How Do You Make a Character Table?
Date: 10/15/2008 at 01:27:16 From: Amy Subject: How do you make a character table (representation theory) How do you make a character table? The course is Representation Theory and we're studying representations of finite groups, specifically character theory. The example in the text book is of the Symmetric group with 3 elements S_3. It is | 1 3 2 S_3 | 1 (12) (123) ________________________________ trivial | 1 1 1 alternating | 1 -1 1 standard | 2 0 -1 I don't get where the inside numbers come from. I get that the inside numbers are the character of the representative. My understanding of this is that I take the trace (sum of the diagonal elements) of the matrix that the element gets sent to in the homomorphism that defines the representation. I'm not seeing how this idea gives me the numbers in this table. The book seems to act as if I should know these without any reference to matrices. My understanding of this is that I take the trace (sum of the diagonal elements) of the matrix that the element gets sent to in the homomorphism that defines the representation. I'm not seeing how this idea gives me the numbers in this table. The book seems to act as if I should know these without any reference to matrices.
Date: 10/16/2008 at 05:11:40 From: Doctor Jacques Subject: Re: How do you make a character table (representation theory) Hi Amy, Usually, you don't need to use the actual matrices of the representations to compute the characters. You can fill most of the character table simply by using general properties of characters (this is the subject matter of character theory). For example: * The number of irreducible characters is equal to the number of conjugacy classes (it is the dimension of the center of the group algebra). * There are orthogonality relations between characters; specifically, between the rows and the columns of the table. (As the character values are in general complex, these are usually not sufficient to determine the table completely). * The characters of an Abelian group have degree 1; their values are roots of unity, and the character group is isomorphic to the original group. This means that you can write them down easily. * If the group G contains a normal subgroup H, any character of the quotient group G/H gives a character of G by composition with the canonical homomorphism G -> G/H. If the quotient happens to be Abelian, you can use the previous remark. In particular, the number of characters of degree 1 is the index of the derived group. * The degrees of the irreducible characters divide the order of the group, and the sum of the squares of these degrees is equal to the order of the group. * Characters come in conjugate pairs, because it is not possible to distinguish i from -i. In particular, if there is only one character of a given degree, its values must be real. * If c_1 is a character of degree 1, and c is any character, there is also a character c_1*c (corresponding to the tensor product of the representations). * In any representation, the character of an element is the complex conjugate of the character of the inverse; in particular, if the element and its inverse belong to the same conjugacy class, the character is real. In most "simple" cases, you can fill the character table completely by using all those properties. Let us look, for example, at S_3. There are 3 conjugacy classes, and therefore there are three irreducible characters. The degrees of these characters are divisors of 6, and the sum of their squares equals 6. It is easy to see in this case that the only possibility is (1,1,2). As the trivial character is always present, we can already fill part of the character table: 1 3 2 () (1,2) (1,2,3) ---------------------------- X1 1 1 1 X2 1 a b X3 2 c d S_3 contains a normal subgroup of order 3, and the quotient subgroup is isomorphic to C_2 (an Abelian group). This means that we must have 2 characters of degree 1, containing 1 (the degree) in the column of the kernel (the third column). The first one is the trivial character, and the second one must be X2. This means that we have b = 1, and, as (1,2) has order 2 in the quotient group, the value of X2 at (1,2) is a primitive square root of 1, i.e., a = -1. The table looks now like this: 1 3 2 () (1,2) (1,2,3) ---------------------------- X1 1 1 1 X2 1 -1 1 X3 2 c d Now, orthogonality of the columns gives: 1*1 + 1*(-1) + 2*c = 0 1*1 + 1*1 + 2*d = 0 from which we obtain c = 0 and d = -1, and this completes the table. For a slightly more elaborated example, let us now consider A_4. There are 4 conjugacy classes, represented by (), (1,2)(3,4), (1,2,3), and (1,3,2). This means that we should have 4 irreducible characters. A_4 contains a normal subgroup N isomorphic to C2 x C2; this subgroup consists of the classes represented by () and (1,2)(3,4). The quotient group is C_3; we should therefore find three characters of degree 1, which take the value 1 in the classes corresponding to N: I A B C 1 3 4 4 () (1,2)(3,4) (1,2,3) (1,3,2) ------------------------------------- X1 1 1 1 1 X2 1 1 ? ? X3 1 1 ? ? X4 ? ? ? ? (The first row is used to label the classes, this makes it easier to refer to entries in the table). Now, class B has order 3 in the quotient group, therefore X2(B) must be a primitive cube root of unity. Let us denote such a cube root by w = (1+sqrt(-3))/2. The other cube root is w^2, and w and w^2 are conjugate. As the classes B and C are inverses of each other, X2(B) and X2(C) are complex conjugates. We can take X2(B) = w and X2(C) = w^2. As characters come in conjugate pairs, we have X3(B) = w^2 and X3(C) = w. The table now looks like this: I A B C 1 3 4 4 () (1,2)(3,4) (1,2,3) (1,3,2) ------------------------------------- X1 1 1 1 1 X2 1 1 w w^2 X3 1 1 w^2 w X4 ? ? ? ? As the sum of the squares of the characters must equal 12, we find that X4 has degree 3 = X4(I). Now, multiplying X4 by X2 (of degree 1) must give a character of degree 3, which must be X4, since it is the only character of degree 3. In particular, we must have: X4(B) = X4(B)*X2(B) = X4(B)*w and this implies that X4(B) = 0; by the same argument, X4(C) = 0. Orthogonality of the first two columns gives: 1*1 + 1*1 + 1*1 + 3*X4(A) = 0 which gives X4(A) = -1. The whole table is: I A B C 1 3 4 4 () (1,2)(3,4) (1,2,3) (1,3,2) ------------------------------------- X1 1 1 1 1 X2 1 1 w w^2 X3 1 1 w^2 w X4 3 -1 0 0 (we could also have found X4(B) and X4(C) by orthogonality). Note that the representation X4 has a natural geometric interpretation, because A4 is the group of rotations of a regular tetrahedron. If we take the center of the tetrahedron as origin, and draw unit vectors to three vertices, the fourth vertex has coordinates (-1,-1,-1) (because the sum of the vectors from the center must be 0). Each element of A4 induces a permutation of the vertices, which is a linear transformation of R^3. You can represent those transformations by 3*3 matrices, and check that the traces of those matrices correspond to the values in the table above. In summary, the use of the above techniques should allow you to construct character tables by hand, at least for simple cases (the ones you should expect to find as exercises in a textbook...). Note that there are general algorithms for computing character tables, although these algorithms are more suited to a computer implementation. These algorithms do not need the explicit matrices of the representations (of course, they need some information about the group, like the conjugacy classes). I hope this helps. Please feel free to write back if you require further assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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