Finding the vector fields when lie derivative is zero

Question

I am trying to figure out two things:

for the following:

$R^{2}$ is given by the coordinates $x^{1}$ and $x^{2}$ and we have a tensor $S$ such that is of type $(2,0)$ in same $R^{2}$ given by:

$S = dx^{1} ⊗ dx^{1} + dx^{2} ⊗ dx^{2}\,\,\,$,where $⊗$ is tensor product.https://en.wikipedia.org/wiki/Tensor_product

1. The vector fields of $X$ such that lie derivative $L_{X}{S}=0$ here $X$ is vector field and $S$ is tensor field. https://en.wikipedia.org/wiki/Tensor

2. figuring out if there exist an $X$ vector field for which that this flow $H(t)=(H^{1}(t),H^{2}(t))^{T}$ is in a bounded set for any fixed initial value.

So I am new to this part of tensor-theory, I am having problem finding the solution for the whole problem, but I think I found the solution if we consider a simpler$S = dx^{1} ⊗ dx^{2} $ with same tensor $(2,0)$but not sure if its correct, here is what I worked out to find the vector fields:

lets assume:

$X=X^{1}(x^{1},x^{2})\partial_{1}+X^{2}(x^{1},x^{2})\partial_{2}$

is a vector filed in $R^{2}$ where $X^{1} , X^{2}$ are smooth fnuctions and so
$L_{X}{S}$ is a tensor of type $(2,0)$ which means we can do this:

$L_{X}(S = dx1 ⊗ dx2) =L_{X}dx^1 ⊗ dx^2+ dx^1 ⊗L_{X}dx^2$

let us have:

$Y=Y^{1}(x^{1},x^{2})\partial_{1}+Y^{2}(x^{1},x^{2})\partial_{2}$

and

$Z=Z^{1}(x^{1},x^{2})\partial_{1}+Z^{2}(x^{1},x^{2})\partial_{2}$

be a any vector fields in $R^{2}$ then $L_{X}{S}$ can be found from $Y$ and $Z$ and the above formula, thus we get :

$L_{X}(S =dx^{1}⊗dx^{2})(Y,Z)=(L_{X}dx^{1})(Y)dx^{2}(Z)+dx^{1}(Y)(L_{X}dx^{2})(Z)=$

$=(L_{X}dx^{1})(Y)Z^{2}+Y^{1}(L_{X}dx^{2})(Z)$

and after a long boring computation we get that : $(L_{X}dx^{2})(Z)=Z^{1}\partial_{1}X^{2}+Z^{2}\partial_{2}X^{2}$ which can be solved for $L_{X}{S}$ giving :

$Y^{1}Z^{2}(\partial_{1}X^{1}+\partial_{2}X^{2})+Y^{2}Z^{2}\partial_{2}X^{1}+Y^{1}Z^{1}\partial_{1}X^{2}$

so to find the vector fields now we just need to let the RHS of this expression vanish for any vectorfields $Y,Z$ by letting them be anywhere over the fields $\partial_{i}$ parallel to the axis I think its reasonable to see that $L_{X}{S}=0$ is a system of PDE:

$\partial_{1}X^{1}+\partial_{2}X^{2}=0 $

$\partial_{2}X^{1}=0 $

$\partial_{1}X^{2}=0$

and solving this gives us the vector fields?

My main problem is especially with the second part, but also I would like to get a confirmation that my method makes sense and it can be applied to the bigger problem and any additional hints are greatly appreciated!

Answer 1

The method you describe is essentially correct (though the indices on the latter two components of the p.d.e. system are incorrect), but NB the answers to (1)-(2) very much depend on the choice of $S$. Indeed, for a generic $2$-tensor $S$ on a $2$-manifold the only vector field satisfying $\mathcal L_X S = 0$ is the zero vector field.

Working in coordinates gives a general coordinate formula for the Lie derivative:

$$(\mathcal L_X S)_{ab} = X^c \partial_c S_{ab} + \partial_a X^c S_{cb} + \partial_b X^c S_{ac} .$$

Our tensor $S = (dx^1)^2 + (dx^2)^2$ has constant coefficients with respect to the coordinate frame, so the first term on the right-hand side disappears, and specializing $(a, b)$ to $(1, 1), (1, 2), (2, 2)$ immediately gives the analogous p.d.e. system for this case. (Since $S$ is symmetric, the equations we get from taking $(a, b)$ to be $(1, 2)$ and $(2, 1)$ are the same, so in this sense it's easier to work with this $S$ than the nonsymmetric tensor $dx^1 \otimes dx^2$.) The system is

\begin{align*} \partial_1 X^1 &= 0 \\ \partial_1 X^2 + \partial_2 X^1 &= 0 \\ \partial_2 X^2 &= 0 . \end{align*}

Notice that, e.g., the first equation tells you that $X^1$ is a function $X^1(x^2)$ of $x^2$ alone.

Remark If $S$ is (as in this case) a Riemannian metric, the equation $\mathcal L_X S = 0$ is the Killing equation, and by definition its solutions are precisely the vector fields whose flows (locally) preserve that metric.