Journal of Lie Theory Volume 13 (2003) 457–464 \begin{matrix}c \\ ◯ \end{matrix} 2003 Heldermann Verlag

On the Volume of Unit Vector Fields on a Compact Semisimple Lie Group

Marcos Salvai^{∗}

Communicated by K. Strambach

Abstract. Let G be a compact connected semisimple Lie group endowed with a bi-invariant Riemannian metric. We prove that maximal singular unit vector fields on G are minimal, that is, they are critical points of the volume functional on unit vector fields on G. Besides, we give a lower bound for the number of nonequivalent minimal unit vector fields on G. Mathematics Subject Classification 2000: 22E46, 53C30, 53C35, 53C42.

1. Introduction

Let M be a compact Riemannian manifold. The volume of a unit vector field on M is by definition the volume of the submanifold that it determines in T^{1}M, the unit tangent bundle of M endowed with the canonical (Sasaki) metric. A unit vector field on M is said to be minimal if it is a critical point of the volume functional on unit vector fields on M. Gil-Medrano and Llinares-Fuster proved in [2] that a unit vector field is minimal if and only if it determines a minimal submanifold of T^{1}M (in this approach, M does not need to be compact). The characterization by Gluck and Ziller of Hopf vector fields as the unit vector fields on S^{3} with minimum volume [4], motivated the study of minimal unit vector fields (see the abundant bibliography on this subject for instance in the references of [1]).

2. Minimality of some left or right invariant unit vector fields

Let G be a compact connected semisimple Lie group endowed with a bi-invariant Riemannian metric. In this note we deal with minimal unit vector fields on G. On other Lie groups with left invariant metrics, they have been studied in [3, 6, 9].

Let 𝔤 ≅ T_{e}G be the Lie algebra of G and let S = T_{e}^{1}G denote the unit sphere in 𝔤. A unit vector v ∈ 𝔤 is said to be maximal singular if the Ad(G)-orbit of v has dimension strictly smaller than that of any other Ad(G)-orbit through a unit vector in some neighborhood of v. A unit vector field on G is said to

ISSN 0949–5932 / $2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag be maximal singular if it is left or right invariant and its value at the identity is maximal singular.

∗ Partially supported by foncyt, ciem (conicet) and secyt (unc).

Remark 2.1. Let U and V be unit vector fields on a compact Riemannian manifold M such that U ∘ ψ = Ψ ∘ V , where ψ and Ψ are isometries of M and T^{1}M, respectively. Clearly, U and V have the same volume and if one of them is minimal, so is the other one. In particular, this happens if U, V are the following pairs of unit vector fields on a compact Lie group G as above.

(a) If U and V are left invariant and U_{e} = Ad (g) V_{e} for some g ∈ G: Take ψ = L_{g} ∘ R_{g^{−1}} and Ψ = dψ, where L_{k} and R_{k} denote left and right multiplication by k, respectively.

(b) If U is left invariant, V is right invariant and U_{e} = V_{e}: Take ψ = Inv : G → G, Inv (h) = h^{−1} (notice that G is a symmetric space and Inv is the geodesic symmetry at e) and Ψ = −d Inv (the opposite of the identity on T^{1}M is an isometry, by definition of the Sasaki metric).

Theorem 2.2. Maximal singular unit vector fields on G are minimal.

