3) 3 3 3 3 + t3 (4) + t(?1) + t(?2) + t(?3) + t(?4) ? 3t(?4) t(?3) t(?1) ? 3t(4) t(3) t(1)

? 3t(?4) t(2) t(3) ? 3t(4) t(?2) t(?3) + 3t(?4) t(?2) t(1) + 3t(4) t(2) t(?1) + 3t(1) t(2) t(?3) + 3t(?1) t(?2) t(3) + t(?2) t(?1) t(2) t(1) + t(?3) t(?2) t(3) t(2) + t(?4) t(?1) t(4) t(1) + t(?4) t(?2) t(4) t(2) + t(?3) t(?1) t(3) t(1) + t(?3) t(?4) t(3) t(4)
2 2 2 2 + t2 (?4) t(?3) t(?2) + t(4) t(3) t(2) + t(?1) t(?2) t(?4) + t(1) t(2) t(4) + t(1) t(?2) t(?3) 2 2 2 2 + t(?1) t2 (2) t(3) + t(?4) t(?3) t(1) + t(4) t(3) t(?1) + t(?4) t(2) t(?3) + t(4) t(?2) t(3) 2 2 2 2 + t2 (?1) t(?3) t(2) + t(1) t(3) t(?2) + t(?4) t(1) t(2) + t(4) t(?1) t(?2) + t(?4) t(3) t(?2) 2 2 2 + t(4) t(?3) t2 (2) + t(1) t(3) t(?4) + t(?1) t(?3) t(4) + t(?1) t(?4) t(3)

(18)

2 2 2 + t(1) t(4) t2 (?3) ? 2t(?3) t(?2) t(?1) ? 2t(3) t(2) t(1) ? 2t(?4) t(?1) t(2) 2 2 2 2 2 2 ? 2t2 (4) t(1) t(?2) + t(?1) t(?2) t(?3) + t(1) t(2) t(3) + t(?4) t(?1) t(2) 2 2 2 + t(4) t2 (1) t(?2) ? t(?4) t(?2) t(2) t(1) ? t(4) t(2) t(?2) t(?1) 2 2 2 ? t(?3) t2 (1) t(?1) t(2) ? t(3) t(?1) t(1) t(?2) ? t(?3) t(2) t(?2) t(1) ? t(3) t(?2) t(2) t(?1) 2 3 3 ? t(?4) t(?2) t(?1) t2 (1) ? t(4) t(2) t(1) t(?1) ? t(?1) t(?2) t(1) ? t(?1) t(2) t(1) 3 ? t3 (?1) t(?2) t(2) ? t(1) t(?2) t(2) ? t(?4) t(?3) t(?2) t(?1) t(2) ? t(4) t(3) t(2) t(1) t(?2) 2 2 2 ? t(?1) t(1) t(2) t(?4) t(3) ? t(?1) t(1) t(?2) t(4) t(?3) + t(?2) t2 (?1) t(1) t(2) + t(?1) t(?2) t(2) t(1) .

Proof. The following argument is an adaptation of an existence argument given in [11], which we use not only to show existence of Q, but to derive the explicit formulation of Q as well. Indeed, with respect to Fact 10, let A1 = B1 = x1 A2 = B2 = x2
1 A4 = B4 = x? 2 1 A7 = B7 = x1 x? 2 1 A5 = B5 = x1 x2 A8 = B8 = x? 2 x1

1 1 A3 = B3 = x? A6 = B6 = x2 x1 A9 = B9 = x2 x? 1 1 . 1 ?1 Then we see that ? has exactly two entries with tr(x1 x2 x? 1 x2 ). After rewriting all matrix entries in terms of our generators of [X], we have 1 ?1 2 ?1 ?1 0 = det(?) = P1 · tr(x1 x2 x? 1 x2 ) + P2 · tr(x1 x2 x1 x2 ) + P3 ,

where P1 , P2 , P3 are polynomials in terms of
1 ?1 ?1 ?1 ?1 ?1 ? = {tr(x1 ), tr(x? R 1 ), tr(x2 ), tr(x2 ), tr(x1 x2 ), tr(x1 x2 ), tr(x1 x2 ), tr(x1 x2 )}.

? , which we have If P1 = 0 then we have a non-trivial relation among the elements of R ? with the aid already seen cannot exist. Alternatively, one can evaluate the elements of R

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES17

of a computer algebra system to verify that P1 = 0. Then by direct calculation, using Mathematica, we verify that P2 = ?P · P1 . Hence it follows that
2 ?P3 = P1 (t2 (5) ? P t(5) ) = P1 (t(5) ? (t(5) + t(?5) )t(5) ) = ?P1 t(5) t(?5) ,

and so we have shown the existence of Q = t(5) t(?5) . Lastly, we simplify P3 /P1 , with the aid of Mathematica, which turns out to be equation (18). 5. Outer Automorphisms Given any α ∈ Aut(F2 ), we de?ne aα ∈ End( [X]) by extending the following mapping aα (tr(w)) = tr(α(w)). If α ∈ Inn(F2 ), then there exists u ∈ F2 so for all w ∈ F2 , α(w) = uwu?1 , which implies aα (tr(w)) = tr(uwu?1 ) = tr(w). Thus Out(F2 ) acts on [X]. By results of Nielsen (see [9], [12]), Out(F2 ) is generated by the following mappings (19) (20) (21) τ= ι= η= x1 → x2 x2 → x1
1 x1 → x? 1 x2 → x2

x1 → x1 x2 x2 → x2

Let D be the subgroup generated by τ and ι, and let D be the corresponding group ring. Then [X] is a D-module. Lemma 12. The action of D preserves R, and D ?xes P and Q. Proof. First we note that it su?ces to check {ι, τ } on {t(±i) , 1 ≤ i ≤ 4}, since the former generates D and the latter generates R. Secondly we observe that both ι and τ are idempotent.

18

SEAN LAWTON

Indeed, ι maps the generators of R as follows: t(1) → t(?1) → t(1) t(3) → t(?4) → t(3) t(2) → t(2) t(?2) → t(?2) t(4) → t(?3) → t(4) . Likewise, τ maps the generators of R by: t(1) → t(2) → t(1) t(?1) → t(?2) → t(?1) t(3) → t(3) t(?3) → t(?3) t(4) → t(?4) → t(4) . Hence both map into R. For the second part of the lemma, it su?ces to observe ι(t(±5) ) = t(?5) = τ (t(±5) ), because in [X], P = t(?5) + t(5) and Q = t(5) t(?5) . Observing ι(t(5) ) = τ (t(5) ) = t(?5) = P ? t(5) , it is apparent that D does not act as a permutation group on the entire coordinate ring of X. However, when restricted to R there is Theorem 13. D restricted to R is group isomorphic to the dihedral group, D4 , of order 8. Moreover, the algebraically independent generators are characterized as those which D acts on as a permutation group. Proof. Let S = Sym(±1, ±2, ±3, ±4) be the symmetric group of all permutations on the eight letters ±i for 1 ≤ i ≤ 4. Then we have worked out, in the proof of Lemma 12, that τ acts on the subscripts of t(±i) as the permutation (1, 2)(?1, ?2)(4, ?4) and likewise, ι acts as the permutation (1, ?1)(3, ?4)(?3, 4). Since D is generated by these elements, we certainly have a well de?ned injection D → S . The Cayley table for D is:

