Clifford Bundles, Dirac Operators, and Index Interfaces: Examples and Counterexamples

Under the unitary Fourier transform [F2], $D$ is multiplication by $2\pi ic(\xi )$. By [F1], its maximal multiplier domain is exactly $\{u\in L^2:\mathrm{\mid }\xi \mathrm{\mid }\widehat{u}\in L^2\}$. If $u\in H^1$, [F3] and the distributional derivative identity in [F2] put $\mathrm{\mid }\xi \mathrm{\mid }\widehat{u}$ in $L^2$. Conversely, if $\mathrm{\mid }\xi \mathrm{\mid }\widehat{u}\in L^2$, then each $2\pi i{\xi }_j\widehat{u}$ is in $L^2$; its inverse Plancherel transform is the distributional derivative ${\mathrm{\partial }}_{j}u$, by the Fourier differentiation identity and injectivity of the Fourier transform on tempered distributions. Hence $u\in H^1$ by [F3]. A multiplication operator by a measurable finite matrix is closed, since convergence of $u_k$ and $c(\xi )u_k$ in $L^2$ has an almost-everywhere convergent subsequence and the pointwise multiplication identity passes to the limit. Schwartz sections are dense in $L^2$, so $D$ is densely defined and closed.

If $h$ is orthogonal to the range, then distributionally $D^{\ast }h=0$. Fourier transformation and [F2] give $c(\xi )\widehat{h}(\xi )=0$ almost everywhere. By [F1], $c(\xi )$ is invertible for every $\xi \mathrm{\ne }0$, and the singleton $\{0\}$ has Lebesgue measure zero; hence $\widehat{h}=0$ almost everywhere and $h=0$. Thus the range is dense.

Fix a unit vector $v\in S_0$ and define $\widehat{g}(\xi )={\mathbf{1}}_{\{0<\mathrm{\mid }\xi \mathrm{\mid }<1\}}\mathrm{\mid }\xi {\mathrm{\mid }}^{-(n\mathrm{/}2-1)}v$. Polar coordinates give $\mathrm{\parallel }\widehat{g}{\mathrm{\parallel }}_2^2=C_n{\int }_0^1{r}^{n-1-2(n\mathrm{/}2-1)}dr=C_n{\int }_0^{1}rdr<\mathrm{\infty }$, so [F2] gives $g\in L^2$. If $g=Du$ for some $u\in H^1$, then for almost every $\xi \mathrm{\ne }0$, [F1] forces $\mathrm{\mid }\widehat{u}(\xi )\mathrm{\mid }=(2\pi {)}^{-1}\mathrm{\mid }\xi {\mathrm{\mid }}^{-1}\mathrm{\mid }\widehat{g}(\xi )\mathrm{\mid }$. Consequently its squared norm near zero is a positive constant times ${\int }_0^1{r}^{n-1-2(n\mathrm{/}2)}dr={\int }_0^1{r}^{-1}dr=\mathrm{\infty }$, contradicting $u\in L^2$. Thus $g$ is not in the range, so the dense range is proper and not closed. A Fredholm operator must have closed range, hence $D$ is not Fredholm.

The construction works also for $n=1$, where $\widehat{g}(\xi )={\mathbf{1}}_{\{0<\mathrm{\mid }\xi \mathrm{\mid }<1\}}\mathrm{\mid }\xi {\mathrm{\mid }}^{1\mathrm{/}2}v$ and the displayed $L^2$ integral is still finite; the logarithmic divergence after division by $\mathrm{\mid }\xi \mathrm{\mid }$ is unchanged. The zero-frequency set has measure zero, but arbitrarily small nonzero frequencies cause the obstruction. Countable choice is used through the Fourier and Sobolev interfaces [F2]–[F3] and the almost-everywhere subsequence argument in step 1.1.

By [F1] with $n=2$, $c_1(T{\mathbb{C}\mathbb{P}}^2)=3x$. The coefficient reduction ${\rho }_2$ is a ring homomorphism, so ${\rho }_2(3x)={\rho }_2(x)$ because $3\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}2)$.

