
<p>Math 651 <br>Groups <br>Homework 1 - Algebras and <br>Due 2/22/2013 </p><p>1) Consider the Lie Group SU(2), the group of 2 × 2 complex matrices A with A<sup style="top: -0.6172em;">T </sup>A = I and det(A) = 1. The underlying set is </p><p></p><ul style="display: flex;"><li style="flex:1">ꢀ ꢁ </li><li style="flex:1">ꢂ</li><li style="flex:1">ꢄ</li></ul><p>ꢃ</p><p>ꢃꢃ</p><p>z −w </p><p></p><ul style="display: flex;"><li style="flex:1">2</li><li style="flex:1">2</li></ul><p></p><p>|z| + |w| = 1 </p><p>(1) (2) </p><p></p><ul style="display: flex;"><li style="flex:1">w</li><li style="flex:1">z</li></ul><p></p><p>3</p><p>with the standard S topology. The usual basis for su(2) is </p><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>0 i </p><p>i 0 </p><p>0 −1 </p><p>i</p><p>0</p><p></p><ul style="display: flex;"><li style="flex:1">X = </li><li style="flex:1">Y = </li><li style="flex:1">Z = </li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">1</li><li style="flex:1">0</li><li style="flex:1">0 −i </li></ul><p>(which are each i times the Pauli matrices: X = iσ<sub style="top: 0.1494em;">x</sub>, etc.). a) Show this is the algebra of purely imaginary quaternions, under the commutator bracket. </p><p>b) Extend X to a left-invariant field and Y to a right-invariant field, and show by computation that the Lie bracket between them is zero. </p><p>c) Extending X, Y , Z to global left-invariant vector fields, give SU(2) the metric g(X, X) = g(Y, Y ) = g(Z, Z) = 1 and all other inner products zero. Show this is a bi-invariant metric. </p><p>d) Pick ꢀ > 0 and set g(X, X) = ꢀ<sup style="top: -0.3615em;">2</sup>, leaving g(Y, Y ) = g(Z, Z) = 1. <br>Show this is left-invariant but not bi-invariant. </p><p>√</p><p>2) The realification of an n×n complex matrix A+ −1B is its assignment it to the 2n × 2n matrix </p><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>A −B </p><p>(3) </p><p></p><ul style="display: flex;"><li style="flex:1">B</li><li style="flex:1">A</li></ul><p></p><p>Any n × n quaternionic matrix can be written A + Bk where A and B are complex matrices. Its complexification is the 2n × 2n complex matrix </p><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>A −B </p><p>(4) </p><p></p><ul style="display: flex;"><li style="flex:1">B</li><li style="flex:1">A</li></ul><p></p><p>a) Show that the realification of complex matrices and complexification of quaternionic matrices are algebra homomorphisms. In the </p><p>1</p><p>quaternionic case, show that quaternionic conjugation corresponds to complex conjugation and the transposition of the off-diagonal blocks (but not the transpositions of the blocks themselves), and that quaternionic transposition corresponds to transposition of the blocks themselves (so that quaternionic conjugate-transpose is the same as complex conjugate-transpose of the complexified matrix). b) The matrix groups Sp(n) are the n × n quaternion-valued matrices <br>Q that satisfy Q<sup style="top: -0.6172em;">T </sup>Q = I. Show that Sp(1) ≈ SU(2). c) We defined the the algebras sp(2n, R) and sp(2n, C) to be the 2 × 2 real (resp. complex) matrices X that satisfy </p><p>X<sup style="top: -0.4114em;">T </sup>J + JX = 0 </p><p>(5) where </p><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>0</p><p>I</p><ul style="display: flex;"><li style="flex:1">J = </li><li style="flex:1">.