EULER FACTORIZATIONS IN CLASSICAL GROUPS. A.M. DuPre. Rutgers University. May 14, 1992

Transcription

1 EULE FACTOIZATIONS IN CLASSICAL GOUPS A.M. DuPre utgers University May 4, 99 Abstract. Several dierent factorizations of elements of the rotation, unitary, indenite orthogonal, symplectic and unitary symplectic groups into elements in these groups which act only on two coordinates, leaving the others xed, are constructively derived, as well as a similar factorization of the matrices which conjugate one of these elements into a maximal torus. These various factorizations are then used to derive a rational parametrization of the respective groups. Introduction G is a matrix with elements in the real or complex eld. The set of nonsingular n n matrices G with elements in I for which G t IG = I forms a group called the n{dimensional real orthogonal group and is denoted by O(n). It follows immediately from the denition that det G =. The subset of matrices in O(n) with det = forms a normal subgroup of index two called the n{ dimensional rotation or special orthogonal group and is denoted by SO(n). The set of matrices U having complex entries and satisfying U IU = I is a group called the n{dimensional unitary group and denoted by U(n). Clearly j det Uj =. The subgroup of matrices U with det u = is called the n{dimensional unitary unimodular or special unitary group and is denoted by SU(n). The set of unitary matrices satisfying S t JS = J?I where J = is called the unitary symplectic group and denoted by U Sp(n). Also I the generalized Lorentz groups, also called the orthogonal groups corresponding to indenite quadratic forms, are studied. The purpose of this paper is to give a reasonably clear introduction to the process of factorizing the members of the various groups of matrices mentioned above into products of planar ones of the same sort. An error in [Mu] is corrected, the same result being obtained in another, quite similar, way. Also the Lorentz groups and their generalizations, the orthogonal groups belonging to indenite quadratic forms are introduced and a general factorization theorem is obtained. I wish to thank professor F.E. Johnston for his unusually astute criticism while I was doing the research for this paper. 99 Mathematics Subject Classication. Primary 5A,9-4. Key words and phrases. Euler factorization, planar matrices, unirational variety. Typeset by AMS-TEX

2 A.M. DUPE. otation Matrices Lemma.. Let SO(). Then can be written in the following form Proof. Let = = cos ' sin ' where? ' < :? sin ' cos '. Now? +? = follows from the orthogonality of. Since a rotation matrix may be constructed having an arbitrary unit vector in its rst row, the point ( ) may lie anywhere on the unit circle in the {plane. But this point may be parametrically represented in the form (cos ' sin '), where? ' <. Since? +? =, we see that is either sin '. If it were + sin ', then using the orthogonality of columns one and two, we obtain sin ' cos ' + sin ' =, which forces =? cos ', contradicting the fact that det =. So =? sin ' = cos ', proving the lemma. We may apply the method in the above proof to show that if O() and det =?, then can be written as cos ' sin ' = where? ' < : sin '? cos ' Lemma.. Let SO() with = =. Then = = and =, assuming the form shown below. = cos ' sin ' 4? sin ' cos ' 5 and?? ' < : Proof. Let = 4 5 and : Since rows one and three and rows two and three are orthogonal, we obtain that = =, which forces = =, because 6=. Also we have? =, so that =. But, and hence =. Now det = det A det, where A is the matrix. Notice that A SO(), since t = we conclude the proof. A t A Theorem.. Every SO() can be written in the form. Applying lemma : to A, cos 4 ' sin ' cos sin? sin ' cos ' cos ' sin '? sin cos? sin ' cos ' 5

3 EULE FACTOIZATIONS IN CLASSICAL GOUPS these three matrices being written (' ) () (' ) and? ' ' <?= = : Proof. We may determine a number ' such that =, where cos '? sin ' 5 = 4 sin ' cos ' and the three matrices are t (' ). To see that this is possible, write 5 = cos ' + sin ' =? sin ' + cos ' : This may be interpreted geometrically as a rotation in the {plane, taking the point ( ) into the point ( ). Now unless both and are zero, we may take ' to be the angle in radians measured from the vector ( ) to the positive {axis. That is, we rotate the vector ( ) through an angle ', thus making =. If both and are zero, just choose ' =. In order to obtain the possible range on ', we observe that the only restriction on ( ) is that? +?, which follows from the fact that the last row of is a unit vector. Consequently, it may lie anywhere on the unit disc, which shows that any angle? ' < may be taken on by ( ). We next determine a number so that =, where now As before we write cos? sin 5 = sin cos? sin cos 4 cos sin : First notice that = =. Since we made, the point ( ) lies in the upper half of the {plane. Now then, and cannot both be zero, since then would have a last row of zeros, which is impossible, because = t (' ), and t are each nonsingular. Thus the range of is?= =. Having determined thus, assumes the form in the hypothesis of lemma, an application of which proves the theorem. The above factorization is called the Brauer factorization. The fact that so factorizes may be given a geometrical interpretation by choosing an orthonormal basis in I and associating each rotation matrix to a linear transformation in the usual manner. Under this interpretation, the linear transformations which are thus represented are rotations about an axis through the origin. A 9 rotation about the x{axis which takes the positive z{ axis into the positive y{axis is termed a clockwise rotation about the x{axis through an 5

