Download Math 319 Problem Set 3: Complex numbers and Quaternions Lie

Math 319
Problem Set 3: Complex numbers and Quaternions
Lie Groups
This problem set introduces two number systems that will play important roles
in our work this semester, the complex numbers C and the quaternions H. (For some
history, see the photocopied excerpt from Conway and Smith.) It also gives more
practice with the definition of a group.
We begin with the complex numbers, which will give us another way to think
about the groups O(2, R) and SO(2, R) from Problem Set 1.
C = {a + bi : a, b ∈ R},
where addition and multiplication are defined by i2 = −1 and the usual rules in R.
That is,
(a + bi) + (a0 + b0 i) = (a + a0 ) + (b + b0 )i
(a + bi)(a0 + b0 i) = (aa0 − bb0 ) + (ab0 + ba0 )i.
You can assume that addition and multiplication are associative and commutative
and satisfy the distributive law (straightforward to verify). Another “operation” on
C is complex conjugation, where the complex conjugate of z = a + bi is
z = a − bi.
Notice that z = z if and only if z is a real number. Also, for any z, zz = a2 + b2 is
a real number, and zz is nonzero if z is nonzero. Also notice that nonzero complex
numbers have multiplicative inverses, z −1 satisfying zz −1 = z −1 z = 1, namely
µ ¶
1
a
1
b
−1
z= 2
z = =
−
i.
z
zz
a + b2 a2 + b2
1. Verify that z = z, z1 + z2 = z 1 + z 2 , z1 z2 = z 1 z 2 , and 1/z = 1/z.
2. a. Show the complex numbers form a group under addition. You have to check
three things:
• Show that C is closed under addition — the sum of two complex numbers is
again a complex number.
• Show that there is an identity for addition in C. [Explain why 0 is the right
choice for the identity for addition, i.e. the analog of I for composition of
transformations or In for multiplication of matrices.]
• Show that every complex number has an additive inverse which is again a
complex number. [What is the inverse of a complex number for addition?]
1
b. Show the nonzero complex numbers form a group under multiplication. You have
to check three things:
• Show that C is closed under multiplication — the product of two nonzero
complex numbers is again a nonzero complex number.
• Show that there is an identity for multiplication among the nonzero complex
numbers. [Explain why 1 is the right choice for the identity for multiplication,
i.e. the analog of I for composition of transformations or In for multiplication
of matrices.]
• Show that every nonzero complex number has a multiplicative inverse which
is again a nonzero complex number. [See the introduction to this problem set.]
A standard way to draw a picture of a complex number is to represent a + bi by the
point (or, equivalently, the vector) (a, b) ∈ R2 . In this representation, the horizontal
axis is the real axis, and the vertical
is imaginary. The length of the vector corre√
sponding to z = a + bi is then zz. Notice that a complex number of “length” 1 is
thus one of the form cos(α) + i sin(α) for some real number α.
3. Recall that the Taylor series (centered at 0) for the exponential function is
ex = 1 + x +
x2
xn
+ ··· +
+ ···
2!
n!
and for the sine and cosine
sin(x) = x −
x3 x5
x2m+1
+
− · · · + (−1)m
+ ···
3!
5!
(2m + 1)!
and
x2m
x2 x4
+
− · · · + (−1)m
+ ···
2!
4!
(2m)!
a. Substitute x = iα into the series for ex and, comparing to the series for sine and
cosine, show that eiα = cos(α) + i sin(α).
b. Use (a) to show e0 = e2πi = 1. What is eπi ?
c. Use the rules for multiplying complex numbers and trig identities to verify that
eiα eiβ = eiα+iβ = ei(α+β) .
cos(x) = 1 −
4. In a natural way, each complex number eiα can be identified with a point of the
unit circle S 1 in R2 . [The unit circle is sometimes called a 1-sphere — hence the
exponent 1 in S 1 — because it is 1-dimensional, a curve.] Similarly, each point of the
unit circle can be identified with a complex number of the form eiα :
S 1 = {eiα : 0 ≤ α < 2π} = {z ∈ C : |z| = 1} = {z ∈ C : zz = 1}
a. Use the description you prefer (and previous results, as needed) to show that S 1
is a group under the usual multiplication of complex numbers. That is, show
2
• S 1 is closed under multiplication: if z1 and z2 belong to S 1 , then so does their
product z1 z2 .
• The complex number 1 is in S 1 .
• If z is in S 1 then its multiplicative inverse z −1 is also in S 1 .
5. Fix the real number α. Define a function Tα : C → C by
Tα (z) = eiα z.
a. Show that Tα preserves “lengths” of complex numbers. That is, assume w = Tα (z),
and then show ww = z z. (Such a transformation is an isometry of C.)
b. Interpreting z as a vector in R2 , show that Tα is a linear transformation. That is,
show Tα (z1 + z2 ) = Tα (z1 ) + Tα (z2 ) and Tα (rz) = rTα (z) for z1 , z2 , z ∈ C and r ∈ R.
c. What is the matrix of Tα with respect to the standard basis of R2 ?
