Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 2 / 2

For $H=\mathrm{diag}(r,s,-r,-s)$, put $L_1(H)=r$, $L_2(H)=s$. Then $L_3=-L_1$, $L_4=-L_2$. The matrix unit identityreads off the root spaces directly. The complete list isThus the C2 root system isAll eight root spaces have dimension one. The zero-weight space is the two-dimensional Cartan subalgebra $\mathbb{C}(E_{11}-E_{33})\oplus \mathbb{C}(E_{22}-E_{44})$; zero itself is not a root.

For the Adjoint representation, the root-space decomposition gives the eight nonzero weightseach of multiplicity one, and weight zero of multiplicity two from the Cartan subalgebra. Thus its weight diagram is the root diagram with a double zero weight, and its total dimension is $8+2=10$.

In the defining representation, the diagonal element of the Cartan subalgebra acts on the four standard basis vectors with eigenvalues $r,s,-r,-s$. ThereforeTheir weight diagram is the four-point cross below.

The tensor-product weight diagram is obtained by adding every ordered pair of defining weights. Each $\pm 2L_i$ occurs once, while each of the four weights $\pm L_1\pm L_2$ occurs twice, from the two orders of the summands. Zero occurs four times, from the ordered pairs $(L_i,-L_i)$ and $(-L_i,L_i)$ for $i=1,2$. HenceThe dimensions sum to $4+8+4=16$, as required for $V\otimes V$.

The printed phrase “not reducible” is a misprint: the requested nonzero complementary submodules establish that this tensor square is reducible. The natural reason is that the flip $\tau (v\otimes w)=w\otimes v$ commutes with the Lie algebra representation. Its two eigenspaces are the symmetric square and the exterior square, givingIn the symmetric square, the four weights $\pm 2L_i$ and four weights $\pm L_1\pm L_2$ each occur once, while zero occurs twice. In the exterior square, the four weights $\pm L_1\pm L_2$ each occur once and zero occurs twice; the $\pm 2L_i$ are absent because repeated wedge factors vanish. These are the two requested complementary weight diagrams.

More precisely, the tensor-square decomposition of the defining symplectic representation identifies ${\mathrm{Sym}}^{2}V$ with the irreducible adjoint module $L(2{\omega }_1)$. The identification is an intertwiner sending a symmetric tensor $vw$ to the element of the symplectic Lie algebra $x\mapsto \omega (v,x)w+\omega (w,x)v$. For the exterior square, symplectic contraction of an exterior square has a five-dimensional kernel, the primitive exterior square, and the invariant tensor $\Omega =e_1\wedge e_3+e_2\wedge e_4$ supplies a complementary line. Thus the full irreducible decomposition isHere ${\omega }_1=L_1$ and ${\omega }_2=L_1+L_2$. The two zero weights in the exterior square split into one primitive zero weight and the invariant line.