Proof. By the previous remark, it suffices to prove the assertion for left invariant maximal singular unit vector fields on G. The product G×G acts on T^{1}G by isometries as follows: (g, h) u = dL_{g}dR_{h^{−1}}u. Let v be a unit vector in 𝔤. The orbit (G × G) v is a submanifold of T^{1}G whose dimension equals dim G+ dim Ad(G) v. Let us consider the action ρ of G on (G × G) v defined by ρ (g, u) = dL_{g}u. Notice that any orbit of ρ is the submanifold of T^{1}M determined by the left invariant vector field taking the value Ad(g) v at the identity, for some g ∈ G. Hence, all orbits of ρ in (G × G) v have the same dimension dim G and also the same volume, by Remark 2.1 (a). Therefore, by the main result of Hsiang and Lawson in [8], they are all minimal in (G × G) v, and so the mean curvature vector field H_{u} of the orbit ρ (G) u is perpendicular to (G × G) v for all u ∈ Ad (g) v. Let u ∈ Ad (G) v. Now, T_{u} (G × G) v ⊂ T_{u}T^{1}G includes the horizontal space, since parallel transport is realized by the action of the group (G is a symmetric space). Therefore, H_{u} (u) is vertical and we have an Ad (G)-invariant vector field V (u) = H_{u} (u) on Ad (G) v, tangent to S. If v is maximal singular, V vanishes identically, since otherwise, for all s ∼ 0, {Exp_{u}sV (u) | u ∈ Ad (G) v} would be an orbit of Ad (G) close to Ad (G) v with the same dimension, but different from it if s ≠ 0 (Exp_{u} : T_{u}S → S denotes the geodesic exponential function of S at u). By G-invariance, H_{v} ≡ 0 and hence the left invariant vector field taking the value v at the identity is minimal.

Remark 2.5 provides another proof of the Theorem.

The following criterion for a unit vector field on a unimodular Lie group to be minimal, obtained by Tsukada and Vanhecke, will be useful below.

Criterion. [9, Proposition 2.2] Let G be a unimodular Lie group endowed with a left invariant metric. Let f : S → ℝ be defined by

f (v) = vol (V) /vol (G) ,

where V is a left invariant vector field with V (e) = v. Then a left invariant unit vector field V is minimal if and only if df_{V (e)} = 0.

For the basic facts from the theory of compact semisimple Lie groups we refer the reader to [7]. Let 𝔱 be a maximal abelian subalgebra of 𝔤 and Δ the corresponding root system. Let C be a Weyl chamber and Φ = {α_{1}, …, α_{n}} the associated basis of Δ. For each 0 ≤ k < n let S_{k} ≅ S^{k} be the unit sphere in the (k + 1)-dimensional wall W = {w ∈ 𝔱 | α_{i} (w) = 0 for all i > k + 1} in 𝔱.

Proposition 2.3. If v ∈ S_{k} is a critical point of the restriction of f to S_{k}, then a left or right invariant vector field V on G with V_{e} = v is minimal.

Proof. We apply the criterion above. First we show that d f|_{T_{v}(𝔱∩S)} = 0. Let H_{i} be the vector dual to α_{i}, Ĥ_{i} = H_{i}/ ∥H_{i}∥ and let ρ_{i} be the reflection of 𝔱 fixing Ker (α_{i}). Let us recall that the action of the Weyl group

{g ∈ G | Ad (g) 𝔱 = 𝔱} / {g ∈ G | Ad (g) u = u for all u ∈ 𝔱}

on 𝔱 is generated by the reflections with respect to the kernels of the roots in Φ, hence any such a reflection ρ may be written as ρ = Ad (g)|_{𝔱} for some g ∈ G. Now, the orthogonal complement of T_{v}S_{k} in T_{v} (S ∩ 𝔱) is the subspace spanned by {H_{i} | i > k + 1} (empty if k = n). For k + 1 < i ≤ n we compute

df_{v} (−Ĥ_{i}) = \frac{d}{dt}|_{0} f ((cos t) v − (sin t) Ĥ_{i})
= \frac{d}{dt}|_{0} f (ρ_{i} ((cos t) v + (sin t) Ĥ_{i}))
= \frac{d}{dt}|_{0} f (Ad (g_{i}) ((cos t) v + (sin t) Ĥ_{i}))
= \frac{d}{dt}|_{0} f ((cos t) v + (sin t) Ĥ_{i}) = df_{v} \binom{}{Ĥ_{i}} .

Hence, d f|_{T_{v}(𝔱∩S)} = 0, since (df|_{S_{k}})_{v} = 0 by hypothesis. Now we prove that df_{v} = 0. The following decomposition of 𝔤 is well-known.

