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Solutions of Qualifying Exams I, 2013 Fall

1. (Algebra)Consider the algebra M2(k) of 2 2 matrices over a field k.Recall that an idempotent in an algebra is an element esuch that e2 =e.

(a) Show that an idempotent eM2(k) different from 0 and 1 is conjugateto

e1 :=

1 00 0

by an element ofGL2(k).

(b) Find the stabilizer inGL2(k) ofe1M2(k) under the conjugation action.

(c) In casek = Fp is the prime field with p elements, compute the number ofidempotents inM2(k). (Count 0 and 1 in.)

Solution. (a) Since e= 0, 1, the image and the kernel ofe are both one-dimensional. Let v1 be a nonzero element in the image, so v1 = e(v0) forsome v0k2. Then

e(v1) =e(e(v0)) = e2(v0) =e(v0) =v1.

Pick a nonzero element v2 in the kernel of e, and we get a basis of k2 in

which e takes the form e1.

(b) For a general element

g =

a bc d

to be in the stabilizer, it must satisfy ge1 = e1g. Writing the equation infour entries out, one sees that it means b = c = 0 (and a, d arbitrary). Sothe centralizer is the subgroup of diagonal matrices.

(c) By (a) and (b), the set of rank 1 idempotents is in bijection with GL2(Fp)/T(Fp),whose cardinality is

(p2

1)(p2

p)

(p 1)(p 1) = (p + 1)p.So the total number of idempotents is equal to p2 +p + 2.

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2. (Algebraic Geometry) (a) Find an everywhere regular differential

n-form on the affine n-space An

.(b) Prove that the canonical bundle of the projective n-dimensional spacePn

isO(n 1).

Solution (Sketch). Part (a) is really a hint for Part (b). Letting x1, x2, . . . , xnbe affine (An) coordinates, put := dx1dx2 dxngiving (a) . Denotingthe corresponding homogenous Pn coordinates t0, t1, . . . , tn, with xi :=ti/t0for i = 1, 2, . . . , nextend to Pn writingdxi = dti/t0 ti/t20dt0 and wedg-ing to discover that the divisor of poles of is (n+ 1)H where H is thehyperplane at infinity (t0= 0) and then conclude (appropriately).

3. (Complex Analysis)(Bols Theorem of 1949). Let Wbe a domain inCand Wbe a relatively compact nonempty subdomain ofW. Let >0 andG be the set of all (a,b,c,d)C such that max (|a 1|, |b|, |c|, |d 1|)< .Assume that cz+d= 0 and az+bcz+d W for z W and (a,b,c,d)G. Letm2 be an integer. Prove that there exists a positive integer (dependingon m) with the property that for any holomorphic function on W suchthat

(z) =

az+b

cz+ d

(cz+ d)2m

(ad bc)mfor zWand (a,b,c,d)G, the -th derivative (z) =()(z) of(z) onW satisfies the equation

(z) =

az+ b

cz+d

(ad bc)m(cz+ d)2(m)

for zWand (a,b,c,d)G. Express in terms ofm.Hint: Use Cauchys integral formula for derivatives.

Solution. Let

Az=az+ b

cz+ d

for A G. We take a positive integer which we will determine later asa function of n. We use Cauchys integral formula for derivatives to takethe -th derivative (z) of(z). For z Wwe use U(z) to denote an open

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neighborhood of z in W and use U(z) to denote its boundary. The -th

derivative of at z Wis given by the formula

(z) = !

21

U(z)

()d

( z)+1

and

(Az) = !

21

U(Az)

()d

( Az)+1 when AzW .

It follows fromU(z)AU(Az),

U(z)AU(Az),with the change of variable A, that

U(Az)

()d

( Az)+1 =AU(Az)

(A)d(A)

(A Az)+1 .

From the following straightforward direct computation of the discrete versionof the formula for the derivative of fractional linear transformation

A Az= a+ bc+ d

az+ bcz+d

=(a+b)(cz+ d) (az+b)(c+ d)

(c+ d)(cz+ d)=

(acz+ bcz+ ad+ bd) (acz+adz+bc+ bd)(c+d)(cz+d)

=(ad bc)( z)(c+ d)(cz+ d)

we obtainAU(Az)

(A)d(A)

(A Az)+1 =U(z)

a+bc+d

adbc(c+d)2

d

(adbc)+1(z)+1(c+d)+1(cz+d)+1

=U(z)

() (adbc)m

(c+d)2m adbc(c+d)2d(adbc)+1(z)+1(c+d)+1(cz+d)+1

= (cz+ d)+1

(ad bc)mU(z)

()d

( z)+1(c+d)12m.

