On the existence of bi-invariant Finsler metrics on Lie groups

Abstract

In this paper, we study the geometry of Lie groups with bi-invariant Finsler metrics. We first show that every compact Lie group admits a bi-invariant Finsler metric. Then, we prove that every compact connected Lie group is a symmetric Finsler space with respect to the bi-invariant absolute homogeneous Finsler metric. Finally, we show that if G is a Lie group endowed with a bi-invariant Finsler metric, then, there exists a bi-invariant Riemanninan metric on G such that its Levi-Civita connection coincides the connection of F.

Introduction

Lie groups are the most beautiful and most important manifolds. On the one hand, these spaces contain many prominent examples which are of great importance for various branches of mathematics, like homogeneous spaces, symmetric spaces, and Grassmannians. On the other hand, these spaces have much in common, and there exists a rich theory. The study of invariant structures on Lie groups and homogeneous spaces is an important problem in differential geometry. Milnor’s research on the properties of invariant Riemannian metrics on a Lie group obtained many interesting and significant results. He computed the connections of these metrics and obtained the formula for geodesics and curvatures. Lie groups are, in a sense, the nicest examples of manifolds and are good spaces on which to test conjectures[1]. Therefore, it is important to study invariant Finsler metrics. In[2], the authors studied invariant Finsler metrics on homogeneous spaces and gave some descriptions of these metrics. Also in[3] and[4], we have studied the homogeneous Finsler spaces and the homogeneous geodesics in homogeneous Finsler spaces.

Among the invariant metrics, the bi-invariant ones are the simplest kind. They have nice and simple geometric properties but still form a large enough class to be of interest. In[5], we have studied the geometry of Lie groups with bi-invariant Randers metrics, and in[6], we have studied the naturally reductive Randers metrics on homogeneous manifolds.

In this paper, we study the geometry of Lie groups with bi-invariant Finsler metrics. We first show that every compact Lie group admits a non-Riemannian bi-invariant Finsler metric. Then, we prove that every compact connected Lie group is a symmetric Finsler space with respect to the bi-invariant absolutely homogeneous Finsler metric. Finally, we show that If G is a Lie group endowed with a bi-invariant Finsler metric, then there exists a bi-invariant Riemanninan metric on G such that its Levi-Civita connection coincides the connection of F.

Bi-invariant Finsler metrics on Lie groups

A Finsler metric on a manifold M is a continuous function, F : T M →[ 0, ∞) differentiable on T M∖{0} and satisfying three conditions[7]:

• (a) F(y) = 0 if and only if y = 0.

• (b) F(λ y) = λ F(y) for any y ∈ T _x M and λ > 0.

• (c) For any non-zero y ∈ T _x M, the symmetric bilinear form g _y  : T _x M × T _x M → R given by:

  $${\mathrm{g}}_y(u,v)=\frac{1}{2}\frac{{\partial }^2}{\mathrm{\partial s\partial t}}\left[F^2\right(y+\mathit{\text{su}}+\mathit{\text{tv}}\left)\right]{|}_{s=t=0}$$

is positive definite.

For each y ∈ T _x M − {0}, define:

$$C_y(u,v,w)=\frac{1}{4}\frac{{\partial }^3}{\partial s\partial t\partial r}\left[F^2\right(y+\mathit{\text{su}}+\mathit{\text{tv}}+\mathit{\text{rw}}\left)\right]{\mid }_{s=t=r=0}.$$

C is called the Cartan torsion.

Let G be a connected Lie group with Lie algebra$\mathfrak{g}=T_{e}G$. We may identify the tangent bundle TG with$G\times \mathfrak{g}$ by means of the diffeomorphism that sends (g, X) to (L _g )_∗X ∈ T _g G.

Definition 1.

A Finsler function F : T G → R₊ will be called G-invariant (left-invariant) if F is constant on all G-orbits in$\mathit{\text{TG}}=G\times \mathfrak{g}$; that is, F(g, X) = F(e, X) for all g ∈ G and$X\in \mathfrak{g}$.

The G-invariant Finsler functions on TG may be identified with the Minkowski norms on$\mathfrak{g}$. If F : T G → R₊ is an G-invariant Finsler function, then, we may define$\tilde{F}:\mathfrak{g}\to R_{+}$ by$\tilde{F}\left(X\right)=F(e,X)$, where e denotes the identity in G. Conversely, if we are given a Minkowski norm$\tilde{F}:\mathfrak{g}\to R_{+}$, then$\tilde{F}$ arises from an G-invariant Finsler function F : T G → R₊ given by$F(g,X)=\tilde{F}\left(X\right)$ for all$(g,X)\in G\times \mathfrak{g}$.

