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The Dirac operator on the 2-sphere

The objective of this post is to explore MathJax, a JavaScript display engine for LaTeX. Being my first post writen with this tool, I want to present a short but fun example: I will give a description of the explicit computation of the spin-Dirac operator (of the unique complex spinor bundle!) on the 2-sphere \(S^2\) equipped with the standar round metric. A more detailed treatment can be found in my expository paper.


The Levi-Civita connection

Let us consider a 2-sphere of radius \(r>0\),

\[\begin{equation*} S^{2}(r):= \{(x_1,x_2,x_3) \: | \: x_1^2+x_2^2+x_3^2=r^{2}\}\subset\mathbb{R}^3 \end{equation*}\]

equipped with the induced Riemanian metric from \(\mathbb{R}^3\). With respect to a local parametrization given by polar coordinates

\[\begin{align*} x_1(r,\theta,\phi)=&\cos\phi\sin\theta,\\ x_2(r,\theta,\phi)=&\sin\phi\sin\theta,\\ x_3(r,\theta,\phi)=&\cos\theta, \end{align*}\]

where \(0<\theta<\pi\) and \(0<\phi<2\pi\), the metric can be written as

\[\begin{align*} g^{TS^2(r)}=r^2d\theta^2+r^2\sin^{2}\theta d\phi^2. \end{align*}\]

A straight forward computation shows that for the Levi-Civita connection we have

\[\begin{align*} \nabla_{\partial_\theta}\partial_{\theta}=&0,\\ \nabla_{\partial_\phi}\partial_\phi=&-\sin\theta\cos\theta\partial_\theta,\\ \nabla_{\partial_\theta}\partial_\phi=&\cot\theta\partial_\phi,\\ \nabla_{\partial_\phi}\partial_\theta=&\cot\theta\partial_\phi. \end{align*}\]

Indeed, the Christoffel symbols are

\[\begin{align*} \Gamma_{\phi\phi}^{\theta}=&\frac{1}{2r^2}(-\partial_\theta (r^2\sin^2\theta))=-\sin\theta\cos\theta,\\ \Gamma_{\theta\phi}^{\phi}=&\frac{1}{2r^2\sin^2\theta}(\partial_\theta (r^2\sin^2\theta))=\cot\theta. \end{align*}\]

The curvature form

Let us consider the following local orthonotmal basis for \(TS^2(r)\),

\[\begin{align*} e_1:=&\frac{\partial_\theta}{r},\\ e_2:=&\frac{\partial_\phi}{r\sin\theta}, \end{align*}\]

with associated dual basis

\[\begin{align*} e^1:=&rd\theta,\\ e^2:=&r\sin\theta d\theta. \end{align*}\]

With respect to this basis the volume form is \(vol_{S^2(r)}=e^1\wedge e^2\). To be precise we consider the orientation such that \[ vol_{\mathbb{R}^3}=rdr\wedge vol_{S^2(r)}. \] For further reference we compute the exterior derivative

\[\begin{align*} de^1=&0,\\ de^2=& d(r\sin\theta d\phi)=r\cos\theta d\theta\wedge d\phi=\frac{\cot\theta}{r} e^1\wedge e^2. \end{align*}\]

We now calculate the components \(\omega_{ij}\) of the connection \(1\)-form associated with this basis, which are defined by the relations

\[\begin{align*} \nabla e_j=:\omega_{ij}\otimes e_i, \end{align*}\]

where the sum over repeated indices is understood. From the expression of the Levi-Civita connection we verify

\[\begin{align*} \nabla_{e_1}e_1=&0,\\ \nabla_{e_1}e_2=&\partial_\theta\left(\frac{1}{r^2\sin\theta}\right)\partial_\phi+\left(\frac{1}{r^2\sin\theta}\right)\nabla_{\partial_\theta}\partial_\phi=-\frac{\cos\theta}{r^2}\partial_\phi+\frac{\cos\theta}{r^2}\partial_\phi =0,\\ \nabla_{e_2}e_1=&\left(\frac{1}{r^2\sin\theta}\right)\nabla_{\partial_\phi}\partial_\theta=\left(\frac{1}{r^2\sin\theta}\right)\cot\theta\partial_\phi=\frac{\cot\theta}{r}e_2,\\ \nabla_{e_2}e_2=&\left(\frac{1}{r\sin\theta}\right)^2\nabla_{\partial_\phi}\partial_\phi=-\left(\frac{1}{r\sin\theta}\right)^2\sin\theta\cos\theta\partial_\theta=-\frac{\cot\theta}{r} e_1.\\ \end{align*}\]


\[\begin{align*} \omega_{12}=-\omega_{21}=-\frac{\cot\theta}{r} e^2. \end{align*}\]