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The extra factor (c+ d)12m inside the integrand on the extreme right-

hand side becomes 1 and can be dropped if 12m = 0, that is, if= 2m + 1. Thus, if = 2m + 1, then

(Az) = (cz+d)+1

(ad bc)m (z).

That is,

(z) =

az+ b

cz+ d

(ad bc)m(cz+ d)2(m)

,

because = 2m + 1 implies + 1 = 2( m).

4. (Algebraic Topology)(a) Show that the Euler characteristic of anycontractible space is 1.

(b) LetB be a connected CW complex made of finitely many cells so that itsEuler characteristic is defined. LetE B be a covering map whose fibersare discrete, finite sets of cardinality N. Show the Euler characteristic ofEis N times the Euler characteristic ofB .

(c) Let G be a finite group with cardinality>2. Show that BG (the classi-fying space ofG) cannot have homology groups whose direct sum has finiterank.

Solution. (a) The homology of a point with coefficients in a fieldkisH0= k,Hi = 0 for i > 0. Hence its Euler characteristic is(1)i dim Hi = 1. All

contractible spaces are homotopy equivalent so their Euler characteristic isthat of the point.

(b) For any open cover{Ui}, we know that the chain complex of singularchains living in Ui for some i has equivalent homology to the chain complexof all chains. Taking the cover of B by trivializing neighborhoods Ui, thechain complex of chains living in Ui receives a map from chains in E livingin1(Ui). The latter is simply|G|direct sums of the former, and the chainmap between them is the add every component map. This shows the ranks

of homology ofE is Ntimes the rank of homology ofB .(c) Strictly speaking, this problem cannot be solved based on easy machinery(as far as I know). A much more reasonable problem would be: Prove BGisnot homotopy equivalent to anything made up of only finitely many cells. Idid not take off points for people not distinguishing between this condition,

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and the condition stated in the problem itself. We know BG = EG/G, but

EG is contractible. So (EG) = 1. If BG has finite homology, (BG) =1/|G|, which cannot be an integer unless|G|= 1.

5. (Differential Geometry) Let H ={(x, y) R2 : y > 0} be theupper half plane. Let g be the Riemannian metric onHgiven by

g=(dx)2 + (dy)2

y2 .

(H, g) is known as the half-plane model of the hyperbolic plane.

(a) Let () = (cos , sin ) and () = (cos + 1, sin ) for (0, ) be twopaths inH. Compute the angleA at their intersection point shown in Figure1, measured by the metric g .

Figure 1: AngleA between the two curves and in the upper half plane

H.

(b) By computing the Levi-Civita connection

xi

xj=

2k=1

kij

xk

ofg or otherwise (where (x1, x2) = (x, y)), show that the path , after arc-length reparametrization, is a geodesic with respect to the metric g .

Solution. (a) The intersection point is (1/2, 3/2): solving for() = (cos , sin ) = (cos + 1, sin ) =()

we obtain = /3,= 2/3.

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The angle A satisfies

cos A= (/3), (2/3)g||(/3)||g|| (2/3)||g

= (3/2, 1/2), (3/2, 1/2)g||(3/2, 1/2)||g||(

3/2, 1/2)||g

= 1

21y2

1y2

=12

and so A = 2/3.(b) Using the formula

ijk =1

2gil(gjl,k + gjl,j gjk,l)

one obtains

ijk =1

y (ijk,2+ kij,2 jki,2).

After arc-length reparametrization, the tangent vectors of the path are

v() = ()||()||g = ( sin2 , sin cos ).

Then

v()v() =v () +

11 12

21 22

v()

where

11= ( sin )111+ (cos )121 =cot ;12= ( sin )112+ (cos )122 = 1;21= (

sin )211+ (cos )

221 =

1;

22= ( sin )212+ (cos )222 =cot .

Thus one hasv()v() = 0.

