1. 18.757 Homework 1 Solutions
1. Write S n−1 for the unit sphere in Rn , and O(n) for the
group of n × n real orthogonal matrices. Write
S k (C)n = complex polynomial fns on Rn , homog of degree k.
Write
V = C(S n−1 )
for the continuous complex-valued functions on the sphere,
and Veven (respectively Vodd ) for the subspace of even (respectively odd) functions.
a) Show that restriction to the sphere defines inclusions
S 2m (Cn ) ,→ Veven ,
S 2m+1 (Cn ) ,→ Vodd ,
for all m > 0. Write W k for the image of S k (Cn ); we
discussed in class the fact that W k is an O(n)-invariant
subspace of V .
If f is homogeneous of degree k on Rn , then
f (x) = f (|x| · (x/|x|)) = |x|d f (x/|x|)
(0 6= x ∈ Rn ).
Since x/|x| ∈ S n−1 , the vanishing of the restriction of f to S n−1 implies
the vanishing of f on Rn −{0}. Obviously a polynomial vanishing on the
open set Rn −{0} must vanish, proving the injectivity of the restriction
map.
It’s also the case that a polynomial homogeneous of degree d satisfies
f (−x) = (−1)d f (x),
and so is even if d is even and odd if d is odd.
b) Show that W k ⊂ W k+2 .
The inclusion map is given on the level of polynomials by multiplication by r2 = x21 + · · · + x2n , which after restriction to the sphere is the
identity map.
c) Show that the sum over k of W k is a dense subspace of
V ; that is, that for any continuous function f on the
sphere, and any > 0, there is an m > 0 and a function
h ∈ W 2m + W 2m+1
such that |h − f | < .
This is the Stone-Weierstrass theorem: the algebra of functions S(Cm )|S n−1
has a unit, separates points (this is already done by the linear polynomials), and is closed under complex conjugation. The last requirement is necessary to get the conclusion that the algebra is uniformly
dense in the complex-valued continuous functions. To see that the last
hypothesis is necessary, consider the algebra C[z] of complex-valued
1
2
polynomials on the (x, y) plane, with z = x + iy. This algebra has a
unit and separates points on the unit circle; but the uniform closure
consists of all continuous functions whose negative Fourier coefficients
are zero. In particular, the function x = cos(θ) cannot be uniformly
approximated by polynomials in z.
d) Write
∂2
∂2
+
·
·
·
+
∂x21
∂x2n
for the Laplace operator. Obviously ∆ defines a linear
map
∆ : S k (Cn ) → S k−2 (Cn ),
(almost obviously) respecting the action of O(n). Define
∆=
H k = ker(∆|S k (Cn ) ),
the space of harmonic polynomials of degree k. Similarly, we have a linear map
r2 : S k−2 (Cn ) → S k (Cn ),
also respecting the action of O(n). Prove that
S k ' im(r2 ) ⊕ H k .
(Hint: one approach is to recall that if T : E → F is
a linear map between finite-dimensional inner product
spaces, then
F = im(T ) ⊕ ker(T ∗ ).)
To define an inner product on S(Cn ), consider the algebra homomorphism
∂
.
∂ : S(Cn ) → constant coefficient differential operators, ∂(xi ) =
∂xi
The thing to notice about this homomorphism is that
∂(r2 ) = ∆.
Define
hp, qi = [∂(q)p](0).
It’s easy to compute that the basis of monomials xα is orthogonal, with
hxα , xβ i = α!δα,β
(using standard notation for multi-indices α ∈ Nn ). Furthermore
h∆p, qi = [∂(q)∂(r2 )p](0) = [∂(r2 q)p](0) = hp, r2 qi.
This says that (r2 )∗ = ∆, and now the hint applies.
It follows from the problem that
W k ' H k ⊕ H k−2 ⊕ H k−4 · · · ,
3
an O(n)-invariant direct sum decomposition with [k/2] terms,
and that
n+k−1
n+k−3
k
dim H =
−
.
k
k−2
2. (With thanks to Inna Entova-Aizenbud.)
a) Prove (as claimed in class) that for n = 2, H k (restricted
to the unit circle) is spanned by cos(kθ) and sin(kθ).
By classical complex analysis, the real and imaginary parts of holomorphic functions are always harmonic. (One way to see this is to
write
∂2
∂2
∂
∂
∂
∂
+
=
+i
−i
;
∂x2 ∂y 2
∂x
∂y
∂x
∂y
one of these factors (I forget which) is the Cauchy-Riemann operator.)
The real and imaginary parts of z k are (in polar coordinates) rk cos(kθ)
and rk sin(kθ), which are therefore harmonic polynomials. Of course
they restrict to cos(kθ) and sin(kθ) on the unit circle.
The formula written before this problem shows that H k has dimension 2 if k > 1, and dimension 1 if k = 0; so we have found all of
Hk.
b) I said in class that H k consists of even functions if k is
even and odd functions if k is odd. But cos(kθ) is always
even, and sin(kθ) is always odd. Explain.
The point antipodal to θ on the unit circle is θ + π; so the notion of
even (respectively odd) discussed in class is f (θ + π) = f (θ) (respectively f (θ + π) = −f (θ)). It’s easy to check that these conditions are
related to k in the indicated way.
The map θ 7→ −θ corresponds to the action of complex conjugation
on R2 ' C. This is an element of the non-identity component of O(2),
and is not central in O(2). The “conjugation-even” functions on the
circle, which include all the cos(kθ), are not preserved by the subgroup
SO(2) of O(2).