Let $ M $ be a compact connected manifold. The degree of symmetry of $ M $, denoted $ N(M) $, is the maximum of the dimensions of the isometry groups of all possible Riemannian structures on $ M $. See for example


Two metrics are considered to be equivalent if they are isometric up to a constant multiple. (see comment from Robert Bryant)

I'm interested in manifolds $ M $ for which there is a unique up to equivalence metric with isometry group of dimension $ N(M) $. My guess is that there is always such a unique metric for manifolds of the form $ G/H $ for $ G $ a compact connected simple Lie group and $ H $ a closed subgroup. Moreover I would imagine that this unique up to equivalence metric is just the pushforward of the unique up to scaling biinvariant metric on the compact connected simple Lie group $ G $.

I believe all spheres $ S^n, n \geq 2 $ have this property. And the unique maximum symmetry metric is the round metric.

What about the manifold $ M=\mathbb{C}P^n $ of real dimension $ 2n $? Is it the case that $$ N(\mathbb{C}P^n) =n(n+2) $$ And moreover is it true that every metric on $ \mathbb{C}P^n $ whose isometry group has maximum dimension must be equivalent to the Fubini-Study metric?

Edit: Ok so the guess about spaces $ G/H $ was a little ambitious and as Robert Bryant points out it is wrong. My new guess is that a unique up to equivalence metric exists for any irreducible compact symmetric space $ M $. (Edit: I originally left out "irreducible" which Robert Bryant pointed out in the comments makes this obviously false).

Ok I made this guess into a new a new question

Unique maximum symmetry metric on irreducible compact symmetric space

  • 1
    $\begingroup$ You are leaving out the diffeomorphism group. There are many (even Kähler) metrics on $\mathbb{CP}^n$ that are isometric to 'the' Fubini-Study metric but are not equal to a (constant) scalar multiple of it. $\endgroup$ Nov 3 at 22:28
  • $\begingroup$ @RobertBryant Is there a good way I could rephrase my question to fix that problem? For example just requiring that they are isometric to Fubini-Study not necessarily a constant scalar multiple? $\endgroup$ Nov 3 at 22:32
  • 1
    $\begingroup$ A plausible guess might be that any metric on $\mathbb{CP}^n$ whose isometry group has dimension $n(n{+}2)$ must be isometric to a constant scalar multiple of the Fubini-Study metric. While it's possible (when $n=1$) for a non-constant scalar multiple of the Fubini-Study metric to be isometric to a constant scalar multiple of the Fubini-Study metric, most of the time, when you have a non-constant scalar multiple of the Fubini-Study metric, the result is not even Kähler. $\endgroup$ Nov 3 at 22:47
  • 1
    $\begingroup$ Unfortunately, 'unique up to isometry' is not what you want, since, on a compact manifold at least, a metric $g$ is not isometric to $cg$ for any constant $c\not=1$, while the non-zero constant multiples of $g$ all have the same isometry group. I would recommend to say something like, 'Two metrics are considered to be equivalent if they are isometric up to a constant multiple.', and then, after that, say 'unique up to equivalence'. $\endgroup$ Nov 4 at 12:30
  • 3
    $\begingroup$ Also, for your 'new guess', you need to restrict to irreducible compact symmetric spaces in order to avoid the obvious counterexamples: $M=\mathbb{R}^n/\Lambda$ where $\Lambda\subset\mathbb{R}^n$ is a lattice. These compact symmetric spaces have $N(M)=n$ realized by the translations in $\mathbb{R}^n$, but there is an $(n(n+1)/2-1)$-parameter family of inequivalent ones. Also, if $M$ is a non-trivial product of irreducible symmetric spaces, you'll have a positive dimensional family of inequivalent $G$-invariant metrics. $\endgroup$ Nov 4 at 12:42

2 Answers 2


There's an easy counterexample to your guess: Let $M^6 = \mathrm{SU}(3)/\mathbb{T}^2$, where $\mathbb{T}^2\subset\mathrm{SU}(3)$ is the maximal torus (for example, the diagonal subgroup). In that case, there is a 3-parameter family of non-isometric metrics on $M^6$ that are invariant under $\mathrm{SU}(3)$, so they are not unique up to a constant (or even non-constant) scalar factor.

I imagine that $M^6$ does not carry a metric whose isometry group has dimension greater than $8=\dim\mathrm{SU}(3)$, but I don't have a proof handy.