4 4 A.M. DUPE angle =, similar denitions being understood for rotations about the y{ and z{axes respectively. Then the above factorization says that every rotation in three dimensions about an axis through the origin can be accomplished by performing three consecutive rotations. First a counterclockwise rotation about the x-axis through an angle ', a second clockwise rotation about the y{axis through an angle, and third, a rotation through an angle ' counterclockwise about the z{axis. There is, besides the Brauer, another useful factorization for three dimensional rotation matrices, called the Euler factorization, which we now proceed to develop. Theorem.. If SO() then it may be factored in the form cos sin 4 cos ' sin ' 5 4? sin ' cos '?sin cos 5 4 cos ' sin '? sin ' cos ' 5 as before calling the matrices (' ) () (' ), and the range of the parameters is? ' ' < and. Proof. Determine ' so that =, where t (' ) =, and cos ' sin ' =? sin ' cos ' The range of ' is seen to be? '. Next, determine so that =, where now we have t () =, and since the point ( ) lies in the right half of the {plane, the range of is. Notice that is now measured from the vector ( ) to the positive {axis. easoning as before in lemma :, we conclude that =, and assumes the form below, and this proves the theorem. = = 4 cos ' sin ' 5? sin ' cos ' Geometrically, this says that any three{dimensional rotation may be accomplished by rotating counterclockwise through an angle ' about the x{axis, clockwise through an angle about the y{axis, and nally rotating again about the x{axis, this time counterclockwise through an angle '. The two angles ' ' are called longitude angles and the latitude angle of the rotation. This economy of axes is obviously mechanically desirable. We shall next concern ourselves with four{dimensional rotations and their Brauer and Euler factorizations. As a simple extension of lemma : to matrices in SO(4), we remark without proofthat if SO(4) such that 4 = 4 = 4 =, then is actually of the form, and is a member of SO(). Theorem.. Let SO(4). Then = (' ) ( ) (' ) 4 ( ) 4 ( ) 4 (' ) :

5 EULE FACTOIZATIONS IN CLASSICAL GOUPS 5 The product of these last three matrices is shown below cos sin cos sin = cos ' sin '? sin cos? sin cos? sin ' cos ' where? ' i < and?= i =. Proof. First determine ' so that 4 = 4 4, where we have t 4(' ) = and? '. If both 4 and 4 4 are zero, then take ' =. We shall not repeat this last step in what follows and it should be understood as always taking place. Next determine so that 4 = and 4 4, where t 4( ) =?= =. Notice that 4 = 4 =, which fact shall henceforth not be mentioned if it is obvious. Lastly, choose the number which makes 4 = 4 4, where we have t 4( ) = and?= =. By the remark preceding the theorem, 4 = 4 = 4 = 4 4, and by theorem :, it readily follows that = (' ) ( ) (' ), and thus the complete Brauer factorization of is as stated in the theorem. As before, the ' i are called the longitude and the i the latitude angles. In order to obtain a geometrical interpretation for this factorization, we select an orthonormal basis fe e e e 4 g in I 4 and then observe that the matrix () xes the two dimensional space spanned by the vectors e e 4. We say then that () is a rotation about the e e 4 {plane through an angle. Notice that although a general rotation in three dimensions must x a one-dimensional subspace, a general four{dimensional rotation need not x any subspace except the zero{dimensional subspace, the origin, which is always xed by any linear transformation. For example, () 4 () represents such. The above theorem tells us that any four{dimensional rotation may be accomplished by rotating about each of the six planes determined by the six possible pairs of basis vectors. We need not use all six planes, as is shown by the next theorem. Theorem.4. Any SO(4) may be factored as with? ' i < i. = (' ) ( ) (' ) 4 ( ) ( ) (' ) Proof. Determine ' making 4 = 4, where t (' ) = and? ' <. Next determine so that 4 4 =, where we have t ( ) =. Since the point ( 4 4 ) is in the upper half of the 4 4 {plane, and because is the angle between the vector ( 4 4 ) and the positive 4 {axis, the range of is. Now choose so that 4 = 4 4, where t 4( ) =, and again. has a one in the 4 4 {place and zeros elsewhere in the last row and last column. The matrix formed from the rst three rows and columns is now a member of SO(). Using a variant of the Euler factorization, we have = (' ) ( ) (' ), which concludes the proof. In his book [Mu], Murnaghan has a mistake on pp.8 9. He proposes to reduce a four{dimensional rotation matrix to the form = D() t (' 4) t (' ) t 4 (' ) t 4 (' )

