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)


\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].


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].


\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}


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.


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. ]

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