definition of symmetry-rank of a Riemannian manifold

Question

I want to know why and how the following definitions are equivalent?

Definition 1: The symmetry-rank of a Riemannian manifold is the dimension of a maximal subspace of commuting Killing fields.

Definition 2: The symmetry-rank of a Riemannian manifold is the rank of its isometry group.

Is that possible that (give an explicit example) a Riemannian manifold $(M,g)$ have symmetry-rank $k=0, 1, 2, ..., \dim(M)$ for various metric $g$?

Answer 1

1. Petersen (on page 237 of his textbook (2006 Ed.)) defines rank of a Riemannian manifold correctly, using parallel Jacobi fields. The advantage of this definition is that it works equally well regardless of curvature sign and compactness.

He only informally says that rank can be also computed using commutative subgroups of the isometry group in the case of symmetric spaces. However, one cannot simply say that the rank of a symmetric space is the maximum of dimensions of abelian subgroups in the isometry group: For instance, for the hyperbolic $n$-space the isometry group contains an abelian subgroup of dimension $n-1$ (and this is the maximal dimension in this case). However, by everybody's definition, hyperbolic spaces have rank 1. Nevertheless, in the context of compact symmetric spaces, indeed, rank equals the maximal dimension of abelian subgroups of the isometry group, i.e. the dimension of maximal commutative subalgebra of Killing fields.

To compare this with the standard notions of rank for semisimple Lie groups: Suppose that $X$ is a symmetric space of noncompact type. Then its isometry group is a Lie group $G$, its rank is usually defined as the dimension of the Cartan subalgebras in its Lie algebra ${\mathfrak g}$. Equivalently, this is the dimension of a maximal group of transvections of the associated symmetric space (all such subgroups are conjugate, their Lie algebras are the Cartan subalgebras of ${\mathfrak g}$). Equivalently, it is the dimension of a maximal flat in the symmetric space. Equivalently, it is the dimension of a maximal ${\mathbb R}$-split torus in $G$, when the latter is regarded as a real-algebraic group. (There is also a notion of rank in the theory of algebraic groups, that you do not have to care about. The split torus I mentioned above is isomorphic to $({\mathbb R}^\times)^n$, not to a product of circles!)

If you are interested in the case of general symmetric spaces (curvature of mixed sign), their rank is the sum of ranks of deRham factors, where the rank of a flat factor is its dimension.

This formula can be easily worked out algebraically since the isometry group is commensurable to the product of isometry groups of the factors.

2. As for the linked paper by Grove and Searle, from the given examples it is clear they did not mean the rank in the sense of the linked Wikipedia page. In fact, they did not give a definition of rank at all, they assume reader's familiarity with the notion. From the context: They are interested in isometry groups of compact manifolds (in fact, of nonnegative curvature). Such groups $G$ are compact, their rank (in the context) is the dimension of a maximal torus in $G$, i.e. the dimension of a maximal commutative subalgebra in the algebra of Killing fields of the manifold.

3. Earlier in the book, p. 195, Petersen also discusses the notion of rank for compact Lie groups and, accordingly, symmetry-rank for compact Riemannian manifolds; these are the same as I mention in Part 2. The definitions in your question miss the critical compactness assumption. I should add that the notion of "symmetry-rank" can be regarded as common only in the community of geometers interested in manifolds of positive (nonnegative) curvature: Grove, Ziller, Wilking, et al. This is why they assume compactness: One can show that if $M$ is a complete manifold of nonnegative sectional curvature then its isometry group is (locally) isomorphic to the product of a compact Lie group with ${\mathbb R}^k$.

4. Rank in the sense of the Wikipedia article is in context of abstract groups (not Lie groups) and is defined as the infimum of cardinalities of generating sets. This definition is designed for (and used by) the theory of finitely generated groups, not Lie groups. In the sense of this definition, every Lie group of positive dimension has the rank of the cardinality of continuum, which is totally useless for you as a geometer.

There are other notions of rank in group theory, e.g. in theory of abstract abelian groups.

5. As for examples you are asking, you can find them among metrics on tori $T^n$. For $k\ne n-1$, take $T^{k}$ and equip it with an arbitrary flat metric $g_k$; $T^{n-k}$ to be the torus with some generic Riemannian metric $g_{n-k}$ that does not admit any nontrivial symmetries (or only finitely many symmetries). Now, take $M= T^k \times T^{n-k}$ with the "product-metric" $g_k\oplus g_{n-k}$. Then the symmetry rank of this manifold will be $k$.

For $k=n-1$ modify this example as follows. Take $T^1$ to be the circle; let $f: T^1\to {\mathbb R}_+$ be some smooth nonconstant function. Take $g_k$ as before to be a flat metric and then equip $T^n=T^1\times T^{n-1}$ with the "warped product metric" using the function $f$. Locally, this metric will be of the form $$ g=dt^2 + f(t)g_k. $$ Thus, for each $t\in T^1$ you will be using the scalar multiple $f(t)g_{n-1}$ as the metric on the fiber $T^{n-1}\times \{t\}$. The symmetry group of the resulting Riemannian metric $g$ will be locally isomorphic to $T^{n-1}$, i.e. have rank $n-1$. (Actually, if you take $f$ to be invariant only under the identity isometry $id: T^1\to T^1$, then the entire symmetry group of $(T^n,g)$ will be isomorphic to that of $(T^{n-1},g_{n-1})$.)