Moduli Space of Bi-Invariant Metrics

In this work, we focus on describing the space of bi-invariant metrics in a Lie group up to isometry. I.e, that is, metrics invariant under both left and right translations. We show that $\mathfrak{BI}$, the moduli space of bi-invariant metrics, is an orbifold. Moreover we give an explicit description of this orbifold, and of $\mathfrak{EBI}$, the space of bi-invariant metrics equivalent under isometries and scalar multiplies.


Introduction
In the study of Riemannian manifolds, it is of interest to show how complicated is the collection of Riemannian geometries satisfying a certain property.For example, given a manifold M we could explore the class of all Riemannian geometries defined on M , such that the curvature induced by these geometries has a fixed lower bound; we could also consider the class of all geometries that turn M into an Einstein manifold.Moreover, one can ask if given two geometries with the desired property, we can deform one into the other via geometries satisfying the given property.This can be quantified in a precise way, by considering the moduli space of all Riemannian metrics satisfying the desired property.This set consist of all the Riemannian metrics up to isometry, equipped with the C ∞ -Whitney topology (see [13]).
For a Lie group G, a natural interesting family of Riemannian geometries consist of the ones that have G contained in their isometry group, i.e. left or right invariant Riemannian metrics.A smaller subfamily consist on the Riemannian geometries that are both left and right invariant, that is bi-invariant metrics.These Riemannian metrics satisfy non-negative lower curvature bounds (see [7]), and have important applications when studying spaces with a lower curvature bound (see [9]).
In this work we study the topology of the moduli space of bi-invariant metrics of a compact Lie group, obtaining the following theorem., are semi-simple subalgebras, such that any two factors of s are non isomorphic to each other, each b i decomposes as the sum of m i isomorphic factors, no simple factor of s is isomorphic to any simple factor of any b i , and a is an abelian subalgebra.Then the moduli space of bi-invairant metrics BI(g) is a contractible orbifold homeomorphic to Here BI(s) is the space of all possible bi-invariant metrics on s, and SP mi (R) is the m i th symmetric product of R.
As mentioned before bi-invariant metrics are useful in the construction of manifolds with lower curvature bounds.For example given a Lie group acting by isometries, Cheeger [3] gave a deformation procedure that will preserve non-negative curvature bounds, by fixing a bi-invariant metric.With this deformation one can construct new spaces with positive Ricci curvature [10].Moreover, for closed manifolds of cohomogeneity one, Grove and Ziller used a fixed bi-invariant metric to show the existence of invariant Riemannian metrics with non-negative curvature [4].
Thus it is relevant to study how bi-invariant metrics relate to each other.Due to the strong geometric properties of these metrics, it is natural that the homotopy information of the moduli space of bi-invariant metrics is simple.This is in contrast to the moduli-space of left invariant metrics as observed by Kodama, Takahara and Tamaru [6].
Nonetheless, we believe it is useful to have a full description of the topology of this moduli space.In particular observe that the conclusion in Theorem 1.0.1 states that the moduli space of bi-invariant metrics of a compact Lie group is an orbifold, and moreover, we can read the orbifold stratification from the Lie algebra decomposition.
It is important to note that the characterization of Theorem 1.0.1 only relies on the group admitting a bi-invariant metric and the decomposition of the semisimple factor in the Lie algebra, not on the compactness of the Lie group.The compactness stated in Theorem 1.0.1 is stated to guarantee the existence of a biinvariant metric.We recall that there are examples of non-compact and non-abelian simple Lie groups, such as SL(2, C), that do not admit bi-invariant metrics.
