## Lie groups with bi-invariant Riemannian metric

This is a sequel to the previous post on the exponential maps of Lie groups. In this note we record some results which are derived using Ken’s theorems. In particular we impose a topological restriction on Lie groups with bi-invariant metric.

Proposition 1 If ${G}$ is a Lie group with a bi-invariant metric, let ${\nabla}$ be its Levi-Civita connection, then for any left-invariant vector field ${X, Y}$

$\displaystyle \nabla_XY=\frac{1}{2}[X, Y]$

where ${[\cdot, \cdot]}$ is the Lie bracket.

Proof: Note that ${\nabla_XX=0}$ at ${e}$, because the integral curve of ${X}$ is exactly ${\gamma(t)=\exp(tX)}$, and by Ken’s theorem, this is a geodesic, so ${\nabla_X X=\nabla_{\gamma'}\gamma'=0}$. In general, for ${g\in G}$, as ${L_g}$ is an isometry, we have

$\displaystyle \nabla_{X_g}X= dL_g(\nabla_{X_e}X)=0.$

For left-invariant ${X, Y}$,

$\displaystyle 0=\nabla_{X+Y}{(X+Y)}= \nabla_XY+ \nabla_YX.$

Together with the torsion-free condition ${\nabla_XY-\nabla_YX=[X, Y]}$, we can get the result. $\Box$

For any vectors ${X, Y\in T_eG}$, there is a natural Lie bracket on ${T_eG}$ defined as follows. Extend ${X, Y}$ to be the unique left-invariant vector fields on ${G}$, still denoted by ${X, Y}$. We can then define the vector field ${[X,Y]}$ on ${G}$. We then define the restriction of this vector field on ${T_eG}$ to be ${[X, Y]}$ (same notation as the vector field Lie bracket!). Thus we will always identify left-invariant vector fields with vectors in ${T_eG}$, which are by definition isomorphic as Lie algebra.

Proposition 2 If ${G}$ is a Lie group with bi-invariant metric, then for ${X, Y, Z\in T_eG}$,

$\displaystyle \langle [X,Y],Z\rangle=\langle [Y,Z],X\rangle=\langle [Z,X],Y\rangle.$

We first set up some notations. Recall that ${Ad_g\in Aut(T_eG)}$ is defined as

$\displaystyle Ad_g= dL_g dR_{g^{-1}}.$

Regarding ${Ad}$ as a map (indeed group homomorphism) ${Ad: G\rightarrow Aut(T_eG)}$, noting that the tangent space of ${Aut(T_eG)}$ at ${id}$ is ${End(T_eG)}$ [${T_eG}$ is just a finite dimensional vector space, by fixing a basis, ${Aut(T_eG)}$ is just the space of invertible matrices, which is open in the space of all matrices, thus its tangent space is the space of all matrices, intrinsically(without basis) this is ${End(T_eG)}$ ], by taking the differential of ${Ad}$ at ${e}$, we have a map ${ad=d_e(Ad)}$

$\displaystyle \begin{array}{rcl} ad: T_eG&\rightarrow &End(T_eG), \\ X&\mapsto &ad_X \end{array}$

so that for any ${X\in T_eG}$, there is a linear map

$\displaystyle \begin{array}{rcl} ad_X: T_eG&\rightarrow &T_eG\\ Y&\mapsto &ad_XY \end{array}$

Proposition 3 For ${X,Y\in T_eG}$,

$\displaystyle ad_XY=[X, Y].$

We briefly recall that the geometric meaning of ${[X, Y]}$ is as follows:

$\displaystyle [X, Y]= \left.\frac{d}{dt}\right|_{t=0} {(\phi_{-t})}_* Y$

where ${\phi_t}$ is the flow of ${X}$.

