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## Representation Theory

I think I will write out the beginnings of a "book" that I'd been meaning to get around to starting at some point. Better that someone might actually read it here than trust to them finding it on my web page.

Any requests, let me know by PM (or if this is inappropriate for this forum). Chapter one will appear some time soon. In the meantime here are some group theory notes to be getting on with:

http://www.maths.bris.ac.uk/~maxmg/docs/groups.pdf
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 Recognitions: Homework Help Science Advisor Chapter 0 Preliminaries and Notational Conventions I will be using the latex features to typeset lots of mathematics, however I will not overly abuse that since the images take time to reload into a page. Thus I will adopt the "psuedo-tex" standards for things. In particular x_r means subscript, ie $x_r$ and x^r means superscript [itex]x^r[/tex], hopefully I will choose whichever format is best and not cause any confusion. 0.1 Groups Recall the bad definition of a group, G, is a set of elements with a rule for composing any two elements to obtain a third element of G that satisfies some rules. Note we already have "closure" since I have declared that composing two elements of G gives an element of G. We use juxtaposition to indicate composition, ie xy means x composed with y. The other rules are associatvity: (xy)z=x(yz) for all x,y,z in G identity: there is an element e satisfying ex=xe=x for all x in G inverses: given any x in G there is some y in G satisfying xy=yx=e Examples: the integers, rationals, reals or complex numbers under addition. the non-zero rationals, reals or complex numbers under multiplication the mxn matrices over any of the above with addition the nxn invertible matrices over any of the above under matrix multiplication. Non-examples: ie these are NOT groups the rationals under multiplication since 0 has no multiplicative inverse. the invertible nxn matrices under addition (exercise find two invertible matrices that add together to give a non-invertible matrix. What I've just done is give an abstract group definition, and this is probably the one you first saw when you did a bit of algebra for the first time. Personally I prefer to give the concrete definition but I'm going to assume that you already know what a group is. If not you could try reading my notes linked to in the first post on concrete groups. Accepting that you already know what a group is.... Definition: a group is abelian (aka commutative) if xy=yx for any pairs of elements x and y in G. Two elements commute if xy=yx Exercise: which of the examples are abelian and which are not? We actually need very little group theory for the most of what we do, and any we need later will be introduced at the relevant time. The only things we need to make sure we agree on are some symbols. List of symbols G is a group, |G| will be its order. We are only interested in the cases when G is finite ie |G| is a natural number. C_G(x) means the conjugacy class in G of an element x. It is the set of all elements of the form yxy^{-1} Z(G) is the centre of G; the set of all elements that commute with every element of G. Exercise show x is in Z(G) iff C_G(x)={x} if x is in G we will denote by ord(x) its order; that is the smallest strictly positive integer n such that x^n=e Recall (though we shan't prove it) that the order of any element divides the order of the group. Finally, the only other thing we need to know right now about groups is what a homomorphism is. Definition. If G and H are two groups are homomorphism from G to H is a map p:G-->H that sends each element in g to some element in p(g) in a way that respects the group structures: p(xy)=p(x)p(y) and p(x^{-1})=p(x)^{-1} these say that if we consider the operations of composition or inverting, well, I can do them in G and send the result to H or send the elements to H then do the operations but that both of these yield the same result. 0.2 Vector Spaces. Recall that a field, F, is a set with two operations + and * such that under + F is an abelian group with identity 0 and that F\{0} is an abelian group under * with identity 1. We often refer to F\{0} (ie F without the element 0) as the multiplicative group and write it is $$F^{\times}$$ Examples: The rational, real or complex numbers. Further examples: Let p be a prime number and consider the set of residues mod p, ie let F={0,1,...,p-1} then F is a field. Proof: that it is a group under addition is obvious, and most of the fact that its non-zero elements are a group under multiplication is too, the only non-obvious thing is multiplicative inverses. Recally Euclid's algorithm and highest common factors. If x is any element of 1,2,..,p-1 then it is coprime (considering it as an integer) to p, so there are integers y and z such that xy+pz=1 (again in the integers). So just take the residue class of y for the inverse of x. F here is a "field of characteristic p", and since it has p elements in it, it is labelled F_p Definition: If F is any field it contains the element 1, and thus it contains 1+1, 1+1+1, etc. If no amount of adding up of 1's ever gives 0 then we say the field has characteristic zero. If we ever manage to get 0 adding up 1's let n be the first time we get 0. We say F has characteristic n. Lemma: if F has characteristic n>0 then n is a prime. Proof: if n is not prime let n=ab where a and b is a non-trivial factorization (ie a and b are both different from 1 and n). Consider the elements gotten by adding up a 1's and b 1's in F. These multiply together to give 0, but we know F\{0} is a group and this is a contrdiction (not closed), so n can have no non-trivial factorizations, hence it is a prime. Lemma: If p is a prime then there is a field of characteristic p Proof: take the resdiue classes mod p. Lemma: there is more than one field of any given prime characteristic. Proof: we omit the proof, preferring instead to give an example that mimics the construction of C from R. For ease we consider the case F_2; the general case is no harder. F_2 as a set is just {0,1}. Consider the polynomial f(x)=x^2+x+1. This has no root in F_2, ie f(0)=1 and f(1)=1, just like x^2+1 has no root in R. So, just like the R to C case we add a symbol, usually an alpha, but we'll use a instead, and consider F=F_2[a] F_2[a] is the set of all elements of the form x+ay just like C=R[i] is the set of all elements x+iy. F_2[a] has 4 elements. 0,1,a,a+1 these are indeed a group under addition as we can see by inspection recalling that 2=0 mod 2. What about multiplication? well we need to work out a^2 (a times a), but we know that a satisfies a^2+a+1=0, or a^2=a+1 (recall also that 1=-1 mod 2). Similarlry a(a+1)=a^2+a=1, and (a+1)(a+1) =a^2+2a+1 multiplying out =a^2+1 since 2=0 =a from the identity a^2+a+1=0 so it is indeed a field with 4 elements. The general idea over F_p is to find a polynomial with no roots, which is easy f(x)=x^p-x-1 will do the trick. Why? Because by Fermat's little theorem x^p=x mod p for all x, so that x^p-x=0 for all x, thus f(s)=1 for any s in {0,1,..,p-1}. Now we add some element a that formally satisfes f(a)=0 and form F_p[a] as before. We'll end this for now, since it is getting too far from representation theory. Though I will perhaps have to do a little more about fields in char p when we come to look at them in more detail to prove to you that i'm not making any restrictions that are too strong ie imposing so many rules that nothing can possibly satisfy them all.