A 5,41,, ta" +WW/xy ILLurem
In W hm Low
Pf' m / We 911 ,
Wbmw aghubm ha???) [a
0.ch quadlm/ farm/2M az O A? a [u Vaduz
0% 4L0 WP $3: W4F(3lra) .9 JZOMZS) use-".4
A/bWW/d no qfnw? W
aw; 215w 35% W649. Heu WM
0?» Wm [WM a mat WW Paper (Swim-.4
41mm
Lie Groups Solutions, Problem Set # 8
Section 5.2:
2: In all cases we look at 1 X , where G and X T G. For R, |(, X )| =
|(1, 1 X )| = | 1 |X |, so the volume form is d /| |. For C , the standard volume
form is dxdy = dzdz /4i. The dz terms picks up a fac
Lie Groups Solutions, Problem Set # 7
Section 4.2:
6: (a) A linear transformation M GL(E ) sends subspaces of E to subspaces of E .
It is clear that this action is transitive on Grm (E ), making Grm (E ) a homogeneous
space for GL(E ). The only question i
Lie Groups Solutions, Problem Set # 6
The problems that say verify, namely problems 1, 2, 3, 4, 5 and 9 of section 3.3,
did not need to be written up, and are not included in these solutions.
Section 3.2:
9: This is a compact version of problem 6, but the
Lie Groups Solutions, Problem Set # 5
Section 3.1:
1: (a) Given a real vector space E , let F = E C, and let C be complex conjugation.
Conversely, given a pair (F, C ) with F of complex dimension n, we can view F as
a real vector space of real dimension 2
Lie Groups Solutions, Problem Set # 4
Section 2.5:
2: If F is g stable, then X v F for all X g, v F . Likewise, X 2 v = X (X v) F ,
and by induction X n v F . Since F is a vector space, exp(X )v =
X n v/n! F ,
so exp(g) sends F to itself. Thus the group g
Lie Groups Solutions, Problem Set # 3
Section 2.2:
1: (a) If G is the group of invertible block-triangular matrices, then g is the vector
space of all block-triangular matrices (with the sizes of the blocks xed). It is easy
to see that the exponential of
Lie Groups Solutions, Problem Set # 2
Section 1.3:
5: This problem is surprisingly subtle, and took me a while to solve. If a = exp(X )
and b = exp(Y ), showing that a1 = exp(X ) is in exp(n) is easy, but showing that
ab exp(n) is much harder.
First note
53m? [ache Mmthwum+
LMLM alngi- (13 V it A row/L Vwlor' SFW
to} 00 VKV % 2 Lu, a swisgmmnéc bfllriw 74mm
w(x,v) LCOQLX)
w Mm WP a:V>\/*
er |> CoCX)= ybwbwy
{23¢- w «Ia M1442ng (:6 [a MMWFP/LIAEW.
4 ,1; 5MO%3MMC:?C mic" hijMe/mé:
{M W W M 0.0» W 4w
Lie Groups Solutions, Problem Set # 1
Section 1.1:
4: This problem can be done either with power series or with dierential equations.
Ill do part (a) with dierential equations and part (b) with power series. For both
parts, note that Xe1 = x e1 = x e2 , 2
Lie Groups, Problem Set # 8
Due Thursday, November 8
In this problem set were going to study some homogeneous spaces.
1) Let V be an n-dimensional vector space over a eld K = R or C. The Grassmanian Grn,k is the set of k -dimensional subspaces of V . Grn,
Lie Groups Solutions, (Optional) Problem Set # 11
Section 6.3:
Problem 6.3.1: (a) Since the irreducible characters span the space of class functions, the number of such characters is the dimension of this space. But a class
function is determined by its v
Lie Groups, Problem Set # 1 Solutions
1) Show that the function f (t) = exp(at)[cos(bt)+ i sin(bt)] has f (t) = (a + bi)f (t)
and f (0) = 1. In other words, that exp(a + bi)t) = f (t).
Solution: By the product rule, f (t) = a exp(at)[cos(bt)+i sin(bt)]+ex
Lie Groups, Problem Set # 2 Solutions
This weeks problems were all from the book, namely Section 2.1, problems 5, 6
and 9, and 2.2, problems 4, 5 and 7.
Problem 2.1.5: (a) First note that ij = k = ji, so for any complex number ,
j = j and j = j . If q1 =