Fourier–Hardy index and the sign of the cotangent plane symbol

Under [A1], all calculus and index constructions below use the declared AC hypothesis. The exponentials are an orthonormal complete system. Completeness can be checked without an additional spectral theorem: convolution of a continuous periodic function with the Fejér kernel is a trigonometric polynomial and converges uniformly, since the nonnegative kernel has integral one and its mass outside each neighborhood of zero tends to zero; continuous functions are dense in $L^2$ by approximation of interval indicators and simple functions. Thus an $L^2$ function orthogonal to every exponential is zero. Consequently $P_{+}$ is an orthogonal projection of norm at most one, including when $V=0$. On Fourier basis vectors, $A(e^{inx}v)=e^{i(n+1)x}v$ for $n\ge 0$ and $A(e^{inx}v)=e^{inx}v$ for $n<0$. These images are exactly the distinct nonzero modes, so $A$ is an isometry onto their closed span, its kernel is zero, and its cokernel is the constants. Every identification commutes with $G$, proving ${\mathrm{Ind}}_G(A)=-[V]$.

To verify the actual classical symbol, let $f$ be a smooth cutoff supported in a coordinate interval of length less than $2\pi$, and extend it by zero to $\mathbb{R}$. With $\widehat{f}(\eta )=\int e^{-ix\eta }f(x)dx$, local application to $f(x)e^{ix\xi }$ gives $e^{-ix\xi }{P}_{+}(f{e}^{i(\cdot )\xi })(x)=(2\pi {)}^{-1}{\sum }_{n\ge 0}\widehat{f}(n-\xi )e^{ix(n-\xi )}$. Every derivative of $\widehat{f}$ decays faster than every power, by repeated integration by parts. For $\xi \to -\mathrm{\infty }$ the sum and all its $x,\xi$ derivatives are rapidly decreasing, since $n-\xi \ge n+\mathrm{\mid }\xi \mathrm{\mid }$. For $\xi \to +\mathrm{\infty }$ use full Fourier reconstruction, whose sum is $f(x)$ on the chosen chart, and subtract the sum over $n<0$; that sum has the same rapid estimates. Uniform bounds on the remaining compact $\xi$ interval follow by convergence of the differentiated sums. Hence this localized complete symbol has leading term $f(x)$ on the positive ray and zero on the negative ray, with smooth rapidly decreasing remainders away from a bounded frequency interval. Off the diagonal the localized kernels are smooth by the same integration-by-parts estimates. This establishes $P_{+}$ in the calculus [F1], with the asserted two-ray symbol. Multiplication by $e^{ix}$ gives the displayed elliptic symbol of $A$.

Choose a smooth real function $\phi$ on $[0,2\pi ]$, equal to $0$ near $0$, equal to $2\pi$ near $2\pi$, and increasing only in a proper interior interval $I$. The loops $u_t(x)=\exp (i((1-t)x+t\phi (x)))$ are smooth periodic degree-one loops, for $0\le t\le 1$, joining $e^{ix}$ to a loop equal to $1$ outside $I$. All derivatives match at the endpoints because $\phi$ is constant there. On the positive cotangent ray replace $e^{ix}$ by $u_t(x)$, leaving the negative symbol $1$. These are elliptic symbol homotopies; [F2] preserves their index. At $t=1$ the symbol is the identity outside the compact subinterval where $\phi$ varies, and its class is extended from $T^{\ast }I$. A coordinate on $I$ gives the following compact representative on ${\mathbb{R}}^2$: $a(x,\xi )=1+(e^{i\phi (x)}-1)h(\xi )$, where $h=0$ on $\xi \le -1$, $h=1$ on $\xi \ge 1$, and $\phi$ is extended by its endpoint constants. Its possible zeros lie in a compact rectangle, and its boundary values are $e^{i\phi (x)}$ on the top and $1$ on the other three sides.

Traverse that rectangle counterclockwise in the coordinate orientation $dx\wedge d\xi$. On the top $x$ decreases, so the change of argument is $-2\pi$. The boundary map has winding $-1$, whereas $x+i\xi$ has winding $+1$. Their class identity can also be seen directly: map the rectangle to a disk; the ratio of its boundary function to $\bar{z}$ has winding zero and hence a continuous logarithm (integrate its logarithmic derivative and use the zero total integral). Exponentiating scalar multiples of that logarithm gives a homotopy to $\bar{z}$. The sum of the $z$ and $\bar{z}$ relative symbols is zero: the boundary matrices $\mathrm{diag}(z,\bar{z})$ contract through $\left(\begin{array}{cc}tz& -\sqrt{1-t^2}\\ \sqrt{1-t^2}& t\bar{z}\end{array}\right)$, $1\ge t\ge 0$, to a constant invertible matrix, and then to the identity. Thus $[a]=-b$, consistently with [F3]. Tensoring the argument by $V$ proves the class assertion. Taking $G=\{1\}$ and $V=\mathbb{C}$ in steps 1.1 and 3.1 gives the scalar compact representative index $-1$; since its class is $-b$, additivity and excision give $a_{\{1\}}(b)=1$, where $a$ denotes the compact-support analytic index. Finally $d\theta =d\xi \wedge dx$ reverses the chosen complex orientation, giving precisely the orientation statement in the example.