We start in Section 2 by collecting results that characterize Lie groups and Lie algebras that possess bi-invariant metrics.Specifically, we discover that the decomposition of these Lie algebras yields comprehensive insights into the behavior of the automorphism group on the space of bi-invariant metrics.This enables us to describe the space of bi-invariant metrics in Section 3. We do this using the factorization of the Lie algebra and a factorization of the action of the automorphism group of the Lie algebra.Likewise, we employ the same procedure to obtain the space of equivalent bi-invariant metrics, encompassing scalar multiples as well.In Section 4 we present explicitly the moduli spaces of bi-invariant metrics for Lie groups of dimension at most 6.We end by restating in Section 5 some results from [7].
to inner products on the Lie algebra g.Thus, the space M(g) (defined in [6] by Kodama, Takahara and Tamaru) consisting of inner products on the Lie algebra g corresponds to the space of left-invariant metrics on G.
2.1.Bi-invariant metrics.A Riemannian metric on G is called bi-invariant if it is invariant under both left and right translation.
We begin by recalling some facts about the adjoint representation.Each g ∈ G defines an inner automorphism ψ g : G → G, ψ g (x) = gxg −1 .Using this, we can define a group homomorphism Ad : G → Aut(g), called the adjoint representation of G, where Ad(g) : g → g is the differential of ψ g at the identity element of G.The differential of Ad : G → Aut(g) gives us a representation between the Lie algebras ad : g → Der(g), and thus we obtain the following commutative diagram.
ad exp e

Ad
Where "e" denotes the exponential map of Aut(g).Next, we state a couple results regarding bi-invariant metrics that can be found in [7].
Lemma 2.1.1.A left-invariant metric in a Lie group G is right-invariant if and only if Ad(g) : g → g is an isometry for any element g ∈ G.
For compact groups we have the following theorem that can be found in [2].Theorem 2.1.4.Any compact Lie group G admits a bi-invariant metric.
For a Lie algebra with a bi-invariant metric we have the following splitting theorem that can be found in [7].
Splitting Theorem 2.1.5.Let g be a Lie algebra with a bi-invariant metric.Then we have g = a 1 ⊕• • •⊕a k an orthogonal direct sum of simple ideals and commutative ideals without proper ideals, where the simply connected Lie group G associated with g can be expressed as the product A 1 × • • • × A k of normal subgroups.Furthermore, for each A i we have two options: (1) If a i is commutative, then it has dimension 1 and A i ∼ = R.
(2) If a i is non-commutative, then the center of a i must be trivial, A i has strictly positive Ricci curvature and A i is compact.
Remark 2.1.6.With the Splitting Theorem 2.1.5,we can say that any Lie algebra with a bi-invariant metric decomposes as g = s ⊕ Z(g), where s is a semi-simple Lie algebra.
For a connected Lie group that admits a bi-invariant metric, Milnor [7] gives the following description.Lemma 2.1.7.A connected Lie group admits a bi-invariant metric if and only if it is isomorphic as a group to the cartesian product K × R l , where K is compact.

2.2.
Curvature of Bi-invariant metrics.Given a left-invariant metric on G we have its associated Levi-Civita connection.We consider the curvature tensor R of the left invariant metric, and define the following curvature operator: for x, y ∈ g we define κ(x, y) = ⟨R xy (x), y⟩.For a bi-invariant metric, Milnor [7] shows that this curvature operator is nonnegative.

Moduli Space of Bi-Invariant Metrics
In this section, we examine the moduli space of bi-invariant metrics of a Lie group G.An advantage of this approach is the fact that to find left invariant metrics with non-negative sectional and scalar curvatures on a Lie group G, under certain conditions (see [7, p.297]), it will be sufficient to find a subgroup with a bi-invariant metric.In this way, knowing how many non isomorphic bi-invariant metrics the subgroup has, we have a rough measure of how many, non isomorphic left invariant metrics with non-negative sectional and scalar curvatures the group G possesses.
3.1.Isometry Classes of Bi-Invariant Metrics.We denote the collection of inner products on a Lie algebra g by M(g).Identifying an inner product on M(g) with a matrix, we endow M(g) with the relative topology as a subspace of a matrix space.Definition 3.1.1.For a Lie algebra g, we define the set of left invariant metrics that are bi-invariant in g, as follows We want to study the moduli space of bi-invariant metrics BI(g) consisting of the isometry classes of BI(g) and the space EBI(g) of conformally equivalent classes BI(g).To do this we introduce the following definition.
