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Jeffrey Hellrung Thursday, April 20, 2006 Math 246A, Homework 02
- Let χ be the continuous homomorphism from R to T that we constructed in the lecture.
(a) Prove that lim φ→ 0
ℑ(χ(φ)) φ exists. (b) Use (a) to prove that χ is differentiable as a map R → R^2. This is a baby version of Hilbert’s fifth problem, to understand when continuity of a group homomorphism, which is all we used to introduce χ, automatically gives differentiability. If you find a slick solution to (a) and (b) using some high-powered analysis tool, try again with fewer high-powered tools. (c) The homomorphism property of χ implies χ(x/n)n^ = χ(x) and hence
lim n→∞ χ(x/n)n^ = χ(x).
Approximate χ(x/n) by using (a) and show that using this approximation we still get the same limit.
Solution
(a) Let φ 0 ∈ (0, 1 /4] ⊂ R, and set φk = φ 0 / 2 k^ for k ≥ 1. Define z 0 = χ(φ 0 ), and note that by the properties of χ, ℜ(z 0 ) ≥ 0 and ℑ(z 0 ) > 0. Set
zk+1 =
1 + zk |1 + zk| for k ≥ 0, such that then zk = χ(φk). Set
ak = ℑ(zk) φk
and note that ℑ(zk+1) =
ℑ(zk) |zk + 1|
hence ak+ ak
φk φk+
ℑ(zk+1) zk
|zk + 1|
so the ak’s are monotone increasing. Now consider the sequence
bk =
ℑ(zk) φkℜ(zk)
We note that ℜ(zk+1) =
ℜ(zk)(1 + 1/ℜ(zk)) |1 + zk|
hence bk+ bk
φk φk+
ℜ(zk)ℑ(zk+1) ℜ(zk+1)ℑ(zk)
1 + 1/ℜ(zk)
so the bk’s are monotone decreasing. Further, ak < bk, and ak/bk → 1 as k → ∞, so it follows that limk→∞ ak = limk→∞ bk exists for each choice of φ 0. A priori, the limit could depend on the choice of φ 0 , and we intend to show this not to be the case. [unfinished]
(b) We have, for φ ∈ R,
χ(φ + h) − χ(φ) h = χ(φ)
χ(h) − 1 h = χ(φ)
ℜ(χ(h)) − 1 h
ℑ(χ(h)) h
Now from part (a), lim h→ 0
ℑ(χ(h)) h exists, while
ℜ(χ(h)) − 1 h
ℜ(χ(h))^2 − 1 h(ℜ(χ(h)) + 1)
ℑ(χ(h))^2 h(ℜ(χ(h)) + 1)
ℑ(χ(h)) ℜ(χ(h)) + 1
ℑ(χ(h)) h
as h → 0, showing that χ is differentiable at φ (indeed, if we have that ℑ(χ(h))/h → 2 π as h → 0, this shows that (^) dφd χ(φ) = 2πiχ(φ)). (c)
- A skew field extension of the real numbers is a real vector space with a multiplication that satisfies associative and distributive laws (but not necessarily commutative law) and has a (unique) identity element 1 with 1 x = x1 = x for all x, and each nonzero element x has an inverse y with
xy = yx = 1.
