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I have a homogeneous Riemannian manifold X with isometry group Iso. Is Iso a maximal group? By maximal group, I mean that there does not exist another group G such that:

  1. Iso is a proper subgroup of G,
  2. G acts transitively on X by diffeomorphisms, and
  3. G has compact stabilizers Gx.

I know that if such a group exists, then there is a G-invariant metric on X. So, in other words, the question is: if I have a Riemannian metric d on X with isometry group Iso which is homogeneous, can there exist another Riemannian metric d' on X with isometry group Iso', such that Iso' contains properly the group Iso?

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    $\begingroup$ You had better specify that your groups are connected, otherwise there are easy counterexamples, i.e., the compact Lie group $G =(\mathbf{S}^1)^r$ is homogeneous both for the regular representation of $G$ on itself, but also for the obvious action by the group $G'$ that is the extension of the symmetric group $\mathfrak{S}_r$ by $G$. $\endgroup$ – Jason Starr Jun 24 '14 at 19:12
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    $\begingroup$ Of course. The odd-dimensional sphere $S^{2n+1}$ with the round metric is $O(2n+2)/O(2n+1)$. The subgroup $U(n+1)$ of $O(2n+2)$ acts transitively, and there are $U(n+1)$-invariant metrics which are not round. $\endgroup$ – Claudio Gorodski Jun 24 '14 at 20:57
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    $\begingroup$ Maby I'll raise the question more precisely. I have riemannian metric d on manifold X with isotopy group Iso. Does there exists another riemannian metric d' on X witch isotopy group Iso', such that Iso is properly contained in Iso'. $\endgroup$ – Adam Jun 24 '14 at 22:12
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    $\begingroup$ Look for example at Hermitian symmetric spaces. $\endgroup$ – Vít Tuček Jun 24 '14 at 22:17
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    $\begingroup$ Consider product of two round circles of distinct length. Making the lengths equal increases the group. (Or make it spheres to make everything orientation preserving as well.) $\endgroup$ – Alex Degtyarev Jun 25 '14 at 0:15
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Let $G$ be a compact Lie group with left-invariant metric $h$. For general $h$, the group of isometries is $G$. However, when $h$ is bi-invariant, the group of isometries is $G\times G$. I think this gives an example of non-maximal group of isometries.

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(This should have been a comment, but got too long, sorry)

Let us call $Iso(G)$ the full isometry group of the homogeneous space $G/H$, i.e., the largest group that acts transitively and effectively on $G/H$.

  1. Expanding on Claudio's comment, e.g., the sphere $S^{4n+3}$ can be written as $SO(4n+5)/SO(4n+4)$, $SU(2n+1)/SU(2n)$ or $Sp(n+1)/Sp(n)$, and there are (arbitrarily) small deformations of the round metric -- through so-called Berger metrics -- that make the (identity component of the) full isometry group drop from $SO(4n+5)$ to $SU(2n+1)$ or $Sp(n)$. Notice that $Iso(G/H)$ actually drops in dimension under this deformation of homogeneous metrics, not only looses some component as in item 3 below. For one more example on $\mathbb C P^n$, see my answer here;

  2. The full isometry group $Iso(G/H)$ of every homogeneous space $G/H$ with positive sectional curvature was computed by Shankar, see table given in Figure 3.

  3. Elaborating on Alex's comment, if we take $(M,g)$ any Riemannian manifold and consider $(M\times M,g\oplus\lambda g)$, then for $\lambda=1$ there is always an extra isometry, namely the one that exchanges both factors $M$.

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It is possible that Iso is not maximal. For example, take smooth manifold diffeomorphic to $S^2\subset \mathbb{R}^3$ such that it is a warped product metric that is not round (but symmetric with respect to rotations around the $z$-axis. Then Iso could be $S^1$ or at least some proper subgroup of $SO(3)$, which clearly acts by diffeomorphisms.

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  • $\begingroup$ But this is not a homogeneous Riemannian manifold. $\endgroup$ – ThiKu Jun 25 '14 at 0:53

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