6 6 A.M. DUPE where D() is a matrix of the form ( ) 4 ( ) or of the form I a t ( ) t 4( ), where I is the matrix (?) i+j ij ij being the Kronecker. If one attempts to apply the procedure described in his book, one soon nds out that something is out of sequence. In other words, as straightfoward as the previous algorithms appear to be, it is possible to work oneself into a corner from which there is no escape. For example, the procedure in the book fails for the following rotation matrix. 6?? 7 4 5???? It is possible to show that the factorization attempted in [Mu] can actually be achieved. Theorem.5. If SO(4), then can be factored in the following form. = ( ) 4 ( ) (' ) (' ) 4 (' ) 4 (' 4 ) = D() (' ) : : : 4 (' 4 ) Proof. We shall factor t as t 4 (' 4) t 4 (' ) t (' ) t (' )D t (), which proves the theorem. First determine so that =, where ( ) = and? <. Then determine in such a way that + = 4 and 4 ( ) =. Next choose ' so that 4 = 4 (' ) =. Then we may apply a Brauer factorization to the matrix in the last three rows and columns of to obtain that = t 4(' 4 ) t 4(' ) t (' ), which proves the theorem. We now pass to the general factorization theorem for n{dimensional rotation matrices. Theorem.6. If SO(n), then there is a Brauer factorization = (' ) ( ) (' ) : : : n (' n? ) where the ' i are longitude angles,? ' i <, and the i are latitude angles?= i = Proof. We proceed by induction on the order of. Our theorem is clearly true for SO(). Let it be true for all SO(n?) and suppose that SO(n). Determine angles so that, in turn, where ' n? (n?)(n?) : : : (n?)(n?)?(n?) n = n n n = n n : : : n (n?) n? = n(n?) n t n(' n? ) = t n( (n?)(n?) ) = : : : : : : (n?) t n?n ( (n?)(n?) =(n?) ) = (n?) :

7 EULE FACTOIZATIONS IN CLASSICAL GOUPS 7 As we sucessively determine these angles, we see that the rst angle is unrestricted and in each following case, the point we must rotate always lies in the upper half plane, which restricts the angles i to the range?= i =. Notice too that once an entry is made zero, it is left zero by the planar matrices following which operate upon it. The resulting matrix (n?) has zeros in its last row except for n n(n?). Applying an obvious generalization of lemma : gives us that (n?) = where SO(n? ), and an application of the induction hypothesis proves the theorem. Theorem.7. If SO(n) then there is a generalized Euler factorization of in the form = (' ) ( ) (' ) : : : (' n? ) where the ranges on the longitude angles ' i are? ' i <, and the range for the latitude angles i j for i = : : : (n?)(n?) i not of the form (k?)(k?) for k = 4 : : : n, and j of the form (k?)(k?) for some k = 4 : : : n, is?= i = j Proof. The proof will again be by induction on the order of. There is only one possible factorization in case SO() and this clearly satises the conditions of the theorem with the only angle appearing being a longitude angle. Assume that the theorem is true if we have SO(n? ) and let SO(n). Choose angles so that, respectively, we have ' n? (n?)(n?) : : : (n?)(n?) n = n n = n : : : n (n?) n? = n(n?) where now and nally n (n?) = n n(n?) t (' n? ) = ( (n?)(n?) ) = : : : (n?) t n? ( ) = (n?)(n?)? (n?) (n?) t n ( (n?)(n?) ) = (n?) : The rst angle is unrestricted and in each sucessive determination except the last, the point to be rotated lies in the right half of the plane so the range on the rst angle ' n? is? ' n? <, and the ranges of the other angles except the last are?= i =. In determining the range of the last angle, we see that it still lies in the right half plane but it must now be rotated to the vertical axis and so the angle for the last determination is bound by (n?)(n?). n? then assumes the form of SO(n? ), which we factor according to the induction hypothesis.this completes the proof of the theorem and shows that a general n{dimensional rotation may be performed by rotating about only n? dierent (n? ){dimensional hyperspaces., where is a member

