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Chapter 18: Problem 22

Let \(X(G)\) be the character ring of a finite group \(G\), generated over \(Z\) by the simple characters over \(\mathbf{C}\). Show that an element \(f \in X(G)\) is an effective irreducible character if and only if \((f, f)_{G}=1\) and \(f(1) \geq 0 .\)

Short Answer

Expert verified
An element \(f \in X(G)\) is an effective irreducible character if and only if \((f, f)_{G} = 1\) and \(f(1) \geq 0\). This is shown by proving that if \(f\) is an effective irreducible character, then \((f, f)_{G}=1\) and \(f(1) \geq 0\), and conversely, if \((f, f)_{G}=1\) and \(f(1) \geq 0\), then \(f\) is an effective irreducible character.

Step by step solution

01

An irreducible character of a finite group \(G\) is a linear character that cannot be expressed as the sum of two or more non-trivial linear characters of \(G\). An effective irreducible character is an irreducible character with non-negative degrees at the identity element. #Step 2: Scalar product of characters#

The scalar product \((\chi, \psi)_{G}\) of two characters \(\chi\) and \(\psi\) of a finite group \(G\) is defined as \[ (\chi, \psi)_{G} = \frac{1}{|G|}\sum_{g \in G}\chi(g)\overline{\psi(g)} .\] #Step 3: Show that if \(f\) is an effective irreducible character, then \((f, f)_{G}=1\) and \(f(1) \geq 0\)#
02

If \(f\) is an effective irreducible character, then by definition \(f(1) \geq 0\). Now, we'll show that \((f, f)_{G} = 1\) for the irreducible character \(f\). Recall that for an irreducible character, the scalar product with itself is always 1, so we have \[ (f, f)_{G} =\frac{1}{|G|}\sum_{g \in G}f(g)\overline{f(g)}= \frac{1}{|G|}\sum_{g \in G}|f(g)|^2 = 1 .\] #Step 4: Show that if \((f, f)_{G}=1\) and \(f(1) \geq 0\), then \(f\) is an effective irreducible character#

Suppose \((f, f)_{G}=1\) and \(f(1) \geq 0\). We'll show that \(f\) is an effective irreducible character. Note that from the given condition, we have \[ 1=(f, f)_{G} =\frac{1}{|G|}\sum_{g \in G}|f(g)|^2 .\] The scalar product of a character with itself is always non-negative, so this implies that \(f(g) = 0\) for all but one element \(g \in G\). Since \(f(1) \geq 0\), it implies that \(f\) has a non-zero value at the identity element, and zero elsewhere. Thus, \(f\) is an irreducible character. And since \(f(1) \geq 0\), it's an effective irreducible character. Therefore, we have shown that an element \(f \in X(G)\) is an effective irreducible character if and only if \((f, f)_{G} = 1\) and \(f(1) \geq 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Character Ring
The character ring, often denoted as \(X(G)\), is a fascinating algebraic structure associated with a finite group \(G\). It consists of functions called characters, which map elements from the group to complex numbers, particularly preserving certain operations from group theory.

These characters are not random; they obey special rules that make them align with the group structure. For instance, every group element's character is a sum of values called the character values, weighted by their respective occurrences within the group.

One key property is that these characters can be added and multiplied together to form new characters, thus giving rise to the term 'ring'. Also, the ring contains a special subset of characters called the irreducible characters. These are akin to prime numbers in the world of integers; they cannot be broken down into more basic characters that are nontrivial. Effectively, they build the foundations for all characters in the ring.
Finite Group
A finite group is a mathematical ensemble where we have a set of elements together with an operation that combines any two elements to form a third element, satisfying certain conditions such as closure, associativity, the existence of an identity element, and the existence of inverse elements for each element in the set.

This group has a finite number of elements, hence its name. The order of the group is the number of elements it contains. In group theory and fields like particle physics, symmetry and transformation operations are often described by such groups.

Finite groups are relatively easy to study and have been extensively cataloged. They serve as the underlying symmetry groups for many geometrical shapes and are key in solving equations and understanding the symmetries in mathematical and physical systems.
Scalar Product of Characters
The scalar product of characters, sometimes referred to as the inner product, is a measure of similarity between two characters of a finite group. It's calculated using the formula:\[ (\chi, \psi)_{G} = \frac{1}{|G|}\sum_{g \in G}\chi(g)\overline{\psi(g)} .\]

Where \(|G|\) represents the order of the group, or the total number of elements, and \(\chi\) and \(\psi\) are characters of the group. This scalar product is an integral part of character theory, as it can help determine if characters are irreducible.

If the scalar product of a character with itself is 1, the character is irreducible and cannot be decomposed into simpler characters. Moreover, the scalar product also aids in establishing orthogonality relations between characters, critical for breaking down group representations into irreducible components.

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Most popular questions from this chapter

The groups \(S_{4}\) and \(A_{4}\). Let \(S_{4}\) be the symmetric group on 4 elements. (a) Show that there are 5 conjugacy classes. (b) Show that \(A_{4}\) has a unique subgroup of order 4, which is not cyclic, and which is normal in \(S_{4}\). Show that the factor group is isomorphic to \(S_{3}\), so the representations of Exercise 1 give rise to representations of \(S_{4}\). (c) Using the relation \(\sum d_{i}^{2}=\\#\left(S_{4}\right)=24\), conclude that there are only two other irreducible characters of \(S_{4}\), each of dimension 3 . (d) Let \(X^{4}+a_{2} X^{2}+a_{1} X+a_{0}\) be an irreducible polynomial over a field \(k\), with Galois group \(S_{4}\). Show that the roots generate a 3 -dimensional vector space \(V\) over \(k\), and that the representation of \(S_{4}\) on this space is irreducible, so we obtain one of the two missing representations. (e) Let \(\rho\) be the representation of (d). Define \(\rho^{\prime}\) by $$ \begin{array}{l} \rho^{\prime}(\sigma)=\rho(\sigma) \text { if } \sigma \text { is even: } \\ \rho^{\prime}(a)=-\rho(a) \text { if } \sigma \text { is odd. } \end{array} $$ Show that \(\rho^{\prime}\) is also irreducible, remains irreducible after tensoring with \(k^{\mathrm{a}}\), and is non-isomorphic to \(\rho\). This concludes the description of all irreducible representations of \(S_{4}\) (f) Show that the 3 -dimensional irreducible representations of \(S_{4}\) provide an irreducible representation of \(A_{4}\) (g) Show that all irreducible representations of \(A_{4}\) are given by the representations in (f) and three others which are one-dimensional.

The group \(S_{3}\). Let \(S_{3}\) be the symmetric group on 3 elements. (a) Show that there are three conjugacy classes. (b) There are two characters of dimension 1, on \(S_{3} / A_{3}\). (c) Let \(d_{i}(i=1,2,3)\) be the dimensions of the irreducible characters. Since \(\sum d_{i}^{2}=6\), the third irreducible character has dimension 2. Show that the third representation can be realized by considering a cubic equation \(X^{3}+a X+b=0\), whose Galois group is \(S_{3}\) over a field \(k .\) Let \(V\) be the \(k\) vector space generated by the roots. Show that this space is 2 -dimensional and gives the desired representation, which remains irreducible after tensoring with \(k^{2}\). (d) Let \(G=S_{3}\). Write down an idempotent for each one of the simple components of \(\mathbf{C}[G] .\) What is the multiplicity of each irreducible representation of \(G\) in the regular representation on \(\mathrm{C}[G] ?\)

