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Berkeley MATH 261A - The exponential map

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4 The exponential mapIf A is a matrix, we can define exp(A) by the usual power series. We shouldcheck this converges: this follows if we define the norm of a matrix to besupx6=0(|Ax|)/|x|. Then |AB| ≤ |A||B| and |A+B| ≤ |A|+|B| so the usual esti-mates show that the exponential series of a matrix converges. The exponentialis a map from the Lie algebra Mn(R) of the Lie group GLn(R) to GLn(R). (Thesame proof shows that the exponential map converges for bounded operators ona Banach space. The exponential map also exists for unbounded self-adjointoperators on a Hilbert space, but this is harder to prove and uses the spectraltheorem.) The exponential map satisfies exp(A+B) = exp(A) exp(B) wheneverA and B commute (same proof as for reals) but this does NOT usually hold if Aand B do not commute. Another useful identity is det(exp(A)) = exp(trace(A))(conjugate A to an upper triangular matrix).To calculate the exponential of a matrix explicitly one can use the Lagrangeinterpolation formula as in the following exercises.Exercise 49 Show that if the numbers λiare n distinct numbers, and Biarenumbers, thenB1(A − λ2)(A − λ3) · · ·(λ1− λ2)(λ1− λ3) · · ·+ B2(A − λ1)(A − λ3) · · ·(λ2− λ1)(λ2− λ3) · · ·+ · · ·is a polynomial of degree less than n taking values Biat λi.Exercise 50 Show that if the matrix A has distinct eigenvalues λ1,λ2,... thenexp(A) is given byexp(λ1)(A − λ2)(A − λ3) · · ·(λ1− λ2)(λ1− λ3) · · ·+ exp(λ2)(A − λ1)(A − λ3) · · ·(λ2− λ1)(λ2− λ3) · · ·+ · · ·(In this formula exp can be replaced by any holomorphic function.)Exercise 51 Find expa bc d. The work can be reduced a little by writingthe matrix as a sum of a multiple of the identity and a matrix of trace 0.In particular for every element of the Lie algebra we get a 1-parametersubgroup exp(tA) of the Lie group. We look at some examples of 1-parametersubgroups.Example 52 If A is nilpotent, then exp(tA) is a copy of the real line, and itselements consist of unipotent matrices. In this case the exponential series isjust a polynomial, as is its inverse log(1 + x), so the exponential map is anisomorphism b etween nilpotent matrices and unipotent ones.Example 53 If the matrix A is semisimple with all eigenvalues real, then it canbe diagonalized, and the image of the exponential map is a copy of the positivereal numbers. In particular it is again injective.28Example 54 If the matrix A is0 1−1 0(semisimple with imaginary eigen-values) then the image of the exponential map is the circle group of rotations.In particular the exponential map is no longer injective.Example 55 A 1-parameter subgroup need not have closed image: consideran irrational line in the torus S1× S1, considered as (say) diagonal matrices inGL2(C).)In general a 1-parameter subgroup may combine features of all the examplesabove.Exercise 56 Show that if A is in the Lie algebra of the orthogonal group (soA + At= 0) then exp(A) is in the orthogonal group.One way to construct a Lie group from a Lie algebra is to fix a representationof the Lie algebra on a vector space V , and define the Lie group to be the groupgenerated by the elements exp(a) for a in the Lie algebra. It is useful to do thisover fields other than the real numbers; for example, we might want to do itover finite fields to construct the finite simple groups of Lie type. The problemis that the exponential series does not seem to make sense. We can get aroundthis in two steps as follows. First of all, if we work over (say) the rationalnumbers, the exponential series still makes sense on nilpotent elements of theLie algebra, as the series is then just a finite polynomial. The other problemis that the exponential series contains coefficients of 1/n!, that make no senseif n ≤ p for p the characteristic of the field. Chevalley solved this problem asfollows. The elements an/n! are elements of the universal enveloping algebraover the rationals. If we take the universal enveloping algebra over the integersand reduce it mod p we cannot then divide anby n!. However we can first dothe division by n! asnd then reduce mod p: in other words we take the subringof the universal enveoping algebra over the rationals generated by the elementsan/n! for a nilp otent, and then reduce this subring mo d p. Then this has welldefined exp onential maps for nilpotent elements of the Lie algebra.Another way to define exponentials without dividing by a prime p is to usethe Artin-Hasse exponentialexp x1+xpp+xp2p2+ · · ·!Exercise 57 Show that a formal power series f (x) = 1 + · · · with ratio-nal coefficients has coefficients with denominators prime to p if and only iff(xp)/f(x)p≡ 1 mod p. Use this to show that the Artin-Hasse power series hascoefficients with denominators prime to p.Example 58 The exponential map need not be onto, even if the Lie group isconnected. As an example, we will work out the image of the exponential mapfor the connected group SL2(R). The Lie algebra is the 2 by 2 matrices of trace0, so the eigenvalues are of the form λ, −λ for λ > 0, or iλ, −iλ for λ > 0,or 0, 0. In the first case exp(A) is diagonalizable with two positive distincteigenvalue with product 1. In the second case A is diagonalizable with two29eigenvalues of absolute value 1 and product 1. In the third case A is unipotent(both eigenvalues 1) but need not b e diagonalizable. If we check through theconjugacy classes of SL2(R) we see that we are missing the following classes:matrices with two distinct negative eigenvalues (in other words trace less than−2), and non-diagonalizable matrices with both eigenvalues −1. So the image ofthe exponential map is not even dense (or open or closed): it omits all matricesof trace less than −2.There is an alternative more abstract definition of the exponential map thatgoes roughly as follows. For any element a of the Lie algebra of a group G ,we show that there is a unique 1-parameter subgroup R 7→ G whose derivativeat the origin is a. Then exp(a) is defined to be the value of this 1-parametersubgroup at 1. This definition has the advantage that it works for all Lie groups,and in particular shows that the expnential map does not depend on a choiceof representation of the Lie group as a matrix group. The disadvantage is thatone has to prove existence and uniqueness of 1-parameter subgroups, which areessentially geodesics for a s uitbale connection on G.Theorem 59


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