Branching from B3 to A3 2026-10-25
For B3 root system simple roots , choose the D3 root system simple roots in chain order . The Dynkin labels project by . This separates the vector weights into the six-dimensional exterior square of the defining SU(4) representation and a singlet representation. The spinor cube vertices split into its defining representation and its dual representation.
Past exam of the groups and representations course of the University of Oxford 2018 4 a Solution 2026-10-25
Choose an orthonormal basis in the real Euclidean space containing the root system, and take simple rootsTheir squared lengths are . Use the Cartan matrix convention , with the denominator indexed by the column. Direct inner products giveThe Dynkin diagram of the B3 root system is a three-node chain with a single bond from node 1 to node 2 and a double bond from node 2 to the short node 3; the arrow points toward node 3. The A3 root system has a chain of three nodes joined by single bonds. A convention putting the row length in the denominator transposes the first displayed Cartan matrix.
For the Extended Dynkin diagram, the highest root is . Add . Its only nonzero off-diagonal inner product among these simple roots is . Deleting the short node 3 leaves the chain , with three equal-length roots, which is .
More concretely, the long roots form the D3 root system, which is isomorphic to the A3 root system. The corresponding root spaces and Cartan subalgebra generatefor the compact real forms. At the matrix level the inclusion fixes one coordinate in seven-dimensional space. This establishes the required Lie subalgebra inclusion; the associated global groups require the distinction explained in d.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 102 3 a Solution Created 2026-10-03 Updated 2026-10-05
Write an element of the diagonal Cartan subalgebra asThe root system of this Special orthogonal Lie algebra isThus the maps requested are , with the two signs independent. This is the D3 root system, with twelve roots.
For a direct check, the defining matrix identity gives the block form with and . If denotes a matrix unit, the root space for is spanned by ; those for and are spanned respectively by and . Commuting these with gives exactly the displayed maps.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 102 3 b i Solution Created 2026-09-24 Updated 2026-09-25