6. Again fix the real number α. This time define a function Sα : C → C by
Sα (z) = eiα z.
a. Show that Sα preserves lengths of complex numbers.
b. Interpreting z as a vector in R2 , show that Sα is a linear transformation.
c. What is the matrix of Sα with respect to the standard basis of R2 ?
7. We’ll also have occasion to use matrices with complex entries: M(n, C) is the set
of all n by n matrices with entries in C. If the matrix A ∈ M(n, C) has entries aij ,
write A for the matrix with entries aij . (Similarly, the complex conjugate of a vector
v = (v1 , . . . , vn ) is v = (v1 , . . . , vn ).) Complex conjugation has a simple relationship
to matrix operations on A, B ∈ M(n, C). For each of the following, you may make
your argument for the case when n = 2, but to make the generalization to arbitrary
n easier later, write your two by two matrices in the form
·
¸
a11 a12
A=
a21 a22
a.
b.
c.
d.
e.
Show A + B = A + B.
Show that AB = A B.
Show AT = (A)T .
Show tr(A) = tr(A).
Show det(A) = det (A).
3
8. This is getting a bit ahead of our story, but there’s a very important group of
matrices with complex entries called the unitary group U (n, C).
T
U (n, C) = {M ∈ M(n, C) : M M = In }
Assume the results of (7) for the following.
a. Show that U (n, C) is a group under matrix multiplication.
b. Show that SU (n, C) is a subgroup of U (n, C), where
SU (n, C) = {M ∈ U (n, C) : det(M ) = 1 }.
c*. Show that if M ∈ SU (2, C), then
·
M=
a −b
b
a
¸
for complex numbers a and b satisfying aa + bb = 1.
[Hint: Set up notation
·
¸
a c
M=
,
b d
and write out the equations involving a, b, c, d that correspond to the requirements
T
M M = I2 and det(M ) = 1. Use the equations to write c and c in terms of a and b,
but consider two cases separately: Case (i) a = 0, and Case (ii) a 6= 0.]
Remark. In Part 5 we’ll motivate this definition more fully and do more with it.
Looking ahead, physicists use four groups to make predictions about the “electron
shells” and “energy levels” of a hydrogen atom: SO(3, R), SO(4, R) and SU (2, C).
(See Linearity, Symmetry, and Prediction in the Hydrogen Atom by Singer.)
9*. The quaternions H (H is for Hamilton — see Conway and Smith and also
Yaglom) are a generalization of the complex numbers:
H = {a + bi + cj + dk : a, b, c, d ∈ R},
where i2 = j 2 = k 2 = −1, ij = k = −ji, jk = i = −kj, and ki = j = −ik. Otherwise,
addition and multiplication follow the usual rules for R: for z = a + bi + cj + dk and
z 0 = a0 + b0 i + c0 j + d0 k,
z + z 0 = (a + a0 ) + (b + b0 )i + (c + c0 )j + (d + d0 )k
zz 0 = (aa0 − bb0 − cc0 − dd0 ) + (ab0 + ba0 + cd0 − dc0 )i
+(ac0 + ca0 + da0 − ad0 )j + (ad0 + da0 + bc0 − cb0 )k
Addition and multiplication are associative and satisfy the distributive law. Addition
is commutative, but multiplication is not. An analogue of complex conjugation is also
defined for H: z = a − bi − bj − bk.
4
a. Show z + z 0 = z + z 0 and z = z for all z, z 0 ∈ H.
b. Calculate zz and find a formula for the multiplicative inverse 1/z for z ∈ H, z 6= 0.
[Hint: imitate the complex case.]
c. Show H is a group under addition.
d. Show the nonzero elements of H form a group under multiplication.
e. Since multiplication in H isn’t commutative, perhaps it’s not surprising that the
relationship between multiplication and complex conjugation is different for H than
for C. Show zz 0 = z 0 z for all z, z 0 ∈ H. [Hint: you can simplify an otherwise messy
calculation by writing z = a + w for w = bi + cj + dk and observing z = a − w. Then
you just have to check that w0 w = ww0 .]
f. Show {z ∈ H : zz = 1} is a group under multiplication ( so it is a subgroup of the
multiplicative group of nonzero quaternions).
Remark: When z ∈ H is identified with the point (a, b, c, d) ∈ R4 , the group in (f)
is often called the 3-sphere S 3 in R4 , by analogy to the 1-sphere S 1 in R2 . The set
S 2 is the surface of a “regular” unit sphere in R3 ; it does not have a natural group
structure.
5