𝔤 = Ker (ad_{v}) ⊕ Image (ad_{v})  (1)
= ⋃ Ad (g) (𝔱) ⊕ \{ \frac{d}{dt}|_{0} Ad (exp tX) v | X ∈ 𝔤\} ,

where the union is over G_{v} = {g ∈ G | Ad (g) v = v} . Let u be a unit vector in T_{v} (𝔱 ∩ S) and let g ∈ G_{v}. We already know that df_{v} (u) = 0. Therefore

df_{v} (Ad (g) u) = \frac{d}{dt}|_{0} f (Ad (g) ((cos t) v + (sin t) u))
= \frac{d}{dt}|_{0} f ((cos t) v + (sin t) u) = df_{v} (u) = 0.

Similarly, if X ∈ 𝔤, we have df_{v} ( \frac{d}{dt} |_{0} Ad (exp tX) v) = 0. By (1), the preceding clearly implies that df_{v} vanishes identically. Thus, V is minimal by the criterion of Tsukada and Vanhecke.

Given α ∈ Φ, let v_{α} be the unique unit vector in 𝔱 satisfying α (v_{α}) > 0 and β (v_{α}) = 0 for all β ∈ Φ − {α} . The vector v_{α} is a vertex of the spherical simplex \overline{C} ∩ S (the unit vectors in the closure of the Weyl chamber C) and is maximal singular. Moreover, any maximal singular unit vector in 𝔤 is in the Ad(G)-orbit of exactly one of these vertices. Let V_{α} denote the left invariant vector field on G taking the value v_{α} at the identity. As a corollary of Proposition 2.3, we have the following result.

Theorem 2.4. If α ≠ β ∈ Φ and the maximal singular unit vector fields V_{α} and V_{β} have the same volume, then there exists v in the shortest geodesic arc in S joining v_{α} with v_{β}, v ≠ v_{α}, v_{β}, such that the left invariant vector field V with V (e) = v is minimal.

Proof. Let u (s), 0 ≤ s ≤ 1 be a parametrization of the shortest geodesic arc in S joining v_{α} with v_{β}, which is an edge of the spherical simplex \overline{C} ∩S. Suppose that the basis of the root system is {α_{1}, …, α_{n}} , with α = α_{1} and β = α_{2}. Then u (s) is a regular curve in S_{1} as in Proposition 2.3, which in this case is a circle. Since f (v_{α}) = f (v_{β}) by hypothesis, there exists s_{o} with (f ∘ u)′ (s_{o}) = 0. Hence, taking v = u (s_{o}) , we have that (d f|_{S_{1}})_{v} = 0. Therefore, the statement follows from Proposition 2.3.

Remark 2.5. With similar arguments we can give another proof of Theorem 2.2: Suppose that α = α_{1} ∈ Φ. By definition of v_{α}, we have that S_{0} has only two elements ±v_{α}. Hence (d f|_{S_{0}})_{v_{α}} = 0 trivially. Thus, V_{α} is a minimal unit vector field by Proposition 2.3.

3. Counting nonequivalent minimal unit vector fields

In order to have a clear statement of Theorem 3.2 below, we say that two unit vector fields U, V on a Riemannian manifold M are equivalent if there exists ϕ in the identity component of the isometry group of M such that U ∘ ϕ = dϕ ∘ V.

Remark 3.1. Two left invariant unit vector fields U and V on G are equivalent if and only if U_{e} = Ad (g) V_{e} for some g ∈ G, since the identity component of the isometry group of G consists of {L_{g} ∘ R_{h} | g, h ∈ G}. In particular, by Remark 2.1 (a), equivalent unit vector fields have the same volume and if one of them is minimal, so is the other one. Moreover, no right invariant unit vector field is equivalent to a left invariant one, since the center of G is finite.

If Φ = {α_{1}, …, α_{n}} is as above a basis of the root system, then V = {V_{α_{1}}, …, V_{α_{n}}} consists of nonequivalent left invariant unit vector fields. Suppose that {V_{1}, …, V_{k_{1}}}, {V_{k_{1}+1}, …, V_{k_{2}}} , etc., are ℓ disjoint subsets of V consisting of vector fields with the same volume. Next, we give a lower bound for the number of nonequivalent minimal unit vector fields on G in terms of the numbers k_{j}.