8 8 A.M. DUPE In these theorems it is clear that the order in which we have chosen the planar matrices of the factorization may be altered to obtain a class of factorizations. The reason for choosing these particular factorizations is that they are amenable to an induction proof and are in a certain way natural generalizations of the three{dimensional case. We may actually give the elements of the planar rotation matrices as algebraic functions of the elements of the original matrix. In fact, the algebraic functions involve radicals of degree at most two. We show how this is done in the three{dimensional case and omit the obvious generalization to the n{dimensional case. Theorem.8. Let SO(). Then may be factored in the form p? ( ) 6 4? p? ( ) 7 4? ? 7 where = p ( ) + ( ), unless, of course has a pair of zeros in the last row, in which case we must replace an appropriate one of the matrices by the identity matrix. Proof. If we wish to rotate the point (x y) to the point (x y ), then the angle is simply tan? (y=x) in case the point (x y ) lies on the positive x{axis and is cot? (y=x) in case it lies on the positive y{axis. But sin (tan? (y=x)) = x px + y cos (tan? (y=x)) = y px + y : 5 For the cot?, we have similar formulas. An application of these formulas to the determination of the angles in theorem : proves theorem :8. These factorizations are useful in determining the volume element of the rotation group, its homology ring and its homotopy groups. One immediate application is the calculation of the dimension as a Lie group of SO(n). In fact, we have the Theorem.9. The total number of planar matrices in an Euler or a Brauer factorization of SO(n) is n(n? )=. Proof. If we are factoring such an, we shell o n? planar matrices in order to reduce it to a form to which we can apply the induction hypothesis. Adding up all these gives us the number in the statement of the theorem. The angles in the various planar matrices into which we were factoring are often called parameters and the Euler and Brauer factorizations referred to as parametrizations.if the range of the parameter is unrestricted, the planar matrix represents what are called one parameter subgroups, which are then isomorphic to a circle group, denoted by T, and also called the {torus. If the range of the parameter is less than full, we are really dealing with a quotient of the circle group by a two element subgroup, which gives an intuitive reason for expecting that this two element group will show up in homology and homotopy. It is an important fact in algebraic geometry that the varieties of linear algebraic groups are rational, i.e., it is possible to parametrize them with parameters so that the functions

9 EULE FACTOIZATIONS IN CLASSICAL GOUPS 9 appearing in this parametrization are rational functions of the parameters. Now the circle group can be parametrized rationally,as follows: t t?! ( + t t? + ) t : We can use this to get an explicit rational parametrization of SO(n), using the results of this chapter. Because of the limited range of variability of some of the angles, we will have some of the parameters in which it is rational being correspondingly restricted. Since there is also a rational parametrization of SO(n) called the Cayley parametrization, it would be interesting to examine the connection between the two. We will not do this here, but leave this for another time.. Unitary Matrices Lemma.. Let U U(). Then U can be written e U = i ' cos e i? sin e?i e i ' sin e i cos e?i where = and? <. Proof. Let det u = e i'. Then U = e i ' U, and det U =. If U a b = then a(u )? d?b = and U = c Thus if we equate entries, we obtain U = d a b?c a a b c d, and aa + b b =, so we may let? b a jaj = cos jbj = sin, and since both sin and cos are positive, we get o =. Let be the arguments of a b respectively. Finally, we have shown that a = cos e i and b = sin e i, which proves the lemma. : Lemma.. Let U SU(). Then U can be written in the form e i cos? sin e U = i e?i sin cos e?i where =? <. Proof. Multiply to obtain U = cos e i sin e i? sin e?i cos e?i which is the most general form for a member of SU(). Lemma.. If U SU(), then U can be factored as e i cos? sin e?i U = e?i sin e i cos where o =? <. Proof. Let = + as in lemma : and multiply.