Prove that the only finite dimensional skew field extension that is not a field is the quaternion algebra (up to isomorphism). Hint: Prove that the reals are embedded into the skew field by x 7 → x1 and commute with all elements of the skew field. Prove that any element y not identified with a real has the property that the linear span of 1 and y is a field isomorphic to C. Consider a fixed embedding of C into the skew field arising in this fashion. Prove that the skew field is a complex vector space via multiplication from the right by those elements in the skew field identified with C. (This proves, in particular, that the real dimension of the skew field is even, though we will not need this fact explicitly.) Consider left multiplication by the element which is identified with i. Consider the Jordan normal of this linear map on the above complex vector space. Discuss the possible eigenvalues and multiplicities. You can use that there are no (commutative) subfields of the skew field properly containing C. Solution
- Let F be any field, let P be the vector space of polynomials with coefficients in this field, and consider the map P 7 → P ′^ given by ∑n
k=
akzk^7 →
∑^ n
k=
kakzk−^1
(where the factor k inside the summation stands for the k-fold sum of the identity element of the field). Prove linearity (over F), the product rule, and the chain rule for this algebraic derivative. Solution Denote by d : P → P the considered map, and note that, by definition,
d
( (^) n ∑
k=
akzk
∑^ n
k=
akd
zk
(b) The claim is that every rational function can be expressed as a linear combination of a polynomial, powers of inverse linear functions, powers of linear over irreducible quadratic functions, and powers of inverse irreducible quadratic functions, where the latter 3 are of the form
1 x − a
x x^2 − 2 ℜ(r)x + |r|^2
, and
x^2 − 2 ℜ(r)x + |r|^2
respectively, a ∈ R and r ∈ C. We start by recalling that any rational function can be expressed as the ratio of two polynomials, and by polynomial division, this may be reexpressed as the sum of a polynomial and a ratio of two polynomials, the numerator having a lesser degree than the denominator. We are thus reduced to proving the claim for p/q when deg p < deg q. We do induction on deg q. The claim clearly holds for deg q = 1 (in which case p is a constant), so we assume deg q > 1. By part (a), q either has a linear or irreducible quadratic factor. Suppose q(x) has a linear factor x − a and deg p ≥ 1. Let
q(x) = (x − a)q′(x), p(x) = (x − a)p′(x) + c,
with c ∈ R. It follows that p(x) q(x)
p′(x) q′(x)
c q(x)
Since deg q′^ = deg q − 1 < deg q, we may apply the induction hypothesis to the term p′/q′, so we are reduced to the case of proving the claim for 1/q. If q has no linear factors, since deg q > 1, it must have an irreducible quadratic factor x^2 − 2 ℜ(r)x + |r|^2. Note that if deg p ≤ 1, we are reduced to proving the claim for x/q(x) (whenever q has only irreducible quadratic factors) in addition to 1/q(x), while if deg p ≥ 2, we can use polynomial division as above to again reduce to the case of x/q(x) and 1/q(x). We now consider the following 4 distinct cases:
- q has 2 distinct linear factors, x − a and x − b. From above, we may assume p(x) ≡ 1. Let
q(x) = (x − a)(x − b)q′(x).
We claim there exists A, B ∈ R such that
1 q(x)
A
(x − b)q′(x)
B
(x − a)q′(x)
indeed, multiplying out gives 1 = A(x − a) + B(x − b), hence A, B satisfies the system ( 1 1 −a −b
A
B
which has determinant a − b 6 = 0 (by the assumption that the factors were distinct). Now we may apply the induction hypothesis to each of the terms in the decomposition of 1/q.
- q has precisely 1 linear factor x − a and 1 irreducible quadratic factor x^2 − 2 ℜ(r)x + |r|^2. From above, we may assume p(x) ≡ 1. Let
q(x) = (x − a)(x^2 − 2 ℜ(r)x + |r|^2 )q′(x).
We claim there exists A, B, C ∈ R such that
1 q(x)
Ax + B (x^2 − 2 ℜ(r)x + |r|^2 )q′(x)
C
(x − a)q′(x)
indeed, multiplying out gives
1 = (Ax + B)(x − a) + C(x^2 − 2 ℜ(r) + |r|^2 ),
hence A, B, C satisfies the system
−a 1 − 2 ℜ(r) 0 −a |r|^2
A
B
C
which has determinant a^2 − 2 ℜ(r)a + |r|^2 6 = 0 (since a cannot be a root of x^2 − 2 ℜ(r)x + |r|^2 ). Now we apply the induction hypothesis to each of the terms in the decomposition of 1/q.
- q has no linear factors and 2 distinct irreducible quadratic factors, x^2 − 2 ℜ(r)x + |r|^2 and x^2 − 2 ℜ(s)x + |s|^2. From above, we may assume p(x) ≡ 1 or p(x) = x. Let
q(x) = (x^2 − 2 ℜ(r)x + |r|^2 )(x^2 − 2 ℜ(s)x + |s|^2 )q′(x).