Since the Levi-Civita connection is compatible with metric then \(\omega_{ij}=-\omega_{ji}\). In addition, since it is also torsion-free then the components satisfy the structure equations

\[\begin{align*} de^i+\omega_{ij}\wedge e^j=0. \end{align*}\]

Note for example for \(i=2\),

\[\begin{align*} de^2+\omega_{21}\wedge e^1=\frac{\cot\theta}{r}e^1\wedge e^2+\frac{\cot\theta}{r}e^2\wedge e^1=0. \end{align*}\]

From the components of the connection 1-form we can compute the components \(\Omega_{ij}\) of the curvature using the relation

\[\begin{align*} \Omega_{ij}=d\omega_{ij}+\omega_{ik}\wedge\omega_{kj}. \end{align*}\]

In this particular case we have

\[\begin{align*} \Omega_{12}=&d\omega_{12}=\frac{\csc^2\theta}{r^2} e^1\wedge e^2-\frac{\cot\theta}{r} de^2=\frac{\csc^2\theta}{r^2} e^1\wedge e^2-\frac{\cot^2\theta}{r^2} e^1\wedge e^2=\frac{1}{r^2}e^1\wedge e^2. \end{align*}\]

The Gauß-Bonnet Theorem

If we integrate the 2-form \(\Omega_{12}/2\pi\) over \(S^2(r)\) we obtain

\[\begin{equation}\label{Eq:GB} \int_{S^{2}(r)}\frac{\Omega_{12}}{2\pi}=\frac{1}{2\pi r^2}\int_{S^{2}(r)}e^1\wedge e^2=\frac{4\pi r^2}{2\pi r^2}=2, \end{equation}\]

which verifies the Gauß-Bonnet theorem since the Euler characteristic \[\chi(S^{2}(r))=2\] for any \(r>0\).

Spin strucrure

A topological condition for the existence of spin structures is the vanishing of the second Stiefel-Whitney class \(w_2(TS^2)\in H^2(S^2;\mathbb{Z}/2\mathbb{Z})\). This characteristic class is the \(\mathbb{Z}/2\mathbb{Z}\)-reduction of the Euler class of \(TS^2\). By the Gauß-Bonnet theorem we know that the integral of this Euler class equals the Euler characteristic \(\chi(S^2)=2\), which modulo \(\mathbb{Z}/2\mathbb{Z}\) is zero. This shows that \(S^2\) is a spin manifold. Moreover, the spin structures are classified by the group \(H^1(S^2;\mathbb{Z}/2\mathbb{Z})=0\), so we conclude that \(S^2\) has only one spin structure. This result is actually valid for all spheres.

The spinor bundle

Now we construct the spinor bundle \(\Sigma(S^2)\) as an associated bundle \(\Sigma(S^2)=Spin(S^2)\times_{\rho_2}\Sigma_2\) where, \(Spin(S^2)\) the principal \(Spin\)-bundle correspinding to the unique spin structure, \(\rho_2:S^1\longrightarrow \text{Aut}(\Sigma_2)\) is the spin representation and \(\Sigma_2=\mathbb{C}^2\) is the spinor space.

To begin with we describe the spin representation of the Clifford algebra \(Cl(2)\) on the spinor vector space \(\Sigma_2=\mathbb{C}^{2}\). To to so its enough to describe the action on basis elements. Recall the definition of the Pauli matrices,

\[\begin{align*} \sigma_1:=\left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right),\quad\quad \sigma_2:=\left( \begin{array}{cc} 0 & -i\\ i & 0 \end{array} \right),\quad\quad \sigma_3:=\left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array} \right). \end{align*}\]

These matrices satisfy the relations

\[\begin{align*} \sigma_j^\dagger=&\sigma_j,\\ \sigma_j^2=&1,\\ \sigma_1\sigma_2=&i\sigma_3,\\ \sigma_j\sigma_k+\sigma_k\sigma_j=&2\delta_{jk}\quad\text{for $j=1,2,3$}. \end{align*}\]

Let \(\{v_1,v_2\}\) be the standard orthonormal basis of \(\mathbb{R}^2\), we define the Clifford action \(\rho_2(v_j):=-i\sigma_j\) for \(j=1,2\). It then follows

\[\begin{align*} \rho_2(v_j)\rho_2(v_k)+\rho_2(v_k)\rho_2(v_j)=-2\delta_{jk}. \end{align*}\]

