# Maximum symmetry metric on $\mathbb{C}P^n$

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

• 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. Nov 3 at 22:28
• @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? Nov 3 at 22:32
• 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. Nov 3 at 22:47
• 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'. Nov 4 at 12:30
• 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. Nov 4 at 12:42

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.

• 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? Nov 3 at 23:06
• 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 $O(6)$. Nov 3 at 23:40
• @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. Nov 24 at 14:24
• 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}.
– mme
Nov 25 at 17:15
• Oh I see its $Sp_n/Sp_{n-1} \cong S^{4n-1}$ but then you mod out by $U_1$ to get $Sp_n/(Sp_{n-1}\times U_1) \cong \mathbb{CP}^{2n-1}$. 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 $$G$$.

• 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. Nov 25 at 17:23
• 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. Nov 28 at 0:15
• @IanGershonTeixeira: A counterexample to what? My examples in point 1. above already have $G$ simple and $H$ semisimple (simple, in fact). Nov 28 at 0:35
• 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.) Nov 28 at 0:45
• @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). Nov 28 at 0:55