Similarly, a Finsler metric is right-invariant if each R _a  : G → G is an isometry.

Definition 2.

A Finsler metric on G that is both left-invariant and right-invariant is called bi-invariant.

Let G be a compact Lie group. Fix a base ω₁, ω₂, …, ω _n in T _e G and put ω = ω₁∧ … ∧ω _n where n = d i m G. Extend ω to a left-invariant differential form ω on G by putting Ω _g  = (L _g )^∗ω. The form ω never vanishes. The form determines an orientation of G. Recall that a chart with coordinates x¹, …, xⁿ is called positively if d x¹∧ … ∧d xⁿ = f Ω where f is a positive function defined on the coordinate neighborhood. Clearly, the atlas consisting of all positively oriented charts determines an orientation on G. Indeed, if d y¹∧ … ∧d yⁿ = h Ω and h > 0, then:

$$d{y}^1\wedge \dots \wedge d{y}^n=\frac{h}{f}d{x}^1\wedge \dots \wedge d{x}^n.$$

On the other hand,

$$d{y}^1\wedge \mathrm{...}\wedge d{y}^n=\frac{\partial (y^1,\dots ,y^n)}{\partial (x^1,\dots ,x^n)}d{x}^1\wedge \dots \wedge d{x}^n,$$

so

$$\frac{\partial (y^1,\dots ,y^n)}{\partial (x^1,\dots ,x^n)}=\frac{h}{f}>0.$$

Now, for any a ∈ G, we can easily see that$R_a^{\ast }\mathrm{\Omega }$ is left-invariant. It follows that$R_a^{\ast }\mathrm{\Omega }=f\left(a\right)\mathrm{\Omega }$. We can easily see that f(a b) = f(a)f(b) that is f:G → R − {0} is a continuous homomorphism of G into the multiplicative group of real numbers. Since f(G) is compact connected subgroup, the conclusion f(G) = 1 holds. Therefore,$R_a^{\ast }\mathrm{\Omega }=\mathrm{\Omega .}$ So ω is bi-invariant volume element on G.

Theorem 1.

Every compact Lie group admits a bi-invariant Finsler metric.

Proof

Let ω be the bi-invariant volume element. Let F _e be a Minkowski norm on$T_{e}G=\mathfrak{g}$. Then, define the function${\tilde{F}}_e$ on T _e G by:

$${\tilde{F}}_e^2\left(X\right)={\int }_G{F}_e^2\left(A{d}_{g}X\right)\mathrm{\Omega .}$$

Let K₁, …, K _N be a covering of G by cubes, and that let ϕ₁, ..., ϕ _N be a corresponding partition of unity, and Ω = Ω_12 … nd x¹∧ … ∧d xⁿ. Now we can write:

$$\begin{array}{c}{\tilde{F}}_e^2\left(X\right)={\int }_G{F}_e^2\left(A{d}_{g}X\right)\mathrm{\Omega }=\sum _{i=1}^{N}n!{\int }_{K_i}{F}_e^2\left({\varphi }_{i}A{d}_{g}X\right){\mathrm{\Omega }}_{12\mathrm{...n}}d{x}^1\wedge \dots \wedge d{x}^n.\end{array}$$

By definition of the orientation, Ω_{12 ... n} > 0, and ϕ _i (x) is positive in K _i ; furthermore,$F_e^2\left(A{d}_{g}X\right)>0$ if X ≠ 0. Since all summands in the above expression for${\int }_G{F}_e^2\left(A{d}_{g}X\right)\mathrm{\Omega }$ are positive, we come to${\int }_G{F}_e^2\left(A{d}_{g}X\right)\mathrm{\Omega }>0$. So,${\tilde{F}}_e$ is well-defined function${\tilde{F}}_e:T_{e}G\to [0,\infty ),{\tilde{F}}_e\left(X\right)>0\mathit{\text{ifX}}\ne 0$. We can easily see that${\tilde{F}}_e\left(X\right)=0$ if and only if X = 0. Clearly${\tilde{F}}_e\left(\mathrm{\lambda X}\right)=\lambda {\tilde{F}}_e\left(X\right)$ for any X ∈ T _e G, λ > 0.