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6. (Real Analysis)For any positive integer nletMnbe a positive number

such that the seriesn=1 Mn of positive numbers is convergent and its limitis M. Let a < b be real numbers and fn(x) be a real-valued continuousfunction on [a, b] for any positive integer n such that its derivative fn(x)exists for everya < x < b with|fn(x)| Mn fora < x < b. Assume that theseries

n=1 fn(a) of real numbers converges. Prove that

(a) the series

n=1 fn(x) converges to some real-valued function f(x) foreveryaxb,

(b) f(x) exists for every a < x < b , and

(c)

|f(x)

| M fora < x < b.

Hint for(b): For fixed x(a, b) consider the series of functionsn=1

fn(y) fn(x)y x

of the variable y and its uniform convergence.

Solution. (a) Fix x(a, b]. For q > p1, by the Mean Value Theoremapplied to the function

qn=p fn on [a, x] we can find a < p,q < xsuch that

qn=p

fn(x) q

n=p

fn(a) = (x a)q

n=p

fn(p,q) ,

which implies thatq

n=p

fn(x)

qn=p

fn(a)

+ (x a)

qn=p

fn(p,q)

qn=p

fn(a)

+ (x a)q

n=p

Mn.

Since both seriesn=1 fn(a) andn=1 Mn are convergent and thereforeCauchy, for any >0 we can find a positive integer N1 such that

qn=p

fn(a)

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for q > pN1 and we can find a positive integer N2 such that

qn=p

Mn < 2(x a)

for q > pN2. Thus for nmax(N1, N2) we haveq

n=p

fn(x)

< and the series

n=1 fn(x) is Cauchy. Hence the series

n=1 fn(x) converges

to some real-valued function f(x) for every a

x

b.

(b) Before the proof of the statement in (b), we would like to state thatthe uniform limit of continuous functions is continuous. That is, ifhn(x) is asequence of functions on a metric spaceEwhich converges to a function h(x)on Euniformly on Eand if for some x0 Eand for every n the functionhn(x) is continuous at x = x0, then h(x) is continuous at x0. This resultsfrom the so-called 3 argument as follows. Given any > 0. The uniformconvergence ofhnh onEimplies that there exists some positive integerNsuch that|hN(x) h(x)|< for allxE. SincehNis continuous atx = x0,there exists some > 0 such that|hN(x) hN(x0)| < for dE(x, x0) < (wheredE(, ) is the metric of the metric space E). Thus for dE(x, x0)< we have

|h(x) h (x0)| |h(x) hN(x)|+|hN(x) hN(x0)|+|hN(x0) h (x0)|< 3,

which implies the continuity ofh at x = x0.

We now prove the statement in (b). Takex0 (a, b). We introduce thefunctiongn,x0(x) on [a, b] which is defined by

gn,x0(x) =

fn(x)fn(x0)xx0 forx=x0

gn,x0(x0) =fn(x0) .

It follows from the continuity offn on [a, b] and the existence offn(x0) that

gn,x0 is a continuous function on [a, b].

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Whenx[a, b] with x=x0, by the Mean Value Theoremfn(x) fn(x0)

x x0 =fn(x)

for some x strictly between x0 andx and as a consequence

|gn,x0(x)|=|fn(x0)| Mn.

Whenx = x0,|gn,x0(x)|=|fn(x0)| Mn.

Thus|gn,x0(x)| Mn for x [a, b]. Fromn=1 Mn M

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Solutions of Qualifying Exams II, 2013 Fall

1. (Algebra)Find all the field automorphisms of the real numbers R.

Hint: Show that any automorphism maps a positive number to a positivenumber, and deduce from this that it is continuous.

Solution. Ift > 0, there exists an element s= 0 such that t = s2. If isany field automorphism ofR, then

(t) =(s2) = ((s))2 >0.

It follows that preserves the order on R: Ift < t, then

(t) =(t+ (t t)) =(t) +(t t)> (t).Any real number is determined by the set (Dedekinds cut) of rational

numbers that are less than , and any field automorphism fixes each rationalnumber. Therefore is the identity automorphism.

2. (Algebraic Geometry)What is the maximum number of ramificationpoints that a mapping of finite degree from one smooth projective curve overCof genus 1 to another (smooth projective curve of genus 1) can have? Givean explanation for your answer.

