1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Is $M$ a Riemannian manifold? $\endgroup$
    – Ben McKay
    Oct 4, 2016 at 6:18
  • 1
    $\begingroup$ Question 2 is just the implicit function theorem applied to the linear equation. You know that there is a pair of eigenvalues of opposite sign, so there is an eigenvector for the positive eigenvalue, and you check the implicit function theorem for a unit length eigenvector. $\endgroup$
    – Ben McKay
    Oct 4, 2016 at 6:22
  • 1
    $\begingroup$ I think "everybody" should be "anybody". $\endgroup$
    – Ben McKay
    Oct 4, 2016 at 6:23
  • $\begingroup$ For question 1, multiply a constant matrix $h_0$ by a bump function. $\endgroup$
    – Ben McKay
    Oct 4, 2016 at 6:24
  • $\begingroup$ Dear @BenMcKay, Yes $M$ is a Riemannian manifold. Can you explain your thinking? $\endgroup$
    – C.F.G
    Oct 4, 2016 at 9:55

1 Answer 1

1
$\begingroup$

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 $f=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.