I read in some papers the following method:


Let $h$ be a $(1,1)$-tensor field on 3-dimensional Riemannian manifold $(M,g)$ and $p\in M$. Then there exists a smooth local orthonormal basis of the form $\{e\_1,e\_2,e\_3\}$ in a
neighborhood of $p$. Now, let $U\_1$ be the open subset of $M$ where $h\neq 0$ and let $U\_2$ be the
open subset of points $p\in M$ such that $h = 0$ in a neighbourhood of $p$. $U\_1\cup U\_2$ is an
open dense subset of $M$. On $U\_1$ we put $he\_1 = \lambda e\_1$ where
$\lambda $ is a non-vanishing smooth function.


**Remark:** we know that $he\_3=0$ and $tr h=0$.


I have three question:


**Questions:**


1. Can anybody construct a simple example with the above properties such that $U\_1\neq \emptyset ,\,\, U\_2\neq \emptyset$?
2. why can we put $he\_1 = \lambda e\_1$?
3. Why do we need to construct an open dense subset?


Thanks.



Why can we construct the following method?
For question 1, take a smooth function $f(x,y,z)$ positive for $x > 0$ and zero for $x < 0$. Let 
$$
h\_0 
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 0 
\end{pmatrix}
$$
and then let $h=fh\_0$.
Take the standard Euclidean metric on $\mathbb{R}^3$. 
Let $e\_1, e\_2, e\_3$ be the standard basis. Clearly $U\_1=(x>0)$ while $U\_2=(x<0)$.
Take $\lambda=f$.


For question 2, if we can prove local existence, uniqueness and smoothness of $e\_1$ (up to sign) so that $he\_1=\lambda e\_1$ with $\lambda > 0$ on $U\_1$ then we clearly have global existence, uniqueness and smoothness (at least one a 2-1 covering space). So we can work in local coordinates $x,y,z$. We take any orthonormal basis $e\_1', e\_2', e\_3'$. We want to solve $e\_1=a e\_1' + b e\_2' + ce\_3'$ so that $1=a^2+b^2+c^2$ and so that $h e\_1 = \lambda e\_1$ and with $\lambda>0$. Since $h \ne 0$ and has trace zero, and a zero eigenvalue, $h$ has a pair of opposite sign eigenvalues. We calculate that the characteristic polynomial of $h$, in a variable $t$, is $\lambda^2 t - t^3$. We take the coefficient of $t$, and take its positive square root, to get $\lambda$ a smooth function on $U\_1$. We have equations for $a,b,c$, and we turn them into an operator 
$$
F(x,y,z,a,b,c)=(h(ae\_1'+be\_2'+ce\_3')-\lambda(ae\_1'+be\_2'+ce\_3'),a^2+b^2+c^2-1).
$$
We compute the matrix
$$
\frac{\partial F}{\partial a,b,c}=\begin{pmatrix}he\_1'-\lambda e\_1' & he\_2'-\lambda e\_2' & he\_3'-\lambda e\_3' \\ 2a & 2b & 2c \end{pmatrix}.
$$
Clearly $F(a,b,c)=0$ just at $ae\_1'+be\_2'+ce'\_3=\pm e\_1$.
If the rank of this matrix were to drop at such a point, a vector in the kernel would be precisely an eigenvector of $h$ with eigenvalue $\lambda$ perpendicular to $e\_1$, not possible.
By the implicit function theorem, the solutions $\pm e\_1$ of $F(a,b,c)=0$ are smooth functions, locally, and locally unique up to $\pm$.


For question 3, try 
$$
h=
\begin{pmatrix}
f & 1 & 0 \\
0 & -f & 1 \\
0 & 0 & 0
\end{pmatrix}
$$
so that $h$ is not diagonalizable at points where $f=0$.


