Orthogonal-equivariant compact normalization of the doubled-real Bott symbol

Under [A1], use the controlled symbol classes and realizations from [F2] throughout. Exterior multiplication satisfies ${\epsilon }_z^2=0$ and ${\epsilon }_z{\epsilon }_z^{\ast }+{\epsilon }_z^{\ast }{\epsilon }_z=\mathrm{\mid }z{\mathrm{\mid }}^{2}I$: verify first for an orthonormal basis and use bilinearity. Hence the stated square identity and compact-support class hold. Under $V=V_1\oplus V_2$, the exterior-algebra identification gives $k_V(z_1,z_2)=k_{V_1}(z_1)\widehat{\otimes }1+1\widehat{\otimes }{k}_{V_2}(z_2)$. This is exactly the graded symbol product, with no extra sign. Thus $b_{V_1\oplus V_2}=b_{V_1}⊠b_{V_2}$, equivariantly for any group preserving the decomposition. For a real orthogonal bundle the real metric extends complex-bilinearly to identify its complexification with its complex dual. The same doubled-real Bott convention can therefore be written with the dual, while such an identification is not asserted for arbitrary complex bundles.

We first fix the nonequivariant line sign explicitly. On $S^1=\mathbb{R}\mathrm{/}(2\pi \mathbb{Z})$ let $P_{+}$ project onto $e^{inx}$, $n\ge 0$, and put $A=e^{ix}{P}_{+}+1-P_{+}$. The Fejér kernels show that trigonometric polynomials are dense in $L^2$: they approximate continuous functions uniformly, and continuous functions approximate interval indicators and hence simple functions in $L^2$. Thus the Fourier vectors are a complete orthonormal basis. The map $A$ sends the $n$th vector to the $(n+1)$st for $n\ge 0$ and to itself for $n<0$. Its images are exactly all nonzero modes, orthonormally, so $\ker A=0$, its range is closed and its cokernel is the constants. Its index is $-1$. To check its actual symbol, localize with a smooth function $f$ in an interval of length less than $2\pi$. The local symbol of $P_{+}$ is $(2\pi {)}^{-1}{\sum }_{n\ge 0}\widehat{f}(n-\xi )e^{ix(n-\xi )}$. For $\xi \to -\mathrm{\infty }$ this and every derivative decrease faster than every power, by rapid decay of the Fourier transform of $f$. For $\xi \to +\mathrm{\infty }$ subtract the rapidly decreasing sum over $n<0$ from the full Fourier reconstruction $f(x)$. The resulting estimates, also off the diagonal, prove that $P_{+}$ is classical of order zero with leading symbol $1$ on $\xi >0$ and $0$ on $\xi <0$. Thus $A$ has symbol $e^{ix}$ on the positive ray and $1$ on the negative ray.

The compact de Rham operator has index its cohomological Euler representation, as follows directly from [F1]. On a compact invariant Riemannian manifold let $D=d+d^{\ast }$, acting from even to odd complex forms, with the usual group action on forms. Integration by parts gives $D^2=\mathrm{\Delta }=d{d}^{\ast }+d^{\ast }d$ and $⟨\mathrm{\Delta }u,u⟩=\mathrm{\parallel }du{\mathrm{\parallel }}^2+\mathrm{\parallel }{d}^{\ast }u{\mathrm{\parallel }}^2$. The principal symbol of $D$ is $i({\epsilon }_{\xi }-{\epsilon }_{\xi }^{\ast })$, so it is elliptic. The parametrix and regularity give smooth finite-dimensional kernels and closed range; since the self-adjoint elliptic $\mathrm{\Delta }$ has cokernel its kernel, its restriction to the orthogonal complement of its kernel is invertible, with inverse $G$ improving Sobolev order by two. For closed smooth $\alpha$ write $\alpha =h+\mathrm{\Delta }G\alpha$. Commutation $d\mathrm{\Delta }=\mathrm{\Delta }d$ shows $dG\alpha$ is harmonic; it is exact, so its norm vanishes by integration by parts against its primitive. Hence $\alpha =h+d{d}^{\ast }G\alpha$. A harmonic exact form is zero by the same norm calculation. Thus harmonic forms identify with cohomology, equivariantly because all operations commute with the compact action. This proves the stated index comparison here without invoking a scalar Hodge assertion for a matrix complex.

