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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?

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Not sure what you were trying to do with the formatting at the end, so I guessed. Hope you don't mind. –  Qiaochu Yuan Jun 4 '11 at 23:49
    
I don't mind at all. Sometimes I have some trouble previewing the text of the question. –  Claudio Gorodski Jun 4 '11 at 23:55
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I assume that you mean that the multiplicity is always even in the bi-invariant case, not that it's always 2. –  Robert Bryant Jun 12 '11 at 18:20
    
Robert: You are completely right, many thanks for pointing this out. I've already edited the question. –  Claudio Gorodski Jun 13 '11 at 1:14
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2 Answers

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:

  • Peter W. Michor: Some Geometric Evolution Equations Arising as Geodesic Equations on Groups of Diffeomorphism, Including the Hamiltonian Approach. IN: Phase space analysis of Partial Differential Equations. Series: Progress in Non Linear Differential Equations and Their Applications, Vol. 69. Birkhauser Verlag 2006. Pages 133-215. (pdf).

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 $$ y_{tt}= [\ad(y)^\top+\ad(y),\ad(u)^\top]u - \ad(u)^\top y_t -\ad(y_t)^\top u + \ad(u)y_t\,. $$

Continued:

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).

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Since we are dealing with symmetric spaces, in particular with compact semi-simple Lie groups, we already know that the space can be equipped with a bi-invariant metric. As you mentioned before, we know that compact semi-simple Lie groups with a bi-invariant metric form a symmetric space. However, the results on which you base your assumption were derived by heavily depending on the symmetric space structure. This seems to be a total loop for me (in the methodological sense!).

If you are referring to the general question below, I would say that I do not see the point in proving something like that. This is not meant as an offence, but I actually do not understand where this question is supposed to be going. All considerations above were made for symmetric spaces, that require the compact Lie Group to be already equipped with a bi-invaraint metric.

I know this is not quite an answer, but I don't know how to comment, so if someone of the moderators feels this should be converted in a comment, I'd not necessarily oppose to doing that.

EDIT: I am currently working on the problem. Thanks to Claudio, it has become clearer to me what actually should be shown.

EDIT 2: So far, in the case of a connected, compact, semi-simple Lie group, I have found out nothing interesting. It is known that these groups are symmetric spaces, or, more precisely, give rise to symmetric spaces. Namely, consider an (open) subgroup $H_0$ of the stabilzer $H(\sigma)$ of $G$, where $\sigma$ is an involutive automorphism of $G$. Then, it is known that the quotient space $G/H_0$ is a symmetric space when one uses the definition of a symmetric space via Lie theory. The question I am now asking myself is how the curvature tensor behaves in general. There is a long formula for the curvature tensor of $G$ in "Differential Geometric Structures" by Walter A. Poor. Then there is a remark about the Levi-Civita connection which takes a particular simple form when the metric is bi-invaraint, namely $\nabla_X Y \cdot_{G} Z = \frac{1}{2}[X,Y]\cdot_{G}Z$. Then the author writes: The converse is true if $G$ is connected and the curvatue tensor takes the form $R(W,X)Y=-\frac{1}{4}[[W,X],Y]$. I am not quite sure what the author wants to say here. Does he mean that then we may choose a bi-invaraint metric and do so in order to obtain this more elegant formula? (This is what I believe actually and if so, this seems like a dead end for a (possible) proof of the above statement - Any comments clarifying this linguistic problem are welcome.). I will further be occupied with the problem of Claudio.

EDIT 3: In fact, what I've posted before seems like a dead end. Now I want to check if there's something interesting in Morse-theory.

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Peter: in principle compact semisimple Lie groups do not carry metrics, and we have to assign a metric in order to talk about conjugate points. Yes, first results about multiplicities of conjugate points are derived starting with a bi-invariant metric on $G$. On the other hand, the question itself starts with a metric which is known to be only left-invariant but such that the above results about conjugate points are assumed to hold and then asks whether this metric has to be bi-invariant. –  Claudio Gorodski Dec 4 '11 at 11:43
    
Thanks Claudio, now the question becomes more transparent for me. I was at first a bit confused as my "answer" probably depicts. But I think your question is important - I will ponder about it further. Just to clarify: You have the derived results, they are necessray for a bi-invariant metric. Now you want to test if they are sufficient for a left-invariant metric on a (semi-simple, compact) Lie group to be a bi-invariant metric on precisely that group. Am I right there? –  gggg gggg Dec 4 '11 at 12:59
    
@Claudio: Just to clarify: I don't see in the definition of a symmetric space the necessity for the underlying manifold (symmetric space in the Riemannian sense) to be equipped with a bi-invariant metric. However, it can be shown that the curvature tensor takes a paricular simple form. In Jost's "Riemannian geometry and Geometric Analysis" chapter 5. There, no bi-invariance is needed in order to calculate the curvature tensor so I am a bit confused. So far, everything seems pretty much like a dead end. suggestion: Couldn't you test the hypothesis via a computer software for some sample groups –  gggg gggg Dec 5 '11 at 8:15
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