Let $$G$$ be Lie group, we denote by $$\mathfrak g$$ it’s Lie algebra. Let $$L_x,R_x$$ be the differentials of $$y \mapsto x\,y$$ and $$y\mapsto y\,x^{-1}$$ at the identity $$e \in G$$.

A bi-invariant metric on $$G$$ is a Riemannian metric on $$G$$ such that $$\langle X , Y \rangle_x = \langle L_y R_z X , L_y R_z Y \rangle_{yxz^{-1}}$$ for all $$x,y,z \in G$$ and $$X,Y \in T_x G$$. We begin by characterizing such metrics.

Proposition 1 There is a correspondence between bi-invariant metrics on $$G$$ and Ad-invariant scalar products on it’s Lie algebra.

Proof: Given a scalar product $$\langle\ ,\ \rangle_e$$ on $$\mathfrak g$$ we define (using the same notation) on $$T_g G$$ a scalar product $$\langle X , Y \rangle_g = \langle L_{g^{-1}} X, L_{g^{-1}} Y \rangle_e$$, this is a left-invariant metric on $$G$$, now given $$X,Y \in T_xG$$ then given two vectors tangent at $$x$$,

\begin{aligned}\langle R_y X, R_y Y \rangle_{xy^{-1}} &= \langle Ad_y L_{x^{-1}} X, Ad_y L_{x^{-1}}Y\rangle_e \\ &= \langle L_{x^{-1}} X, L_{x^{-1}}Y\rangle_e \\ &= \langle X, Y \rangle_x\end{aligned}.

Conversely, if we start with a bi-invariant metric, it is easy to see that it’s restriction to the Lie Algebra is a Ad-invariant. $$\Box$$

It turns out that for connected Lie groups, bi-invariant metrics on $$G$$ are actually the same as ad-invariant scalar products on it’s Lie Algebra –those who satisfy $$\langle [X,Y],Z\rangle + \langle Y,[X,Z]\rangle = 0$$, indeed

$\frac d {dt} \langle Ad_{exp(tX)} Y , Ad_{exp(tX)}Z \rangle \rvert_{t = s} = \langle [X,Ad_{exp(sX)}Y], Z \rangle + \langle Ad_{exp(sX)}Y, [X,Z]\rangle$

Since the image of the exponential map, generates the connected component of the identity, one invariance gives the other.

## A Little of Riemannian Geometry

The Riemannian connection on a manifold is uniquely determined by the formula

$\begin{array}{rcl} \langle \nabla_X Y,Z \rangle &=& \frac 1 2 \{ X\langle Y,Z\rangle + Y\langle X,Z\rangle - Z\langle X,Y\rangle \\ && - \langle [Y,Z],X \rangle - \langle [Y,Z],X \rangle -\langle [Y,X],Z \rangle \} \end{array}$

In the case of a Lie group with a bi-invariant metric, since everything is constant along $$G$$ the terms of the form $$X\langle Y,Z\rangle$$ vanish, and using the fact that the bracket of two left-invariant vector coincides with the left-invariant vector coming from $$[X_e,Y_e]\in \mathfrak g$$, the formula above implies that for left-invariant vector fields $$X,Y$$ we have

$\nabla_X Y = \frac 1 2 [X,Y]\ .$

Note that by the previous formula one parameter subgroups are geodesics, conversely given any geodesic $$\gamma$$, such that $$\gamma(0) = x, \dot \gamma(0) = L_xX$$ for $$X \in \mathfrak g$$, then $$\alpha(t) = x \exp( tX)$$ is a path such that, $$\dot \alpha(t) =L_x X (\exp(tX)) = X (\alpha (t))$$ then by uniqueness $$\alpha = \gamma$$, so geodesics starting at the identity are in 1-1 correspondence with one parameter subgroups, in particular those are defined for all $$t \in \mathbb R$$ so the existence of a bi-invariant metric implies that $$G$$ is a complete Riemannian manifold.