Solution (Sketch). By the Riemann-Hurwitz formula, if we have a mapping

fof finite degree d from one smooth projective (irreducible, say) curve ontoanother the Euler characteristic of the source curve is d times the Euler char-acteristic of the target minusa certain nonnegative number e, and moreovere is zero if and only if the mapping is unramified. Now compute: the Eulercharacterstic of our source and target curves is, by hypothesis, 0 and so thise is zero, and therefore the mapping is unramified.

3. (Complex Analysis)Let and be two complex numbers such that

Im

> 0. Let G be the closed parallelogram consisting of all z C such

that z= + for some 0, 1. Let G be the boundary ofG andLet G

0

=G G be the interior ofG. Let P1, , Pk, Q1, , Q be pointsinG0 and letm1, , mk, n1, , n be positive integers. Letfbe a functionon G such that

f(z)

j=1(z Qj)njkp=1(z Pp)mp

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is continuous and nowhere zero on G and is holomorphic on G0. Let (z)

and(z) be two polynomials onC. Assume thatf(z+ ) =e(z)

f(z) if bothzand z+ are in G. Assume also thatf(z+) = e(z)f(z) if both z andz+ are in G. Express

kp=1 mp

j=1 nj in terms of and and the

coefficients of(z) and(z).

Solution. Let A = 0, B = , C = + , and D = . Since Im

> 0,

it follows that going from A to B, to C, to D and then back to A is in thecounterclockwise direction. By the argument principle

k

p=1mp

j=1nj =

1

2

1 G

d log f

= 1

21

AB

d log f+BC

d log f+CD

d log f+DA

d log f

= 1

21

AB

d log fCD

d log f+

BC

d log fAD

d log f

=

1

21

AB

d(z) +

AD

d(z)

=

1

21 (() +(0) +() (0)) .

Thus, the answer is

kp=1

mp

j=1

nj = 1

21 (() +(0) +() (0)) .

4. (Algebraic Topology)(a) Fix a basis for H1 of the two-torus (withinteger coefficients). Show that for every element x SL(2,Z), there is anautomorphism of the two-torus such that the induced map on H1 acts byx.

Hint: SL(2,Z) also acts on the universal cover of the torus.

(b) Fix an embedding j : D2 S1 S3. Remove its interior from S3 toobtain a manifold Xwith boundary T2. Let f be an automorphism of the

two-torus and consider the glued spaceXf := (D

2 S1) fX.IfX is homotopy equivalent to D2 S1, compute the homology groups ofXf.

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Solution. (a) Given g SL(2,Z) SL(2,R) let x : R2 R2 be the in-duced action. Sinceg is in SL(2,Z) it respects the relationship of whethertwo vectors in R2 differ by integer coordinates. So the map on the torus[(x1, x2)][g(x1, x2)] is well-defined. This clearly sends a homology gener-ating pair given by the curves (x1, 0) and (0, x2) to the expected images viag.

(b) There is an ambiguity in the problem about how f glues X and D2 S1 together; so I gave full credit regardless of whether you identified thisambiguity or not. Note Xf = (D

2 S1) S1S1X. Write U=D2 S1 andV =X. The Mayer-Vietoris sequence gives

H0(U

V) H0(U)

H0(V) H0(U

V)

H1(U V) H1(U) H1(V) H1(U V)

H2(U V) H2(U) H2(V) H2(U V)

H3(U V) H3(U) H3(V) H3(U V)

but because we know the homology ofD2

S1

S1 andS1

S1, we can fill

in various groups in the long exact sequence:

Z Z Z H0(U V)

Z2 g

Z Z H1(U V)

Z 0 0 H2(U V)

0 0 0 H3(U V)

Sinceg is an isomorphism, we know H1 must inject intoZ, but the inclusionmap H0(U V)H0(U) H0(V) is an injection, so H1(U V) = 0.

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We know H0 is either equal to Zfrom the long exact sequence above, or

by observing that Xf is path-connected.Iffinduces an isomorphism, we seeH2must be zero; this was the intent of

the problem, but you can get a different answer based on how you interpretedthe gluing by f.

Finally,H3is also isomorphic to Z by the exactness of the above sequence.

5. (Differential Geometry) LetM=U(n)/O(n) forn1, whereU(n)is the group ofnn unitary matrices and O(n) is the group ofnn orthogonalmatrices. M is a real manifold called the Lagrangian Grassmannian.

