Real Clifford Modules, KO Theory, and ABS Orientations: Examples and Counterexamples

For ${\mathrm{C}\mathrm{l}}_{1,0}$, send its positive generator to $(1,-1)\in \mathbb{R}\oplus \mathbb{R}$; its square is $(1,1)$. The image contains the linearly independent vectors $(1,1)$ and $(1,-1)$, so the homomorphism is onto. Every word in one generator reduces using $e^2=1$ to a linear combination of $1,e$, so the source has dimension at most two; the target has dimension two, hence the map is an isomorphism. For ${\mathrm{C}\mathrm{l}}_{0,1}$, send its negative generator to $i\in \mathbb{C}$; $i^2=-1$, and $1,i$ span the target. Again every word reduces to $1,e$, so the surjection between spaces of dimension at most two and two is an isomorphism.

Put $A=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $B_{+}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, and $B_{-}=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Direct multiplication gives $A^2=B_{+}^2=I$, $B_{-}^2=-I$, and $A{B}_{+}=-B_{+}A$, $A{B}_{-}=-B_{-}A$. Thus $A,B_{+}$ define a homomorphism from ${\mathrm{C}\mathrm{l}}_{2,0}$ and $A,B_{-}$ one from ${\mathrm{C}\mathrm{l}}_{1,1}$. In the positive case the four matrices are $I,A,B_{+},B_{-}$; in the mixed case they are $I,A,B_{-},B_{+}$. For any real matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, its unique expansion in $I,A,B_{+},B_{-}$ has coefficients $(a+d)\mathrm{/}2,(a-d)\mathrm{/}2,(b+c)\mathrm{/}2,(b-c)\mathrm{/}2$, so these four matrices are a basis and each displayed list spans. The defining relations reduce every word to a linear combination of $1,e_1,e_2,e_1{e}_2$, so each source has dimension at most four; surjectivity onto the four-dimensional matrix algebra makes both maps isomorphisms.

In $\mathbb{H}$, the elements $i,j$ square to $-1$ and anticommute. They therefore define a homomorphism from ${\mathrm{C}\mathrm{l}}_{0,2}$; its image contains $1,i,j,ij=k$, which is a basis of $\mathbb{H}$. Every word reduces to a linear combination of $1,e_1,e_2,e_1{e}_2$, so the source has dimension at most four. The surjection onto the four-dimensional $\mathbb{H}$ is an isomorphism.

For one generator $e$, the tensor algebra is $\mathbb{R}[t]$ and the Clifford relation is $t^2=Q(e)$. Division by either monic quadratic leaves a unique remainder $a+bt$. The positive evaluation map sends this remainder to $(a+b,a-b)$, which determines $(a,b)$, so $\mathbb{R}[t]\mathrm{/}(t^2-1)\cong \mathbb{R}\oplus \mathbb{R}$. The negative evaluation map sends it to $a+bi$, which determines $(a,b)$, so $\mathbb{R}[t]\mathrm{/}(t^2+1)\cong \mathbb{C}$.

The element $(1,0)$ is a nontrivial idempotent in $\mathbb{R}\oplus \mathbb{R}$. If $z=a+bi\in \mathbb{C}$ satisfies $z^2=z$, comparing real and imaginary parts gives $2ab=b$ and $a^2-b^2=a$. If $b=0$, then $a=0$ or $a=1$. If $b\mathrm{\ne }0$, then $a=1\mathrm{/}2$ and $b^2=-1\mathrm{/}4$, impossible because a real square is nonnegative. Thus $\mathbb{C}$ has no idempotents other than $0,1$. An algebra isomorphism preserves idempotents and $0,1$, so these algebras are not isomorphic.

Put $m=2$ in [F1]. Modulo two and $a^3=0$, $(1+a{)}^3=1+3a+3a^2+a^3=1+a+a^2$. Its degree-one part is $w_1(T{\mathbb{R}\mathbb{P}}^2)=a\mathrm{\ne }0$. The exact SW definition supplier is the existing item with the in its ID; no nonexistent substitute is used.

By [F2] the tangent bundle is not orientable. A Spin lift would supply the orientation required by [F3], a contradiction. Thus $T{\mathbb{R}\mathbb{P}}^2$ has no Spin structure, refuting the asserted universal existence and exhibiting the intended first-class obstruction. AC is carried from the characteristic-class and orientability suppliers.

The eigenvalues of $H$ are $-1$ and $+1$, so $H$ is self-adjoint and gapped. The projection onto its negative eigenspace is $P=(I-H)\mathrm{/}2=\mathrm{diag}(1,0)$. Direct multiplication gives $P^2=P=P^{\ast }$; its image is the one-dimensional span of $(1,0)$.

Therefore $\dim E_{<0}(H)=1$, while the stated zero-band reference has negative eigenspace dimension zero. Substitution in the defined rank difference gives $\nu (H)=1-0=1$.

