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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

https://www.ams.org/journals/tran/1969-146-00/S0002-9947-1969-0250340-1/S0002-9947-1969-0250340-1.pdf

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

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $G$.

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    $\begingroup$ 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
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    $\begingroup$ @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

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