The CW structure in [F4] has cellular chains $C_4\cong \mathbb{Z}$, $C_2\cong \mathbb{Z}$, and $C_0\cong \mathbb{Z}$, with all other groups zero. Every cellular boundary is zero because its target is zero, so cellular and singular homology give $H_2({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})\cong \mathbb{Z}$ and $H_1=0$. By UCT with integral coefficients, evaluation identifies $H^2({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})$ with $\mathrm{Hom}(\mathbb{Z},\mathbb{Z})$; since $x$ is an integral cohomology generator by [F2], it evaluates as $+1$ or $-1$ on a homology generator. UCT with ${\mathbb{F}}_2$ coefficients identifies $H^2({\mathbb{C}\mathbb{P}}^2;{\mathbb{F}}_2)$ with $\mathrm{Hom}(\mathbb{Z},{\mathbb{F}}_2)$, and naturality of coefficient reduction makes ${\rho }_2(x)$ evaluate as $1$. Hence ${\rho }_2(x)\mathrm{\ne }0$.

Applying [F3] to the complex tangent bundle and using steps 1.1–1.2 gives $w_2(T{\mathbb{C}\mathbb{P}}^2)={\rho }_2(x)\mathrm{\ne }0$. The manifold is complex-oriented, hence its tangent bundle is oriented; [F5] therefore implies that it admits no Spin lift. The coefficient group has characteristic two and no sign convention changes the nonzero class.

First classify the characters of $\mathbb{T}=\mathbb{R}\mathrm{/}\mathbb{Z}$. If $\chi :\mathbb{T}\to S^1$ is a continuous character, let $f(t)=\chi ([t])$. By continuity at $0$, choose $\delta >0$ so that on $\mathrm{\mid }t\mathrm{\mid }<2\delta$ the values of $f$ lie in the arc with a continuous argument $\theta (t)\in (-1\mathrm{/}8,1\mathrm{/}8)$, normalized by $f(t)=e^{2\pi i\theta (t)}$. Whenever $\mathrm{\mid }s\mathrm{\mid },\mathrm{\mid }t\mathrm{\mid },\mathrm{\mid }s+t\mathrm{\mid }<\delta$, the character identity gives $\theta (s+t)-\theta (s)-\theta (t)\in \mathbb{Z}$; this difference has absolute value less than $3\mathrm{/}8$, so it is zero. Define $\mathrm{\Theta }(u)=N\theta (u\mathrm{/}N)$ for any positive integer $N$ with $\mathrm{\mid }u\mathrm{/}N\mathrm{\mid }<\delta$. If $N,N^{\mathrm{\prime }}$ both work, repeated local additivity gives $N\theta (u\mathrm{/}N)=N{N}^{\mathrm{\prime }}\theta (u\mathrm{/}(N{N}^{\mathrm{\prime }}))=N^{\mathrm{\prime }}\theta (u\mathrm{/}{N}^{\mathrm{\prime }})$, so $\mathrm{\Theta }$ is well-defined. Choosing $N$ large enough for $s,t,s+t$ shows $\mathrm{\Theta }(s+t)=\mathrm{\Theta }(s)+\mathrm{\Theta }(t)$. Near zero $\mathrm{\Theta }=\theta$, so $\mathrm{\Theta }$ is continuous, and the continuous additive equation gives $\mathrm{\Theta }(t)=\alpha t$ for some $\alpha \in \mathbb{R}$. For arbitrary $t$, choose $N$ with $\mathrm{\mid }t\mathrm{/}N\mathrm{\mid }<\delta$; then $f(t)=f(t\mathrm{/}N{)}^N=e^{2\pi i\alpha t}$. Since $f(1)=1$, $\alpha \in \mathbb{Z}$. Conversely each $k\in \mathbb{Z}$ defines ${\chi }_k([t])=e^{2\pi ikt}$, so $\widehat{\mathbb{T}}=\mathbb{Z}$. Finite-product duality now identifies the frequencies of ${\mathbb{T}}^{2m}$ with ${\mathbb{Z}}^{2m}$, and compact-group Plancherel makes these characters a complete orthonormal basis.