In a graded $C_1$-module, the odd generator $e$ maps $M^0$ to $M^1$ and $M^1$ to $M^0$ by [F2]. Since $e^2=-I$, these restrictions are inverse up to sign, so $\dim M^0=\dim M^1$. Thus every restricted class has equal two coordinates. The explicit module $\mathbb{R}\oplus \mathbb{R}$ with $e(x,y)=(-y,x)$ satisfies $e^2=-I$, so its restriction is $(1,1)$; hence the image in the Grothendieck group is exactly $\mathbb{Z}(1,1)$.

The homomorphism ${\mathbb{Z}}^2\to \mathbb{Z}$, $(a,b)\mapsto a-b$, is surjective and has kernel $\mathbb{Z}(1,1)$. By [F1], its quotient is $A_0$, so the induced map $A_0\to \mathbb{Z}$ is an isomorphism and sends a graded module to its even dimension minus its odd dimension.

In this rank the ungraded fiber Clifford algebra is a quaternionic matrix algebra. Its spinor commutant is ${\mathbb{H}}^{\mathrm{o}\mathrm{p}}$, and the multiplicity objects in [F1] are right-quaternionic bundles. Hence its map is $[E]\mapsto {\mathrm{\Phi }}_V(E{\otimes }_{\mathbb{H}}\mathrm{\Sigma })$, not the ordinary real tensor product by a real line coefficient.

At a point in rank four, a quaternionic line produces the actual quaternionic Hopf generator of ${\widetilde{KO}}^0(S^4)$. Regarding that multiplicity line merely as a four-dimensional real space changes the balanced tensor operation; it does not turn the theorem into a real-line-source formula. The same Clifford residue and balanced evaluation apply in every rank $8m+4$, proving the stated source distinction.

In the first four residues [F1] identifies the integer unit with the real line, the first order-two generator with the Möbius clutching and the second with the stable one-turn orthogonal clutching. Degree $-3$ is zero. Thus the first four entries are $\mathbb{Z},\mathbb{Z}\mathrm{/}2,\mathbb{Z}\mathrm{/}2,0$, including their geometrically specified generators.

The rank-four quaternionic spinor map in [F1] identifies $K{O}^{-4-j}(\mathrm{p}\mathrm{t})$ with $KS{p}^{-j}(\mathrm{p}\mathrm{t})$ for $0\le j\le 3$. The latter are $\mathbb{Z},0,0,0$, supplying exactly the remaining entries; the quaternionic line gives the integer generator. Multiplication by the actual rank-eight symbol then repeats these residues, so the example does not infer a real theorem from complex Bott or from a bare list of abstract groups.

Put $e=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $\gamma =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ and $s=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Direct multiplication gives $e^2=s^2=I$, ${\gamma }^2=-I$ and pairwise anticommutation. On ${\mathbb{C}}^2\otimes {\mathbb{C}}^N$ let $B$ be a real symmetric involution, $M_B=s\otimes B$, and $H_B(k)=ik\gamma \otimes I+M_B$. The kinetic term and mass are Hermitian and anticommute, so $H_B(k{)}^2=(k^2+1)I$. Thus every $k$ has gap at least one.

Take $T=K$ and $C=(e\otimes I)K$. Both square to $1$ and commute. Complex conjugation changes $ik\gamma$ to $-ik\gamma$ and fixes $M_B$, giving $T{H}_B(k)T^{-1}=H_B(-k)$. Conjugating additionally by $e$ reverses both $\gamma$ and $s$, giving $C{H}_B(k)C^{-1}=-H_B(-k)$. Their product $e\otimes I$ is the self-adjoint chiral involution. These are precisely the momentum-reversing BDI symmetries of the one-dimensional Dirac family.

The admissible real symmetric masses anticommuting with both $e\otimes I$ and $\gamma \otimes I$ are exactly $s\otimes B$. Indeed anticommutation with $e$ makes the diagonal blocks zero, symmetry makes the other blocks $B,B^{\mathsf{T}}$, and anticommutation with $\gamma$ forces $B=B^{\mathsf{T}}$. Squaring the mass gives $B^2=I$. Therefore the reduced multiplicity problem is the actual real index-zero mass problem of [F1], with invariant $\nu =\dim E_{-}(B)-\dim E_{-}(B_0)$. Rank is constant along an involution homotopy because the continuous idempotent $(I-B)\mathrm{/}2$ has integer-valued continuous trace. Fixed-rank involutions form a connected real Grassmannian, so this invariant also distinguishes exactly the stable components.

For $\nu \ge 0$ take $N=\nu$, $B=-I_N$, $B_0=+I_N$; for $\nu <0$ take $N=-\nu$, $B=+I_N$, $B_0=-I_N$. Use the zero-dimensional empty system at $\nu =0$. The formula of step 2.2 gives invariant exactly $\nu$ in both cases. For $N=1$, the negative mass against the positive reference gives $\nu =1$, which [F1] sends to the unit of $K{O}^0(\mathrm{p}\mathrm{t})$. Direct sum adds the integers and reversing the difference negates them. The entire continuous $k$ family in step 1.1 has the same invariant; no gap-preserving continuous integer jump is claimed.