For any Lie group $G$ there exist many left-invariant Riemannian metrics, namely, one just takes any inner product on the tangent space at the identity $T_eG$ and then left translate it to the other tangent spaces. If $G$ is compact, one can do better. Namely, one can start with an inner product at $T_eG\cong\mathfrak g$ which is also invariant under the adjoint representation $\mathrm{Ad}:G\to GL(\mathfrak g)$ (obtained by averaging an arbitrary inner product with respect to a Haar measure) and it turns out that the resulting Riemannnian metric on $G$ is also right-invariant. Since it is invariant
under left and right translations, it is called bi-invariant.
The Riemannian geometry of bi-invariant metrics is very nice. For instance, geodesics
through the identity coincide with one-parameter groups, so the Riemannian exponential map coincides with the Lie group exponential. The Riemann curvature tensor has a simple formula
$R(X,Y)X=-[[X,Y],X]$ for unit vectors $X$, $Y\in\mathfrak g$, from which follows that the sectional curvature is nonnegative. Actually, compact semisimple Lie groups equipped with bi-invariant Riemannian metrics are symmetric spaces of compact type (the geodesic symmetry at the identity is the inversion map $g\mapsto g^{-1}$) and hence its geometric and topological invariants are amenable to explicit computations.
Fix a bi-invariant metric in a compact semisimple Lie group $G$. As for any other symmetric space, the Jacobi equation along a fixed geodesic $\gamma$ (say starting at $e$) has constant coefficients with respect to a parallel orthonormal frame (since the sectional curvature is parallel).
Invoking the real root space decomposition of the Lie algebra $\mathfrak g = \mathfrak t+\sum_{\alpha\in\Delta^+} \mathfrak g_\alpha$ with respect to a maximal torus with Lie algebra $\mathfrak t$, we have $R ( H , X_\alpha)H = \mathrm{ad}_H^2 X_\alpha = -\alpha(H)^2 X_\alpha$ and hence a typical Jacobi field $J$ along $\gamma(t)=\exp tH$ (for a unit vector $H\in\mathfrak t$) vanishing at $t=0$ is
of the form $J(t)=\sin(\alpha(H)t)X_\alpha(t)$, where $X_\alpha(t)$ is the parallel
vector field along $\gamma$ with $X_\alpha(0)=X_\alpha\in\mathfrak g_\alpha$.
It follows that $\gamma(t)$ is a conjugate point to $\gamma(0)=e$ along $\gamma$ if $\alpha(H)t\in \pi\mathbf Z$, and then the contribution to the multiplicity is $\dim\mathfrak g_\alpha=2$. We see that the total multiplicity of the conjugate point $\gamma(t)$ is twice the number of roots $\alpha\in\Delta^+$ such that $\alpha(H)t\in\mathbf Z$, hence it is even. In other words, in bi-invariant metrics conjugate points always have even multiplicity (in particular, due to the Morse index theorem also the index of geodesics is always even).
My question is whether this property characterizes bi-invariant metrics among left-invariant ones. Namely, assume we have a left-invariant Riemannian metric on $G$ such that for every point $g\in G$ conjugate to the identity element along a geodesic the multiplicity is even. Is it true that the metric must be bi-invariant?
Best Answer
A left-invariant Riemannian metric on Lie group is a special case of homogeneous Riemannian manifold, and its differential geometry (geodesics and curvature) can be described in a quite compact form. I am most familiar with the description in 28.2 and 28.3 of here of covariant derivative and curvature.
But on a Lie group itself there is an explicit description of Jacobi fields available for right invariant metrics (even on infinite dimensional Lie groups) in section 3 of:
I shall now describe the results (which go back to Milnor and Arnold). For detailed computations, see the paper.
Let $G$ be a Lie group with Lie algebra $\def\g{\mathfrak g}\g$. Let $\def\x{\times}\mu:G\x G\to G$ be the multiplication, let $\mu_x$ be left
translation and $\mu^y$ be right translation, given by $\mu_x(y)=\mu^y(x)=xy=\mu(x,y)$.
Let $\langle \;,\;\rangle:\g\x\g\to\Bbb R$ be a positive
definite inner product. Then
$$\def\i{^{-1}} G_x(\xi,\eta)=\langle T(\mu^{x\i})\cdot\xi, T(\mu^{x\i})\cdot\eta)\rangle $$ is a right invariant Riemannian metric on $G$, and any right invariant Riemannian metric is of this form, for some $\langle \;,\;\rangle$.
Let $g:[a,b]\to G$ be a smooth curve.
In terms of the right logarithmic derivative $u:[a,b]\to \g$ of $g:[a,b]\to G$, given by
$u(t):= T_{g(t)}(\mu^{g(t)\i}) g_t(t)$, the geodesic equation has the expression $$ \def\ad{\text{ad}} \partial_t u = u_t = - \ad(u)^{\top}u\,, $$ where $\ad(X)^{\top}:\g\to\g$ is the adjoint of $\ad(X)$ with respect to the inner product $\langle \;,\; \rangle$ on $\g$, i.e., $\langle \ad(X)^\top Y,Z\rangle = \langle Y, [X,Z]\rangle$.
A curve $y:[a,b]\to \g$ is the right trivialized version of a Jacobi field along the geodesic $g(t)$ described by $u(t)$ as above iff $$ y_{tt}= [\ad(y)^\top+\ad(y),\ad(u)^\top]u - \ad(u)^\top y_t -\ad(y_t)^\top u + \ad(u)y_t\,. $$
For connected $G$, the right invariant metric is biinvariant iff $\ad(X)^\top = -\ad(X)$. Then the geodesic equation and the Jacobi equation reduces to $$ u_t = \ad(u)u = 0,\qquad y_{tt} = \ad(u)y_t $$ Now we can look at examples. If $G=SU(2)$ then $\g=\mathfrak{sl}(3,\mathbb R)$ and we can take an arbitrary inner product on it. (Maybe, I will continue if I have more time in the next few days).