Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 1 / 6

A reduced crystallographic abstract root system in a finite-dimensional real inner product space $E$ is a finite set $R$ such that: it spans $E$ and does not contain zero; the only scalar multiples of $\alpha \in R$ belonging to $R$ are $\pm \alpha$; every root reflectionpermutes $R$; and each Cartan integer $2(\beta ,\alpha )/(\alpha ,\alpha )$ is integral. These are the crystallographic and reduced conventions appropriate to roots of a complex semisimple Lie algebra. Without crystallographic integrality, the numerical restrictions in the following subparts would not hold.

If $\beta =-\alpha$, then both Cartan integers are $-2$. This opposite-root case satisfies the printed inner-product hypothesis and must be included.

Otherwise reducedness makes the roots nonproportional. Writing the angle as $\theta$, strict Cauchy-Schwarz inequality givesBoth factors are negative integers, so their product is $1,2$ or $3$. The full list is thereforeThe nonproportional pairs correspond respectively to angles $120^{\circ }$, $135^{\circ }$ and $150^{\circ }$; reversing the order of the roots reverses the unequal pairs.

There is an exception to the printed assertion: for $\beta =-\alpha$, the inner product is negative but $\alpha +\beta =0$ is not a root. This already occurs in the rank-one root system $R=\{\alpha ,-\alpha \}$.

For the intended assertion, assume $\beta \ne -\alpha$. By the preceding list, at least one of $n_{\alpha ,\beta },n_{\beta ,\alpha }$ equals $-1$. If the first does, the reflection axiom givesIf the second does, use $s_{\beta }(\alpha )=\alpha +\beta$ instead. Thus the precise reusable conclusion isThis proves that obtuse nonopposite roots have a root sum directly from the reflection and integrality axioms, without appealing to the classification of root systems.

Choose a linear functional $\ell :E\to \mathbb{R}$ which is nonzero on every root; one exists because there are only finitely many root hyperplanes to avoid. Declare a root positive when $\ell (\alpha )>0$, and negative when $\ell (\alpha )<0$. This orders the roots by sign, compatibly with addition whenever the sum is a root. One can refine it to a lexicographic total order by completing $\ell$ to a coordinate system. Since negation permutes the roots,A simple root is a positive root not expressible as the sum of two positive roots. If distinct simple roots $\alpha ,\beta$ had $\alpha -\beta$ as a root, it would be either positive or negative. In the first case $\alpha =\beta +(\alpha -\beta )$ decomposes $\alpha$ into two positive roots; in the second $\beta =\alpha +(\beta -\alpha )$ decomposes $\beta$. Both contradict simplicity. If they are equal, their difference is zero and is also not a root. Hence the difference of two simple roots is never a root.

For use in the final subpart, the simple roots form a basis, and every positive root is a nonnegative integer combination of them. Here are the needed reasons. Repeatedly decompose a nonsimple positive root into two positive roots; their $\ell$ values decrease, and the finite root set makes this terminate. This gives the integer combinations. Distinct simple roots have nonpositive inner product: a positive inner product would make $\alpha$ and $-\beta$ obtuse, so the qualified root-sum lemma would make $\alpha -\beta$ a root, just ruled out. Finally, a nontrivial linear relation between simple roots can be split into positive and negative coefficients, giving the same vector $u$ as positive combinations of disjoint sets. Its squared norm, computed using those two expressions, is nonpositive, whereas a nonzero positive combination has positive $\ell$ value. This contradiction proves linear independence. Their span is $E$ because they generate all roots.

The single bonds give equal simple-root lengths and inner products, after normalization,The positive roots of this A3 root system areThe first two sums are roots by the qualified obtuse-root lemma. The last is a root because $(\alpha +\beta ,\gamma )=-1$. All have nonnegative simple-root coordinates, hence are positive.

To prove completeness, first note that every root is Weyl-conjugate to a simple root. For a nonsimple positive root $\delta =\sum c_i{\alpha }_i$, some $(\delta ,{\alpha }_i)>0$, since $(\delta ,\delta )=\sum c_i(\delta ,{\alpha }_i)>0$. Reflection subtracts a positive integer multiple of ${\alpha }_i$, decreasing its integer height. The reflected root stays positive: it leaves the coefficients at all other simple roots unchanged, at least one of which is positive; a root has coordinates all of one sign. Repeating this height decrease reaches a simple root. Thus every root has squared length two in this connected single-bond system.

Write an arbitrary positive root as $a\alpha +b\beta +c\gamma$, with nonnegative integers $a,b,c$. Its squared length condition isThus $a,c\in \{0,1\}$. For $(a,c)=(0,0)$ the equation gives $b=1$; for $(1,0)$ or $(0,1)$ it gives $b=0$ or $1$; for $(1,1)$ it gives $b=1$. These are exactly the six coefficient triples in the displayed list. This proves completeness from the axioms, not merely from recognizing the Dynkin diagram.