Lie algebras with inner products.We say that: (1) They are isometric if there exists a Lie algebra isomorphism ϕ : (2) They are conformally equivalent, if there exists a Lie algebra isomorphism ϕ : g 1 → g 2 and a real number λ > 0 such that It is clear that both are equivalence relations on BI(g).Now we are going to prove that an equivalence class of a bi-invariant metric contains only bi-invariant metrics.
Definition 3.1.4.For a Lie algebra g with a bi-invariant metric we define: (1) BI(g) as the space of isometry classes of BI(g) (2) EBI(g) as the space of conformally equivalent classes of BI(g) We conclude from the definition of both equivalence relations the following corollary.
Corollary 3.1.5.For a Lie algebra g with a bi-invariant metric, we have: (1) BI(g) is the orbit space BI(g)/Aut(g) under the left action (2) EBI(g) is the orbit space BI(g)/R × Aut(g) where The simplest case is when we have an abelian Lie algebra.Corollary 3.1.7.Let g an abelian Lie algebra, then we have that BI Proof.In an abelian Lie algebra any metric is bi-invariant, in addition we have GL(g) = Aut(g).This implies that two metrics in g are always related under an automorphism, thus Simplifying to the compact and semi-simple case.Using the Splitting Theorem 2.1.5and Lemma 3.2.1 we can describe the group Aut(g) for a Lie algebra with a bi-invariant metric.From now on, every time we write g = s ⊕ Z(g) we refer to a Lie algebra that admits a bi-invariant metric as indicated in Remark 2.1.6.
Before starting the study of Aut(g), we are going to prove the following lemma.
where each a i is simple, so a i is non abelian and contains no nonzero proper ideals.This implies that [a i , a i ] = a i .Therefore given x ∈ g we have For a Lie algebra with a bi-invariant metric g ∼ = s ⊕ Z(g) we have Aut(g) ∼ = Aut(s) ⊕ Aut(Z(g)).
Proof.For ϕ ∈ Aut(g), we will see that ϕ(s) ⊂ s and ϕ(Z(g)) ⊂ Z(g).Take x ∈ s, so by Lemma 3.2.1 we know that For the Lie algebra g ∼ = s ⊕ Z(g), the spaces BI(g) and EBI(g) are homeomorphic to BI(s) and EBI(s).
Proof.Let us consider a metric ⟨•, •⟩ g on g ∼ = s⊕Z(g).This metric can be expressed as a sum of metrics ⟨ Thus we can identify the moduli space Since Z(g) is abelian, by Corollary 3.1.7we have In conclusion we have that for g ∼ = s ⊕ Z(g) the moduli spaces BI(g) and EBI(g) are homeomorphic to BI(s) and EBI(s) respectively.□ With this lemma, we can conclude that the study of moduli spaces BI(g) and EBI(g) reduces to the case of compact and semi-simple Lie groups.We know from Theorem 2.1.4that any compact Lie group admits a bi-invariant metric.The following lemma will be useful for examining what happens when the group is compact and simple.Lemma 3.2.4.If we have a compact and simple Lie group, then the bi-invariant metric is unique up to a positive scalar multiple.
The reader interested in a proof, may consult [2].Remark 3.2.5.In [2, Theorem 2.35], it is verified that the Killing form, B(X, Y ) = tz(ad(X)ad(Y )) is Ad invariant.Furthermore, if the group is compact, connected, and semi-simple, then B is negative definite and −B plays the role of a bi-invariant metric.In the case of Lemma 3.2.4,the unique metric up to scalar multiple is represented by the Killing form.That is, for any bi-invariant metric ⟨•, •⟩ on g there exists α > 0 such that −αB = ⟨•, •⟩.
