Note: I missed this class as I was attending a confrence of the Association of Symbolic Logic in San Diego. I plan to write this lecture up by borrowing some notes soon. Those of you who are interested in Group Theory should check out the Automorphism Tower Problem at some time, which Simon Thomas presented his proof of in San Diego.
3/25/99
Def: If E is a Real Euclidean space and F
is a subset of E, then F is called a root system
is it satisfies the following axioms:
1) F in nonempty, F
spans E and F does not contain 0.
2) If a is in F
then ka is in F iff
k is in {-1, 1}.
3) If a and b
are in F then sa(b)
is in F.
4) If a and b
are in F then <b,
a> is in Z.
where sa(b)
= b - <b, a>a
and <b, a> = 2(b,
a)/(a, a).
Note: sa(b)
is the reflection of b in the hyperplane perpendicular
to a (This hyperplane is defined as Pa
= {x in E : (x, a) = 0}).
Def: The rank of F is the dimension
of the euclidean space of which F is a subset
of.
Note: Here there should be some pictures of root systems. I have not yet decided how I wish to store them so at least for now, we will not attempt to draw the basic examples.
Note: If v and w are in E then (v, w), the inner product, is given by the cosine formula (v, w) = |v||w|cos q. Thus (v, w)2 = (v, v)(w, w)cos2 q and cos2 q = ((v, w)/(v, v))((v, w)/(w, w)). This gives us (2(v, w)/(v, v))(2(v, w)/(w, w)) = 4cos2 q = <w,v><v,w>. If v and w are in a root system, both <w, v> and <v, w> are integers, so we know 4cos2 q is an integer as well. Because of the bounds on the cosine function, as well as the squaring of the function we know 4cos2 q is in {0, 1, 2, 3, 4}. Thus cos2 q is in {0, 1/4, 1/2, 3/4, 1}. If q is acute then it is in {00, 300, 450, 600, 900} = {0, p/6, p/4, p/3, p/2}.
Note: Suppose a and b are in F. We have determined that 4cos2 q = <a, b ><b, a> is in {0, 1, 2, 3, 4}, so now we can examine the possibile values for <a, b > and <b, a>. We can assume WLOG that |b| is greater than or equal to |a|.
Lemma: Given the precceding assumptions, the following hold
1) |<b, a>|
is greater than or equal to |<a, b
>|.
2) <a, b
> and <b, a> have
the same sign.
3) <a, b
> = 0 iff <b, a>
= 0.
Proof: As |b| is greater than or equal to |a|, we know (b, b) is greater than or equal to (a, a) so (a, b)/(b, b) is less than or equal to (a, b)/(a, a) which gives us 1. As (a, a) and (b, b) are positive, (a, b)/(b, b) is positive iff (a, b)/(a, a) is which gives us 2. <a, b> = 0 iff (a, b)/(b, b) = 0 iff (a, b) = 0 iff (a, b)/(a, a) = 0 iff <b, a> = 0 giving us 3. n
Lemma: Given the precceding assumptions, we know either b equals plus or minus a or |<a, b >| = 1.
Proof: <a, b ><b, a> is in {0, 1, 2, 3, 4}. Neither <a, b > nor <b, a> can equal zero as then they would both be zero. If their product is 1,2 or 3 then we are done as a product of integers equaling any of these means the smaller of the two has absolute value of one. The only other option is that <a, b ><b, a> = 4, which means 4cos2 q = 4, so q is in {0, p} and b equals plus or minus a. n
Note: We can thus create the following table for q.
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Proof: Assume |b| is greater than or equal to |a|. Then |<a, b>| = 1. By the third axiom of root systems, we know if a and b are in F then sb(a) = a - <a, b>b is in F. If (a, b) > 0 then <a, b> = 1 so a - b is in F. If (a, b) < 0 then <a, b> = -1 so a + b is in F. n
Proposition: If a and b are in F then the a root string b - pa, b - (p-1)a, ... , b, ... , b + qa is unbroken.
Proof: If there is a break, we can find integers a and b in [-p, q] so that b - aa is a root, b - (a + 1)a is not a root, b + ba is a root, and b - (b - 1)a is not a root. If (b - aa, a) > 0 then b - (a + 1)a is a root by the lemma, a contradiction. Likewise if (b - ba, a) < 0 then b - (b - 1)a is root, another contradiction. We get (b, a) - (aa, a) < or = 0 and (b, a) - (ba, a) > or = 0, or -(b, a) + (ba, a) < or = 0. By adding we get (ba, a) - (aa, a) < or = 0 or (b - a)(a, a) < or = 0. As both b - a and (a, a) are strictly positive we have reached a contradiction. n
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