Homotope the loop $e^{ix}$ to $e^{i\phi (x)}$, where $\phi$ rises smoothly from $0$ to $2\pi$ inside a proper coordinate interval and is constant near the ends; the homotopy is $e^{i((1-t)x+t\phi (x))}$. It is smooth periodic, including at the cut, and preserves symbol invertibility on both rays. The final symbol is extended from the interval. On its cotangent plane a representative is $1+(e^{i\phi (x)}-1)h(\xi )$, where $h=0$ for $\xi \le -1$ and $h=1$ for $\xi \ge 1$. Its only possible zeros lie in a compact rectangle. On the rectangle it is $e^{i\phi (x)}$ on the top and $1$ elsewhere. With orientation $dx\wedge d\xi$, the top is traversed right to left, so its winding is $-1$. More explicitly its boundary loop divided by $\bar{z}$ on a disk has winding zero and a continuous logarithm, so a logarithm homotopy identifies it with $\bar{z}$. The loops $z\oplus \bar{z}$ contract to a constant invertible matrix through $\left(\begin{array}{cc}tz& -\sqrt{1-t^2}\\ \sqrt{1-t^2}& t\bar{z}\end{array}\right)$ for $1\ge t\ge 0$. Consequently this relative class is $-b_{\mathbb{R}}$. By symbol homotopy, excision and step 1.2, $a_{\{1\}}(b_{\mathbb{R}})=1$. This establishes the sign independently of de Rham dimensions.

We next prove a symbol identity on $S^m\subset {\mathbb{R}}^m\oplus \mathbb{R}$, for $m=1,2$, equivariantly under $O(m)$ acting on the first summand. The de Rham symbol is equivalent to the Koszul symbol of the real covector $\xi$: on exterior degree $k$, multiplication by $i^k$ conjugates ${\epsilon }_{\xi }+{\epsilon }_{\xi }^{\ast }$ to $i({\epsilon }_{\xi }-{\epsilon }_{\xi }^{\ast })$. Use the round metric to identify exterior covectors and vectors. Let $f(q,t)=t$ be the smooth invariant height function and deform this real Koszul symbol to exterior multiplication by $\xi +isdf$, plus its adjoint, for $0\le s\le 1$. This is a globally smooth equivariant symbol on the same bundles. Real and imaginary parts are orthogonal in the Hermitian norm, so its square is $(\mathrm{\mid }\xi {\mathrm{\mid }}^2+s^2\mathrm{\mid }df{\mathrm{\mid }}^2)I$. At $s=0$ its noninvertibility is the zero section; for $s>0$ it is precisely the zero covectors over the two critical points of $f$, the north and south poles. Thus the full homotopy has one common compact noninvertibility set, the zero section. No comparison of only the principal symbols is being substituted for this compact-support homotopy.