Then, ⟨•, •⟩ restricted to each a i is bi-invariant in a i , and for all u ∈ a i , v ∈ a j with i ̸ = j, it holds that ⟨u, v⟩ = 0 Proof.Let's recall that the Lie bracket in s decomposes as [ where each [•, •] i is the bracket of the summand a i .Now, let x, y, and z be in This implies that ⟨•, •⟩ restricted to a i is skew-adjoint, and by Lemma 2.1.2,it is bi-invariant in a i .
Finally, let u ∈ a i , v ∈ a j with i ̸ = j.Using Lemma 3.2.1,since a i is simple, we have As previously mentioned in Lemma 3.2.3, the study of our moduli spaces is reduced to the case of compact, semi-simple groups.The following result gives us a description of the space BI(s), when s semi-simple.Theorem 3.2.7.If S is a compact, connected, and semi-simple Lie group with Lie algebra decomposition s = a 1 ⊕ • • • ⊕ a k , where a i is simple, it holds that BI(s) can be identified with we know from Lemma 3.2.4 that each a i admits a unique bi-invariant metric up to positive scalar multiples, and without loss of generality by Remark 3.2.5 we can assume that it is given by the Killing form −B i .In this way, for each α i > 0, the metric As a result, ⟨•, •⟩ s decomposes uniquely as a bi-invariant metric in each summand, so we have that Thus, we can identify BI(s) with {(α 1 , ..., α k )|α i > 0}.□ Remark 3.2.8.From Theorem 3.2.7,we can conclude that in a compact and semisimple group S with s = a Remark 3.2.9.Note that in this case, the space BI(s) = {(α 1 , ..., α k )|α i > 0} has the structure of a group with the product of real numbers entry by entry.This fact has strong consequences later.

3.3.
Action of the automorphisms group on BI.By Lemma 2.1.1 we recall that a bi-invariant metric on a Lie group G is invariant under Ad(g) for all g ∈ G. Thus any metric in BI(g) remains fixed under the action (Ad(g This leads us to studying the action of the automorphisms of g that are distinct from the ones given as Ad(g).In order to give a complete description of the moduli spaces of bi-invariant metrics, we need to know more about the group of automorphisms of a semi-simple Lie algebra.Definition 3.3.1.For G a connected Lie group, we define the group of inner automorphisms of the Lie algebra g as It is easy to verify that Inn(g) does not depend on G.In [12, Chapter II] we can find the following result.Remark 3.3.3.The inner automorphisms of g are induced by inner automorphisms of G, since for each g ∈ G, Ad(g) is the differential of some inner automorphism of G.When g is semi-simple, then Inn(g) = Aut 0 (g), that is, the connected component of the identity in Aut(g) is comprised solely of internal automorphisms.
In this way, we know that if S is compact and semi-simple then for its Lie algebra s, the action of Inn(s) on BI(s) is trivial, so we must only study how the outer automorphisms of the Lie algebra act on BI(s).Definition 3.3.4.For a Lie algebra g, we define the group of outer automorphisms as Out(g) = Aut(g)/Inn(g).
We recall that for a Lie group G, the quotient G/G 0 is a group where each element corresponds to a connected component of G.In [8, Corollary 2] it is verified that for a connected and semi-simple Lie group we have that Aut(G)/Aut 0 (G) is finite.In this way, we conclude that if s is a semi-simple Lie algebra, then Aut(s) has a finite number of connected components and therefore Out(s) is a finite group.Proof.Take the bi-invariant metric given by −B, where B is the Killing form of g.Let ϕ ∈ Out(g).As the metric ϕ * (−B) is bi-invariant, then by Lemma 3.2.4ϕ * (−B) = α(−B) for some α > 0. Since Out(g) has finite order, then ϕ n = Id for some n ∈ N, and thus (ϕ n ) * (−B) = α n (−B) = −B.This implies that α = 1 and therefore the action of Out(g) is trivial on BI(g).□ Remark 3.3.6.We have that for a compact and simple group G, the action of Aut(g) on BI(g) is trivial.With this we conclude that BI(g) = BI(g) = R + and EBI(g) = {⟨•, •⟩ 0 }.Now we study the semi-simple case.To do this, we review the following results.