Theorem 3.2. Let n be the rank of G and let k_{1}, …, k_{ℓ} be the cardinalities of disjoint sets of nonequivalent maximal singular left invariant unit vector fields having the same volume. Then the number of nonequivalent minimal unit vector fields on G is not smaller than 2n + ∑_{i=1}^{ℓ} k_{i} (k_{i} − 1).

Proof. By Theorem 2.4, the number of nonequivalent minimal left invariant unit vector fields on G is not smaller than n + ∑_{i=1}^{ℓ} \binom{k_{i}}{2}. By Remark 3.1, this bound doubles if one considers the corresponding right invariant vector fields.

4. An expression for the volume of a left invariant unit vector field

We show later that in some cases, a lower bound as in Theorem 3.2 can be obtained without actually computing the volume of maximal singular unit vector fields, but just using the symmetries of the associated root system. Nevertheless, if it has no symmetries (or if it has, but one wants to try to improve the lower bound), Proposition 4.1 below gives a simple formula for the volume of maximal singular unit vector fields in terms of the root system (see the examples).

Let Δ^{+} be the set of positive roots with respect to Φ. Given a root ϕ ∈ Δ^{+}, there exist nonnegative integers m_{β} (ϕ) , β ∈ Φ, such that ϕ = ∑_{β∈Φ} m_{β} (ϕ) β.

Proposition 4.1. The volume of a left or right invariant unit vector field V on G with V_{e} = v is given by

vol (G) ∏_{ϕ∈Δ^{+}} (1 + \frac{1}{4}ϕ (v)^{2}) .

In particular, the volume of the maximal singular unit vector field V_{α} is given by

vol (G) ∏_{ϕ∈Δ^{+}} (1 + \frac{1}{4}α (v_{α})^{2} m_{α} (ϕ)^{2}) .

Proof. It is well-known that the volume of a unit vector field V on a compact Riemannian manifold M is given by

∫_{M} \sqrt{det} (I + (∇V)^{∗} (∇V)) ,

where integration is taken with respect to the Riemannian volume form on M. Let V be a left invariant unit vector field on G with V_{e} = v. The Levi-Civita connection of the bi-invariant metric on G satisfies that ∇_{W} V = \frac{1}{2} [W, V] for any left invariant vector field W. Since ad v is skew-symmetric with respect to any bi-invariant metric, we have then

vol (V) = vol (G) \sqrt{det} (I + (∇V)^{∗} (∇V))
= vol (G) \sqrt{det} (I + \frac{1}{4} (ad v)^{∗} (ad v))
= vol (G) \sqrt{det} (I − \frac{1}{4} (ad v)^{2}) .

Let 𝔪 be the orthogonal complement of 𝔱 in 𝔤 with respect to the Killing form. There exists an orthonormal basis {x_{ϕ}|ϕ ∈ Δ^{+}} ∪ {y_{ϕ}|ϕ ∈ Δ^{+}} of 𝔪 such that

ad_{v}x_{ϕ} = ϕ (v) y_{ϕ} and ad_{v}y_{ϕ} = −ϕ (v) x_{ϕ}

for all v ∈ 𝔱 and ϕ ∈ Δ^{+}. The matrix of − (ad v)^{2}|_{𝔪} with respect to that basis consists then of blocks ϕ (v)^{2} I, ϕ ∈ Δ^{+}, where I is the 2 × 2 identity matrix. Moreover, clearly, ad_{v} = 0 on 𝔱. Therefore, the stated formula for vol(V) is valid. The expression for vol(V_{α}) follows, since by definition of v_{α} we have that ϕ (v_{α}) = m_{α} (ϕ) α (v_{α}).

5. Examples

In the following, compact semisimple Lie groups are supposed to be endowed with a bi-invariant Riemannian metric. A reference for the symmetries of the Dynkin diagrams referred to below is for instance [7, Ch. X, Theorem 3.29]. For α_{k} belonging to a basis of a root system we denote v_{k} = v_{α_{k}} and V_{k} = V_{α_{k}}.

