0
$\begingroup$

I've the following doubt:

Let $(M^n,g)$ a Riemannian Manifold, where $g=dr^2+\gamma^2(r)d\omega^2$ is given in geodesic spherical coordinates. Suppose that the radial sectional curvatures of $M^n$ are $\geq0$. If we consider a radial geodesic starting in a point $p$, along it the Jacobi equation reduces to $$\gamma''+K(r)\gamma=0$$ due to rotational symmetry

Note: $K$ is the sectional curvature

how can I show this statement?(About the jacobi equation)

Thanks!

$\endgroup$

1 Answer 1

4
$\begingroup$

The Jacobi equation for a field J(t) along a radial unit speed geodesic $c(t)$ is $$ J'' + R_cJ = 0 $$ where $R_c$ is the tensor given by $R_c J = R(J, c')c'$. At every point the tensor $R_c$ takes $c'(t)$ to zero, and one checks that $R_c$ is symmetric from $\langle R(X,Y)Z, W \rangle = \langle R(W, Z)Y, X \rangle$. It follows that at any point $q = c(t)$, $R_c$ preserves the orthogonal complement to the span of $c'(t)$ and is symmetric, hence diagonalizable, on this subspace. Then rotational symmetry of your metric guarantees that all the eigenvalues of $R_c$ on this subspace are equal, that is, $R_c$ is a constant multiple of the identity, say $R_c = \lambda$, on this orthogonal complement. $\lambda$ is seen to be the radial sectional curvature from $$ K(J, c'(t)) = \langle R(J, c'(t))c'(t), J(t) \rangle = \langle R_c J, J \rangle = \lambda $$ whenever $||J(t)|| = 1$. Thus the Jacobi equation reduces to $$ J'' + K(r) J = 0 $$ for Jacobi fields perpendicular to $c$. None of this assumes that $M$ has nonnegative sectional curvature.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .