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Given:

1) a Finsler metric $F_p : T_p G \rightarrow \mathbb{R}$ on $SU(N+1)$

2) a $U(N)$ subgroup of $SU(N+1)$ which is a stabilizer subgroup of some point on $CP^N$ (complex projective space) in the standard action of $SU(N+1)$ on $CP^N.$ This action is given by realizing $CP^N$ as a quotient manifold on $\mathbb{C}^{N+1}$. The appropriate equivalence relation on $\mathbb{C}^{N+1}$ is given by $u \sim v$ if $u = Zv$ for some $Z \in \mathbb{C}/\{0\}$. $SU(N+1)$ acts on $\mathbb{C}^{N+1}$ by matrix multiplication. The standard action I'm speaking of is now given by setting $U \in SU(N+1)$ acts on $CP^N$ by the formula $U \circ [v] = [Uv]$.

It is true (and well known) that $CP^N \cong SU(N+1)/U(N)$. But what exactly is invariance the condition on $F$ for it to push forward through this isomorphism to a Finsler metric on $CP^N$ that is invariant under the action of $SU(N+1)$?

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  • $\begingroup$ D0 you mean what you wrote at the end, i.e., 'invariant under $SU(N)$?', instead of 'invariant under $SU(N{+}1)$'? If so, is the $SU(N)$ you meant the obvious subgroup of the $U(N)$ that is given, i.e., the stabilizer of the identity coset in $SU(N{+}1)/U(N)$? If not, and you want the condition on $F$ that it push forward to define a Finsler metric on $CP^N$ that is invariant under $SU(N{+}1)$, then the requirement is that $F$ be invariant under both left $SU(N{+}1)$ action and right $U(N)$ action. The resulting push forward Finsler metric will be a multiple of the Fubini-Study metric, though. $\endgroup$
    – Robert Bryant
    Commented Dec 5, 2013 at 1:43
  • $\begingroup$ Thanks. I've made an edit to fix the error you spotted. How does one know that any such metric will be a multiple of the FS metric? Are you saying that there are no $SU(N+1)$ invariant Finsler metrics on $CP^N$ which are not multiples of the FS metric? If so, how does one know this? Also, do you have a ref for the required invariance properties? Thanks again. $\endgroup$
    – Benjamin
    Commented Dec 5, 2013 at 2:22

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Because $\mathrm{SU}(N{+}1)$ acts transitively on the projectivized tangent space of $\mathbb{CP}^N$ (a fact that has nothing to do with metrics on $\mathbb{CP}^N$), any $\mathrm{SU}(N{+}1)$-invariant Finsler structure is determined by knowing its value on a single tangent line at a single point of $\mathbb{CP}^N$. In particular, since the Fubini-Study metric is $\mathrm{SU}(N{+}1)$-invariant, any $\mathrm{SU}(N{+}1)$-invariant Finsler structure on $\mathbb{CP}^N$ must be a constant multiple of the Fubini-Study metric.

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  • $\begingroup$ Does this argument extend without modification to homogenous functions on $T \mathbb{C} P^N$? I mean to ask if Finsler metrics can be replaced with homogenous lagrangians on $\mathbb{C} P^N$, $\mathcal{L}: T \mathbb{C} P^N \rightarrow \mathbb{R}$ such that $\mathcal{L}(p, \lambda v) = \lambda \mathcal{L}(p, v)$ for all $ \lambda > 0$? It seems not to but I only ask so as to better understand the above argument. $\endgroup$
    – Benjamin
    Commented Mar 3, 2014 at 10:02
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    $\begingroup$ Yes, it does. The point is that $\mathrm{SU}(N{+}1)$ acts transitively on the unit tangent bundle of $\mathbb{CP}^N$ (in the Fubini-Study metric), so any $\mathrm{SU}(N{+}1)$-invariant function $\mathcal{L}:T\mathbb{CP}^N\to\mathbb{R}$ with the homogeneity property that you assume (even when you only require it to hold for $\lambda>0$) must be a constant multiple of the length function associated to the Fubini-Study metric. $\endgroup$
    – Robert Bryant
    Commented Mar 3, 2014 at 12:09
  • $\begingroup$ Ok. I was not clear if positive homogeneity was the only property of a Finler metric being used but I follow now. Is it correct to say then that $\mathcal{L}(|\psi \rangle, | \delta \psi \rangle ) = \frac{\langle \psi | \delta \psi \rangle}{(-i) \langle \psi | \psi \rangle^{1/2} }$ must be a constant multiple of the FS metric? I've used the notation of QM here similarly to the way it is used for the FS metric. But I'm still not clear if what I've written is actually a valid Lagrangian on $\mathbb{C P^N}$ which counts as invariant in the same sense as the FS metric. $\endgroup$
    – Benjamin
    Commented Mar 3, 2014 at 12:43
  • $\begingroup$ Unfortunately, I'm not familiar with this notation at all, so I can't say whether it represents the same thing that I have been saying. Sorry not to be of more help. $\endgroup$
    – Robert Bryant
    Commented Mar 3, 2014 at 13:44
  • $\begingroup$ Thanks for all your help, it is appreciated! In case of interest I was referring to the notation used here: en.wikipedia.org/wiki/…. $\endgroup$
    – Benjamin
    Commented Mar 3, 2014 at 13:57

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