10 A.M. DUPE Lemma.4. If U SU() and U = U =, then U = U = and ju j =. Proof. Similar to lemma :. Notation. D( ) is and U ( ) is 4 ei e i e i 5 4 cos? sin e?i sin e i cos 5 : From the above, the meaning of U ( ) and U ( ) should be clear. Theorem.. If U SU(), then we may write U as shown below. U = D( )U ( )U ( )U ( ) where the ranges on the variables are? i i < i = + + (mod ) : Proof. Let U = UU( ). Then U = U cos? U sin e i. Set U =. Then U =U = tan e i, so if we let = tan? (ju j=ju j) and = arg U? arg U, where =, then we have U =. Of course, if U is zero, then just let =, and pick for any number? <. We shall assume, in what follows, that this exceptional case has been disposed of. Next we determine, in the same way, and so that U =, where U U ( ) = U.Then clearly, U obtain U U = z = U =, and we may apply lemma :4 to, where U U() and jzj =. We can write U = e i U, where U is a unimodular unitary matrix. Now factor U according to lemma : and we obtain the result stated in the theorem. The restriction + + (mod ) comes from the fact that U is supposed to be unimodular. As a result of this theorem, we see that every three=dimensional unitary matrix can be written in the form asserted in theorem with the exception of the removal of the restriction + + = (mod ). We have a corresponding factorization to the Euler, for unitary matrices. Theorem.. If U is a unitary matrix of dimension three, then it can be factored into the form below. U = D( )U ( )U ( )U ( ) and the ranges are i = and? i i <. Proof. First determine so that U =, where UU ( ) = U =. The argument from here is similar to the previous theorem.

11 EULE FACTOIZATIONS IN CLASSICAL GOUPS Theorem.. If U is in U(n) then it can be factored as U = D( : : : n )U ( )U ( )U ( ) : : : U n ( j j ) where the ranges are? i i < i =. Proof. As earlier, the proof will be by induction on the order of U. So let the theorem be true for all U U(n? ) and let U U(n). First choose numbers j j so that U n(j) j = for j = : : : n?, where we have assumed U (j) U j ( n?j n?j ) = U (j+). As these angles are chosen, it is clear that the ranges are? k < and k =. Once an entry is made zero, the matrices operating on it after its determination leave it zero. The matrix U (n?) has zeros in every entry in the rst row except the rst, which has absolute value one. It follows from the fact that U (n?) is unitary that the entries in the rst column are also zero except the rst. The matrix in the last n? rows and columns is a unitary matrix to which we may apply the induction hypothesis and factor. This concludes the proof of the theorem. There is a factorization of unitary matrices corresponding to the Euler factorization for rotation matrices. Only n? \planar" matrices of the form U ij ( ) need be used. This is shown in the theorem below. Theorem.4. If U U(n), then it can be factored as U = D( : : : n )U ( )U ( )U ( ) : : : U ( k k ) where the ranges are? i < i =. Proof. We prove the theorem by induction on the order of U. Assume the theorem true for U U(n? ) and let U U(n). Let j j j?? j? : : : be chosen which make U n = U n = : : :, where we have U (j) Uj+( n?j n?j ) = U (j+), for j = : : :. The ranges are easily seen to be? i i < i = : The remainder of the proof is similar to that of theorem : after we apply the induction hypothesis to U (n?). Further factorization of a general unitary matrix is possible if we factor each U ij according to lemma :. The order of the various matrices in the product may clearly be changed and a whole class of factorizations obtained. We merely remark that a geometrical interpretation can be given to this factorization, using complex geometry.. Symplectic Matrices If we consider the bilinear form x y? x y and all nonsingular transformsations on two variables which leave this form invariant. Then this set of transformations forms a group, called the two{dimensional symplectic group. The word symplectic means twisted together, which clearly describes the above bilinear form. Formulated matrix-theoretically, we are considering all nonsingular n{dimensional matrices S with real or complex entries such that S t JS = J, where J =?I I, and I is the n n identity matrix. If we now let