Proposition 3.3.7.If u is an ideal in a Lie algebra g and ϕ ∈ Aut(g), then ϕ(u) is an ideal in g.
Proof.Let u ∈ u and x ∈ g.We observe that Thus ϕ(u) is an ideal in g. □ Corollary 3.3.8.Let s = a 1 ⊕• • •⊕a k be a semi-simple Lie algebra and ϕ ∈ Aut(g).
Then ϕ(a i ) = a j for some j.
Proof.We know that ϕ(a i ) is a simple ideal of g.Furthermore, for all j, ϕ(a i ) ∩ a j is an ideal of a j .But since a j is simple, then ϕ(a i ) ∩ a j = 0 or ϕ(a i ) ∩ a j = a j .In the second case, we have that a j ⊂ ϕ(a i ) but since ϕ(a i ) is simple, then ϕ(a i ) = a j .Also, if ϕ(a i ) ∩ a j = 0 for all j, then ϕ(a i ) = 0 which is a contradiction.□ The next step will be to study the semi-simple case where all the factors are non-isomorphic to each other.Proof.Let ϕ ∈ Aut(s).Since all the summands are not isomorphic to each other by the previous corollary we have that ϕ(a i ) = a i .Thus we conclude that ϕ = ϕ 1 + • • • + ϕ k , where each ϕ i ∈ Aut(a i ), which finishes the proof.□ Lemma 3.3.10.Let G be a compact, connected, and semi-simple Lie group with Lie algebra g = a 1 ⊕ • • • ⊕ a k , where the summands are not isomorphic to each other and are simple.Then the action of Out(g) on BI(g) is trivial.
) be the bi-invariant metric given by the sum of the Killing forms B i in each a i .We define the mapping From the previous lemma, we know that In this way, we conclude that In Remark 3.2.9we had identified BI(g) with the multiplicative group of vectors with k positive real entries.From what we have seen above, we conclude that Φ is a group homomorphism and the image of Out(g) is a finite subgroup of BI(g).But this group does not have any non-trivial finite subgroups, so we conclude that Φ is trivial and the action of Out(g) on BI(g) is also trivial.□ This small detail allows us to relate Aut(g) to the symmetric group of permutations.We start with the following definition.Definition 3.4.2.Given a set X and S n the symmetric group of permutations of n elements, we define the left action of S n on In conclusion, we have that each automorphism of h 1 ⊕ • • • ⊕ h k is given by automorphims ϕ i ∈ Aut(h i ) for i = 1, . . ., k, and a permutation σ ∈ S k , where it holds that, if x i ∈ h i , then ϕ i (x i ) ∈ h σ(i) .Now we will see what happens when each h i is the Lie algebra of a simple and compact group.Theorem 3.4.4.Let H be a connected, simple, and compact Lie group and consider the product of H k-times, G = H × • • • × H. Then the action of Aut(g) on BI(g) coincides with the action of the symmetric group S k .Observe that the inner product ⟨•, •⟩ can be decomposed as a sum of inner products: Proof.Let ϕ ∈ Aut(g) and let us see how this automorphism acts on a bi-invariant metric ⟨•, •⟩ 0 ∈ BI(g).We know that ⟨ , where B is the Killing form on H and each α i > 0. Using Observation 3.2.8we know that the decomposition h ⊕ • • • ⊕ h is orthogonal for any bi-invariant metric.From this we obtain But as we saw in Lemma 3.3.5 the action of the automorphism group of a compact and simple group on the space BI(h) is trivial.With this we have that Thus we have that the action of Aut(g) on BI(g) is given by the action of the symmetric group S k .□ Definition 3.4.5.We consider the action of the symmetric group of permutations where X is a topological space, given by (σ, (x 1 , ..., x n )) → (x σ(1) , ..., x σ(n) ).
We define the nth symmetric product of X as SP n (X) = X n /S n .