On the other hand, it is true that any Riemannian metric on $\mathbb{CP}^n$ whose isometry group has dimension at least $n(n{+}2)$ must be isometric to a constant scalar multiple of the Fubini-Study metric. Here is one argument:

Suppose that a connected, compact group $G$ acts effectively and smoothly on $\mathbb{CP}^n$. Then, by averaging, there exists a $G$-invariant metric $g$. Moreover, since $H^2_{dR}(\mathbb{CP}^n,\mathbb{R})\simeq\mathbb{R}$, it follows from the Hodge Theorem that there is a $g$-harmonic $2$-form $\omega$ that represents a generator of $H^2_{dR}(\mathbb{CP}^n,\mathbb{R})$, and it is unique up to constant multiples. Since $G$ is connected, it follows that it must leave $\omega$ fixed. Moreover, because of the structure of the cohomology ring of $\mathbb{CP}^n$, the top-degree form $\omega^n$ must represent a generator of $H^{2n}_{dR}(\mathbb{CP}^n,\mathbb{R})$. In particular, $\omega^n$ does not vanish identically.

Thus, there is a point $p\in\mathbb{CP}^n$ such that $\omega_p\in \Lambda^2(T^*_pM)$ is a 2-form of full rank. Consider the stabilizer $G_p\subset G$ of $p$. Since $G$ acts by isometries and $\mathbb{CP}^n$ is connected, $G_p$ injects into $\mathrm{O}(T_pM)$ by identifying $g\in G_p$ with $g'(p):T_pM\to T_pM$. Moreover, $G_p$ leaves $\omega_p$ fixed. Thus, $G_p$ must lie inside a subgroup of $\mathrm{O}(T_pM)$ that fixes a complex structure $J:T_pM\to T_pM$ and hence must have dimension at most $\dim \mathrm{U}(n) = n^2$. Now, we have $$ \dim G = \dim G_p + \dim G/G_p = \dim G_p + \dim G{\cdot}p \le n^2 + 2n = n(n{+}2). $$ If equality holds, then $\dim G_p = n^2$ and $\dim G{\cdot}p = 2n = \dim \mathbb{CP}^n$. Thus, the orbit $G{\cdot}p$ is both open and closed in $\mathbb{CP}^n$, so $G$ acts transitively on $\mathbb{CP}^n$. It follows that $\omega$ is everywhere of full rank and, after scaling $\omega$ so that it has comass 1, we have that $\omega(u,v) = g(Ju,v)$ for a unique almost-complex structure $J$ on $\mathbb{CP}^n$ that is preservd by $G_p$, which has the same dimension as the connected group $\mathrm{U}(g_p,J_p)\simeq \mathrm{U}(n)$. Thus, $G_p = \mathrm{U}(g_p,J_p)$. Since $G_p$ contains $-I\in\mathrm{U}(g_p,J_p)$, it follows that there is an element of $G$ that fixes $p$ and reverses all $g$-geodesics through $p$. Since $G$ acts transitively on $\mathbb{CP}^n$, it follows that $(\mathbb{CP}^n,g)$ is a Riemannian symmetric space. Using the classification, it follows that $G\simeq \mathrm{SU}(n{+}1)/Z$ (where $Z\simeq\mathbb{Z}_{n+1}$ is the center of $\mathrm{SU}(n{+}1)$) and that the metric $g$ is, up to isometry, a constant multiple of the standard Fubini-Study metric.

  • $\begingroup$ Ok so my guess was a little ambitious. What about the title question (reworked as suggested by you in the comments)? If $ g $ is a metric on $ \mathbb{CP}^n $ with isometry group of dimension $ n(n+2) $ then must $ g $ be isometric to a constant scalar multiple of the Fubini-Study metric? And if it's true for $ \mathbb{CP}^n $ and spheres then maybe it's true for all compact symmetric spaces? $\endgroup$ Nov 3 at 23:06
  • $\begingroup$ Is it easy to see thay none of the $SU(3)$-invariant metrics in your answer "accidentally" has a larger isometry group? E.g., if one considers the same problem on $SU(3)/SU(2)$, there is a 2-parameter family of invariant metrics, but for a 1-parameter sub-family, the isometry group is $O(6)$. $\endgroup$ Nov 3 at 23:40
  • 1
    $\begingroup$ @RamiroLafuente: Of course you are right. That remark was wrong-headed. I'll remove it. Anyway, I have realized that there is a much simpler argument for the maximum symmetry of $\mathbb{CP}^n$ that works for all $n$, so I'll replace that entire segment. $\endgroup$ Nov 24 at 14:24
  • 1
    $\begingroup$ It acts on quaternionic space H^n, transitively on its unit sphere. It commutes with the left action of H (it is the group of H-linear isometries) and this descends from a transitive action on S^{4n-1} to a transitive action on HP^{n-1}. In between, it factors through a transitive action on CP^{2n-1}. $\endgroup$
    – mme
    Nov 25 at 17:15
  • 1
    $\begingroup$ Oh I see its $ Sp_n/Sp_{n-1} \cong S^{4n-1} $ but then you mod out by $ U_1 $ to get $ Sp_n/(Sp_{n-1}\times U_1) \cong \mathbb{CP}^{2n-1} $. $\endgroup$ Nov 25 at 17:25