(a) Compute and state the dimension ofM.

(b) Construct a Riemannian metric which is invariant under the left actionofU(n) onM.

(c) Let be the corresponding Levi-Civita connection on the tangent bundleT M, andX, Y, Zbe any U(n)-invariant vector fields onM. Using the givenidentity (which you are not required to prove)

XY =12

[X, Y],

show that the Riemannian curvature tensor R ofsatisfies the formulaR(X, Y)Z=

1

4[Z, [X, Y]].

Solution. (a)T[I]M= u(n)/o(n)=Sym2(Rn)

where Sym2(Rn) denotes the space of real n n symmetric matrices. Thus

dim M=n(n + 1)

2 .

(b) Define a metric on Sym2(Rn) by

A, B= tr(ABt) = tr(AB).g

O(n) acts on T[I]M=Sym2(Rn) byg

A= gAg1. Then

g A, g B= tr(g ABg1) =A, B.Hence this metric is invariant under the action ofO(n). By translating themetric to tangent spaces at other points by the action ofU(n), this gives awell-defined invariant metric on U(n)/O(n).

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(c)

XY =1

2[X, Y].

Then

R(X, Y)Z=XYZ YXZ [X,Y]Z=

1

4([X, [Y, Z]] [Y, [X, Z]]) 1

2[[X, Y], Z]

=1

4[Z, [X, Y]]

where the last equality follows from Jacobi identity.

6. (Real Analysis)Show that there is no function f: R R whose setof continuous points is precisely the set Q of all rational numbers.

Solution. For fixed >0 let C() be the set of points xR such that forsome >0 we have|f(x) f(x)|< for all x, x(x , x + ). ClearlyC() is open since for every x C(), we have (x , x+) C(). NowletCdenote the set of continuous points off .From the definitions, we havethat

C=n=1

C(1/n).

Now suppose that C=Q. Then

RQ=n=1

Xn,

where Xn = RC(1/n). Since C(1/n) is open, Xn is closed. Also Q iscountable, say Q={q1, q2, . . . }. Let Yn ={qn}.Then

R=

n=1

Xn

n=1

Yn

,

i.e. we have writtenR as a countable union of closed sets. Then by Bairestheorem, some Xn or Yn has nonempty interior. Clearly it cannot be one oftheYn.So there existsXncontaining an interval (a, b).But this is impossiblebecause Xn R Q and every interval contains a rational number. Thus,we obtain a contradiction, which shows that C=Q.

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Solutions of Qualifying Exams III, 2013 Fall

1. (Algebra)Consider the function fields K= C(x) and L = C(y) of onevariable, and regard L as a finite extension ofKvia the C-algebra inclusion

x(y5 1)24y5

Show that the extension L/Kis Galois and determine its Galois group.

Solution. Consider the intermediate extension K = C(y5). Then clearly[L: K] = 5 and [K: K] = 2, therefore [L: K] = 10.

Thus, to prove that L/K is Galois it is enough to find 10 field automor-phisms ofL over K. Choose a primitive 5th root of 1, say = e2i/5. ForiZ/5 and s {1}, the C-algebra automorphismi,s ofLdefined by

yiys

leavesx, hence K, fixed.

There can be many ways to determine the group, heres one.

Looking at the law of composition of these automorphisms, one sees thatthe subgroup Gal(L/K)

Z/5, (which is necessarily normal, being of index

2) is not central, for conjugation by 0,1 acts as1 on it.So the group is the dihedral group of 10 elements.

2. (Algebraic Geometry) Is every smooth projective curve of genus0 defined over the field of complex numbers isomorphic to a conic in theprojective plane? Give an explanation for your answer.

Solution (Sketch). Yes. Apply the Riemann-Roch theorem which guar-antees the existence of a nonconstant meromorphic function with a simplepole at exactly one point. Argue that this meromorphic function identifiesthe curve with P1, and using that fact, embed the curve as a conic in theplane in any convenient way, e.g., If t0, t1 are projective (P

1) coordinates,let z0 = t

20, z1 = t0t1 z2 = t

21 be the map to P

2. The conic, then, wouldbe z0z2 = z

21 . (Alternatively: one can consider the complete linear system

attached to the anticanonical divisor.)