By [1], when M is a differentiable manifold, then SP n (M ) has an orbifold structure, and if M = R then SP n (R) is homeomorphic to the product R×(R + ∪{0}) n−1 .Theorem 3.4.6.If H is a compact, connected, simple Lie group and G is the product of H k-copies of H, i.e.G = H × • • • × H, then the moduli space BI(g) is homeomorphic to the kth symmetric product of R, With this we conclude that the isometry class of a bi-invariant metric is the set [(α 1 , ..., α k )] = {(α σ(1) , ..., α σ(k) )|σ ∈ S k }.It follows that □ Remark 3.4.7.In this case, BJ(g) is homeomorphic to R × (R + ∪ {0}) k−1 , which is homotopic to a point and thus contractible.Remark 3.4.8.In the case of EBI(g), recall that by Remark 3.1.6we have that the actions of Aut(g) and R × on BI(g) commute, so we can first consider the space BI(g)/R × and later factor it with Aut(g).Since BI(g) is formed by vectors of strictly positive k real numbers, we can think of the space BI(g)/R × as S k−1 + = S k−1 ∩ BI(g), and in this way EBI(g) = S k−1 + /Aut(g).Having said this, we can state the following corollary.given by and each b i is semi-simple and decomposes as the sum of m i isomorphic factors but none of these factors is isomorphic to any a j .Then BI(g) is homeomorphic to Proof.Let ϕ ∈ Aut(g).By Corollary 3.3.8,we have ϕ = ψ + φ 1 + • • • + φ l where ψ ∈ Aut(s) and each φ i ∈ Aut(b i ).We know that the action of ψ on BI(s) is trivial and the action of φ i on BI(b i ) is through a permutation σ ∈ S mi .Therefore, upon factoring BI(g) with Aut(g), we have that BI(g) is homeomorphic to the product And for the space EBI(g) we have the following theorem.
Theorem 3.5.2.Let G be a connected, compact, and semi-simple Lie group with and each b i is semi-simple and decomposes into a sum of m i isomorphic factors, but none of these factors are isomorphic to any a j .Then EBI(g) is homeomorphic to In both cases, the spaces BI(g) and EBI(g) are products of contractible spaces and therefore they are contractible as well.Also note that for a Lie group G admitting a bi-invariant metric, the description of these spaces depends on the decomposition of the Lie algebra into semi-simple components, and in particular on the number of simple components which are isomorphic.
Figure 1.Since BI(so(4)) = (R + ) 2 , each metric is represented by a pair of positive real numbers.Therefore, we can visualize the space BI(so(4)) as the positive quadrant of R 2 , where each point (α 1 , α 2 ) represents a bi-invariant metric.Taking the quotient of BI(so(4)) by S 2 , each (α 1 , α 2 ) is identified with (α 2 , α 1 ).This implies that BI(so(4)) is the shaded area with the boundary given by the line y = x.The dotted line passing through the shaded area represents an element of EBI(so(4)), so this space is represented by the circular segment extending from the x-axis to the line x = y.

Curvature
Using the language developed so far, we can restate some results from Milnor's previous work in [7] as follows.Consequently, we have the following corollary.

Theorem 3 . 3 . 2 .
Let G be a connected semi-simple Lie group.Then the connected component of the identity of the automorphism group Aut 0 (G) in Aut(G) coincides with Inn(G).

Lemma 3 . 3 . 5 .
If G is a compact and simple Lie group then the action of Out(g) on BI(g) is trivial.

3. 4 .Theorem 3 . 4 . 1 .
Description of BI and EBI.With what has been developed so far, we can state the following theorem.For a compact, connected, and semi-simple Lie group G with g = a 1 ⊕ • • • ⊕ a k where the summands are simple and pairwise non-isomorphic, it follows that BI(g) = BI(g).Furthermore, EBI(g) = BI(g)/R × corresponds to the set of vectors with k − 1 strictly positive real number entries.Now we are going to study the case G = H × • • • × H, where H is a compact and simple Lie group, with Lie algebra