PomeLemma 11.27.2.594B40F9CD8C If $g$ is not flat, the open set $\{ x:\det A_ x\ne 0\} $ is nonempty.
Lemma 11.27.2594B40F9CD8C
Proof. Suppose that $\det A=0$ everywhere. On the open set where $A\ne 0$, its three eigenvalues are $\lambda _0,-\lambda _0,0$ with $\lambda _0>0$; they are distinct, so $A$ has smooth local eigenframes there.
Step 1: a Bochner identity. On an open set where $A$ has distinct eigenvalues, we compute $\Delta |A|^2$, keeping track of the normalizations.
Choose an oriented local eigenframe $\xi _1,\xi _2,\xi _3$ of $\Lambda ^-$ with $|\xi _ i|^2=2$, and denote the corresponding eigenvalues by $\lambda ,\mu ,\nu $. As skew-adjoint endomorphisms, these forms satisfy the multiplication rules of the quaternions. Write
\[ \nabla \xi _1=c\otimes \xi _2-b\otimes \xi _3,\quad \nabla \xi _2=-c\otimes \xi _1+a\otimes \xi _3,\quad \nabla \xi _3=b\otimes \xi _1-a\otimes \xi _2. \]
Here $a,b,c$ are real one-forms. The Ricci identity and the contracted Bianchi identity, using $\operatorname {Ric}(g)=0$, give
\[ \mathrm da+b\wedge c=\lambda \xi _1,\qquad \mathrm d\lambda =(\lambda -\mu )\xi _3c+(\lambda -\nu )\xi _2b, \]
and their simultaneous cyclic permutations. In the second formula the two-forms act as skew-adjoint endomorphisms on one-forms. Put $D=\mathrm d^*\mathrm d=-\operatorname {div}_ g\nabla $ on functions. Applying $\mathrm d^*$ to the second formula yields
\[ D\lambda =2\lambda ^2+4\mu \nu +2(\nu -\lambda )|b|^2+2(\mu -\lambda )|c|^2. \]
For example, the calculation uses
\begin{align*} \mathrm d^*(\xi _3c)& =-\langle b,\xi _1c\rangle +\langle a,\xi _2c\rangle +\langle a,\xi _3b\rangle -2\nu ,\\ \mathrm d^*(\xi _2b)& =-\langle b,\xi _1c\rangle -\langle a,\xi _2c\rangle -\langle a,\xi _3b\rangle -2\mu , \end{align*}
which follow from the connection and curvature formulas above; substitution and the cyclic formulas for $\mathrm d\mu ,\mathrm d\nu $ cancel the mixed pairings.
Now $\lambda +\mu +\nu =0$ and $|A|^2=\lambda ^2+\mu ^2+\nu ^2$. Multiplying the three scalar equations by their eigenvalues and adding gives
\[ \lambda D\lambda +\mu D\mu +\nu D\nu =18\lambda \mu \nu -2\bigl ((\mu -\nu )^2|a|^2 +(\nu -\lambda )^2|b|^2+(\lambda -\mu )^2|c|^2\bigr ). \]
The connection expression also gives
\[ |\nabla A|^2=|\mathrm d\lambda |^2+|\mathrm d\mu |^2+|\mathrm d\nu |^2 +2\bigl ((\mu -\nu )^2|a|^2+(\nu -\lambda )^2|b|^2 +(\lambda -\mu )^2|c|^2\bigr ). \]
The product rule therefore gives, with $\Delta =\operatorname {div}_ g\nabla =-D$,
D65B8F2D9646
11.27.3
\begin{equation} \Delta |A|^2=2|\nabla A|^2-36\det A=2|\nabla A|^2. \end{equation}
This identity also holds on the interior of $\{ A=0\} $. Both sides are smooth functions, and the two open sets have dense union in $X$, so the identity holds everywhere. Integration over $X$ gives $\nabla A=0$.
Step 2: a parallel operator $A$ forces flatness. If $g$ is not flat, then $A\ne 0$ somewhere, and since $A$ is parallel and $\det A=0$, it is nonzero everywhere and has three distinct constant eigenvalues. Its eigenlines are parallel; local unit sections of them are parallel as well. The induced connection on $\Lambda ^-$ therefore has zero curvature. Under $\mathfrak {so}(4)=\Lambda ^+\oplus \Lambda ^-$, the two summands commute, while the action of $\Lambda ^-$ on itself is faithful. These facts follow directly from the bases $e_1\wedge e_2\pm e_3\wedge e_4$, $e_1\wedge e_3\mp e_2\wedge e_4$, $e_1\wedge e_4\pm e_2\wedge e_3$. Thus zero induced curvature forces $\mathscr R(S\wedge T)$ to lie in $\Lambda ^+$ for every $S,T$. It already lies in $\Lambda ^-$, so $\mathscr R=0$, a contradiction. $\square$
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