The two pole classes have the following exact signs. In the north and south stereographic disk coordinates $x\mapsto (2x\mathrm{/}(1+\mathrm{\mid }x{\mathrm{\mid }}^2),\pm (1-\mathrm{\mid }x{\mathrm{\mid }}^2)\mathrm{/}(1+\mathrm{\mid }x{\mathrm{\mid }}^2))$, the round metric is $c(x{)}^2{\sum }_{j}d{x}_j^2$ with $c(x)=2\mathrm{/}(1+\mathrm{\mid }x{\mathrm{\mid }}^2)$, and $df=\mp 4{\sum }_j{x}_{j}d{x}_j\mathrm{/}(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^2$. If $\eta$ is the canonical cotangent coordinate, the orthonormal coframe coefficients of $\xi +idf$ are $c^{-1}(\eta \mp ia(x)x)$, where $a(x)=4\mathrm{/}(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^2>0$. The real orthonormal frames identify the exterior bundles equivariantly with ${\mathrm{\Lambda }}^{\bullet }{\mathbb{C}}^m$. Positive scalar multiplication by $c$ preserves their class. Excision into disjoint pole neighborhoods is permitted because the final symbol is invertible on their complement; adding elementary symbols and gluing along that invertible map is precisely open extension for compact-support triples. On each neighborhood the homotopy $a_t=(1-t)a+t>0$ changes the symbol to $\eta \mp ix$, keeping its sole zero at $(0,0)$. On the north, multiplication of its vector by $i$ gives $x+i\eta$. On the south it gives $-x+i\eta$, and subsequent multiplication by $-1$ gives $x-i\eta$. Scalar unit complex multiplications are joined to the identity through unit scalars, preserve equivariance and introduce no zeros. With $\vartheta (x,\eta )=(x,-\eta )$ and open extensions $j_0,j_{\mathrm{\infty }}$ from the pole charts, the actual symbol identity is $[\sigma (D_{S^m})]=j_0{b}_{{\mathbb{R}}^m}+j_{\mathrm{\infty }}{\vartheta }^{\ast }{b}_{{\mathbb{R}}^m}$. This proves both the compact support and the two local signs without a continuous-to-smooth sphere gluing premise.

The sphere map interchanging the two hemispheres is $O(m)$-equivariant and carries their chosen tangent coordinates into one another. Excision and diffeomorphism covariance therefore give $a_{O(m)}(\sigma (D_{S^m}))=a_{O(m)}(b_{{\mathbb{R}}^m})+a_{O(m)}({\vartheta }^{\ast }{b}_{{\mathbb{R}}^m})$. For $m=1$ let $ϵ$ be the nontrivial character of $O(1)$. The class identity ${\vartheta }^{\ast }{b}_{\mathbb{R}}=-ϵb_{\mathbb{R}}$ is explicit. On the unit complex circle the direct sum on the left of ${\vartheta }^{\ast }b+ϵb=0$ has domain $1\oplus ϵ$ and, after exchanging target summands, matrix $\left(\begin{array}{cc}0& z\\ \bar{z}& 0\end{array}\right)$. The path $\left(\begin{array}{cc}\sin t& \cos tz\\ \cos t\bar{z}& -\sin t\end{array}\right)$, $0\le t\le \pi \mathrm{/}2$, is equivariant for the involution $z\mapsto -z$ and the matrix representation $\mathrm{diag}(1,-1)$, is invertible since its determinant is $-1$, and ends at a constant equivariant isomorphism. Hence the sum is zero as a relative class. For $m=2$ restricted to $SO(2)$, $\vartheta$ is properly equivariantly homotopic to the identity by $(x,\xi )\mapsto (x,R_{\pi t}\xi )$; on the symbols its square norm remains $\mathrm{\mid }x{\mathrm{\mid }}^2+\mathrm{\mid }\xi {\mathrm{\mid }}^2$. Thus ${\vartheta }^{\ast }b=b$ for this group.