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3. (Complex Analysis)Letf(z) =z+ez forzC and letR, >1.Prove or disprove the statement that f(z) takes the value exactly once inthe open right half-plane Hr ={zC : Re z >0}.Solution. First, let us consider the real function f(x) = x+ex. Since fis continuous, f(0) = 1 and limx f(x) =, by the intermediate valuetheorem, there exists uR such that f(u) =. Now let us show that suchu is unique. LetR > 2 and let be the closed right half disk of radius Rcentered at the origin

{z= x +iyC : x= 0, |y| R}

zC :|z|= R, 2 arg(z)

2

.

Let F(z) =

z and G(z) =

ez. Then for z

, we have|

G(z)|

=|eRez| 1 since Re z 0, while|F(z)| > 1 by construction. Hence byRouches theorem, f(z) = F(z) + G(z) has the same number of zerosinside as F(z), namely 1. Since this is true for all R large enough, weconclude that the point u is unique.

4. (Algebraic Topology)(a) Let XandY be locally contractible, con-nected spaces with fixed basepoints. Let XY be the wedge sum at thebasepoints. Show that 1(X Y) is the free product of1Xwith1Y.(b) Show that 1(X Y) is the direct product of1Xwith 1Y.(c) Note the canonical inclusion f :X

Y

X

Y. Assume thatX and

Yhave abelian fundamental groups. Show that the map f on fundamentalgroups exhibits 1(X Y) as the abelianization of1(X Y).Hint: The Hurewicz map is natural.

Solution. (a) This follows form the Van Kampen theorem: Writing X Yas the union

XYwe have that 1(X Y)=1(X) 1()1(Y) =1(X) 1Y.(b) There is the obvious continuous map

Maps(S1

, X) Maps(S1

, Y)Maps(S1

, X Y)given by sending (t X(t), t Y(t)) (t (X(t), Y(t))). This mapis a continuous so it induces a map

0(Maps(S1, X) Maps(S1, Y))0Maps(S1, X Y)

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where the lefthand side is isomorphic to 0Maps(S1, X)0Maps(S1, Y)).Further, the above map is clearly a bijection, so it induces an injection anda surjection on 0.

(c) The Hurewicz map is natural so we have a commutative diagram

1(X Y) f q

1(X Y)

H1(X Y) f H1(X Y)

where the vertical maps are abelianizations by the Hurewicz theorem. Butthe lower-right corner is equal to H

1(X)

H

1(Y) by the Kunneth theorem

(since X and Y are connected), and the bottom copy of f is the obviousisomorphism onH1. Sinceqis an abelianization by definition, but the bottomarrow and rightmost arrow are both isomorphisms, the top arrow must alsobe an abelianization.

5. (Differential Geometry)(a) Let S1 =R/Zbe a circle and considerthe connection

:= d + 1ddefined on the trivial complex line bundle over S1, whereis the standard co-ordinate on S1 =R/Z descended from R. By solving the differential equation

for flat sectionsf()

f= df+ 1fd= 0

or otherwise, show that there does not exist global flat sections with respecttoover S1.(b) Let T =V / be a torus, where is a lattice and V = R is the realvector space containing . Let Lbe the trivial complex line bundle equippedwith the standard Hermitian metric. By identifying flat U(1) connectionswithU(1) representations of the fundamental group1(T) or otherwise, show

that the space of flat unitary connections onL is the dual torus T= V/,where := Hom(,Z) is the dual lattice and V := Hom(V,R) is the dualvector space.

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Solution. (a) The differential equation

f() +1f() = 0

has a unique solutionf() =Ae

1

up to a constant AC. This is not a well-defined function over S1 becausef(0)=f(1).(b) The space of flat G-connections over Tcan be identified as

Hom(1(T), G)/AdG.

Since 1(T) = and for the abelian group G = U(1) the adjoint action istrivial, we have

Hom(1(T), G)/AdG= Hom(, U(1)) =T.