In particular using Jacobi identity, the tensor curvature becomes

$R(X,Y)Z =\nabla_{[X,Y]} Z - [\nabla_X,\nabla_Y ] Z = \frac 1 4 [[X,Y],Z]$

Choosing orthogonal vector fields, the sectional curvature is

$K(X,Y) = \langle R(X,Y)X, Y \rangle = \frac 1 4 \langle [X,Y],[X,Y] \rangle \geq 0$

For a unitary vector $$X$$ we choose an orthonormal basis $$\{Z_1= X, Z_2 \dots , Z_n\}$$ of $$\mathfrak g$$, And the Ricci curvature becomes

$\text{Ric}_x (X) = \sum_{i=1}^n K(X,Z_i) =-\frac 1 4 \sum_{i=1}^n \langle [X,[X,Z_i],Z_i\rangle = -\frac 1 4 \text{tr}(ad_X ad_X).$

So what does it means to have a bi-invariant metric? it’s easier to start with a ad-invariant scalar product, the most important example of a bi-linear form on a Lie Algebra which is ad-invariant –an the only example in the semisimple case– is the Killing form, defined by

$k(X,Y) = \text{tr}( ad_X ad_Y)$

this won’t be a scalar product, since it’s non degenerate if and only if the algebra is semi simple nor is going to be positive definite, but there is a special case, namely Compact Lie Algebras, where by definition the Killing form is negative definite. One remark is that since the Killing form is ad-invariant, every ideal has a orthogonal complement, so we have a decomposition $$\mathfrak g = Z(\mathfrak g) \oplus [ \mathfrak g , \mathfrak g ]$$, where $$Z(\mathfrak g)$$ is the sum of the 1-dimensional ideals and $$[ \mathfrak g , \mathfrak g ]$$ is semisimple, in particular the killing form is non degenerate in that ideal

Proposition 2: A compact Lie Group admits an bi-invariant metric.

Proof: given any scalar product $$\langle , \rangle_0$$ on $$\mathfrak g$$, choose a Haar measure on $$G$$ then define

$\forall\ X,Y \in {\mathfrak g} \qquad \langle X, Y \rangle = \int_G \langle Ad_g X, Ad_g Y \rangle_0 dg$

this is clearly a Ad-invariant scalar product on $$\mathfrak g$$, so it defines a bi invariant metric on $$G$$. $$\Box$$

Proposition 3: A Lie Group with discrete center is compact iff it’s Lie Algebra is compact.

Proof: If the Group is compact, by the previous proposition there is a ad-invariant scalar product on the Lie algebra on which $$ad_X$$ is skew-symmetric hence it has only pure imaginary eigenvalues, so the Killing form is negative semi-definite, since it is degenerate exactly on $$Z(\mathfrak g) = 0$$, then it must be negative definite. Conversely if the Lie algebra is compact, the Killing form is negative definite on $$\mathfrak g = [\mathfrak g, \mathfrak g]$$ so taking it’s negative, we have a bi-invariant metric on $$G$$, for $$k$$ is negative definite using Bonnet Myers Theorem $$G$$ must be compact. $$\Box$$

As a corollary, the universal cover must be also compact and the fundamental group of $$G$$ must be finite. with small modifications we have the following theorem

Theorem 4: A Lie algebra is compact if and only if it is the Lie algebra of a compact simply connected semisimple Lie Group The Killing Form on a Lie Algebra is negative semidefinite if and only if it is the Lie Algebra of a product of a Euclidean space and a compact simply connected semisimple Lie Group

In particular every compact Lie group must be a quotient, by a discrete subgroup of

$\mathbb T ^n \times K_1 \times \dots \times K_m$

where all the $$K_i$$’s are simply connected compact simple Lie groups. there is a theorem who states that any Lie Groups deformation retracts into a maximal compact subgroup, combining that and the fact that cover maps induce isomorphisms on the fundamental groups $$\pi_n$$ for $$n \geq 2$$ we have

Proposition 5: For any Lie Group $$\pi_2(G)$$ is trivial.

Proof: we only have to check the list of simple simply connected compact Lie groups, these are $$SU(n)$$ and $$SO(n)$$, using the fibrations

$SO(n-1) \rightarrow SO(n) \rightarrow S^n$ $SU(n-1) \rightarrow SU(n) \rightarrow S^{2n-1},$

it’s easy to show that $$\pi_2$$ is trivial, it left as an exercise to check the same for the compact exceptional groups. $$\Box$$