有限群表示理論

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在數學裡，表示理論是以線性變換的群來分析一般抽象群的一種技術。相關的介紹請見群表示，此條目則討論含有有限個元素的群的表示理論。

[编辑] 基本定義

此條目中的所有線性變換都是有限維的，且除了有另外提起外，都假定為複數。G的表示是一個群同構 ρ:G → GL(n,C)，由 G 至一般線性群 GL(n,C) 的映射。因此，要選定一個表示，則只要將群內的每個元素配定一個方陣，其中方陣的相乘和群元素間的運算會是一樣的。

若矩陣是實數的，則稱 ρ 是 G 的一個實表示。換句話說， ρ(G) ⊂ GL(n,R)。

[编辑] 另一種公式化

表示 ρ: G → GL(n,C) 定義了 G 在向量空間 Cⁿ 上的群作用，而且此一作用也可以完全決定 ρ 。因此，要選定一個表示，選定在表示的向量空間上的作用即已足夠。

換言之，群 G 在複向量空間 V 上的作用可以推導出群代數 C[G] 在向量空間 V 上的左作用，反之亦然。因此，表示會等價於左 C[G]-模。

群代數 C[G]是一個在複數上，以 G 作用的 |G| 維代數。（參見彼德-外爾定理中緊緻群的例子。）而實際上， C[G]是 G×G 的一個表示。更具體地來說，若 g₁ 跟 g₂ 是 G 的元素，且 h 是 C[G] 中相對應至 G 的 h 的一個元素，則

(g₁,g₂)[h]=g₁ h g₂^-1。

C[G] 也可以以三種方式來做為 G 的表示：

共軛： g[h] = g h g^-1
左作用： g[h] = g h （正則表示）
右作用： g[h] = h g^-1 （同上）

這些都可以在 G×G 作用中被「找到」。

[编辑] 例子

對許多的群而言，用矩陣來表示完全是一件很自然的事情。例如，一個二面體群 D₄ －正方形的對稱，即可以兩個鏡射矩陣的表示來產生：

$m = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$

$n = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

這裡， m 是由 (x,y) 映射至 (− x,y) 的鏡射，而 n 則是由 (x,y) 映射至 (y,x) 的鏡射。這些矩陣的相乘一共可以產生構成此群的八個矩陣。如上所述，可以以矩陣來表示，或者也可以以在二維向量空間 (x,y) 上的作用來表示。

此一表示是「真實的」－亦即，在矩陣和群的元素之間是一對一對應的，因為不存在在群作用下不變的 (x,y) 的子空間。

[编辑] 表示間的態射

Given two representations ρ: G → GL(n,C) and τ: G → GL(m,C) a morphism between ρ and τ is a linear map T : Cⁿ → C^m so that for all g in G we have the following commuting relation: T ° ρ(g) = τ(g) ° T.

According to Schur's lemma, a non-zero morphism between two irreducible complex representations is invertible, and moreover, is given in matrix form as a scalar multiple of the identity matrix.

This result holds as the complex numbers are algebraically closed. For a counterexample over the real numbers, consider the two dimensional irreducible real representation of the cyclic group C₄ = 〈x〉 given by:

$\rho : x \mapsto \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.$

Then the matrix $\left [ \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right ]$ defines an automorphism of ρ, which is clearly not a scalar multiple of the identity matrix.

[编辑] 子表示和不可約表示

As noted earlier, a representation ρ defines an action on a vector space Cⁿ. It may turn out that Cⁿ has an invariant subspace V ⊂ Cⁿ. The action of G is given by complex matrices and in this in turn defines a new representation σ : G → GL(V). We call σ a subrepresentation of ρ. A representation without subrepresentations is called irreducible.

[编辑] 由舊表示建構新表示

There are number of ways to combine representations to obtain new representations. Each of these methods involves the application of a construction from linear algebra to representation theory.

Given two representations ρ₁, ρ₂ we may construct their direct sum ρ₁ ⊕ ρ₂ by (ρ₁ ⊕ ρ₂) (g)(v,w) = (ρ₁(g)v, ρ₁(g)w).