The following formalism is the analogue of Artin's formalism of \(L\) -series in number theory. Cf. Artin's "Zur Theorie der \(L\) -Reihen mit allgemeinen Gruppencharakteren", Collected papers, and also S. Lang. "L-series of a covering", Proc. Nat Acad. Sc. USA (1956). For the Artin formalism in a context of analysis, see J. Jorgenson and S. Lang, "Artin formalism and heat kernels", J. reine angew. Math. 447 (1994) pp. 165-200. We consider a category with objects \(\\{U\\}\). As usual, we say that a finite group \(\mathrm{G}\) operates on \(U\) if we are given a homomorphism \(\rho: G \rightarrow\) Aut \((U)\). We then say that \(U\) is a G-object, and also that \(\rho\) is a representation of \(G\) in \(U\). We say that \(G\) operates trivially if \(\rho(G)=\) id. For simplicity, we omit the \(\rho\) from the notation. By a G-morphism \(f: U \rightarrow V\) between \(G\) -objects, one means a morphism such that \(f \circ \sigma=\sigma \circ f\) for all \(\sigma \in G\). We shall assume that for each \(G\) -object \(U\) there exists an object \(U / G\) on which \(G\) operates trivially, and a \(G\) -morphism \(\pi_{v, G}: U \rightarrow U / G\) having the following universal property: If \(f: U \rightarrow U^{\prime}\) is a \(G\) -morphism, then there exists a unique morphism $$ f / G: U / G \rightarrow U^{\prime} / G $$ making the following diagram commutative: In particular, if \(H\) is a normal subgroup of \(G\), show that \(G / H\) operates in a natural way on \(U / H\). Let \(k\) be an algebraically closed field of characteristic \(0 .\) We assume given a functor \(E\) from our category to the category of finite dimensional \(k\) -spaces. If \(U\) is an object in our category, and \(f: U \rightarrow U^{\prime}\) is a morphism, then we get a homomorphism $$ E(f)=f_{*}: E(U) \rightarrow E(U) $$ (The reader may keep in mind the special case when we deal with the category of reasonable topological spaces, and \(E\) is the homology functor in a given dimension.) If \(G\) operates on \(U\), then we get an operation of \(G\) on \(E(U)\) by functoriality. Let \(U\) be a \(G\) -object, and \(F: U \rightarrow U\) a \(G\) -morphism. If \(P_{r}(t)=\Pi\left(t-a_{1}\right)\) is the characteristic polynomial of the linear map \(F_{t}: E(U) \rightarrow E(U)\), we define $$ Z_{f}(t)=\prod\left(1-\alpha_{i} t\right) $$ and call this the zeta function of \(F\). If \(F\) is the identity, then \(Z_{P}(t)=(1-t)^{\text {R } W}\) ) where we define \(B(U)\) to be \(\operatorname{dim}, E(U)\) Let \(\chi\) be a simple character of \(G\). Let \(d_{z}\) be the dimension of the simple representation of \(G\) belonging to \(\chi\), and \(n=\operatorname{ord}(G) .\) We define a linear map on \(E(U)\) by letting $$ e_{x}=\frac{d_{x}}{n} \sum_{\text {eff } G} \chi\left(\sigma^{-1}\right) \sigma_{\text {* }} $$ If \(P_{x}(t)=\prod\left(t-\beta_{X}(x)\right)\) is the characteristic polynomial of \(e_{x}=F_{*}\), define $$ L_{p}(t, \chi, U / G)=\prod(1-\beta(x) t) $$ Show that the logarithmic derivative of this function is equal to $$ -\frac{1}{N} \sum_{\mu=1}^{\infty} \operatorname{tr}\left(e_{x}=F_{\%}^{\mu}\right) r^{\mu-1} $$ Define \(L_{p}(t, \chi, U / G)\) for any character \(x\) by linearity. If we write \(V=U / G\) by abuse of notation, then we also write \(L_{p}(t, \chi, U / V)\). Then for any \(\chi, \chi\) ' we have by definition, $$ L_{P}\left(t_{+} x+\chi^{\prime}, U / V\right)=L_{F}(t, \chi, U / V) L_{F}\left(t, X^{\prime}, U / V\right) $$ We make one additional assumption on the situation: Assume that the characteristic polynomial of $$ \frac{1}{n} \sum_{d \in G} \sigma_{*} \cdot F_{*} $$ is equal to the characteristic polynomial of \(F / G\) on \(E(U / G)\). Prove the following statement: (a) If \(G=\\{1\\}\) then $$ L_{F}(t, 1, U / U)=Z_{p}(t) $$ (b) Let \(V=U / G\). Then $$ L_{p}(t, 1, U / V)=Z_{F}(t) $$ (c) Let \(H\) be a subgroup of \(G\) and let \(\psi\) be a character of \(H\). Let \(W=U / H\), and let \(\psi^{\theta}\) be the induced character from \(H\) to \(G\). Then $$ L_{F}(t, \psi, U / W)=L_{F}\left(t, \psi^{G}, U / V\right) $$ (d) Let \(H\) be normal in \(G\). Then \(G / H\) operates on \(U / H=W\). Let \(\psi\) be a character of \(G / H\), and let \(\gamma\) be the character of \(G\) obtained by composing \(\psi\) with the canonical map \(G \rightarrow G / H .\) Let \(\varphi=F / H\) be the morphism indused on $$ U / H=W $$ Then $$ L_{\varphi}(t, \psi, W / V)=L_{F}(t, \chi, U / V) $$ (e) If \(V=U / G\) and \(B(V)=\operatorname{dim}_{k} E(V)\), show that \((1-t)^{n \cdot \eta_{1}}\) divides \((1-t)^{\text {At }}\). Use the regular character to determine a factorization of \((1-t)^{\operatorname{sic}}\).

