1

We consider the vector field $X =h(y)∂x$. Give the necessary conditions for $ X $ to be a Killing field for the metric $g$ with $$g=\left(\begin{array}{cc} 1+y^{2} & x y \\ x y & 1+x^{2} \end{array}\right) $$

$X$ is a Killing field iff $L_X g=0 ,$ a.e: $$ L_X g(\partial_i,\partial_j)=X(g(\partial_i,\partial_j)) -g([X,\partial_i], \partial_j)-g(\partial_i, [X,\partial_j])$$

\begin{cases} L_X g(\partial_1,\partial_1)=X(g(\partial_1,\partial_1)) =0 \\ L_X g(\partial_1,\partial_2)=yh(y)-h'(y)(1+y^2)=0\\ L_X g(\partial_2,\partial_2)=2xh(y)-2h'(y)xy=0 \\ \end{cases} $\iff$

\begin{cases} h(y)= C\sqrt{1+y^2} \\ h(y)=Dy \end{cases}

M-S
  • 272
  • 1
  • 12
  • 2
    Can you show your calculation? – Arctic Char Jun 27 '21 at 15:38
  • I have edited the post! Can you please see where's the problem? – M-S Jun 27 '21 at 17:08
  • 1
    Your computations look correct to me (note however that from the differential equations you should get $h(y)=C\sqrt{1+y^2}$ and $h(y)=Dy$ for arbitrary constants, not that it solves the contradiction...) – Giulio R Jun 27 '21 at 18:13
  • Thank you for answering, but how the existence of constants solves the contradiction ? – M-S Jun 27 '21 at 19:19

0 Answers0