Quantum Computing
(Fall 2013)
Instructor: Shengyu Zhang.
Lecture 3 Hidden Subgroup Problem 1: Group Representations
Studies of Hidden Subgroup Problem for non-Abelian groups often need knowledge of group
representations, which is the subject of this lecture.
1. Linear representation of finite groups
Suppose that π is a vector space over β. The general linear group πΊπΏ(π) is the group of
invertible linear operators from π to itself. Here we only consider the case π = βπ . The
group πΊπΏ(π) can also be identified with the group of invertible matrices in βπ×π .
For a finite group πΊ, a linear representation is a homomorphism π: πΊ β πΊπΏ(π). The
dimension of π is called the degree of π, denoted by ππ . (Homomorphism: The map π is
a homomorphism if π(π₯)π(π¦) = π(π₯π¦), βπ₯, π¦ β πΊ.) Note that this implies that π(π₯ β1 ) =
π(π₯)β1 , and π(1πΊ ) = πΌ where 1πΊ is the identify element of the group πΊ.
Example 1. Trivial representation: π(π₯) = 1, βπ₯ β πΊ. Here the degree ππ = 1.
Example 2. Regular representation: π = π πππ{|π₯βͺ: π₯ β πΊ}. The left regular representation πΏ
is defined by πΏ(π₯)|π¦βͺ = |π₯π¦βͺ, and the right regular representation π
is defined by
π
(π₯)|π¦βͺ = |π¦π₯ β1 βͺ. Here the degree ππΏ = ππ
= |πΊ|.
Two representations π1 and π2 are isomorphic if there is an invertible linear operator
π: π β π s.t. π1 (π₯)π = ππ2 (π₯), βπ₯ β πΊ. Intuitively, it means that the two representations
are the same up to a change of basis.
For a representation π: πΊ β πΊπΏ(π), a subspace π of π is called πΊ-invariant if
π(π₯)π β π, βπ₯ β πΊ.
Claim. If π is πΊ-invariant, then there is another subspace πβ² s.t. π = π β πβ², and πβ²
is also πΊ-invariant.
See [Ser77] for two proofs. By this result, we know that if there is a πΊ-invariant subspace π,
then π = π β πβ² for a subspace πβ², and the representation can also be decomposed into
the direct sum of two subrepresentations, namely π = ππ β ππ β² . In matrix form, π is a
π
block diagonal matrix ( π
ππ β²
).
If π doesnβt have any nontrivial πΊ-invariant subspace, we call π irreducible. We use πΊΜ to
denote the set of irreducible representations of πΊ. We sometimes use βirrepβ as a shorthand
for βirreducible representationβ.
Fact. Every representation is isomorphic to a direct sum of irreducible representations.
An important result about irreducible representations is the following Schurβs lemma.
Theorem (Schurβs lemma). For two irreducible representations π1 and π2 with degree π1
and π2 , respectively, assume that there is a matrix π β βπ1 ×π2 s.t. π1 (π₯)π =
ππ2 (π₯), βπ₯ β πΊ. If π1 is not isomorphic to π2 , then π = 0. If π1 = π2 and π1 = π2 ,
then π = ππΌ for some real number π.
Proof. See [Ser77].
This lemma can be used to prove the following fundamental orthogonality theorem.
Theorem. For any two different irreducible representations π1 and π2 in matrix form, and
any π, π β [π1 ], πβ², πβ² β [π2 ], it holds that
β π1 (π₯ β1 )ππ π2 (π₯)π β² πβ² = 0.
π₯βπΊ
When π1 and π2 take unitary matrices, then the above equality changes to
β π1 (π₯)βππ π2 (π₯)π β² πβ² = 0.
π₯βπΊ
Proof. See [Ser77]
2. Character theory
For a representation π, its character is simply a function π: πΊ β β defined by π(π₯) =
π‘π(π(π₯)), where π‘π is the trace---sum of eigenvalues. Note that π(1) = ππ . A character is
irreducible if it is the character of an irreducible representation.
The following theorem is a simple corollary of the orthogonality of irreducible
representations.
Theorem. Different irreducible characters π and πβ² are orthogonal: βπ₯βπΊ π(π₯)β πβ²(π₯) = 0.
Two elements π₯ and π¦ are called conjugate if βπ β πΊ s.t. ππ₯πβ1 = π¦. Itβs an equivalent
relation and thus partitions πΊ into conjugacy classes. Class functions are constant for members
of the same conjugacy class. It is not hard to see that characters are class functions.