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES19
id id ι τ ιτ τι τ ιτ ιτ ι τ ιτ ι id ι τ ιτ τι τ ιτ ιτ ι τ ιτ ι ι ι id τι ιτ ι τ τ ιτ ι ιτ τ ιτ τ τ ιτ id ι τ ιτ τι τ ιτ ι ιτ ι ιτ ιτ τ τ ιτ τ ιτ ι id ιτ ι ι τι τι τι ιτ ι ι id τ ιτ ι τ τ ιτ ιτ τ ιτ τ ιτ τ ιτ ι ιτ τ ιτ ι id τι ι ιτ ι ιτ ι τι τ ιτ ι τ ιτ ι ιτ id τ τ ιτ ι τ ιτ ι τ ιτ ιτ ι τι ιτ ι τ id

where id → (1) τ → (1, 2)(?1, ?2)(4, ?4) τ ι → (1, ?2, ?1, 2)(3, 4, ?3, ?4) ιτ ι → (1, ?2)(2, ?1)(3, ?3) ι → (1, ?1)(3, ?4)(?3, 4) ιτ → (1, 2, ?1, ?2)(3, ?4, ?3, 4) τ ιτ → (2, ?2)(3, 4)(?3, ?4) τ ιτ ι → (1, ?1)(2, ?2)(3, ?3)(4, ?4).

It is an elementary exercise in group theory (see [7]) to show any group presentable as {a, b |a| = n ≥ 3, |b| = 2, ba = a?1 b} is isomorphic to the dihedral group Dn of order 2n. However, letting a = τ ι and b = ι we see |a| = 4, |b| = 2, D is generated by a and b, and ba = ιτ ι = (τ ι)?1 ι = a?1 b. The last statement in the theorem follows from the fact that {t(±i) | 1 ≤ i ≤ 4} are algebraically independent and D does not act as a permutation group if t(5) were included. Remark 14. The action of D on [X] determines an action on X. Since D acts as a permutation group on R the surjection from Theorem 8, X → 8 , is D-equivariant. In this way X exhibits 8-fold symmetry. As already noted, the group ring D acts on [X]. By brute force computation, one can establish the following succinct expressions for the polynomial relations P and Q. Corollary 15. In D de?ne

?D to be the group “symmetrizer”
σ.
σ∈D

20

SEAN LAWTON

Then P = ?D(p) ? 3 and Q = ?D(q ) + 9 where p and q are given by: 1 p = t(1) t(?1) t(2) t(?2) ? 4t(1) t(?2) t(?4) + 2t(1) t(?1) + 2t(3) t(?3) 8 1 2 2 2 3 2 q = 2t(?2) t2 (?1) t(1) t(2) + 4t(1) t(2) t(3) ? 4t(1) t(?2) t(2) ? 8t(?4) t(?2) t(?1) t(1) ? 4t(4) t(3) t(2) t(1) t(?2) + 8 2 2 2 8t(1) t(3) t2 (?4) + 8t(?4) t(1) t(2) ? 8t(3) t(2) t(1) + 4t(4) t(?3) t(2) + t(?2) t(?1) t(2) t(1) + t(?3) t(?4) t(3) t(4) +
3 4t(?3) t(?1) t(3) t(1) + 4t3 (1) + 4t(3) + 12t(?4) t(?2) t(1) ? 12t(?4) t(2) t(3) ? 12t(1) t(?1) ? 12t(3) t(?3) .