6. (Real Analysis)(Fundamental Solutions of Linear Partial DifferentialEquations with Constant Coefficients). Let be an open interval (M, M) inR with M >0. Let n be a positive integer and L=

n=0 a

d

dx be a lineardifferential operator of order n on R with constant coefficients, where thecoefficientsa0, , an1, an= 0 are complex numbers andxis the coordinateof R. Let L =n=0(1)a ddx . Prove, by using Plancherels identity,that there exists a constant c > 0 which depends only on M and an and isindependent ofa0, a1, , an1 such that for any f L2() a weak solutionu ofLu=f exists withuL2() c fL2(). Give one explicit expressionfor c as a function ofM and an.

Hint: Aweaksolutionu ofLu = fmeans that (f, )L2() = (u, L)L2() for

every infinitely differentiable function on with compact support. For thesolution of this problem you can consider as known and given the followingthree statements.

(I) If there exists a positive number c >0 such that L2()c LL2()for all infinitely differentiable complex-valued functions on withcompact support, then for any f L2() a weak solutionu ofLu = fexists withuL2()c fL2().

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(II) Let P(z) = zm +

m1k=0 bkz

k be a polynomial with leading coefficient

1. IfFis a holomorphic function on C, then

|F(0)|2 12

2=0

PeiFei2 d.(III) For an L2 function f on R which is zero outside = (M, M) its

Fourier transform

f() =

MM

f(x)e2ixdx

as a function ofR can be extended to a holomorphic function

f(+i) = MM f(x)e2ix(+i)dxon Cas a function of+ i.

Solution. This problem is to compute the constant c in Lemma 3.3 onp.225 of the book of Stein and Shakarchi on Real Analysisby going over itsarguments and keeping track of the constants involved in each step.

Introduce the polynomial

Q() =n

k=0(1)k ak(2)k

so that

(#)L () =Q()()

any C0 (R), where denotes taking the Fourier transform. Considerfirst the special case where an =

1(2i)n so that the coefficient of

n in the

polynomial Q() of degree n in is 1. Writing =+1 (with both

and real) and taking the L2 of both sides of (#) over R as functions of.Then

()

Q (+i) (+ i)2 d=

L (+i)2 d.Since from the definition of Fourier transformL (+ i) =

x=(L) (x)e2i(+i)xdx=

x=

(L) (x)e2x

e2ixdx,

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it follows that

L

(+ i) is equal to the value at of the Fourier trans-

form of the function (L) (x)e2x . Thus, by applying Plancherels identityto the function (L) (x)e2x , we get

=

L (+ i)2 d=

x=

(L) (x)e2x2 dxe4||M

|(L) (x)|2 dx,

because the support of(x) (as well as the support of (L) (x)) is in theinterval = (M, M). Thus from () it follows that

()

Q (+ i) (+ i)2

d

e4||M

|(L) (x)

|2 dx.

Setting= sin in (), we get from|| 1 that

()

Q (+ i sin ) (+ i sin )2 de4M

|(L) (x)|2 dx.

Replacingby+ cos in the integrand on the left-hand side of (), we get

()

Q (+ cos + i sin ) (+ cos + i sin )2 de4M

|(L) (x)|2 dx.

By Statement (III) given above the function (+ i) as a function of+iC is holomorphic on C. Since Q (+i) as a function of +i C is apolynomial of degree n with leading coefficient 1, it follows from Statement(II) applied to F(z) = (+ z) andP(z) =Q(+z) that ()2 1

2

2=0

Q (+ cos + i sin ) (+ cos + i sin )2 d.Integrating both sides over (, ) and using (), we get

=

()

2

=

1

2 2

=0 Q (+ cos +i sin ) (+ cos + i sin )

2

d

d

= 1

2

2=0

=

Q (+ cos + i sin ) (+ cos + i sin )2 d d 1

2

2=0

e4M

|(L) (x)|2 dx

d= e4M

|(L) (x)|2 dx.

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By applying Plancherels formula to , we conclude that

()2L2()e4M (L) (x)2L2()under the additional assumption that an =

1(2i)n

. When this additionalassumption is not satisfied, we can apply the argument for the special caseto

1

an(2i)nL

instead of toLto conclude that

()

2L2()

e4M

|an(2)n

|2

(L) (x)

2L2() ,

or ()L2()c (L) (x)L2() ,

with

c= e2M

|an| (2)n .

By Statement (I) given above, when we set

c= e2M

|an

|(2)n

,

we can conclude that for any f L2() a weak solution u ofLu= f existswithuL2()c fL2().

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