On the circle, constants generate $H^0$ and $dx$ generates $H^1$; reflection acts by $1$ and $-1$, respectively. Hence step 1.3 gives $a_{O(1)}(\sigma (D_{S^1}))=1-ϵ$. If $r=a_{O(1)}(b_{\mathbb{R}})=u+vϵ$, step 4.1 gives $(1-ϵ)r=1-ϵ$, or $u-v=1$. Restriction to the trivial group and step 2.1 give $u+v=1$. Thus $u=1,v=0$: this is the full $O(1)$ representation, not just its dimension. On $S^2$, [F4] gives only $H^0$ and $H^2$; the invariant area form has nonzero integral and is therefore not exact, so it generates the latter group. Both representations are trivial under rotations. Thus $a_{SO(2)}(\sigma (D_{S^2}))=2$. Step 4.1 implies $2a_{SO(2)}(b_{{\mathbb{R}}^2})=2$. By [F2], averaging a positive Hermitian form makes every finite-dimensional representation unitary. To see directly that $R(SO(2))$ has no torsion, let $\rho :SO(2)\to U(W)$ be any such continuous representation and set $P_{k}w=(2\pi {)}^{-1}{\int }_0^{2\pi }{e}^{-ikt}\rho (R_t)wdt$ for $k\in \mathbb{Z}$. Changing variables in the integral gives $\rho (R_s)P_{k}w=e^{iks}{P}_{k}w$, so each $P_{k}w$ is a weight vector or zero. For fixed $w$, the Fejér means of the continuous periodic function $t\mapsto \rho (R_t)w$ at $0$ are finite linear combinations of these $P_{k}w$ and converge to $w$: the Fejér kernel is nonnegative with integral one, and its mass outside every neighborhood of zero tends to zero, so uniform continuity proves uniform convergence of its convolutions. Since $W$ is finite-dimensional, the span of all weight spaces is closed and contains every $w$. Distinct weights form a direct sum, as applying the corresponding Fourier projections isolates each weight. Thus $W$ is a direct sum of characters $z\mapsto z^m$, $m\in \mathbb{Z}$, and virtual representations are uniquely determined by their integer multiplicities; multiplication by $2$ is injective in $R(SO(2))$. Therefore $2a_{SO(2)}(b_{{\mathbb{R}}^2})=2$ implies $a_{SO(2)}(b_{{\mathbb{R}}^2})=1$.

Every orthogonal matrix $Q$ on $V$ admits an orthogonal decomposition into invariant real lines with eigenvalue $+1$ or $-1$, and real planes on which it is a rotation: diagonalize its complex unitary complexification and pair conjugate eigenvectors; real eigenvectors span the remaining $\pm 1$ spaces. Thus $Q$ belongs to a block subgroup $H$ which is a product of copies of $O(1)$ and $SO(2)$. On restriction to this entire subgroup, step 1.1 identifies $b_V$ with the product of the line and plane symbols. By [F3] and step 5.1 its index is $1\in R(H)$. Restriction of a Fredholm index simply restricts its finite kernel representations, so the character of $a_{O(n)}(b_V)$ takes the value $1$ at the original arbitrary matrix $Q$. This proves equality of characters on every component of $O(n)$, including orientation-reversing matrices.

Equality of these characters implies equality in $R(O(n))$. Indeed average inner products over $O(n)$ to make representations unitary, and take invariant orthogonal complements to decompose them into irreducibles. Normalized invariant volume on $O(n)$ can be obtained from the bi-invariant metric induced by the trace inner product on its skew-symmetric matrices, giving equal mass to its two components. Averaging on $\mathrm{Hom}(U,W)$ is the orthogonal projection onto intertwining maps, so the integral of $\overline{{\chi }_U}{\chi }_W$ is their dimension; for irreducibles this is $1$ if isomorphic and $0$ otherwise, by Schur's lemma (an eigenvalue of an intertwiner and its invariant kernel prove that lemma over $\mathbb{C}$). Pairing a zero virtual character against every irreducible therefore gives zero multiplicities. Apply this to step 6.1 minus the trivial character to obtain $a_{O(n)}(b_V)=1$. For $n=0$ the open Euclidean set is a point and its symbol is the even line, whose index is directly $1$. Group restriction and coefficient tensoring in [F2] give the general assertion. Finally the complex orientation is the ordered product $d{x}_1\wedge d{\xi }_1\wedge \cdots \wedge d{x}_n\wedge d{\xi }_n$, whereas $(d\theta {)}^n\mathrm{/}n!$ is its $(-1{)}^n$ multiple, proving the exact orientation convention.