We claim that, in either case of p above, there exists A, B, C, D ∈ R such that
p(x) q(x)
Ax + B (x^2 − 2 ℜ(s)x + |s|^2 )q′(x)
Cx + D (x^2 − 2 ℜ(r)x + |r|^2 )q′(x)
indeed, multiplying out gives
p(x) = (Ax + B)(x^2 − 2 ℜ(r)x + |r|^2 ) + (Cx + D)(x^2 − 2 ℜ(s)x + |s|^2 ),
hence A, B, C, D satisfies the system
− 2 ℜ(r) 1 − 2 ℜ(s) 1 |r|^2 − 2 ℜ(r) |s|^2 − 2 ℜ(s) 0 |r|^2 0 |s|^2
A
B
C
D
or
which has determinant (after some algebra)
D =
|r|^2 − |s|^2
ℜ(r)|s|^2 − ℜ(s)|r|^2
|r|^2 − ℜ(r)^2 − |s|^2 + ℜ(s)^2
2 |r|^2 − ℜ(r)^2 + 2|s|^2 − ℜ(s)^2
since |r|^2 > ℜ(r)^2 , |s|^2 > ℜ(s)^2 , and we can’t have both ℜ(r) = ℜ(s) and |r|^2 = |s|^2 (otherwise the quadratic factors would not be distinct). Now we apply the induction hypothesis to each of the terms in the decomposition of 1/q.
- Finally, q has only a single distinct factor (either linear or quadratic). But this is exactly the claim. Therefore, the claim is proved by induction on deg q.
(c) In light of part (b) and linearity, we can reduce the problem of showing that every rational function has an antiderivative (of the desired form) if we can show that the component functions of the expansion have antiderivatives (of the desired form). We thus compute the following antiderivatives.
Solution
(a) (b)
- This is an exercise in elementary metric spaces that will be useful later.
Let (X, d) be a metric space. A sequence of functions fn : X → C is said to converge locally uniformly to a function f : X → C if for each point x 0 ∈ X there is an open neighborhood U (x 0 ) on which fn converges uniformly to f.
(a) Prove that if fn converges locally uniformly to a function f : X → C, then fn converges uniformly to f on each compact set K ⊂ X. (b) Prove that the converse statement of (a) holds if X is an open subset of C. What can you say about general metric spaces X? (c) Prove that if fn : X → C is a sequence of continuous functions which converge locally uniformly to f , then f is continuous.
Solution
(a) We have, for each x ∈ K, an open neighborhood Ux of x on which fn → f uniformly. Since the Ux’s are an open cover of K and K is compact, there exists a finite subset {xi}ki=1 ⊂ K such that the Uxi = Ui’s cover K. Now given ǫ > 0, to each Ui there exists ni such that ‖fn − f ‖u < ǫ on Ui whenever n ≥ ni, by uniform convergence on Ui, hence ‖fn − f ‖ < ǫ on K ⊂
⋃k i=1 Ui^ whenever n ≥ max 1 ≤i≤k ni. It follows that fn → f uniformly on K. (b) Suppose first that X ⊂ C is open and that fn → f uniformly on each compact subset K ⊂ X. Now each x ∈ X is contained in some B(ǫ, x) ⊂ X, hence B(ǫ/ 2 , x) ⊂ X is a compact subset of X, hence fn → f uniformly on B(ǫ/ 2 , x) ⊂ B(ǫ/ 2 , x), an open neighborhood of x. It follows that fn → f locally uniformly. In the general case, I believe we need X to be locally compact. (c) Let x ∈ X and ǫ > 0 be given. Since fn → f locally uniformly, there exists an open neighborhood U of x such that fn → f uniformly on U. Choose N large enough such that ‖fN − f ‖u < ǫ on U. By continuity of fN , there exists a δ > 0 such that B(δ, x) ⊂ U and |fN (y) − fN (x)| < ǫ for y ∈ B(δ, x). It follows then that
|f (y) − f (x)| ≤ |f (y) − fN (y)| + |fN (y) − fN (x)| + |fN (x) − f (x)| < 3 ǫ
for y ∈ B(δ, x). Thus f is continuous at x, and since x ∈ X was arbitrary, we conclude that f is continuous on all of X.
