Math 678, Week 2 Exercises

Bryden Cais

September 14, 2002

Exercise 1. Show that

ζ(k) = −\frac{(2πi)^{k}B_{k}}{2k!} ,

where B_{k} is defined by

\frac{x}{e^{x} − 1} = ∑_{k=0}^{∞} B_{k} \frac{x^{k}}{k!} .

Recall that we proved the formula

f(w) = ∑_{n∈ℤ} \frac{1}{(n + w)^{k}} = ∑_{n=1}^{∞} \frac{(−2πi)^{k}n^{k−1}}{(k − 1)!} e(nw)

in class. We write this as

f(w) = \frac{1}{w^{k}} + ∑_{n=1}^{∞} \{\frac{1}{(n + w)^{k}} + \frac{1}{(−n + w)^{k}} \} = ∑_{n=1}^{∞} \frac{(−2πi)^{k}n^{k−1}}{(k − 1)!} e(nw).

Observe that equality holds wherever the two sides above converge. Moreover, for k = 1 the left hand summand is O(1/n^{2}) and therefore converges (provided we do not rearrange the sum). The right hand side obviously converges to

−2πi ∑_{n=1}^{∞} e(nw) = −2πi\frac{1}{e(−w) − 1}

= \frac{1}{w} ∑_{k=0}^{∞} B_{k} \frac{(−2πiw)^{k}}{k!} ,

while we can write the left hand side as

\frac{1}{w} − 2w ∑_{n=1}^{∞} (\frac{1}{n^{2}} \frac{1}{(1 − w^{2}/n^{2})}) = \frac{1}{w} − 2w ∑_{n=1}^{∞} (\frac{1}{n^{2}} + \frac{w^{2}}{n^{4}} + \frac{w^{4}}{n^{6}} + ⋯)

= \frac{1}{w} − 2w ∑_{k=0}^{∞} ζ(2k + 2)w^{2k},

where the change in order of summation is justified by absolute convergence. Now compare coefficients of w^{2k} to find that

ζ(2k) = −\frac{(−2πi)^{2k}B_{2k}}{2(2k)!} ,

as required.

Exercise 2. Show that σ_{k} is multiplicative; more generally, show that

σ_{k}(m)σ_{k}(n) = ∑_{d|(m,n)} σ_{k}(\frac{mn}{d^{2}}).

Recall that if φ is a multiplicative function then so is

ψ(n) := ∑_{d_{n}} φ(n).

To see this, suppose that (m, n) = 1. We then have

ψ(mn) = ∑_{d|mn} φ(d)

= ∑_{d_{1}|m,d_{2}|n} φ(d_{1}d_{2})

= ∑_{d_{1}|m} ∑_{d_{2}|n} φ(m)φ(n)

= ψ(m)ψ(n),

where we have used the fact that if (m, n) = 1 then the divisors of mn are in bijective correspondence with pairs of divisors d_{1}, d_{2} of m, n respectively. This follows from the euclidian algorithm. It is therefore immediate that

σ_{k}(n) = ∑_{d|n} d^{k}

is multiplicative, since d^{k} is. Let p be prime and suppose that a ≤ b are nonnegative integers. Then we have

σ_{k}(p^{a})σ_{k}(p^{b}) = (∑_{r=0}^{a} p^{kr}) (∑_{s=0}^{b} p^{ks})

= \frac{1 − p^{k(a+1)}}{1 − p^{k}} \frac{1 − p^{k(b+1)}}{1 − p^{k}}

= \frac{1}{1 − p^{k}} (\frac{1 − p^{k(a+1)}}{1 − p^{k}} − \frac{p^{k(b+1)} − p^{k(a+b+2)}}{1 − p^{k}} )

= \frac{1}{1 − p^{k}} (\frac{1 − p^{k(a+1)}}{1 − p^{k}} − p^{k(a+b+1)} \frac{1 − p^{−k(a+1)}}{1 − p^{−k}} )

= \frac{1}{1 − p^{k}} (∑_{j=0}^{a} p^{jk} − p^{(a+b+1)k} ∑_{j=0}^{a} p^{−jk})

= ∑_{j=0}^{a} p^{jk} \frac{1 − p^{(a+b+1−2j)k}}{1 − p^{k}}

= ∑_{d|(p^{a},p^{b})} d^{k}σ_{k} (\frac{p^{a+b}}{d^{2}} ) .

Now let

a = ∏_{i} p_{i}^{a_{i}}, and b = ∏_{i} p_{i}^{b_{i}},

where i ranges over a finite set and the a_{i}, b_{i} are nonnegative integers. Then we have

σ_{k}(a)σ_{k}(b) = ∏_{i} σ_{k}(p_{i}^{a_{i}})σ_{k}(p_{i}^{b_{i}})

since σ_{k} is multiplicative,

= ∏_{i} ∑_{d_{i}|(p_{i}^{a_{i}},p_{i}^{b_{i}})} d_{i}^{k}σ_{k} (\frac{p_{i}^{a_{i}+b_{i}}}{d_{i}^{2}} )

= ∑ ; ∏ d_{i}| ∏(p_{i}^{a_{i}},p_{i}^{b_{i}})(∏ d_{i})^{k} ∏ σ_{k} (\frac{p_{i}^{a_{i}+b_{i}}}{d_{i}^{2}} )

= ∑ ; ∏ d_{i}| ∏(p_{i}^{a_{i}},p_{i}^{b_{i}})(∏ d_{i})^{k}σ_{k} (∏ \frac{p_{i}^{a_{i}+b_{i}}}{d_{i}^{2}} ) ,

where we have used the fact that p_{i} ≠ p_{j} for i ≠ j and therefore that (p_{i}^{a_{i}+b_{i}}/d_{i}^{2}, p_{j}^{a_{j}+b_{j}}/d_{j}^{2}) = 1. Finally, observe that the divisors of (∏(p_{i}^{a_{i}}, ∏ p_{i}^{b_{i}}) are in bijective correspondence with lists d_{i} of divisors of (p_{i}^{a_{i}}, p_{i}^{b_{i}}). Therefore, we see that the above expression is equal to

∑_{d|(a,b)} d^{k}σ_{k} (\frac{ab}{d^{2}})

and we are done.