Since F _e is C^∞ on T _e G − {0}, we see that${\tilde{F}}_e$ is C^∞ on T _e G − {0}. Now for any y ≠ 0, u, v ∈ T _e G by a direct computation, we have:

$${\tilde{g}}_y(u,v)={\int }_G{g}_{A{d}_g\left(Y\right)}\left(A{d}_g\right(u),A{d}_g(v\left)\right)\mathrm{\Omega .}$$

Since$g_{A{d}_g\left(Y\right)}$ is positively definite, hence${\tilde{g}}_y$ is positive definite. So$\tilde{F}\left(e\right)$ is a Minkowski norm on T _e G. In the following, we show that${\tilde{F}}_e$ is A d(G)−invariant i.e.:

$$\forall h\in G,{\tilde{F}}_e\left(A{d}_{\left(h\right)}X\right)={\tilde{F}}_e\left(X\right).$$

Applying the definition of${\tilde{F}}_e$ to the left side of this expression, we find that:

$${\tilde{F}}_e^2\left(A{d}_{h}X\right)={\int }_G{F}_e^2\left(A{d}_{g}A{d}_{h}X\right)\mathrm{\Omega }={\int }_G{F}_e^2\left(A{d}_{\mathit{\text{gh}}}X\right)\mathrm{\Omega .}$$

The diffeomorphism L _h and R _h preserve the orientation because ω is bi-invariant. So, for the diffeomorphism$I_h=R_{x^{-1}}\circ L_x:G\to G$, we have:

$${\int }_{I_h\left(G\right)}{F}_e^2\left(A{d}_{g}X\right)\mathrm{\Omega }={\int }_G{F}_e^2\left(A{d}_{\mathit{\text{gh}}}X\right)R_h^{\ast }\mathrm{\Omega .}$$

Since I _h (G) = G and$R_h^{\ast }\mathrm{\Omega }=\mathrm{\Omega }$, we see that:

$${\tilde{F}}_e^2\left(A{d}_{h}X\right)={\tilde{F}}_e^2\left(X\right).$$

Thus,${\tilde{F}}_e\left(A{d}_{h}X\right)={\tilde{F}}_e\left(X\right)$.

Extend the Minkowski norm in T _e G, thus defined to a left-invariant Finsler metric on G, by putting:

$$\tilde{F}\left(X\right)={\tilde{F}}_e\left({\left(L_{a^{-1}}\right)}_{\ast a}X\right),$$

whenever X ∈ T _a G.

We show that this Finsler metric is bi-invariant. We only need to check the right-invariance, we have:

$$\tilde{F}\left({\left(R_a\right)}_{\ast }X\right)=\tilde{F}\left({\left(L_{a^{-1}}\right)}_{\ast }{\left(R_a\right)}_{\ast }X\right).$$

Furthermore,$A{d}_{a^{-1}}={\left(L_{a^{-1}}\right)}_{\ast }{\left(R_a\right)}_{\ast }$. Consequently, by A d−invariance of$\tilde{F}$:

$$\tilde{F}\left({\left(R_a\right)}_{\ast }X\right)=\tilde{F}\left(A{d}_{a^{-1}}X\right)=\tilde{F}\left(X\right).$$

A connected Riemannian manifold M is said to be symmetric Riemannian space if to each p ∈ M there is an associated isometry σ _p :M → M which is (i) involutive, and (ii) has p as an isolated fixed point. As an example, every compact connected Lie group G is a symmetric Riemannian space with respect to the bi-invariant Riemannian metric[8, 9].

The definition of symmetric Finsler space is a natural generalization of the definition of symmetric Riemannian spaces[3].

Definition 3.

A connected Finsler space (M, F) is said to be symmetric, if to each p ∈ M there is associated an isometry s _p :M → M which is:

• (a) involutive ( $s_p^2$ is the identity).

• (b) has p as an isolated fixed point, that is, there is a neighborhood U of p in which p is the only fixed point of s _p .

As p is an isolated fixed point of σ _p , it follows that (d σ _p ) _p  = −i d, and therefore, symmetric Finsler spaces have reversible metrics and geodesics.

In the following theorem similar to the Riemmanian case, we show that Lie groups with bi-invariant absolutely homogeneous Finsler metrics are symmetric Finsler space.

Theorem 2.

Every compact connected Lie group G is a symmetric Finsler space with respect to the bi-invariant absolutely homogeneous Finsler metric.