</li></ul><p></p><p>(6) </p><p>−I 0 </p><p>The Lie algebra sp(n) of the group Sp(n) defined above is necessarily a real Lie algebra. Via the complexification process, show that sp(n)⊗C ≈ sp(2n, C) (the real algebras sp(2n, R) and sp(n) are real forms of the complex algebra sp(2n, C)). </p><p>3) Let g be a finite-dimensional Lie algebra over C with radical r, and assume r is abelian. In this problem we will use the vanishing of H<sup style="top: -0.3616em;">2 </sup>for semisimple algebras to prove that g = h⊕r. As a side note, this is the essential step in Levi’s theorem (that g = h n r whether or not r is abelian). </p><p>a) Let g = g/r and show that r is a g-module. In the future, denote the canonical projection g → g simply with a bar: x → x¯ ∈ g. </p><p>b) Let σ : g → g be any vector-space splitting. This means that when x ∈ g we have σ(x¯) = x¯. Show that </p><p>g(x, y) , σ([x, y]) − [σ(x), σ(y)] </p><p>(7) </p><p>V</p><p>is an element of <sup style="top: -0.4843em;">2 </sup>g<sup style="top: -0.3653em;">∗</sup>⊗g, which can actually be considered an element </p><p>V</p><p>of <sup style="top: -0.4842em;">2 </sup>g<sup style="top: -0.3652em;">∗ </sup>⊗ r. c) Show that the map g is closed, so g represents a class in H<sup style="top: -0.3615em;">2</sup>(g, r). d) By the vanishing of H<sup style="top: -0.3615em;">2</sup>(g, r), we know g is exact. Show that this results in a map η : g → η so that g(x, y) = η([x, y]), and that σ −η : g → g is a Lie algebra monomorphism, so provides the desired Lie algebra compliment to r. </p><p>2</p><p>4) In class we stated that H<sup style="top: -0.3616em;">2</sup>(g, M) could be identified with the isomorphism classes of abelien extensions of g by M. We proved half of this assertion: that given a representative of a class in H<sup style="top: -0.3615em;">2</sup>, we could construct such a Lie algebra extension, and if a different representative in the same class were chosen, the two extensions, though not the same, were isomorphic. Now assume M is a g-module and </p><p></p><ul style="display: flex;"><li style="flex:1">i</li><li style="flex:1">π</li></ul><p></p><p>(8) <br>0</p><p>M</p><p></p><ul style="display: flex;"><li style="flex:1">h</li><li style="flex:1">g</li></ul><p></p><p>0is an exact sequence of Lie algebras and so that the bracket in h is given by </p><p></p><ul style="display: flex;"><li style="flex:1">[x, i(m)] = i (π(x).m) </li><li style="flex:1">(9) </li></ul><p>for m ∈ M, and that [i(m), i(n)] = 0. Prove there is class in H<sup style="top: -0.3615em;">2</sup>(g, M) that provides this extension. </p><p>5) Let G be a connected Lie algebra, meaning a manifold with a given differentiable structure, along with a group structure. Prove that G has a canonical analytic structure. You may proceed as you wish, but here is one practicable approach: Let Ω ⊂ g be a connected domain containing 0 which is small enough that exp : Ω → U is a diffeomorphism, and let {p<sub style="top: 0.1494em;">i</sub>}<sub style="top: 0.1494em;">i∈Λ </sub>(where Λ ⊆ Z) be a set of points in G. Let U<sub style="top: 0.1494em;">i </sub>= L<sub style="top: 0.1494em;">p </sub>U, and let </p><p>i</p><p>π<sub style="top: 0.1494em;">i </sub>: U<sub style="top: 0.1494em;">i </sub>→ Ω, given by </p><p>π<sub style="top: 0.1494em;">i </sub>, exp<sup style="top: -0.4113em;">−1 </sup>◦ L<sub style="top: 0.2758em;">p </sub></p><p>(10) </p><p>−1 </p><p>i</p><p>be the charts. Then show the following: a) The p<sub style="top: 0.1495em;">i </sub>can be chosen so that {Ω<sub style="top: 0.1495em;">i</sub>}<sub style="top: 0.1495em;">i∈Λ </sub>is a locally finite covering of G b) Set U<sub style="top: 0.1494em;">ij </sub>= U<sub style="top: 0.1494em;">i </sub>∩ U<sub style="top: 0.1494em;">j </sub>whenever i = j and the intersection is non-empty. <br>Then both π<sub style="top: 0.1494em;">i</sub>, π<sub style="top: 0.1494em;">j </sub>: U<sub style="top: 0.1494em;">ij </sub>→ Ω. Using the Cambpell-Baker-Hausdorff formula, show that the transition maps π<sub style="top: 0.1494em;">i </sub>◦ π<sub style="top: 0.2552em;">j</sub><sup style="top: -0.4124em;">−1 </sup>are analytic. </p><p>c) Indicate that the Lie algebra structure (itself determined by the differentiable structure of G) uniquely determines the analytic structure. </p><p>3</p>
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