12 A.M. DUPE A S = C B D may be written, where A B C D are n n matrices, then the condition that S is symplectic S t JS = C t A? A t C D t A? B t C C t B? A t D D t B? B t D which gives the following set of equations to be satisied by A B C D. C t A? A t C = D B? B t D = D t A? B t C = I C t B? A t D =?I : The rst two equations tell us that C t A and D t B are symmetric and the last two equations are dependent, being transposes of one another. Using these equations, we prove the remarkable Theorem.. Every symplectic matrix is unimodular. Proof. Assume that S is symplectic and partitioned as above so that the above equations are satised. If A is nonsingular, then we may write I A B det S = det (A t )? = det? A t C A t A t? A B det D A t C A t : D Now we may multiply [ A B ] on the left by C t and subtract from [ A t C A t D ] without changing the value of the last determinant. det S = det? A t? A B det = det A? A det = : A t C? C t A A t D? C t B B I If A is singular then we shall construct a unimodular symplectic matrix S such that S = S S is also unimodular, which will prove the theorem.now let S I B =, where B is a symmetric matrix to be determined so that A is nonsingular, where A = S I B A B S = : S = B C D I C D I S is clearly symplectic and unimodular. A = A + B C cannot be singular for every choice of B. For, if this were so, choose for B the matrix with zeros everywhere except in one diagonal entry, say the k th. This implies that the k th row of B is a linear combination of the rows of A, which contradicts the fact that S is nonsingular, and we may apply the rst part of this proof to show thatr S= is unimodular. This completes the proof of the theorem. We now consider symplectic matrices which are at the same time unitary and call this group the unitary symplectic group and denote it by U Sp(n). The symplectic group itself will be denoted by Sp(n). The general unitary symplectic matrix S can be written as A? B S = A where A A + B B = I : B A C This may easily be seen by noticing that the inverse of S must be,and also B D D = t?b t = J? S t J, which follows from S t JS = J. Equating these two expressions?c t A t for the inverse and interchanging B andc, we have the assertion.

13 EULE FACTOIZATIONS IN CLASSICAL GOUPS Let U ij ( ) be a matrix of the form introduced in the last section. We denote by Uij ( ) S ij ( ) the symplectic matrix. We shall state and prove the two U ij ( ) main theorems of this section. As in the case of unitary matrices, there are two ways of factoring a unitary symplectic matrix into planar factors, corresponding to the Euler and Brauer factorizations. Theorem.. Every unitary symplectic matrix of order n may be factored in the form S = D ( : : : n )S ( )S ( )S ( ) : : : S n ( k k ) : : : where D is the diagonal matrix with entries e i : : : e i n e i : : : e i n, and the ranges on the variables are? i i < and i = : Proof. The proof is by induction on half the order of S. Let S be of order n and determine numbers j j j? j? : : : which make the rst row of S zero except the rst entry which is e i. It can then be easily shown that the (n+) st row and column are also zero except for a single common nonzero entry which must be e?i. Now apply the induction hypothesis to the matrix obtained by deleting from S the rst and (n + ) st row and column. This concludes the proof. Theorem.. Every s USp(n) can be written as S = D ( : : : n )S ( )S ( )S ( ) : : : S ( k k ) where the ranges are as in theorem. Proof. By induction on half the order of S, we show that we need use only a limited number of dierent planar unitary symplectic matrices. We omit the obvious required determinations. By counting parameters, we see that the n{dimensional unitary group has n parameters and the n{dimensional unitary symplectic group n(n? )=.These parameters are useful in dening an explicit invariant integral on these groups, as very clearly presented in [Ch],[We],[Po]. c d 4. Lorentz Matrices Let I nr be the n{dimensional matrix whose rst r entries are = and whose last n? r entries are?. Then consider the set of all nonsingular square matrices L r with real entries and such that L t I nr L = I nr. This set forms a group called the n{dimensional Lorentz group of type r and is denoted by L r (n).l (4) was introduced by H.A. Lorentz in connection with an electrodynamical transformation.it is clear that if L r L r (n) then L t r L r(n). We now turn our attention to L (). From now on, we denote the members of any particular Lorentz group under discussion by L rather than L r, If no confusion is possible.so a b let L L () L =. Then we may write L t a I L = b c d a? c b d a? c = ab? cd ab? cd b? d = I :