Example 1. The number of nonequivalent minimal unit vector fields on the group SU (n + 1) is not smaller than 2n + ∑_{i=1}^{[n/2]} 2 = 2n + 2 [n/2], where [a] denotes the integral part or a.

It follows from Theorem 3.2, since vol(V_{k}) = vol(V_{n−k+1}) by the Z_{2} symmetry of the Dynkin diagram of SU (n + 1) .

Example 2. For SO (5) and G_{2} (the only groups of rank 2 which remain to analyze), two nonequivalent maximal singular unit vectors have different volumes. Thus, in these cases, Theorem 3.2 yields the lower bound 4 and so it does not improve Theorem 2.2.

The statement for SO (5) follows from Example 4 below for n = 2. On the other hand, a basis of the root system of 𝔤_{2} (contained in ℝ^{2} with the canonical metric) consists of α_{1} = r (2, 0) and α_{2} = r (−3, \sqrt{3}) for some r > 0. The positive roots are α_{1}, α_{2}, α_{1} +α_{2}, 2α_{1} +α_{2}, 3α_{1} +α_{2} and 3α_{1} +2α_{2}. One easily computes v_{1} = (1, \sqrt{3}) /2 and v_{2} = (0, 1) . Hence α_{1} (v_{1}) = r and α_{2} (v_{2}) = \sqrt{3}r. Therefore, by Proposition 4.1,

vol (V_{1}) = vol (G) (1 + r^{2}) (1 + \frac{r^{2}}{4} )^{2} (1 + \frac{9r^{2}}{4} )^{2} ,
vol (V_{2}) = vol (G) (1 + 3r^{2}) (1 + \frac{3r^{2}}{4} )^{4} .

With the aide of the computer one obtains easily that vol(V_{2}) > vol(V_{1}) for any r > 0.

Example 3. The group SO (8) has at least 14 nonequivalent minimal unit vector fields. Besides, for this group, no lower bound as in Theorem 3.2 is sharp.

Let {α_{1}, α_{2}, α_{3}, α_{4}} be a basis of the root system of SO (8), where α_{4} is the distinguished root, that is, there is an action of the group of permutations of three elements by root system isomorphisms (corresponding to the symmetries of the Dynkin diagram), fixing α_{4} and interchanging the rest of the roots. Hence the left invariant vector fields V_{1}, V_{2}, V_{3} have the same volume. Thus, the first assertion follows from Theorem 3.2. Next we check the validity of the second one. In fact, for SO (8) there exists at least one minimal unit vector field not detected by Theorem 3.2 and its proof: Let \overline{C} be the closure of the Weyl chamber associated to the given basis and let S_{2} be as in Proposition 2.3. Then S_{2} ∩ \overline{C} is an equilateral spherical triangle with vertices v_{1}, v_{2}, v_{3}. Now, by the symmetry of the root system, the center v of the spherical triangle is a critical point of f restricted to S_{2}. By Proposition 2.3, a left or right invariant vector field on SO (8) taking the value v at the identity is minimal.

Example 4. Let G = SO (2n + 1) be endowed with the metric −λ^{2}B, where B is the Killing form, and let {α_{1}, …, α_{n}} be a basis of the root system with Dynkin diagram

\begin{matrix}∘^{2} & – & ∘^{2} & – ⋯ – & ∘^{2} & = & ∘^{1} \\ α_{1} & & α_{2} & & α_{n−1} & & α_{n} \end{matrix} .

Then, for 1 ≤ k ≤ n, the volume of V_{k} is given by

vol (G) (1 + \frac{ρ^{2}}{4k} )^{k+2k(n−k)} (1 + \frac{ρ^{2}}{k} )^{k(k−1)/2} ,  (2)

where 2 (2n − 1) ρ^{2} = λ^{2}. If λ and the rank n are given concretely, using this expression, one can compare the values of vol(V_{k}) with the aide of a computer. If n = 3, one sees that for all but two values of λ^{2}, three nonequivalent maximal singular unit vector fields have different volumes, while for the remaining values of λ^{2}, two of them have the same volume.