Proof

Let ψ : G → G denote the inversion map g → g⁻¹. Clearly ψ is involutive and is an isometry of G with e as isolated fixed point. Let X∈T _e G, since ψ(exp(t X)) = exp(−t X) we obtain:

$${\psi }_{\ast }\left(X_e\right)=\frac{d}{\mathit{\text{dt}}}{|}_{t=0}\psi (exp(\mathit{\text{tX}}\left)\right)=\frac{d}{\mathit{\text{dt}}}{|}_{t=0}exp(-\mathit{\text{tX}})=-X.$$

This means that ψ_∗e = −I d, hence ψ is an isometry of G, and e is an isolated fixed point of ψ.

For every g ∈ G, clearly$\psi =R_{g^{-1}}\circ \psi \circ L_{g^{-1}}$. Therefore${\psi }_{\ast g}:\mathfrak{g}\to \mathfrak{g}$ is an isometry for any g ∈ G. Now for g ∈ G, Let σ _g (x) = g x⁻¹g, x ∈ G. The mapping σ _g is an isometry because:

$${\sigma }_g=R_g\circ \psi \circ R_{g^{-1}}.$$

Obviously, σ _g fixing the point g and is involutive${\sigma }_g^2\left(x\right)=g{\left(g{x}^{-1}g\right)}^{-1}g=X.$ Now, it suffices to show that g is isolated fixed point. If$X\in \mathfrak{g}$, then for an arbitrary g ∈ G, we have:

$$\begin{array}{c}{\left({\sigma }_g\right)}_{\ast }X={\left(R_g\right)}_{\ast }{\psi }_{\ast }{\left(R_{g^{-1}}\right)}_{\ast }\left(X\right)={\left(R_g\right)}_{\ast }(-{\left(R_{g^{-1}}\right)}_{\ast }X)=-X,\end{array}$$

and the proof is complete.

Theorem 3.

If G is a Lie group endowed with an absolutely homogeneous bi-invariant Finsler metric, then, the geodesics through the identity of G are exactly one-parameter subgroups.

Proof

Suppose γ(t) is a geodesic in G with e = γ(0). We show that γ(t) is a one-parameter subgroup. With the help of symmetries, we have σ_γ(c)σ _e γ(t) = γ(t + t c). Since σ_γ(c)σ _e (x) = γ(c)x γ(c), so γ(c)γ(t)γ(c) = γ(t + 2c). In particular, γ(2c) = γ(c)², and by an induction, we have γ(n c) = γ(c)ⁿ. Now, if$\frac{c_1}{c_2}\in Q$, c₁ = α n₁, c₂ = α n₂ where n₁ and n₂ are integers, then:

$$\begin{array}{ll}\gamma (c_1+c_2)& =\gamma \left(\right(n_1+n_2\left)\alpha \right)=\gamma {\left(\alpha \right)}^{n_1+n_2}=\gamma {\left(\alpha \right)}^{n_1}\gamma {\left(\alpha \right)}^{n_2}\\ =\gamma \left(c_1\right)\gamma \left(c_2\right).\end{array}$$

Now, by the continuity of γ(t) for arbitrary c₁ and c₂, we have:

$$\gamma (c_1+c_2)=\gamma \left(c_1\right)\gamma \left(c_2\right).$$

Hence, γ(t) is a one-parameter subgroup of G.

Now, we show that one-parameter subgroups are geodesics. Suppose γ(t) is a one-parameter subgroup. Let$\xi =\frac{\mathrm{d\gamma }\left(t\right)}{\mathit{\text{dt}}}{|}_{t=0}$ be the tangent vector of γ(t) at e. There is a geodesic x(t) through e determined by ξ. We have already shown that x(t) is a one-parameter subgroup, so x(t) = γ(t), and γ(t) is a geodesic.

Definition 4.

Let G be a connected Lie group,$\mathfrak{g}=T_{e}G$ its Lie algebra identified with the tangent space at the identity element,$\tilde{F}:\mathfrak{g}\to R_{+}$ a Minkowski norm and F the left-invariant Finsler metric induced by$\tilde{F}$ on G. A geodesic γ:R → G is said to be homogeneous if there is a$Z\in \mathfrak{g}$ such that γ(t) = exp(t Z)γ(0), t∈R holds. A tangent vector X ∈ T _e G − {0} is said to be a geodesic vector if the 1-parameter subgroup t→ exp(t X), t ∈ R, is a geodesic of F.

For results on homogeneous geodesics in homogeneous Finsler manifolds, we refer to[4, 10–12]. The basic formula characterizing geodesic vector in the Finslerian case was derived in[4], Theorem 3.1. For Lie groups with left invariant metrics, we have the following theorem.

Theorem 4.