14 4 A.M. DUPE So we obtain the equations We may now assume that a? c = ab? cd = d? b = : a = cosh b = sinh c = k sinh d = k cosh : Clearly det L = k, which must be. If k =, there are two possibilities: if a = cosh, then d = + cosh, and if a =? cosh, then d =? cosh, both of which follow readily from ab = cd. If k =?, there are also two possibilities: if a = cosh then d =? cosh and if a =? cosh, thend = cosh.the four possible forms for an element L of L () are shown below: cosh sinh sinh cosh? cosh sinh sinh? cosh cosh sinh? sinh? cosh? cosh sinh? sinh cosh If we consider the matrices L () as points in a four{dimensional space, we see that L () falls into four pieces. This follows from the fact that the determinant is a continuous function and that it can never happen that cosh =? cosh, because cosh. The subgroup of L () which consists of either of the rst two forms above is called the unimodular Lorentz group and is denoted by SL () Lemma 4.. Let L L r () r =. Then if L = L = L, it follows that L = L = and L =. Proof. For L L (), we have 4 L L L L L L L? 4 L L L L L L L 5 = 4? L L L L L 5 = I : Since L is nonsingular, L 6=. So L = L =. Also,? L = and L force L =. The proof is similar for L (). Lemma 4.. If L L r () r =, then if L = L = L, then L = L = and L =. Proof. Apply lemma 4: to L t, which is in L r (), by a previous remark. Theorem 4.. Every L L () can be factored as cosh L = 4 cosh sinh 5 4 sinh cos sin 5 4? sin cos k sinh k cosh sinh cosh where? < < + and? <. The center matrix above is L ( ). Proof. Determine so that L = L, where L t () = L. Next let L L ( ) = L and observe that 5 : L = L cosh + L sinh L = L sinh + L cosh

15 It follows from tanh? (?L =L ), which gives EULE FACTOIZATIONS IN CLASSICAL GOUPS 5 L = L = that jl j > jl j. This allows us to choose = r sinh =?L = L r? L and cosh = L = L? L which gives L = and L >. Applying lemma 4:, we conclude that the matrix in the last two rows and columns of L is a Lorentz matrix of the group L (), which may take any one of the four forms listed previously. The proof is completed. Theorem 4.. Let L L (). Then L can be factored in the following form: L = cosh sinh 4 cos sin cos sin sin cos sinh cosh? sin cos where? < and? < < +. Proof. First determine so that L = L, where, as usual, L t ( ) = L. Then? <. Since, as in the previous theorem, jl j < jl j, and also since L, we may choose so that L L? () = L implies that we have L = and jl j. We now apply lemma 4: in case L is positive, and if it is negative, we see that it must be?. In either case, the matrix in the last two rows and columns of L is an orthogonal matrix. Notice that the groups L () and L () are isomorphic, since the permutation matrix which interchanges the rst and third rows will conjugate I into?i, but the group xing?i is just L (). Thus we expect that the two previous theorems apply equally to both these groups. Theorem 4.. Let L L (). Lorentz matrices. L = 5 Then it can be factored into the product of three planar 4 cosh sinh cosh k sinh k cosh 54 sinh cosh 54 sinh sinh cosh sinh cosh where? < i <. Proof. Choose so that L L = L, where LL? ( ) = L and choose so that = L, where L L? ( ) = L. We can do this because of the fact that jl j > jl j follows from L = L =, and jl j > jl j follows from L? L? L =. Apply lemma 4: to L and we see that the matrix in the rst two columns and rows of L is a member of L () and may be any of the four types described earlier. In order to obtain generalized orthogonality relations for the rows and columns of Lorentz matrices similar to those of orthogonal matrices, and in order to gain an explicit expression for the inverse of a general Lorentz matrixn notice that L? = I nr L t I nr, for L L r (n). 5

16 6 A.M. DUPE Then the relations may be immediately read o from L? L = LL? = I. Also, as a remark, the reader may convince herself that the groups L r (n) and L n?r (n) are isomorphic with an obvious change of variables. It is just as easy to see that there are exactly + [n=] non-mutually isomorphic groups L r (n) for xed n. The general Lorentz groups consist of four separated pieces, as was pointed out for the special case of L (). An excellent proof of this last fact is given in [Bo]. We shall not give a proof of this here but the following theorem should at least make this plausible. Theorem 4.4. Let L L r (n) r [n=]. Then L = L rr+ X, where the rst matrix is a planar Lorentz matrix of one of the four dierent forms listed earlier, and X is a product of L ij 's and ij 's. Proof. The matrices which we shall determine rst, hence those which will appear last in the factorization, are the following: r+r+ r+r+ : : : n?n r?r r?r? : : : : We choose the parameters in these matrices in such a way that L r+ = L r+ = : : : L (n?r?) n? = L (n?r) r = L (n?r+) r? = : : : L (n?) = : Suppose also that these numbers have been chosen so that L (n?) and L n(n?) are both positive. This allows us to x the parameter of L n so that L n(n?) = and so that L (n?) is strictly positive, and in fact, equal to one. We do not appeal to a proof by induction, as will become clear. Now one of two things is true: n = r or n = r + k k > In the event that the second case holds, we apply this process just described to the matrix in the last n? columns and rows of L (n?). If n = r, then the matrix in the last n? rows and columns is not a matrix of the sort we are considering because its r is too large. We can, though, apply a similar process to that described above to sucessively make the entries in its last column zero but for the rst and last, which may again be assumed to be positive. The parameter of the matrix L n is then determined so that the top entry in the last column is zero and so that the bottom entry is positive. We then look at the matrix formed from the rst n? columns and rows and consider its r, there being again the two possibilities above. Eventually we arrive at the situation of having a three{dimensional submatrix of a four{dimensional one before us. when it is required that the three{dimensional matrix be factored into a rotation matrix and two Lorentz matrices, it is seen that the two{dimensional Lorentz matrix left in the rst two rows and columns or last two rows and columns can be in any one of the four for Lemma 5.. Let be of the following form 4 a b x c d y x y z 5. A Special Theorem cos sin 5 and b > c: Then = 4? sin cos 5 and < :

