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question: For the complex projective space of $n$-complex dimensions, $$\mathbb{P}^n,$$

what is the symmetry group / isometry group of this complex projective space $\mathbb{P}^n$?

Attempt: Naively, for $\mathbb{P}^n$, we have $$\sum_{i=1}^{n+1} |z_i|^2=1$$ so $$ (\bar z_1, \bar z_2, ...) \cdot (z_1,z_2, ...) \equiv (z_1,z_2, ...)^\dagger\cdot (z_1,z_2, ...)=1, $$ It looks that we have a unitary group $$u \in U(n+1)$$ with $$u^\dagger \cdot u =1$$ symmetry group, so $$ z_i \to u z_i $$ still satisfies the $$ (u(z_1,z_2, ...))^\dagger\cdot (u(z_1,z_2, ...))=(\bar z_1, \bar z_2, ...)u^\dagger \cdot u (z_1,z_2, ...)=1, $$

So the symmetry group / isometry group of the complex projective space $\mathbb{P}^n$ are both the $ U(n+1)$?

Edit: You can choose any metrics. e.g. Fubini-Study metric; or, any other metrics possible. If the answer depends on the metric choices, then give the isommetry for your metric choices.

See also https://en.wikipedia.org/wiki/Complex_projective_space#Differential_geometry

annie marie cœur
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  • Your equation $\sum_{i=1}^{n+1} |z_i|^2=1$ does not make sense in projective space. – Georges Elencwajg Nov 02 '18 at 17:00
  • It is complex projective space - it makes sense – annie marie cœur Nov 02 '18 at 17:45
  • e.g. https://math.stackexchange.com/questions/818193/standard-action-of-su3-on-mathbbc3?rq=1 – annie marie cœur Nov 02 '18 at 17:46
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    maybe related https://math.stackexchange.com/q/2758936/79069 Why is Fubini-Study metric invariant under SU(n+1) action on $ℙ^nℂ$ – wonderich Nov 02 '18 at 18:19
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    The link in your second commment is irrelevant because it is about points in $\mathbb C^3$, not projective space. You don't seem to understand that points in projective space have coordinates only defined up to a non-zero complex scalar. Hence your question does not make sense because any point in $\mathbb P^n$ can be given coordinates $[z_1:...:z_{n+1}]$ such that $\sum_{i=1}^{n+1} |z_i|^2=1$. For example the point $A=[3+4i:-12i]\in \mathbb P^1$ can also be written $A=[\frac {3+4i}{13}:\frac {-12i}{13}]$ and then $\vert \frac {3+4i}{13}\vert^2 + \vert \frac {-12i}{13}\vert^2=1$. – Georges Elencwajg Nov 02 '18 at 22:06
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    The group of symmetries of projective space as an algebraic variety is the projective general linear group. If, however, you equip it with the Fubini-Study metric (which is the unique metric on projective space compatible with the usual metric on the $2n+1$-sphere $S^{2n+1}$ via the quotient map $S^{2n+1} \to \mathbf{P}^n$), then the group of symmetries that preserve this is the projective unitary group. Of course, you could ask for symmetries preserving some other structure as well, and depending upon your choice you may get a different answer. – Stephen Nov 02 '18 at 22:28
  • Do you mean the answer is PU(n+1)? – annie marie cœur Nov 02 '18 at 22:32
  • @Stephen, Do you mean the answer is PU(n+1)? no U(n+1)? – annie marie cœur Nov 02 '18 at 22:32
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    I think the comments are suggesting that you are missing context. We need to know what metric you are trying to preserve (since you are asking for an isometry group (or is that not what you are asking for, since you seem unsure)). As Stephen pointed out there are at least two plausible answers based on what information is given in the question. – jgon Nov 02 '18 at 23:27
  • Thanks very much: Edit - You can choose any metrics. e.g. Fubini-Study metric; or, any other metrics possible. – annie marie cœur Nov 03 '18 at 01:10
  • Maybe also related https://math.stackexchange.com/q/302197/141334 – annie marie cœur May 21 '21 at 14:10

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