The quotient ${\mathbb{R}}^{2m}\mathrm{/}{\mathbb{Z}}^{2m}$ has a global parallel coframe, so the Levi–Civita connection and the periodic spin connection have zero local connection matrix. For each $k\in {\mathbb{Z}}^{2m}$, the Fourier character $e_k(x)=e^{2\pi ik\cdot x}$ satisfies $D(e_k\otimes s)=2\pi i{e}_k\otimes c(k)s$. For $k\mathrm{\ne }0$, $c(k{)}^2=-\mathrm{\mid }k{\mathrm{\mid }}^{2}I$, so $c(k)$ is invertible.

By the completeness in [F3], every $L^2$ spinor is determined by its Fourier coefficients. For $\psi \in H^1$, testing each weak derivative against a Fourier character and integrating over a fundamental cube gives $\widehat{D\psi }(k)=2\pi ic(k)\widehat{\psi }(k)$; the opposite boundary faces cancel by periodicity. Thus $D\psi =0$ forces every nonzero-frequency coefficient to vanish by step 1.2, since $c(k)$ is invertible there; the zero-frequency coefficient is arbitrary. The same argument on the two graded summands shows that $\ker D^{+}$ consists exactly of constant elements of $S^{+}$ and $\ker (D^{+}{)}^{\ast }=\ker D^{-}$ consists exactly of constants in $S^{-}$.

By [F2], ${\dim }_{\mathbb{C}}{S}^{+}={\dim }_{\mathbb{C}}{S}^{-}=2^{m-1}$. Hence the Fredholm index supplied by [F4] is $\dim \ker D^{+}-\dim \ker (D^{+}{)}^{\ast }=2^{m-1}-2^{m-1}=0$. For $m=1$ both dimensions are one; the zero-dimensional torus is outside the stated range $m\ge 1$. The half-spin dimensions use the exact algebra suppliers in [A1].

By [F1]–[F2], $w_2(T{S}^2)={\rho }_2(2x)=0$; the complex orientation orients its tangent bundle. The spin-structure criterion in [F2] supplies a Spin lift.

The sectional curvature is $1$ by [F3]. In an orthonormal frame on a surface, scalar curvature is $R_{1221}+R_{2112}=2$, with the sign convention of [F3]. It is strictly positive everywhere.

Apply [F4] to the Spin structure from step 1.1 and the positive scalar curvature from step 1.2. Every smooth untwisted harmonic spinor is zero. The assertion is for the untwisted operator; a line twist has its own curvature term and is treated separately.

By [F1], $T{S}^2$ admits a Spin lift; fix one. The unit sphere has area $4\pi$, so $\omega =(4\pi {)}^{-1}\sin \theta d\theta \wedge d\varphi$. On north and south charts define unitary connection forms $A_N=-\frac{ik}{2}(1-\cos \theta )d\varphi$ and $A_S=\frac{ik}{2}(1+\cos \theta )d\varphi$. They extend smoothly over their respective poles, and $A_S-A_N=ikd\varphi =g^{-1}dg$ for the unitary transition $g(\varphi )=e^{ik\varphi }$. Hence they define a Hermitian line bundle connection; differentiation gives $F=d{A}_N=d{A}_S=-(ik\mathrm{/}2)\sin \theta d\theta \wedge d\varphi =-2\pi ik\omega$.

By [F3]–[F5], the real Chern class of this connection integrates to $k$, so its integral first Chern class is $c_1(L_k)=kx$; the integral group is torsion-free by [F5], so its image in real cohomology determines it. The line-bundle curvature term in [F4] is $c(e^1)c(e^2)F_{12}=c(e^1)c(e^2)(-ik\mathrm{/}2)=-\frac{k}{2}\mathrm{\Gamma }$, because $\mathrm{\Gamma }=ic(e^1)c(e^2)$. This is nonzero when $k\mathrm{\ne }0$.

In dimension two $\widehat{A}$ has only its degree-zero term $1$, while $\mathrm{tr}\exp (-F\mathrm{/}(2\pi i))=1+k\omega$ because ${\omega }^2=0$. Applying [F4] yields ${\mathrm{ind}}_{\mathbb{C}}(D_{S\otimes L_k}^{+})={\int }_{S^2}k\omega =k$. For $k=0$ both the curvature term and index vanish. The construction is smooth at both poles by the displayed vanishing factors, the orientation is the complex orientation, and the formula uses the same actual spin-module interface as [F4].