We now want to study the restriction of this representation to \(Spin(2)\subset Cl(2)\). Note that every element of \(Spin(2)\) can be written as

\[\begin{align*} \cos t + (\sin t) v_1v_2=-(\sin(t/2)v_1+\cos(t/2)v_2)(\cos(t/2)v_1+\sin(t/2)v_2), \end{align*}\]

for \(t\in[0,2\pi]\), so

\[\begin{align*} \rho_2(\cos t+\sin t v_1v_2)=& \left( \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array} \right)\cos t +\left( \begin{array}{cc} -i & 0\\ 0 & i \end{array} \right)\sin t= \left( \begin{array}{cc} e^{-it} & 0\\ 0 & e^{it} \end{array} \right). \end{align*}\]

This shows that the spin representation restricted to \(Spin(2)=S^1\) is given by

\[\begin{align*} \rho_2(z) = \left( \begin{array}{cc} \bar{z} & 0\\ 0 & z \end{array} \right). \end{align*}\]

We now compute the transition functions of the spinor bundle \(\Sigma(S^2)=Spin(S^2)\times_{\rho_2}\Sigma_2\). These are obtained by composing the transition functions of the Hopf bundle \(\pi:S^3\subset\mathbb{C}^2\longrightarrow S^2\) (which defines the Spin structure on \(S^2\)) with \(\rho_2\), i.e. for \(\pi(z_0,z_1)\in U_N\cap U_S=S^1\) we have

\[\begin{equation*} \rho_2(\pi(z_0,z_1))=\rho_2\left(\frac{z_0}{z_1}\right)= \left( \begin{array}{cc} z_0/z_1 & 0\\ 0 & \bar{z}_0/\bar{z}_1 \end{array} \right). \end{equation*}\]

Here \(U_N\) and \(U_S\) denote the north and south hemisphere of \(S^2\) respectively.

We now sketch the proof to show that the spinor bundle \(\Sigma(S^2)\) is trivial. The argument goes into the direction of clutching functions in K-theory. Let \(\text{Vect}_\mathbb{C}^k(S^2)\) denote the monoid of isomorphims classes of complex vector bundles of rank \(k\) over \(S^2\). An important result in the context of classification of vector bundles states that the map \(\Phi:[S^1,GL(k,\mathbb{C})]\longrightarrow \text{Vect}_\mathbb{C}^k(S^2)\) defined by the transition functions (clutching functions) is a bijection. Here \([S^1,GL(k,\mathbb{C})]\) denotes the space of maps up to homotopy. Moreover, as groups we have an isomorphism \([S^1,GL(k,\mathbb{C})]\cong [S^1,U(k,\mathbb{C})]\cong\mathbb{Z}\), the first one given by the Gram-Schmidt orthogonalization process and the second one because of the known computation of the fundamental group for the unitary group. One also verifies that the determinant \(det:[S^1,U(k,\mathbb{C})]\longrightarrow S^1\) is well defined and the degree of it \(deg\circ det:[S^1,U(k,\mathbb{C})] \longrightarrow\mathbb{Z}\) is actually an isomorphism. This means that the degree of the determinant of the clutching functions characterizes the isomorphism class of the associated vector bundle. This argument shows, in view of

\[\begin{align*} \det\left( \begin{array}{cc} \bar{z} & 0\\ 0 & z \end{array} \right)=\bar{z}z=1, \end{align*}\]

for \(z\in S^1\), that the associated bundle of this clutching function must be isomorphic to the trivial bundle. One can actually define an explicit homotopy for \(t\in[0,\pi/2]\),

\[\begin{align*} \gamma(t):=\left( \begin{array}{cc} \bar{z} & 0\\ 0 & 1 \end{array} \right) \left( \begin{array}{cc} \cos t & \sin t\\ -\sin t & \cos t \end{array} \right) \left( \begin{array}{cc} 1 & 0\\ 0 & z \end{array} \right) \left( \begin{array}{cc} \cos t & -\sin t\\ \sin t & \cos t \end{array} \right). \end{align*}\]

Observe that \(det(\gamma(t))=1\) for all $\(t\in[0,\pi/2]\) and

\[\begin{align*} \gamma(0)= \left( \begin{array}{cc} \bar{z} & 0\\ 0 & z \end{array} \right) \quad\text{and}\quad \gamma(\pi/2)=\left( \begin{array}{cc} 1& 0\\ 0 & 1 \end{array} \right). \end{align*}\]

As a consequence, the sections of \(\Sigma(S^2)\), called spinors, can be regarded as functions \(\psi:S^2\longrightarrow \mathbb{C}^2\).