The tensor representation of ρ₁, ρ₂ is defined by (ρ₁ ⊗ ρ₂) (v ⊗ w) = ρ₁(v) ⊗ ρ₂(w).

Let ρ : G → GL(n,C) be a representation. Then ρ induces a representation ρ^* on the dual of the vector space Hom(Cⁿ,C). Let f : Cⁿ → C be a linear functional. The representation ρ^* is then defined by the rule ρ^* (g) (f) = f(ρ(g)^-1). The representation ρ^* is called either the dual representation or the contragredient representation of ρ.

Further more, if a representation ρ has a subrepresentation σ then the quotient of the representing vector spaces for ρ and σ has a well defined action of G on it. We call the resulting representation the quotient representation of ρ by σ.

[编辑] 應用舒爾引理

Lemma. If f: A⊗B → C is a morphism of representations, then the corresponding linear transformation obtained by dualizing B is: f′: A → C⊗B^* is also a morphism of representations. Similarly, if g: A → B ⊗C is a morphism of representations, dualizing it will give another morphism of representations g′: A⊗C^* → B.

If ρ is an n-dimensional irreducible representation of G with the underlying vector space V, then we can define a G×G morphism of representations, for all g in G and x in V

$f:\mathbb{C}[G]\otimes (1_G\otimes V)\rightarrow (V\otimes 1_G)$

$f(g\otimes x)=\rho(g)[x]$

where 1_G is the trivial representation of G. This defines a G×G morphism of representations, as can be explicitly checked.

Now we use the above lemma and obtain the G×G morphism of representations

$f':\bar{V}\otimes V\rightarrow \overline{\mathbb{C}[G]}$ .

The dual representation of C[G] as a G×G-representation is equivalent to C[G]. An isomorphism is given if we define the contraction 〈g,h〉 = δ_gh, as you may check. So, we end up with a G×G-morphism of representations

$f'':\bar{V}\otimes V\rightarrow \mathbb{C}[G]$ .

It turns out

$f''(x\otimes y)=\sum_{g\in G}\langle x,\rho(g)[y]\rangle g$

for all x in $\bar{V}$ and y in V.

By Schur's lemma, the image of f″ is a G×G irreducible representation, which is therefore n×n dimensional, which also happens to be a subrepresentation of C[G] (f″ is nonzero).

This, of course would be n direct sum equivalent copies V. Note that if ρ₁ and ρ₂ are equivalent G-irreducible representations, the respective images of the intertwiners would give rise to the same G×G-irreducible representation of C[G].

Here, we use the fact that if f is a function over G, then

$\sum_{g\in G}f(g)hgk^{-1}=\sum_{g\in G}f(h^{-1}gk)g$

We convert C[G] into a Hilbert space by introducing the norm where 〈g,h〉 is 1 if g is h and zero otherwise. This is different from the 'contraction' given a couple of paragraphs back, in that this form is sesquilinear. This makes C[G] a unitary representation of G×G. In particular, we now have the concepts of orthogonal complement and orthogonality of subrepresentations.

In particular, if C[G] contains two inequivalent irreducible G×G subrepresentations, then both subrepresentations are orthogonal to each other. To see this, note that for every subspace of a Hilbert space, there exists a unique linear transformation from the Hilbert space to itself which maps points on the subspace to itself while mapping points on its orthogonal complement to zero. This is called the projection map. The projection map associated with the first irreducible representation is an intertwiner. Restricted to the second irreducible representation, it gives an intertwiner from the second irreducible representation to the first. Using Schur's lemma, this has got to be zero.

Now suppose A ⊗ B is a G×G-irreducible representation of C[G].

Note. The complex irreducible representations of G×H are always a direct product of a complex irreducible representation of G and a complex irreducible representation of H. This is not the case for real irreducible representations. As an example, there is a 2 dimensional real irreducible representation of the group C₃ × C₃ which transforms nontrivially under both copies of C₃ but can't be expressed as the direct product of two irreducible representations of C₃.