Proof. We work out P only since the computation for Q is established in the same way but longer. Indeed, 1 ?D(p) = ?D(t(1) t(?1) t(2) t(?2) ) ? 4?D(t(1) t(?2) t(?4) ) + 2?D(t(1) t(?1) ) + 2?D(t(3) t(?3) ) 8 1 = 8t(1) t(?1) t(2) t(?2) ? 4(2t(1) t(2) t(?3) + 2t(?1) t(?2) t(3) 8 + 2t(1) t(?2) t(?4) + 2t(?1) t(2) t(4) ) + 2(4t(1) t(?1) + 4t(2) t(?2) + 4t(3) t(?3) + 4t(4) t(?4) ) =P + 3 . With the help of Mathematica or a tedious hand calculation, the formula for Q is equally veri?ed. In [1] an algorithm is deduced that can be adapted to write minimal generators for [X] when Fr is free of arbitrary rank, which we do is an upcoming paper. It is the hope of the author that exploiting symmetries as above will simplify the calculations involved in describing the ideals for free groups of rank 3 or more. Consequently, this would allow for subsequent advances in determining the de?ning relations of X in general. References
[1] Abeasis, A., and Pittaluga, M., On a minimal set of generators for the invariants of 3 × 3 matrices, Comm. Alg. 17(2) (1989), 487-499 [2] Aslaksen, H., Drensky, V., and Sadikova, L., De?ning relations of invariants of two 3 × 3 matrices, J. Algebra 298 (2006), 41-57 [3] Dolgachev, I.,“Lectures on Invariant Theory,” London Mathematical Lecture Notes Series 296, Cambridge University Press, 2003 [4] Drensky, V., and Formanek, E., “Polynomial Identity Rings,” Advanced Courses in Mathematics CRM Barcelona, Birkh¨ auser Verlag Basel, 2004 [5] Dubnov, J., Sur une g? en? eralisation de l’? equation de Hamilton-Caley et sur les invariants simultan? es de plusieurs a?neurs, Proc. Seminar on Vector and Tensor Analysis, Mechanics Research Inst., Moscow State Univ. 2/3 (1935), 351-367. [6] Eisenbud, D., “Commutative Algebra with a View Toward Algebraic Geometry,” Graduate Texts in Mathematics No. 150, Spring-Verlag New York, 1995 [7] Hungerford, T., “Algebra,” Graduate Texts in Mathematics No. 73, Spring-Verlag New York, 1974

GENERATORS, RELATIONS AND SYMMETRIES IN PAIRS OF 3 × 3 UNIMODULAR MATRICES21

[8] Lawton, S., SL(3, )-Character Varieties and ??2 -Structures on a Trinion, PhD Thesis, University of Maryland, 2006 [9] Magnus, W., Karrass, A., and Solitar, D., “Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations,” Pure and Applied Mathematics Vol. XIII, Interscience Publishers, 1966. [10] Marincuk, A. and Sibirskii, K., Minimal polynomial bases of a?ne invariants of square matrices of order three, Mat. Issled. 6 (1971), 100-113 [11] Nakamoto, K., The structure of the invariant ring of two matrices of degree 3, J. Pure and Applied Alg. 166 (2002), 125-148 [12] Nielsen, J., Die Isomorphismengruppe der freien Gruppen, Math. Ann., 91 (1924), 169-209 [13] Previte, J. and Xia, E., Various letters to the author. [14] Procesi, C., Invariant theory of n × n matrices, Advances in Mathematics 19 (1976), 306-381 [15] Razmyslov, Y., Trace identities of full matrix algebras over a ?eld of characteristic zero. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 38 (1974), 723-756. [16] Rowen, L. H., “Polynomial Identities in Ring Theory,” Academic Press, Inc., New York, 1980 [17] Schwartz, R., Spherical CR Geometry and Dehn Surgery, Ann. of Math. Stud. 165, (2007), in Press [18] Shafarevich, I., “Basic Algebraic Geometry 1,” 2nd edition, Springer-Verlag, Berlin, 1994 [19] Sikora, A., SLn -Character Varieties as Space of Graphs, Trans. Amer. Math. Soc. 353 (2001), no. 7, 2773-2804 [20] Spencer,A. and Rivlin, R., Further results in the theory of matrix polynomials, Arch. Rational Mech. Anal. 4 (1960), 214-230 [21] Teranishi, Y., The ring of invariants of matrices, Nagoya Math. J. 104 (1986), 149-161. [22] Wen, Z.X., Relations polynomiales entre les traces de produits de matrices, C.R. Acad. Sci. Paris 318 (1994) no. 2, 99-104 [23] Wolfram, S., “Mathematica: A System for Doing Mathematics by Computer,” Wolfram Press, 2000 Mathematics Department, Kansas State University, Manhattan, KS 66506 E-mail address: slawton@math.ksu.edu URL: http : //www.math.ksu.edu/ ? slawton



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These matrices have been studied extensively since the discovery of Hoffman and Kruskal [13] who proved that an integral matrix A is totally unimodular if...
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