For two complex-valued functions π and π on πΊ, define two inner products
(π, π) =
1
β
π(π₯)π(π₯ β1 )
|πΊ| π₯βπΊ
1
and β©π, πβͺ = |πΊ| βπ₯βπΊ π(π₯)π(π₯)β
where the superscript * is the complex conjugate. These two inner products are the same if
one of the functions is a character, due to the following fact.
Fact. For a character π of πΊ, π(π₯ β1 ) = π(π₯)β.
β
Proof. π(π₯ β1 ) = tr(π(π₯ β1 )) = βπ ππ (π(π₯ β1 )) = βπ ππ (π(π₯)β1 ) = βπ ππ (π(π₯)) = π(π₯)β ,
where ππ (π)βs are the eigenvalues of π.
3. Regular representation and Fourier transform
Recall the left regular representation πΏ(π₯)|π¦βͺ = |π₯π¦βͺ and the right regular representation
π
(π₯)|π¦βͺ = |π¦π₯ β1 βͺ. The following equalities hold.
πΏ βΌ β¨πβπΊΜ(π β 1ππ ) and π
βΌ β¨πβπΊΜ(1ππ β πβ ).
(1)
In fact, these hold for the same isomorphism, which is the Fourier transform over πΊ.
Formally, define the Fourier transform by the following.
|π₯βͺ β¦ |π₯Μβͺ β
1
β|πΊ|
βπβπΊΜ ππ |π, π(π₯)βͺ, where |π(π₯)βͺ =
1
βππ
π
π
βπ,π=1
π(π₯)π,π |π, πβͺ.
In other words, the Fourier transform is
πΉπΊ = β|π₯Μβͺβ©π₯| .
π₯βπΊ
Note that πΉπΊ depends on the choice of basis for each irrep of dimension greater than 1.
Theorem. The Fourier transform πΉπΊ simultaneously decomposes the left and right regular
representations πΏ and π
into their irreducible components.
Proof. We verify for left regular representation, and the case for π
is similar.
The identity Eq.(1) also implies the following basic fact.
Fact. |πΊ| = βπβπΊΜ ππ2 .
4. Abelian groups
π₯π¦
For a cyclic group β€π , the irreps are ππ¦ : |π₯βͺ β¦ ππ
ππ : |π₯βͺ β¦
1
β π π₯π¦ |π¦βͺ.
βπ π¦ π
for π¦ β [π]. The Fourier transform is
For a finite Abelian group, suppose it is isomorphic to β€π1 × β¦ × β€ππ‘ , then
π₯ π¦1
the irreps are ππ¦1 β¦π¦π‘ : |π₯1 β¦ π₯π‘ βͺ β¦ ππ11
π₯ π¦π‘
β¦ πππ‘π‘
ππ¦1 β¦π¦π‘ : |π₯1 β¦ π₯π‘ βͺ β¦
1
where π₯π , π¦π β [ππ ], and the Fourier transform is
π₯ π¦1
β ππ11
βπ π¦
π₯ π¦
β¦ πππ‘π‘ π‘ |π¦1 β¦ π¦π‘ βͺ.
1 β¦π¦π‘
For Abelian groups, all irreps are one-dimensional. The converse is also true: Any non-Abelian group
has an irrep with degree strictly larger than 1.
One can also note that Abelian groups πΊ have the property that πΊ β
πΊΜ , which non-Abelian groups
do not enjoy.
Note
A good (and concise) reference for group representation is [Ser77]. [CvD10] also has an
appendix for some basic knowledge about group representation, though it takes a matrix
(instead of operator) view. For general introduction of algebra, see [Art10]. For more
extensive introduction of abstract algebra, a standard textbook is [DF03].
Reference
[Art10] Michael Artin, Algebra, 2nd edition, Pearson, 2010.
[CvD10] Andrew M. Childs and Wim van Dam, Quantum algorithms for algebraic problems, Reviews
of Modern Physics, Volume 82, JanuaryβMarch 2010.
[DF03] David S. Dummit and Richard M. Foote, Abstract Algebra, Wiley, 2003.
[Ser77] Jean-Pierre Serre, Linear representations of finite groups, Springer-Verlag, 1977.
Exercise
1. Prove that characters are class functions, namely π(π₯π¦π₯ β1 ) = π(π¦). Also prove another property:
π(π₯ β1 ) = π(π₯)β .
2. Check that the Fourier transform defined is unitary.
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