This representation is also a G-representation (n_A direct sum copies of B where n_A is the dimension of A). If Y is an element of this representation (and hence also of C[G]) and X an element of its dual representation (which is a subrepresentation of the dual representation of C[G]), then

$f''(X\otimes Y)=\sum_{g\in G}\langle X,\rho(g)[Y]\rangle g=\sum_{g\in G} \langle X,Yg^{-1}\rangle g$

$=\sum_{g\in G}\langle X,g^{-1}\rangle gY=\sum_{g\in G}\langle X,g^{-1}\rangle (g,e)[Y]$

where e is the identity of G. I know the f″ defined a couple of paragraphs back is only defined for G-irreducible representations and A ⊗ B isn't a G-irreducible representation in general. But since A ⊗ B is simply the direct sum of copies of Bs and we've shown that each copy all maps to the same G×G-irreducible subrepresentation of C[G], we've just showed that $\bar{B}\otimes B$ as an irreducible G×G-subrepresentation of C[G] is contained in A ⊗ B as another irreducible G×G-subrepresentation of C[G]. Using Schur's lemma again, this means both irreducible representations are the same.

Putting all of this together,

Theorem. C[G] ≅ $\sum_{inequivalent\ G-irreducible\ representations\ V} \bar{V}\otimes V.$

Corollary. If there are p inequivalent G-irreducible representations, V_i, each of dimension n_i, then |G| = n₁² + ... + n_p².

[编辑] 特徵理論

主条目：特徵理論

There is a mapping from G to the complex numbers for each representation called the character given by the trace of the linear transformation upon the representation generated by the element of G in question

χ_ρ(g)=Tr[ρ(g)].

All elements of G belonging to the same conjugacy class have the same character: in other words χ_ρ is a class function on G. This follows from

Tr[ρ(ghg^-1)]=Tr[ρ(g)ρ(h)ρ(g)^-1]=Tr[ρ(h)]

by the cyclic property of the trace of a matrix.

What are the characters of C[G]? Using the property that gh^-1 is only the same as g if h = e, χ_C[G](g) is |G| if g=e and 0 otherwise.

The character of a direct sum of representations is simply the sum of their individual characters.

Putting all of this together,

$\sum_{i=1}^p n_i \chi_{V_i}(g)=|G|\delta_{ge}$

with the Kronecker delta on the RHS.

Repeat this, working with characters of G×G instead of characters, of G which I'll call Δ. Then, Δ_C[G](g,h) is the number of elements k in G satisfying g k h^-1 = k. This is equal to

$\sum_{i=1}^p \chi_{\bar{V}_i}(g)\chi_{V_i}(h)=\sum_{i=1}^p \chi_{V_i}(g^{-1})\chi_{V_i}(h)=\sum_{i=1}^p \chi_{V_i}(g)^*\chi_{V_i}(h)$

where * denotes complex conjugation. After all, C[G] is a unitary representation and any subrepresentation of a finite unitary representation is another unitary representation; and all irreducible representations are (equivalent to) a subrepresentation of C[G].

Consider

$\sum_{h \in G}\Delta_{C[G]}((g,hkh^{-1}))$ .

This is |G| times the number of elements which commute with g; which is |G|² divided by the size of the conjugacy class of g, if g and k belong to the same conjugacy class, but zero otherwise. Therefore, for each conjugacy class C_i of size m_i, the characters are the same for each element of the conjugacy class and so we can just call χ_ρ(C_i) by an abuse of notation). Then,

$\frac{|G|}{|C_i|}\delta_{ij}=\sum_{k=1}^p\chi_{V_k}(C_i)^*\chi_{V_k}(C_j)$ .

Note that

$\sum_{g\in G}\rho(g)$

is a self-intertwiner (or invariant). This linear transformation, when applied to C[G] (as a representation of the second copy of G×G), would give as its image the 1-dimensional subrepresentation generated by

$\sum_{g\in G}g$ ;

which is obviously the trivial representation.

Since we know C[G] contains all irreducible representations up to equivalence and using Schur's lemma, we conclude that

$\sum_{g\in G}\rho(g)$

for irreducible representations is zero if it's not the trivial irreducible representation; and it's of course |G|1 if the irreducible representation is trivial.

