Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 / 6

The irreducible finite-dimensional unitary representations of $SU(2)$ are $E_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$, one for each $n\ge 0$, with dimension $n+1$. Equivalently they are spin-$j$ representations with $j=n/2$. The defining representation has $n=1$, and $n=0$ is trivial. Every finite-dimensional unitary representation is an orthogonal direct sum of these.

Choose self-adjoint angular-momentum generators $J_x,J_y,J_z$ with $[J_x,J_y]=i{J}_z$ cyclically. The actual SU(2) Lie algebra generators are $L_a=-i{J}_a$, so $[L_x,L_y]=L_z$. On an orthonormal weight basis $v_m$, $m=-j,-j+1,\dots ,j$, the complexified action isThe Casimir operator in this normalization is $C=-{\sum }_a{L}_a^2={\sum }_a{J}_a^2$, acting byOther Lie-algebra metric normalizations rescale this scalar. On the diagonal subgroup $\mathrm{diag}(e^{it},e^{-it})$, the weights of $E_n$ are $n,n-2,\dots ,-n$, all with multiplicity one.

The kernel and cokernel of an equivariant Fredholm operator are finite-dimensional representations. Define its equivariant Fredholm index in the representation ring of a compact group byFor $G=SU(2)$ write $H={\widehat{⨁}}_n{E}_n\otimes {\mathbb{C}}^{m_n}$ and $H^{\prime }={\widehat{⨁}}_n{E}_n\otimes {\mathbb{C}}^{m_n^{\prime }}$, with finite multiplicities. By Schur lemma an intertwiner has blocks $I_{E_n}\otimes s_n$, where $s_n:{\mathbb{C}}^{m_n}\to {\mathbb{C}}^{m_n^{\prime }}$. If its rank is $r_n$, the kernel and cokernel multiplicities are $m_n-r_n$ and $m_n^{\prime }-r_n$. ConsequentlyOnly finitely many coefficients can be nonzero, because the global kernel and cokernel are finite-dimensional. Closed range is part of the Fredholm hypothesis and supplies the uniform lower bound on the remaining blocks; multiplicity matching alone would not establish that hypothesis. The ordinary index is obtained by taking dimensions, $\mathrm{ind}S={\sum }_n(n+1)(m_n-m_n^{\prime })$. This calculation applies to any compact group with its finite-dimensional irreducibles in place of the $E_n$.

Use the round unit sphere with its complex orientation and unique spin structure. Its chiral spinor bundles are $S^{+}=\mathcal{O}(-1)$ and $S^{-}={\Lambda }^{0,1}\otimes \mathcal{O}(-1)$, the latter a smooth line bundle of degree $+1$. The Dirac operator isHere Clifford multiplication has $c(v{)}^2=-|v{|}^2$, and $D$ acts self-adjointly on the L2 spinors with first-order Sobolev domain. The geometric spin connection is invariant under the lifted $SU(2)$ action, hence $D$ is equivariant.

We justify the spectral properties needed for its phase. The Peter-Weyl theorem and Frobenius reciprocity for compact groups give the decomposition of sections of an isotropy-weight-$q$ line bundle: $E_n$ occurs once exactly when $n\ge |q|$ and $n\equiv q(\mathrm{mod}2)$, because its torus weights are $n,n-2,\dots ,-n$. The spinor isotropy weights are $+1,-1$, so each chirality contains $E_1,E_3,E_5,\dots$, each once. The horizontal connection Laplacian is the total Casimir minus the vertical generator square, ${\nabla }^{\ast }\nabla =C-1/4$ on these weight-one bundles. Expanding the Dirac square in a normal orthonormal frame gives the diagonal connection-Laplacian terms and the off-diagonal curvature contraction. The Clifford relations contract spin curvature to $\mathrm{Scal}/4$, the Lichnerowicz formula for a twisted Dirac operator with no twisting bundle. Since the scalar curvature here is two,Thus there is no kernel, the eigenvalues of $D$ are $\pm 1,\pm 2,\dots$ with finite multiplicities, and the resolvent is compact by spectral truncation. These round-sphere spectral facts agree with the explicit calculation in Abrikosov's paper.

Define the chiral phase of the round-sphere Dirac operatorIt is unitary because both chiral kernels vanish; equivariance follows from that of $D$. The full phase is $F=D|D{|}^{-1}=\left(\begin{array}{cc}0& V^{\ast }\\ V& 0\end{array}\right)$.

For a smooth scalar $f$, the Leibniz rule gives $[D,M_f]=c(df)$, a bounded bundle endomorphism. The spectral integral for the sign operator gives, in the strong topology,Write $R_{\pm }=(D\mp it{)}^{-1}$. Since $2D(D^2+t^2{)}^{-1}=R_{+}+R_{-}$ and $[R_{\pm },M_f]=-R_{\pm }[D,M_f]R_{\pm }$, the commutator integral is norm convergent: its integrands are compact and bounded by $\Vert [D,M_f]\Vert /(1+t^2)$. A norm limit of their integrals is compact. Thus $[F,M_f]$ is compact, and its off-diagonal blocks say $V{M}_f^{+}-M_f^{-}V$ is compact. Uniform approximation of continuous $f$ by smooth functions extends this to all $f\in C(S^2)$, using the bound $2\Vert f-g{\Vert }_{\infty }$ for the change of commutator.

The chiral phase is an equivariant unitary which intertwines the two multiplication representations modulo compacts. This specifies the grading implicit in the subsequent index question; treating the full self-adjoint phase as an endomorphism would yield a different, zero, compressed index.

Amplify the chiral phase by the identity on the auxiliary ${\mathbb{C}}^2$, and let $P$ act by multiplication on each chirality. The displayed projection has image line $E_z=\mathbb{C}(1,\overline{z})$. On the chart $w=1/z$ near infinity it has frame $(\overline{w},1)$. On the equator the normalized second frame is $z$ times the first, so its clutching function has winding $+1$. Equivalently it is the conjugate of the tautological degree-$-1$ line. Thus $E$ has degree $+1$ with the complex orientation fixed in part (c).

It is equivariant for the conjugate defining action on the auxiliary ${\mathbb{C}}^2$; this representation is equivalent to the defining representation of $SU(2)$. At $z=0$ its isotropy character under $\mathrm{diag}(e^{it},e^{-it})$ is $e^{-it}$. Tensoring with the positive spinor weight $+1$ gives weight zero on $S^{+}\otimes E$, whereas tensoring with the negative weight $-1$ gives weight $-2$ on $S^{-}\otimes E$.

The compressed map isIts parametrix is $P{V}^{\ast }P$: the two products equal the respective identity $P$ modulo compacts, since every matrix entry of $P$ is a continuous function and the phase commutators in part (c) are compact. Hence this is an equivariant Fredholm operator.

Use the precisely stated homogeneous-bundle multiplicity rule from part (c). The domain, of weight zero, decomposes as $E_0\oplus E_2\oplus E_4\oplus \cdots$. The target, of weight $-2$, decomposes as $E_2\oplus E_4\oplus \cdots$. Part (b) subtracts these multiplicities, leaving only the trivial representation. ThereforeThis computes the index from the actual projection and representation weights without invoking an index theorem that merely supplies the requested number. Reversing the chiral direction gives $-1$. If $V$ instead denotes the full self-adjoint phase on $H_{+}\oplus H_{-}$, its compression is self-adjoint Fredholm and has index zero; the nonzero answer uses the standard chiral interpretation made explicit above.