[4] Let G be a connected Lie group with Lie algebra$\mathfrak{g}$, and let F be a left-invariant Finsler metric on G. Then$X\in \mathfrak{g}-\left\{0\right\}$ is a geodesic vector if and only if:

$$g_X(X,[X,Z\left]\right)=0$$

holds for every$Z\in \mathfrak{g}$.

Corollary 1.

If G is a Lie group endowed with a bi-invariant Finsler metric, then, the geodesics through the identity of G are homogeneous geodesics.

Theorem 5.

Let G be a Lie group with a bi-invariant Finsler metric F. Then, there exists a bi-invariant Riemanninan metric on G such that its Levi-Civita connection coincides the connection of F.

Proof

If F is a bi-invariant Finsler metric on G, as:

$$A{d}_g={\left(R_{a^{-1}}\right)}_{\ast a}\circ {\left(L_a\right)}_{\ast e},$$

it is clear that A d _g is an isometry on$\mathfrak{g}$. Now define:

$$I_e=\{x\in \mathfrak{g}|F\left(x\right)=1\}.$$

Then, the adjoint group$\mathit{\text{Ad}}\left(G\right)\subseteq \mathit{\text{GL}}\left(\mathfrak{g}\right)$ leaves I _e invariant. Let G₁ be the subgroup of the general linear group$\mathit{\text{GL}}\left(\mathfrak{g}\right)$ consisting of the elements which leaves I _e invariant. Then, G₁ is a compact Lie group, and A d(G) is a subgroup of G₁. So, A d(G) has compact closure in$\mathit{\text{GL}}\left(\mathfrak{g}\right)$. Therefore, the Lie group G admits a bi-invariant Riemannian metric g.

For (G, F) and (G, g), we have that the geodesics of (G, F) and (G, g) coincide, and hence (G, F) is a Berwald space. Now, we show that the Levi-Civita connection of (G, g) coincides the connection of (G, F).

Let ∇ be the connection of F. For any left invariant vector fields X, Y, Z on G, we have:

$$Y{g}_X(Z,X)=g_X({\nabla }_{Y}Z,X)+g_X(Z,{\nabla }_{Y}X)$$

(1)

Similarly,

$$Z{g}_X(Y,X)=g_X({\nabla }_{Z}Y,X)+g_X(Y,{\nabla }_{Z}X)$$

(2)

$$X{g}_X(Z,X)=g_X({\nabla }_{X}Z,X)+g_X(Z,{\nabla }_{X}X)$$

(3)

All covariant derivatives have$\tilde{X}$ as reference vector.

Subtracting (2) from the summation of (1) and (3), we get:

$$\begin{array}{ll}{g}_X(Z,{\nabla }_{X+Y}X)& +g_X(X-Y,{\nabla }_{Z}X)=Y{g}_X(Z,X)-Z{g}_X(Y,X)\\ +X{g}_X(Z,X)-g_X\left(\right[Y,Z],X)-g_X\left(\right[X,Z],X),\end{array}$$

where we have used the symmetry of the connection, i.e., ∇ _Z X − ∇ _X Z = [ Z, X]. Set Y = X − Z in the above equation, we obtain:

$$\begin{array}{c}2g_X(Z,{\nabla }_{X}X)=2X{g}_X(Z,X)-Z{g}_X(X,X)-2g_X\left(\right[X,Z],X).\end{array}$$

(4)

Since F is left-invariant, d L _x is a linear isometry between the spaces$T_{e}G=\mathfrak{g}$ and T _x G, ∀x ∈ G. Therefore, for any left-invariant vector field X, Z on G, we have:

$$g_X(Z,X)=g_{X_e}(Z_e,X_e)$$

i.e., the functions g _X (Z, X), g _X (X, X) are constant. Therefore, from (4) and Theorem 4, the following is obtained:

$$g_X(Z,{\nabla }_{X}X){\mid }_e=-g_X\left(\right[X,Z],X){\mid }_e=-g_X\left(\right[X,Z],X)=0.$$

Consequently for any left invariant X, we have:

$${\nabla }_{X}X=0.$$

Now using the identity:

$${\nabla }_{X}Y=\frac{1}{2}\left({\nabla }_{X+Y}\right(X+Y)-{\nabla }_{X}X-{\nabla }_{Y}Y+[X,Y\left]\right),$$

we get:

$${\nabla }_{X}Y=\frac{1}{2}[X,Y].$$

Consequently, the assertion of the theorem follows. □

Corollary 2.

If a connected Lie group G admits a bi-invariant Finsler metric, then, it is isomorphic to the cartesian product of a compact group and an additive vector group.