Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 1 6 iii Solution Created 2026-10-03 Updated 2026-10-14
The double-bond arrow points to the middle node in the PDF. Since and , the short root must occupy that middle node and the long root the left node. The right node , joined to by a single bond, is another short root. Thus the diagram is C3 root system type, labelled as follows.
Labelling the C3 Dynkin diagram with the given short and long roots
. Details. The bonds require , and . Writing gives , , and . One choice is thereforeThe opposite choice of the last coordinate reflects the entire realization across the first two coordinate axes' plane and is equally valid.
To enumerate exhaustively, set , and . These are orthonormal, andThe complete C3 root system is , with eighteen roots. Its nine roots with nonnegative coefficients in this simple basis are exactlyThere are six short and three long positive roots, and the other nine roots are their negatives.
The single-bond rank-two diagram is A2 root system type. Three suitable positive simple pairs for its root subsystems areEach pair has squared lengths one and inner product , so its angle is degrees. Its plane meets the full root set in , with respectively , and . These exhaust the roots of the plane, and their coordinates in the pair have one sign. Thus each pair is genuinely a simple basis of an subsystem.
The double-bond diagram is rank-two B2 root system type, equivalently . Three suitable pairs, written short root first, areFor each pair , , and . The remaining positive roots in its plane areTogether with their negatives these are precisely the eight plane roots, and all four positive roots have nonnegative integral pair-coordinates. Hence the displayed pairs are simple bases of the required subsystems. The rank-two subsystems of a C3 root system description explains these as the three coordinate-plane subsystems in the orthonormal realization; the chosen pairs lie in three different diagonal planes.
