Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 18 / 4

Let $v_{1,+}=e_1+i{e}_2$, $v_{1,-}=e_1-i{e}_2$, $v_{2,+}=e_3+i{e}_4$, $v_{2,-}=e_3-i{e}_4$ and $v_0=e_5$. Direct multiplication by the printed rotation blocks gives their torus eigenvaluesThus the differentiated weights of a representation are $i{x}_1,-i{x}_1,i{x}_2,-i{x}_2,0$, with this sign convention fixed by the given matrix blocks.

A wedge of two distinct eigenvectors has the product eigenvalue, so its differentiated weight is the sum. The ten vectors in a basis of the exterior square split into three groups:The eight nonzero weights are distinct, so each has multiplicity one; the zero weight space has dimension two. The multiplicities add to $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, accounting for the entire representation.

For a real inner product space $V={\mathbb{R}}^5$, defineThis map takes values in the Special orthogonal Lie algebra: its matrices are skew-symmetric. It sends $e_i\wedge e_j$ to $E_{ij}-E_{ji}$, so it is an isomorphism between the two ten-dimensional spaces. For $g\in SO(5)$, invariance of the inner product givesThis proves the exterior square realization of the orthogonal adjoint representation, including equivariance, rather than merely matching dimensions.

After complexification, part (a) is therefore the weight decomposition of the Adjoint representation. Its zero weight space is the complex centralizer of $\mathfrak{t}$. The two zero-weight wedges are nonzero multiples of $e_1\wedge e_2$ and $e_3\wedge e_4$, which map under $\Phi$ to the two infinitesimal rotation generators of $T$. HenceIf a larger torus contained $T$, its Lie algebra would commute with $\mathfrak{t}$ and therefore lie in $\mathfrak{t}$. Equality of Lie algebras and connectedness of both tori force equality of the groups. Thus $T$ is a maximal torus.

Let ${\epsilon }_j$ denote the real coordinate functional $x_j$. In the real weight convention, with the common factor $i$ removed from the differentiated torus weights, the root system isThese are the roots of a root system of type B2 root system. In the imaginary-valued compact Lie-algebra convention each displayed functional is multiplied by $i$. Every root space is one-dimensional, as part (a) explicitly shows.

The root hyperplanes are $x_1=0$, $x_2=0$, $x_1=x_2$ and $x_1=-x_2$. Their reflections are coordinate sign changes, coordinate interchange and a signed interchange. They generate all signed permutations of the two coordinates:Geometrically this is the dihedral symmetry group of a square, with order eight. It preserves both the four short roots and the four long roots of the B2 root system.

Representatives in $SO(5)$ can be given explicitly. The matrixhas determinant one and conjugates the first rotation block to its inverse, giving $(x_1,x_2)\mapsto (-x_1,x_2)$. The matrix $D_2=\mathrm{diag}(1,1,1,-1,-1)$ similarly reverses $x_2$. Flipping the fifth coordinate compensates for the determinant of the planar reflection.

The permutation matrixswaps the two coordinate planes. Its permutation is $(13)(24)$, so its determinant is also one. Conjugation gives $(x_1,x_2)\mapsto (x_2,x_1)$, the reflection in $x_1=x_2$. These elements normalize $T$, and $P,D_1$ generate all eight Weyl actions. For example, $P{D}_1$ induces $(x_1,x_2)\mapsto (x_2,-x_1)$, a quarter-turn, and $D_1{D}_2$ induces the half-turn. The representatives supply the specified reflections and their typical products inside the actual special orthogonal group.

Choose positive rootsThe corresponding simple roots are ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_2$. The vector $v_{1,+}\wedge v_{2,+}$ has weight $\lambda ={\epsilon }_1+{\epsilon }_2$. Adding any positive root to this weight produces a weight absent from part (a), so every positive root vector annihilates it. It is consequently a highest-weight vector with highest weight $\lambda$.

A representation of a compact Lie group is completely reducible: Haar averaging makes an inner product invariant, and orthogonal complements of invariant subspaces remain invariant. Therefore the exterior square contains an irreducible summand $L_{\lambda }$ of this highest weight. This can also be seen by projecting the displayed highest-weight vector into irreducible summands; a nonzero projection remains annihilated by all positive root vectors and has the same weight. Its multiplicity is one because the weight space at $\lambda$ is one-dimensional.

Use the Weyl dimension formula, expressly allowed in this part:Here $L_{\lambda }$ is the irreducible highest-weight representation, ${\Phi }^{+}$ is the chosen positive root system, $\rho$ is half its sum, and the pairing is the invariant Euclidean inner product on the real weight plane, with ${\epsilon }_1,{\epsilon }_2$ orthonormal. Using coroots instead gives the same ratio because the length factor cancels for each root. In this convention,The four factors, in the listed positive-root order, are respectivelyTheir product is $10$. Since $\dim {\Lambda }^2{\mathbb{C}}^5=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, this irreducible summand occupies the entire exterior square. HenceThe proof uses the dimension formula only where it is allowed, not in the symmetric-power argument of Question 2. Its B2 adjoint representation weight diagram is exactly the eight roots and the double zero weight obtained above.