17 EULE FACTOIZATIONS IN CLASSICAL GOUPS 7 Proof. We may write down the following set of equations: cx + dy + yz = since rows and columns are orthogonal bx + dy + yz = (b? c)x = subtracting the equations. x = because b > c. ax + cy + xz = since rows and columns are orthogonal. ax + by + xz = (b? c)y = subtracting the equations. y = because b > c. Two possibilities now arise: a b a det =? and z =?: or det c d c b d = + and z = +: We have seen that every member of O() of determinant? may be written as cos sin sin where? <? cos which immediately rules out the rst possibility above. This is because we assumed that b > c. Thus the second possibility holds and we see that sin >. An application of lemma : proves the lemma. Theorem 5.. Let SO() and 6= I. Then may be factored as = t ( ) t ( ) () ( ) ( ) where the ranges on the parameters are? <? = = : Proof. Determine so that =, where we have let = 4 cos sin cos? sin? sin cos sin cos 4 = t ( ) Multiplying this out, we obtain for 4 cos + sin? sin + cos cos + sin? cos + sin Now write?? cos sin =?? sin cos? 5 : 5 :

18 8 A.M. DUPE which shows that we may obtain as stated above. If, of course, both = and =, then let =. The range of is unrestricted, so? <. We next choose so that =, where we have let = ( ) t ( ). Again, in order to see that this is possible, we multiply and obtain for the following matrix: cos + sin 4 cos + sin? sin + cos? sin + cos 5 yielding? cos sin =?? sin cos?? : Since?, the point (?? ) lies in the upper half plane, and thus the range of is?= =. Notice that we also have =, since?? = cos? sin + cos + sin? sin + cos? sin cos + sin + cos cos + sin =? cos + sin?? sin + cos =? = Now not both = and =, since then t =, from which it would follow that ( ) = I, which would also force = I, because symmetry is preserved under conjugation by a rotation matrix. Thus we are denitely assured that >, and an application of lemma 5: proves the theorem. Theorem 5.. If SO() and = I then it can be factored as in theorem 5:. Proof. Since symmetry of a matrix is preserved under conjugation by rotation matrices and since is symmetric, we may choose ' so that =, where (') t (') =. Then we may assume that takes the form 4 a x b z 5 : x z We have xz =. If x = then = (), and if z = then = (), proving the theorem. It is, of course, possible to generalize this theorem to n dimensions, where it gives an explicit way of showing the well-known theorem of semisimple groups, that every element in such a group is conjugate to an element in a maximal torus, by exhibiting an element which performs the conjugation. The only group we have not covered in this paper is the symplectic group. This will be the subject of a forthcoming paper.

19 EULE FACTOIZATIONS IN CLASSICAL GOUPS 9 eferences [[Ch]] C. Chevalley, Theory of Lie Groups, Princeton, Princeton, 946. [[Kl]]A.Kleppner and H.V. McIntosh, The Theory of the Three{dimensional otation Group:Part I, Description and Parametrization, IAS, Baltimore, 958. [[Li]]D.E. Littlewood, The Theory of Group Characters and Matrix epresentations of Groups, Oxford, Oxford, 95. [[Mu]] F.D. Murnaghan, The Orthogonal and Symplectic Groups, Baltimore, 958. [[Mu]] F.D. Murnaghan, The Theory of Group epresentations, Dublin, 98. [[Po]] L. Pontrajagin, Topological Groups, Princeton, Princeton, 946. [[We]] H. Weyl, The Classical Groups: Their Invariants and epresentations, Princeton, Princeton, 99. Newark, N.J.