Given two irreducible representations V_i and V_j, we can construct a G-representation

$\bar{V}_i\otimes V_j$ ,

this time not as a G×G representation but an ordinary G-representation. See direct product of representations. Then,

$\chi_{\bar{V}_i\otimes V_j}(g)=\chi_{\bar{V}_i}(g)\chi_{V_j}(g)=\chi_{V_i}(g)^*\chi_{V_k}(g)$ .

It can be shown that any irreducible representation can be turned into a unitary irreducible representation. So, the direct product of two irreducible representations can also be turned into a unitary representations and now, we have the neat orthogonality property allowing us to decompose the direct product into a direct sum of irreducible representations (we're also using the property that for finite dimensional representations, if you keep taking proper subrepresentations, you'll hit an irreducible representation eventually. There's no infinite strictly decreasing sequence of positive integers). See Maschke's theorem.

If i≠j, then this decomposition does not contain the trivial representation (Otherwise, we'd have a nonzero intertwiner from V_j to V_i contradicting Schur's lemma). If i=j, then it contains exactly one copy of the trivial representation (Schur's lemma states that if A and B are two intertwiners from V_i to itself, since they're both multiples of the identity, A and B are linearly dependent). Therefore,

$\sum_{g\in G}\chi_{V_i}(g)^*\chi_{V_j}(g)=\sum_{k}|C_k|\chi_{V_i}(C_k)^*\chi_{V_j}(C_k)=|G|\delta_{ij}$

Applying a result of linear algebra to both orthogonality relations (|C_i| is always positive), we find that the number of conjugacy classes is greater than or equal to the number of inequivalent irreducible representations; and also at the same time less than or equal to. The conclusion, then, is that the number of conjugacy classes of G is the same as the number of inequivalent irreducible representations of G.

Corollary. If two representations have the same characters, then they are equivalent.

Proof. Characters can be thought of as elements of a q-dimensional vector space where q is the number of conjugacy classes. Using the orthogonality relations derived above, we find that the q characters for the q inequivalent irreducible representations forms a basis set. Also, according to Maschke's theorem, both representations can be expressed as the direct sum of irreducible representations. Since the character of the direct sum of representations is the sum of their characters, from linear algebra, we see they are equivalent.

We know that any irreducible representation can be turned into a unitary representation. It turns out the Hilbert space norm is unique up to multiplication by a positive number. To see this, note that the conjugate representation of the irreducible representation is equivalent to the dual irreducible representation with the Hilbert space norm acting as the intertwiner. Using Schur's lemma, all possible Hilbert space norms can only be a multiple of each other.

Let ρ be an irreducible representation of a finite group G on a vector space V of (finite) dimension n with character χ. It is a fact that χ(g) = n if and only if ρ(g) = id (see for instance Exercise 6.7 from Serre's book below). A consequence of this is that if χ is a non-trivial irreducible character of G such that χ(g) = χ(1) for some g≠1 then G contains a proper non-trivial normal subgroup (the normal subgroup is the kernel of ρ). Conversely, if G contains a proper non-trivial normal subgroup N, then the composition of the natural surjective group homomorphism G → G/N with the regular representation of G/N produces a representation π of G which has kernel N. Taking χ to be the character of some non-trivial subrepresentation of π, we have a character satisfying the hypothesis in the direct statement above. Altogether, whether or not G is simple can be determined immediately by looking at the character table of G.

[编辑] 歷史

The general features of the representation theory of a finite group G, over the complex numbers, were discovered by Ferdinand Georg Frobenius in the years before 1900. Later the modular representation theory of Richard Brauer was developed.

[编辑] 另見

[编辑] 參考文獻

Fulton, William; and Harris, Joe (1991). Representation Theory: A First Course. New York: Springer. ISBN 0-387-97495-4.

The standard graduate level reference for representations of groups in general.

James, Gordon; and Liebeck, Martin (1993). Representations and Characters of Finite Groups. Cambridge: Cambridge University Press. ISBN 0-521-44590-6.

A beautiful and readable introduction; designed for self study.

Jean-Pierre, Serre (1977). Linear Representations of Finite Groups. Springer-Verlag. ISBN 0-387-90190-6.

A very well-written introduction to stated topic: concise and extremely readable.

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