Let {e_{1}, …, e_{n}} be an orthogonal basis of ℝ^{n} with ∥e_{i}∥ = r for all i. The roots

α_{i} = e_{i} − e_{i+1} (1 ≤ i < n) and α_{n} = e_{n}

form a basis of the root system of so (2n + 1) , corresponding to the given Dynkin diagram, whose positive roots are (see [7, p 462])

e_{i} = α_{i} + ⋯ + α_{n}  for 1 ≤ i ≤ n,
e_{i} − e_{j} = α_{i} + ⋯ + α_{j−1}  for 1 ≤ i < j ≤ n,
e_{i} + e_{j} = α_{i} + ⋯ + α_{j−1} + 2α_{j} + ⋯ + 2α_{n} for 1 ≤ i < j ≤ n.

Denoting m_{α_{k}} by m_{k}, one has then for 1 ≤ k ≤ n that

m_{k} (e_{i} − e_{j}) = \{ \begin{matrix}1 & if i ≤ k < j \\ 0 & if not \end{matrix} , m_{k} (e_{i} + e_{j}) = \{ \begin{matrix}1 & if i ≤ k < j \\ 2 & if k ≥ j \\ 0 & if not \end{matrix}

and m_{k} (e_{i}) = \{ \begin{matrix}1 & if k ≥ i \\ 0 & if not \end{matrix} .

Solving the equation for v_{k} one obtains v_{k} = x_{k} (e_{1} + ⋯ + e_{k}) , with kx_{k}^{2}r^{2} = 1, for 1 ≤ k ≤ n. Hence, α_{k} (v_{k})^{2} = x_{k}^{2}r^{4} = r^{2}/k. Now, since for 1 ≤ k ≤ n the cardinalities of the sets

{(i, j) | 1 ≤ i ≤ k < j ≤ n} and {(i, j) | 1 ≤ i < j ≤ k}

are k (n − k) and k (k − 1) /2, respectively, we have by Proposition 4.1 that the volume of V_{k} equals

vol (G) (1 + \frac{r^{2}}{4k} )^{k+2k(n−k)} (1 + \frac{r^{2}}{k} )^{k(k−1)/2} .

Finally, since the opposite of the Killing form corresponds to 2 (2n − 1) r^{2} = 1, formula (2) is true (set ρ^{2} = λ^{2}r^{2}).

Acknowledgments. I would like to thank the referee for making me aware of a bad mistake in the first version, and Carlos Olmos for his helpful comments.

References

[1] Gil-Medrano, O., Relationship between volume and energy of vector fields, Diff. Geom. Appl. 15(2) (2001), 137–152.

[2] Gil-Medrano, O., and E. Llinares-Fuster, Minimal unit vector fields, Tohoku Math. J. 54 (2002), 71–84.

[3] Gil-Medrano, O., J. C. González-Dávila, and L. Vanhecke, Harmonic and minimal invariant unit vector fields on homogeneous Riemannian manifolds, Houston J. Math. 27 (2001), 377–409.

[4] Gluck, H., and W. Ziller, On the volume of a unitvector field on the three sphere, Comm. Math. Helv. 61 (1986), 177–192.

[5] González-Dávila, J. C., and L. Vanhecke, Examples of minimal unit vector fields, Ann. Global Anal. Geom. 18 (2000), 385–404.

[6] —, Energy and volume of unit vector fields on three-dimensional Riemannian manifolds, Diff. Geom. Appl. 16 (2002), 225–244.

[7] Helgason, S., “Differential geometry, Lie groups and symmetric spaces, ” Academic Press, 1978.

[8] Hsiang, W. Y., and H. B. Lawson, Jr., Minimal submanifolds of low cohomogeneity, J. Diff. Geom. 5(1) (1971), 1–38.

[9] Tsukada, K., and L. Vanhecke, Invariant minimal unit vector fields on Lie groups, Period. Math. Hung. 40 (2000), 123–133.

Marcos Salvai

famaf - ciem

Ciudad Universitaria, 5000 Córdoba, Argentina salvai@mate.uncor.edu

Received March 26, 2002 and in final form February 20, 2003