The quaternion group. Let \(Q=\\{\pm 1, \pm x, \pm y, \pm z\\}\) be the quaternion group, with \(x^{2}=y^{2}=z^{2}=-1\) and \(x y=-y x, x z=-2 x, y z=-z y .\) (a) Show that \(Q\) has 5 conjugacy classes. Let \(A=\\{\pm 1\\}\). Then \(Q / A\) is of type \((2,2)\), and hence has 4 simple characters. which can be viewed as simple characters of \(Q\). (b) Show that there is only one more simple character of \(Q\), of dimension 2 . Show that the corresponding representation can be given by a matrix representation such that $$ \rho(x)=\left(\begin{array}{rr} i & 0 \\ 0 & -i \end{array}\right), \rho(y)=\left(\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right), \rho(z)=\left(\begin{array}{ll} 0 & i \\ i & 0 \end{array}\right) $$ (c) Let \(\mathrm{H}\) be the quaternion field, i.e. the algebra over \(\mathrm{R}\) having dimension 4 , with basis \(\\{1, x, y, z\\}\) as in Exercise 3, and the corresponding relations as above. Show that \(\mathrm{C} \otimes_{\mathrm{R}} \mathrm{H}=\mathrm{Mat}_{2}(\mathrm{C})(2 \times 2\) complex matrices). Relate this to (b).

Let \(\mathbf{F}\) be a finite field and let \(G=S L_{2}(\mathbf{F})\). Let \(B\) be the subgroup of \(G\) consisting of all matrices $$ \alpha=\left(\begin{array}{ll} a & b \\ 0 & d \end{array}\right) \in S L_{2}(\mathbf{F}), \text { so } d=a^{-1} $$ Let \(\mu: \mathbf{F}^{*} \rightarrow \mathbf{C}^{*}\) be a homomorphism and let \(\psi_{\mu}: B \rightarrow \mathbf{C}^{*}\) be the homomorphism such that \(\psi_{\mu}(\alpha)=\mu(a)\). Show that the induced character ind \(g\left(\psi_{\mu}\right)\) is simple if \(\mu^{2} \neq 1\)
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