I just wanted to add two points:

  1. A bi-invariant metric on a compact Lie group $G$ does not always induced the maximum symmetric metric on $G/H$. The most familiar examples are spheres: $S^{2n+1} = SU(n+1)/SU(n)$. For $n\geq 3$, the bi-invariant metric on $SU(n+1)$ induces a Berger metric on the sphere. To get the round metric (with strictly larger isometry group), one needs a particular left-invariant metric on $SU(n+1)$. See this article by Kerin and Wraith for details on the $n=2$ case.

  2. Robert Bryant's guess that $M^6 = SU(3)/T^2$ does not carry a Riemmannian metric with isometry group of dimension $9$ or higher is true. Slightly more is true: if $K$ is a compact Lie group acting on $M$ and $\dim K\geq 8$, then $K = SU(3)$. This follows easily from Theorem 1 and 4 in

Hauschild, Volker. “The Euler Characteristic as an Obstruction to Compact Lie Group Actions.” Transactions of the American Mathematical Society 298, no. 2 (1986): 549–78. https://doi.org/10.2307/2000636.

Special case of Theorem 1: If a compact Lie group $K$ acts effectively on a homogeneous space of the form $G/T$ for $G$ a compact Lie group and $T$ a maximal torus, then $\operatorname{rk} K\leq \operatorname{rk} G = \dim T$.

Special case of Theorem 4: Under the same hypothesis of Theorem 1, the order of the Weyl group of $K$ divides that of $G$.

  • 1
    $\begingroup$ Thanks! I could imagine how to prove the special case of $\mathrm{SU}(3)/\mathbb{T}^2$, but the details were messy. It's good to know that it works for all $G/T$, which I wouldn't even have thought of attempting to prove. $\endgroup$ Nov 25 at 17:23
  • $\begingroup$ Do you know a counterexample of the form $ G/H $ where $ G $ simple and $ H $ is semisimple? Just wondering since $ SU_3/\mathbb{T}^2 $ is not of that form. $\endgroup$ Nov 28 at 0:15
  • $\begingroup$ @IanGershonTeixeira: A counterexample to what? My examples in point 1. above already have $G$ simple and $H$ semisimple (simple, in fact). $\endgroup$ Nov 28 at 0:35
  • $\begingroup$ What I said was unclear, let me clarify. In my question I mention that "My guess is that there is always such a unique [maximum symmetry] metric for manifolds of the form $ G/H $ for $ G $ a compact connected simple Lie group and $ H $ a closed subgroup." $ SU_3/T^2 $ is a counterexample to this. $ SU_{n+1}/SU_n \cong S^{2n+1} $ is not a counterexample because spheres do admit a unique maximum symmetry metric (although this is a counterexample to the other guess I made that for simple $ G $ the push forward of the biinvariant metric onto $ G/H $ always induces a maximum symmetry metric.) $\endgroup$ Nov 28 at 0:45
  • 1
    $\begingroup$ @Ian: Well, I have a way of generating guesses, but proving that any of them works will require some effort. Pick any $H\subseteq G$ for which the isotropy action of $H$ on $\mathfrak{g}/\mathfrak{h}$ splits into has precisely two irreducible summands. (This situation has been classified: link.springer.com/article/10.1007/s10455-008-9109-9). Each of these admits $G$-invariant metrics which are not isometric, even up to scaling. I would bet that generically, the isometry group of each of these metrics is $G$ (perhaps up to components and up to cover). $\endgroup$ Nov 28 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.