The de Rham index character of a rotating two-sphere

At $x\in S^2$ and $0\mathrm{\ne }\xi \in T_x^{\ast }{S}^2$, the positive-frequency principal symbol of $d_{\mathbb{C}}$ is $B=i\xi \wedge$. For the round metric let ${\xi }^{\mathrm{♯}}\in T_x{S}^2$ satisfy $⟨{\xi }^{\mathrm{♯}},v⟩=\xi (v)$ and define contraction on a $j$-form $\eta$ by $({\iota }_{{\xi }^{\mathrm{♯}}}\eta )(v_1,\dots ,v_{j-1})=\eta ({\xi }^{\mathrm{♯}},v_1,\dots ,v_{j-1})$, with contraction zero in degree zero. The identity ${\iota }_{{\xi }^{\mathrm{♯}}}(\xi \wedge \eta )+\xi \wedge {\iota }_{{\xi }^{\mathrm{♯}}}\eta =\mathrm{\mid }\xi {\mathrm{\mid }}^2\eta$ gives $Bh+hB=I$ for $h=(i\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-1}{\iota }_{{\xi }^{\mathrm{♯}}}$. Since $B^2=0$, every $B$-closed form is $B$-exact, including the endpoint degrees, so the symbol sequence is exact. Also $d_{\mathbb{C}}^2=0$, and pullback by every rotation commutes with $d_{\mathbb{C}}$. Thus this is an equivariant elliptic complex in [F1]. Writing every complex form uniquely as $\alpha +i\beta$ identifies the complex kernel and image of $d_{\mathbb{C}}$ with the respective real kernel and image tensored with $\mathbb{C}$: closedness and exactness hold componentwise. Thus $H^j(E^{\bullet })=H_{\mathrm{d}\mathrm{R}}^j(S^2;\mathbb{R}){\otimes }_{\mathbb{R}}\mathbb{C}$ equivariantly. The fiber action is ${\rho }_t(x)={\mathrm{\Lambda }}^j((d{t}_x^{-1}{)}^{\ast })\otimes 1$, covering $x\mapsto t\cdot x$, and the section action is $(t^{-1}{)}^{\ast }$. The normalized area form is $\omega =(xdy\wedge dz+ydz\wedge dx+zdx\wedge dy)\mathrm{/}(4\pi )$. In spherical coordinates it is $\sin \vartheta d\vartheta \wedge d\phi \mathrm{/}(4\pi )$, so its integral over $S^2$ is $1$. It is rotation-invariant. By [F3], it cannot be exact; [F2] then makes $[\omega ]$ a generator of $H_{\mathrm{d}\mathrm{R}}^2(S^2)$. Constants generate $H^0$ and are fixed by $T$, while $H^1=0$. Thus the action on both nonzero cohomology groups is trivial and [F1] gives ${\chi }_{{\mathrm{Ind}}_T(E^{\bullet })}(g)=1+1=2$.

Write $\theta =2\pi s$. The equation $g(x,y,z)=(x,y,z)$ and $\theta \mathrm{\notin }2\pi \mathbb{Z}$ force $x=y=0$, so $M^g=\{(0,0,1),(0,0,-1)\}$. At either pole, $dg$ on the tangent plane is the rotation matrix $R_{\theta }$, hence $\mathrm{\mid }{\det }_{\mathbb{R}}(I-dg)\mathrm{\mid }=\det (I-R_{\theta })=(1-\cos \theta {)}^2+{\sin }^2\theta =2-2\cos \theta >0$.

The complex graded fiber at a pole is ${\mathrm{\Lambda }}^{\bullet }(T_p^{\ast }{S}^2{\otimes }_{\mathbb{R}}\mathbb{C})$, with fiber action ${\mathrm{\Lambda }}^j{A}_p$ in degree $j$. The eigenvalues of $A_p$ are $e^{i\theta }$ and $e^{-i\theta }$, so its exterior powers have traces $1$, $e^{i\theta }+e^{-i\theta }$, and $1$ in degrees zero, one, and two. Their alternating sum is $2-2\cos \theta$. Substitution into the definition of $c_p(g)$, using the positive denominator in step 1.2, gives $c_p(g)=1$ at each pole. Thus ${\sum }_{p\in (S^2{)}^g}{c}_p(g)=1+1=2={\chi }_{E^{\bullet }}(g)$ by step 1.1. This proves the fixed-point identity for this example by computing its two sides independently.