Proof: Let ${\alpha(s,t)=c(\gamma^X(s))\gamma^Y(t)}$ where ${c(g)}$ is the conjugate map by ${g}$ and ${\gamma^X(s)=\exp(sX), \gamma^Y(t)=\exp(tY)}$. Then

$\displaystyle Ad_{\gamma^X(s)} Y= \left.\frac{\partial}{\partial t}\right|_{t=0}\alpha(s,t)$

and

$\displaystyle ad_XY=\left.\frac{\partial^2}{\partial s\partial t}\right|_{s=0,t=0}\alpha(s,t).$

So for a function ${f}$,

$\displaystyle (ad_XY)f= \left.\frac{\partial^2}{\partial s \partial t}\right|_{s=0, t=0}f(\alpha(s,t)) =\left.\frac{\partial^2}{\partial s \partial t}\right|_{s=0, t=0}f(\gamma ^X(s) \gamma^Y(t)\gamma ^X(-s))$

Applying chain rule to the function ${(u,v,w)\mapsto f(\gamma ^X(u) \gamma^Y(v)\gamma ^X(-w)}$, we have

$\displaystyle (ad_XY)f= \left.\frac{\partial^2}{\partial s \partial t}\right|_{s=0, t=0}f(\gamma ^X(s) \gamma^Y(t))-\left.\frac{\partial^2}{\partial s \partial t}\right|_{s=0, t=0}f( \gamma^Y(t)\gamma ^X(s)). \ \ \ \ \ (1)$

Consider the first term on RHS, and for fixed ${g=\gamma^X(s)}$, ${g\gamma^Y(t)}$ is the integral curve of ${Y}$ starting at ${g}$, so

$\displaystyle \left.\frac{\partial}{ \partial t}\right|_{ t=0}f(\gamma ^X(s) \gamma^Y(t))= Y(f(\gamma^X(s))).$

Thus by taking the derivative w.r.t. ${s=0}$ of the above,

$\displaystyle \left.\frac{\partial^2}{\partial s \partial t}\right|_{s=0, t=0}f(\gamma ^X(s) \gamma^Y(t))=XY(f).$

For the second term on the RHS of (1), we switch the order of differentiation first and apply the same analysis to get ${\left.\frac{\partial^2}{\partial s \partial t}\right|_{s=0, t=0}f(\gamma^Y(t)\gamma ^X(s) )=YX(f)}$. Therefore ${ad_XY(f)=[X, Y]f}$, we conclude that

$\displaystyle ad_XY=[X, Y].$

$\Box$

We now prove Proposition 2.

Proof: [Proof of Proposition 2]
As ${\langle\cdot, \cdot \rangle}$ is bi-invariant, ${\langle \cdot, \cdot\rangle_{e}}$ is Ad-invariant. So for all ${g}$,

$\displaystyle \langle Ad_g Y, Ad_g Z\rangle=\langle Y,Z\rangle.$

so if ${g=\exp(tX)}$, by taking the derivative of above, and by Proposition 3

$\displaystyle 0=\langle ad_X Y, Z\rangle + \langle Y, ad_X Z\rangle =\langle [X, Y], Z\rangle +\langle Y , [X,Z]\rangle.$

By the skew-symmetry of ${[\cdot, \cdot]}$, rearranging, we can get the result. $\Box$

We have the following important Jacobi identity.

Proposition 4 [Jacobi identity] For any vector field ${X, Y, Z}$ on a manifold, we have

$\displaystyle [X, [Y, Z]]+[Y, [Z, X]]+[Z, [X, Y]]=0.$

Proof: (This of course can be verified directly by examining its action on a function ${f}$. But this seems unnatural and is not easy to understand. We here show that this is just the familiar product rule as in calculus. ) Here we write ${L_XY}$ instead of ${[X, Y]}$. We prove the following product law first:

$\displaystyle L_X[Y, Z]=[L_X Y, Z]+[Y, L_X Z].$

For convenience let the one parameter family of vector field ${d\phi_t(Y)}$ induced by the flow ${\phi_t}$ of ${X}$ be denoted by ${Y_t}$. It is easy to see that we have the invariance of Lie bracket:

$\displaystyle d\phi_t[X, Y]=[d\phi_t (Y), d\phi_t (Z)]=[Y_t, Z_t].$

Under local coordinates,

$\displaystyle [Y_t, Z_t]=(Y_t(Z^j)-Z_t(Y^j))\frac{\partial}{\partial x^j}.$

So using product rule, (with ${\cdot}$ = derivative w.r.t. ${t}$)

$\displaystyle \frac{d}{dt}[Y_t, Z_t] =(\dot{Y_t}(Z_t^j)-Z_t(\dot{Y_t^j}))\frac{\partial}{\partial x^j} + ({Y_t}(\dot{Z_t^j})-\dot{Z_t}({Y_t^j}))\frac{\partial}{\partial x^j}\\ =[\dot Y_t, Z_t] + [Y_t, \dot Z_t].$

