# MATRIX LIE GROUPS AND LIE GROUPS

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1 MATRIX LIE GROUPS AND LIE GROUPS Steven Sy December 7, 2005 I MATRIX LIE GROUPS Definition: A matrix Lie group is a closed subgroup of Thus if is any sequence of matrices in, and for some, then either or is not invertible Example of a Group that is Not a Matrix Lie Group Let where Then there exists! such that "# \$, but "# is invertible Thus is not a matrix Lie group Examples of Matrix Lie Groups % & '(, etc Another example is the Heisenberg group ) described below: Let + *-/ 03 6, / Note: If 9 :, then 9 : - If 9 / 03 / 2 4, then 9; - / < 2 < 03 / <2 4 Then = is a subgroup of > 5 and if 9! with 9 9, then 9, so = is a matrix Lie group, called the Heisenberg group ) 1

2 Facts about the Heisenberg group ) 1 ) 2 Let + *-/ , / *- 3 6, Then is the Lie algebra of ) - 3 Let / / : Then 9 : is a basis for so # 9 / 1 3 / 1 4 Thus ) 5-1 / : 9 and 9 : II LIE GROUPS Definition: is a : < Lie group if = where 1 is a : -manifold 2 = is a group 3 = is a smooth function with respect to the smooth product structure on and the smooth structure on (determined by ) 4 Letting be defined by ;, is a smooth function (with respect to ) 2

3 The Lie Group Let 5 5 ' 5 and ' / Let be the subspace topology on ' induced from Let be the standard product topology on induced from, the standard metric topology, and Note: is second countable and Hausdorff Let & 1 " & and < " " Let 1 " and < " " Let & be defined by unique 1 " such that Let & be defined by unique < " " such that Let & & Then ' is a : -manifold Via the standard product construction, with product atlas, we have that is a : -manifold Note: If, then in each coordinate chart, for an appropriate choice of Define: = locally (ie in each coordinate chart) as = This is well-defined on (independent of choice of coordinate chart) Facts about = 1 = is associative 2 11/ is the identity element 3 ; < < ; Thus = is a group 3

4 Now = and are smooth in each coordinate chart (by inspection), so are smooth Hence = is a Lie group III EVERY MATRIX LIE GROUP IS A LIE GROUP Every matrix Lie group is a smooth embedded submanifold of and hence a Lie group Idea of Proof: For each point in, take small enough neighborhood on which is defined to map neighborhood to Lie algebra, which is a vector subspace of 5 To do this formally, we need some facts Definition: for Proposition 1: If with, then is defined and Definition: [Exercise: See Theorem 27 in Lie Groups, Lie Algebras, and Representations, by Brian Hall] Let 1 Then let & and & Note: By Proposition 1, is open in Proposition 2: Suppose! 1 such that for all, we have is a matrix Lie group with Lie algebra Then there exists [Exercise: See Theorem 227 in Lie Groups, Lie Algebras, and Representations, by Brian Hall] Proposition 3: Every matrix Lie group is a smooth embedded submanifold of and hence a Lie group Proof: Let 1 Let be the subspace topology on 4

5 Let : Let 1 Then # : & Since is open in and multiplication by is open in Thus is an open neighborhood of Note: is a homeomorphism onto, ;, and by Proposition 2, ; ; Then define by ; Then, by Proposition 2, is a well-defined homeomorphism Now!, and is a vector space, so is a vector subspace of Let be a basis for Extend to a basis for Let 5 be defined by Then is a linear isomorphism Furthermore, 5 1 ; is a linear isomorphism Let Then 5 1 ;! 5, so is a smooth embedded submanifold of Then 5 " is a chart for 5

6 Let 9 5 " 9 Then =, where = is standard matrix multiplication, is a Lie group IV NOT EVERY LIE GROUP IS A MATRIX LIE GROUP In fact, we will show even more Namely, not every Lie group is algebraically isomorphic to a matrix Lie group! Nilpotent Matrix Lemma Definition: A matrix 5 is called nilpotent if there exists such that 1 Lemma: If 5 is a nonzero nilpotent matrix, then for all nonzero real numbers, # Proof: Let 1 be a nilpotent matrix, and suppose, by way of contradiction, there exists 5 such that 1 and # Since is nilpotent, there exists such that 1 Let 5 Then # # Let 2 be the th entry of === === Then # 2 === / < / ; ; ; ; Hence there exists polynomials such that 6 /

7 Now let Then #, so # Then, so # Let < # Now, if by way of contradiction, is nonconstant, then is nonconstant and has roots, for all Thus has infinitely many roots, violating the Fundamental -th root Theorem of Algebra Thus is constant, ie there exists 2 5 such that 2 Thus, by (1), for all 5, 2 Letting : 2, we have : Then 1, so 1 Since this holds for all 5, it holds for 1, so 1 Thus # 1, so 1, a contradiction Relate the Heisenberg matrix Lie group ) to the Lie group Define ) by - / 03 / Then is a surjective homomorphism Then + *-/ 1, / " Let By the / st Isomorphism Theorem, 7

8 Definition: is said to be a Lie group homomorphism from Lie group = to Lie group ) = if the following two conditions are met: 1 is smooth 2 = = is a homomorphism Then we write ) Definition: complex representation of (resp) is a Lie group homomorphism Let be a Lie group Let be any (unrelated) Lie algebra Then a finite-dimensional (resp Lie algebra homomorphism " ) If (resp " ) is injective, then we say that (resp ") is faithful Theorem: Let be any finite dimensional representation of ) If!, then )! Proof: Let be any finite dimensional representation of ) Let 5 be defined by Then is a finite dimensional representation of, and Since is Lie algebra homomorphism, : 9 and 9 : 9 1 Let 9 Let be the eigenvalues for and the associated linear operator Let < # 1 for some (generalized eigenspace) Let for some / Then < # < # = 1 1, so Hence is invariant under 8

9 Let Note: < # is nilpotent Now let be an eigenvalue of 9 Since and : commute with 9, they also leave invariant Now 9 : 1, since the trace of a commutator is zero However, 9 # # Thus 1 Now 1, since is an eigenvalue, so 1 Hence, for all, < # Thus is nilpotent for each Fact: === Thus 9 is nilpotent Now! (hypothesis), so for all # Hence, # for all / If, by way of contradiction, 9 1, then 9 is a nonzero nilpotent matrix, so by the Nilpotent Matrix Lemma, # for all 5 with 1, which contradicts / Thus 9 1 Now let ) Then there exists 5 such that Then # 9

10 Thus, so )! Proposition: The Lie group has no faithful finite dimensional representations Proof Suppose is a finite dimensional representation of Then let ) Then is a finite dimensional representation of ) Let Then # Then # #, so Thus! Then, by the above Theorem, )! Since ) is nontrivial, is nontrivial, so is not injective Thus has no faithful finite dimensional representation Now we show that is not isomorphic to any matrix Lie group: Assume, by way of contradiction, that there exists an isomorphism for some matrix Lie group Then is an injective Lie group homomorphism Hence is an injective Lie group homomorphism, so is a faithful finite dimensional representation, which is a contradiction to the above proposition Thus is not isomorphic to any matrix Lie group Note: Since and since ) is a matrix Lie group, we see that matrix Lie groups are not preserved by taking quotients 10

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