The Dirac operator

We now want to construct the Dirac operator \({D}_{S^2(r)}\) associated to the Clifford bundle \(\Sigma(S^2)\). This operator is defined by the formula

\[\begin{align*} {D}_{S^2(r)}:=\rho_2(e_1)\nabla^\Sigma_{e_{1}}+\rho_2(e_2)\nabla^\Sigma_{e_{2}}, \end{align*}\]

where \(\{e_1,e_2\}\) is the orthonormal basis defined above and \(\nabla^\Sigma\) is the spin connection induced by the Levi-Civita connection and the spin representation. This connection can be computed using the component of the connection \(1\)-form as

\[\begin{align*} \nabla^{\Sigma}=\frac{1}{2}\omega_{21}\otimes\rho_2(e_1)\rho_2(e_2)=-\frac{1}{2}\omega_{12}\otimes\sigma_2\sigma_1=-\frac{i}{2}\frac{\cot\theta}{r}e^2\otimes\sigma_3. \end{align*}\]

Applying the Clifford action we get

\[\begin{align*} -i\sigma_1\nabla^\Sigma_{e_1}=&-i\sigma_1e_1,\\ -i\sigma_2\nabla^\Sigma_{e_2}=&-i\sigma_2e_2-\frac{1}{2}\frac{\cot\theta}{r}\sigma_2\sigma_3=-i\sigma_2e_2-\frac{i}{2}\frac{\cot\theta}{r}\sigma_1, \end{align*}\]

so we obtain

\[\begin{align*} {D}_{S^2(r)}=-i\sigma_1\nabla^\Sigma_{e_1}-i\sigma_2\nabla^\Sigma_{e_2}=-i\sigma_1\left(e_1+\frac{1}{2}\frac{\cot\theta}{r}\right)-i\sigma_2e_2. \end{align*}\]

Remark: Note that as \(r\longrightarrow +\infty\), the Dirac operator becomes \(-i\sigma_1 e_1 - i\sigma_2 e_2\), which is the spin-Dirac operator on \(\mathbb{R}^2\).

In terms of the coordinate vector fields we can write

\[\begin{align*} {D}_{S^2(r)}=-i\sigma_1\left(\frac{1}{r}\partial_\theta+\frac{1}{2}\frac{\cot\theta}{r}\right)-\frac{i\sigma_2}{r\sin\theta}\partial_\phi. \end{align*}\]

Lichnerowicz formula

Let \(\Delta^{\Sigma} := (\nabla^{\Sigma})^{*}\nabla^{\Sigma}\) be the spinor Laplacian, then the Lichnerowicz formula takes the form

\[ D^2_{S^2(r)} = \Delta^{\Sigma} + \frac{scal}{4} = \Delta^{\Sigma} + \frac{1}{2r^2}, \]

where \(scal\) is the the scalar curvature and equals \(scal=2/r^2\) for the sphere.

The spectrum

The spin-Dirac operator is a first order, self-adjoint elliptic operator, which implies (as \(S^2\) is compact) that it has a discrete spectrum. The eigenvalues of \(D_{S^2}\) (for \(r=1\)) are given by \(\pm(k+1)\), for \(k\geq 0\), with multiplicities

\[\begin{equation*} 2\left( \begin{array}{c} k+1\\ k \end{array} \right). \end{equation*}\]

This can be deduced from the work of Christian Bär.

The Fredholm index

The chirality operator of associated to the spinor bundle \(\Sigma(S^2)\) is

\[\begin{align*} \Gamma=i(-i\sigma_1)(-i\sigma_2)=\sigma_3. \end{align*}\]

Since the dimension of \(S^2\) is even we can verify that \({D}_{S^2(r)}\sigma_3+\sigma_3{D}_{S^2(r)}=0\). With respect to \(\Gamma\) we can therefore decompose

\[\begin{equation*} D_{S^2(r)} = \left( \begin{array}{cc} 0 & D_{S^2(r)}^{-}\\ D_{S^2(r)}^{+} & 0 \end{array} \right). \end{equation*}\]

Finally, by the Atiyah-Singer Index Theorem, we have \(\text{ind}(D_{S^2(r)}^{+})=0\) as the \(\widehat{A}\)-polynomial is a polynomial in the Pontryagin classes which are of degree \(4j\), for \(j\in\mathbb{N}_0\).