Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 4 / 2 / iv / Solution

Label the weight vertices by $1,\dots ,n,0,\bar{n},\dots ,\bar{1}$, with weights $e_1,\dots ,e_n,0,-e_n,\dots ,-e_1$. The crystal of the defining odd-orthogonal representation is the chainEach arrow of color $i$ subtracts the simple root ${\alpha }_i$. The two arrows through zero form the length-two short-root string.

Here is a complete grid description of the tensor product of crystals, valid for arbitrary $n$. Place the vertex $a\otimes b$ in row $a$, column $b$, in the above chain order; there are $(2n+1{)}^2$ vertices. Let ${\epsilon }_i(a)$ and ${\phi }_i(a)$ count incoming and outgoing steps along the color-$i$ string. For $i<n$, the two strings are $i\to i+1$ and $\overline{i+1}\to \bar{i}$, each of length one. For $i=n$, the string is $n\to 0\to \bar{n}$, with pairs $({\epsilon }_n,{\phi }_n)$ equal to $(0,2),(1,1),(2,0)$. Other vertices have both counts zero.

Using the crystal tensor-product rule, draw every color-$i$ edge byThe first case is a downward arrow to the next row and the second a rightward arrow to the next column; omit an arrow when its indicated factor has no outgoing edge. Thus the rule specifies every vertex and every arrow of the general tensor-square diagram, including the boundary nodes. The figure displays the resulting entire grid for $n=3$, alongside the vector crystal; the generator can draw any positive rank.

The raising rule chooses the first factor when ${\phi }_i(a)\ge {\epsilon }_i(b)$ and the second otherwise. For $n\ge 2$, its only highest vertices areIndeed a highest tensor must have highest first factor, hence $a=1$. For colors $i>1$, the second factor cannot have an incoming edge. For color one, its incoming count must be at most ${\phi }_1(1)=1$. Inspecting the strings leaves precisely $b=1,2,\bar{1}$.

These three connected components encode the Irreducible Lie algebra representationsHere $S_0^{2}V$ is the traceless symmetric square, ${\Lambda }^{2}V$ is the Adjoint representation, and the invariant symmetric form supplies the trivial summand. Their dimensions are $n(2n+3)$, $n(2n+1)$, and one, summing to $(2n+1{)}^2$. In fundamental-weight notation the highest weights are $2{\omega }_1,{\omega }_2,0$ for $n\ge 3$, but $2{\omega }_1,2{\omega }_2,0$ for $n=2$.

For $n=1$ the vector crystal is $1\stackrel{1}{\to }0\stackrel{1}{\to }\bar{1}$. The same tensor rule gives highest vertices $1\otimes 1,1\otimes 0,1\otimes \bar{1}$, of weights $2e_1,e_1,0$, or $4{\omega }_1,2{\omega }_1,0$, respectively. Its summand dimensions are $5,3,1$.