Exercise 3. Show that

n^{k} < σ_{k}(n) < ζ(k)n^{k}

and that these are the sharpest asymptotics one can hope for in the sense that

lim inf \frac{σ_{k}(n)}{n^{k}} = 1

and

lim sup \frac{σ_{k}(n)}{n^{k−1}} = ζ(k).

For any n > 1 one has

n^{k} < ∑_{d|n} d^{k} ≤ n^{k} + (\frac{n}{2})^{k} + (\frac{n}{3})^{k} + … + 1

and therefore that

1 < \frac{σ_{k}(n)}{n^{k}} ≤ 1 + \frac{1}{2^{k}} + \frac{1}{3^{k}} + … + \frac{1}{n^{k}} < ζ(k).  (1)

Observe that if p is prime then

\frac{σ_{k}(p)}{p^{k}} = 1 + \frac{1}{p^{k}} ⟶ 1

as p ⟶ ∞, and

1 + \frac{1}{2^{k}} + … + \frac{1}{m^{k}} < \frac{σ_{k}(m!)}{(m!)^{k}} < ζ(k),

and therefore that σ_{k}(m!)/(m!)^{k} ⟶ ζ(k) as m ⟶ ∞. Combining these results with (1) shows that

lim inf \frac{σ_{k}(n)}{n^{k}} = 1

and

lim sup \frac{σ_{k}(n)}{n^{k}} = ζ(k),

as desired.

Exercise 4. Show that Γ := SL_{2}(ℤ) acts discontinuously on ℍ.

Suppose that Γ does not act discontinuously on ℍ. Then there exists some z ∈ ℍ such that the orbit Γz has a limit point in ℍ. This implies that there exists a cauchy sequence {g_{i}z}_{i=1}^{∞} with g_{i} ∈ Γ and g_{i} ≠ g_{j} if i ≠ j. Write

g_{i} = (\begin{matrix}a_{i} & b_{i} \\ c_{i} & d_{i} \end{matrix})

and recall that we have

ℑ(g_{i}z) = ℑ(z)\frac{1}{|c_{i}z + d_{i}|^{2}} .

Now observe that since a_{i}, b_{i}, c_{i}, d_{i} ∈ ℤ there are only finitely many g_{i} ∈ Γ with bounded entries. Put differently, for any M > 0, there exists some i such that all the entries of g_{i} are larger than M (since there are infinitely many g_{i}). This implies that

lim_{i→∞} ℑ(g_{i}z) = 0,

and therefore that g_{i}z does not tend to a limit in ℍ, contradicting our hypothesis. Therefore, Γ acts discretely on ℍ.

Exercise 5. Let Q be a quadratic form Q(x, y) = ax^{2} + bxy + cy^{2} with a, b, c ∈ ℤ. We have an action of Γ on quadratic forms given by

gx = αx + βy, and gy = γx + δy

for

g = (\begin{matrix}α & β \\ γ & δ \end{matrix}) ∈ Γ

and we say that two forms are equivalent if they are equivalent under this action. What is the connection between this and the fundamental domain we constructed for SL_{2}(ℤ)?

Of course, there are many connections. Observe that since any SL_{2}(ℤ) transformation is invertible, equivalent quadratic forms represent the same integers. Observe that we have

g \frac{x}{y} = \frac{αx + βy}{γx + δy} = \frac{α(x/y) + β}{γ(x/y) + δ}

if y ≠ 0. Setting x/y = z we see that our quadratic form Q gives rise to a quadratic polynomial Q(z) = az^{2} + bz + c and that the action of Γ on Q is now just the usual action of Γ on ℍ ∪ ℙ^{1}(ℚ). Furthermore, the discriminant of Q, defined to be D = b^{2} − 4ac is just the discriminant of Q(z). Since SL_{2}(ℤ) is generated by z ↦ −1/z and z ↦ z + 1, we see that these transformations give us all equivalent quadratic forms.

Now suppose that D < 0. Then the roots of az^{2} + cz + d are complex, so that one root, say τ lies in ℍ. We have just seen that two quadratic forms are equivalent if and only if their respective roots (in ℍ) are Γ equivalent. In particular, two quadratic forms are equivalent if and only if the j invariants of their roots (in ℍ) are equal. Now we have a map from the set of equivalence classes S of quadratic forms to the set class group of the ring of integers of ℚ(τ) given by

ax^{2} + bxy + cx ↦ a(1, τ)

. We have just shown that this map is injective. It is also a surjective map since if (u, v) is any fractional ideal then we can assume that u/v ∈ ℍ has minimal polynomial az^{2} + bz + c. Then ax^{2} + bxy + cy^{2} has discriminant D and maps to a(1, τ). In particular, this shows that the number of equivalence classes of quadratic forms of discriminant D is just the class number of the quadratic field ℚ(τ) where τ is a root in ℍ of az^{2} + bz + c for any form Q = ax^{2} + bxy + cy^{2} in the class.