Thus

$\displaystyle L_X[Y, Z] =\left.\frac{d}{dt}\right|_{t=0}d\phi_t([Y, Z]) =[\dot Y (0), Z(0)] + [Y(0), \dot Z(0)] =[L_X Y , Z] + [Y, L_X Z]$

where ${Y(0)=Y_{t=0}=Y}$. As ${[\cdot,\cdot]}$ is anti-symmetric, by rearranging the above equation,

$\displaystyle \begin{array}{rcl} 0=L_X[Y, Z]-[L_X Y, Z]-[Y, L_X Z] &=&[X, [Y, Z]]-[[X, Y], Z]-[Y, [X, Z]]\\ &=&[X, [Y, Z]]+[Y, [Z, X]]+[Z, [X, Y]]. \end{array}$

$\Box$

Finally, we compute the curvature of ${G}$ with a bi-invariant metric. Recall that the Riemannian curvature tensor is defined as (by definition of ${\nabla^2}$, this is a tensor)

$\displaystyle R(X, Y)Z:= \nabla^2Z(X,Y)-\nabla^2Z(Y,X).$

By an easy computation, this is equal to

$\displaystyle R(X, Y)Z=\nabla_X \nabla_Y Z-\nabla_Y\nabla_X Z-\nabla_{[X,Y]}Z.$

Theorem 5 For left-invariant vector fields ${X,Y,Z }$ on a Lie group ${G}$ with bi-invariant metric, we have

$\displaystyle R(X, Y)Z= -\frac{1}{4}[[X,Y],Z].$

Proof: Note that by invariance of Lie bracket, ${[X,Y]}$ etc are also left-invariant. By Proposition 1,

$\displaystyle \begin{array}{rcl} R(X, Y)Z&=&\nabla_X \nabla_Y Z-\nabla_Y\nabla_X Z-\nabla_{[X,Y]}Z\\ &=&\frac{1}{2}\nabla_X [Y,Z]- \frac{1}{2}\nabla_Y[X,Z]- \frac{1}{2}[[X,Y],Z]\\ &=&(\frac{1}{4}[X, [Y,Z]]- \frac{1}{4}[Y,[X,Z]])- \frac{1}{2}[[X,Y],Z]\\ &=& \frac{1}{4}[[X,Y],Z]- \frac{1}{2}[[X,Y],Z]\\ &=&- \frac{1}{4}[[X,Y],Z] \end{array}$

where we have used the Jacobi identity (Proposition 4) on the forth line. $\Box$

For ${G}$ with bi-invariant metric, for orthonormal ${X, Y\in T_eG}$, by the left-invariance of the metric, if we uniquely extend ${X, Y}$ as left-invariant vector fields on ${G}$, then ${X}$ and ${Y}$ are still orthonormal at each point of ${G}$, thus they span a two-plane ${\Pi_{XY}}$ in each tangent space ${T_gG}$. We define the sectional curvature of ${\Pi_{XY}}$ by

$\displaystyle K(X,Y)= \langle R(X, Y)Y, X\rangle.$

Theorem 6 [Ken’s theorem, 2011] If ${G}$ is a Lie group with a bi-invariant metric. Let ${X,Y\in T_eG}$ be orthonormal and left-invariantly extend them. Then at each point

$\displaystyle K(X, Y)= \frac{1}{4}\left|[X, Y]\right|^2.$

Proof: By Theorem 5,

$\displaystyle \langle R(X, Y)Y, X\rangle= - \frac{1}{4}\langle [[X,Y],Y], X\rangle.$

By Proposition 2,

$\displaystyle \langle [[X,Y],Y], X\rangle=\langle [Y,X],[X,Y]\rangle=-\left|[X,Y]\right|^2.$

$\Box$

Corollary 7 [Ken’s theorem, 2011] If a Lie group ${G}$ has a bi-invariant metric (for example ${G}$ compact by Ken’s theorem), it must have non-negative sectional curvature.

Remark 1 There’re some topological restrictions on such $G$. For example according to Soul theorem of Cheerger and Gromoll, since ${G}$ has non-negative curvature, it is diffeomorphic to the normal bundle of a compact totally geodesic (and thus also non-negatively curved by Gauss equation) submanifold. In particular $G$ must be homotopy equivalent to a compact manifold. [But does this submanifold have anything to do with the group structure I don’t know. ]