Induced metric via $\mathbb C P^n \cong SU(n + 1)/S(U(n) \times U(1))$ I was wondering if the homeomorphism above gives me the Fubini Study metric on $\mathbb C P^n$. More precisely:
Consider $\mathbb C P^n$ equipped with the metric induced by the standard construction $\mathbb C^* \to \mathbb C^{n + 1} \to \mathbb C P^n$. This metric is known as the fubini study metric.
Another way to construct a metric on $\mathbb C P^n$ is the following: Consider $SU(n + 1)$ equipped with the bi-invariant metric $\langle , \rangle$ given by $\langle X,Y \rangle = trace(XY^*)$, where $^ *$ denotes the conjugate transpose and $X,Y \in \mathcal{su}(n + 1)$, the Lie Algebra of $SU(n + 1)$ concsisting of traceless anti-hermitian matrices. Let $S(U(n) \times U(1)) \subset SU(n + 1)$ be the subgroup consisting of elements
\begin{pmatrix}
\det A^{-1} & 0 \\
0 & A \\
\end{pmatrix}
for $A \in U(n)$. I learned that we have $\mathbb C P^n \cong SU(n + 1)/S(U(n) \times U(1)).$ 
My question is if the induced quotient metric coincides with the Fubini Study metric. 
 A: Yes, the two metrics coincide, at least up to scaling.
Rather than cook up an explicit isometry, allow me to use some general theory.
First, notice that $SU(n+1)$ acts on $\mathbb{C}P^n$ in the following way.  Given a point (equivalence class) $[p]\in \mathbb{C}P^n$ and a matrix $A\in SU(n+1)$, lift $[p]$ to any preimage $p$ in $\mathbb{C}^{n+1}$.  Then define $A[p] = [Ap]$.  One can check that this is well defined (which amounts to the fact that $A$ commutes with $\lambda\cdot \operatorname{Id}$) and isometric (since the $SU(n+1)$ action on $\mathbb{C}^{n+1}$ is isometric).
One more important property:  This action is transitive.  Giveny any two points $[p]$ and $[q]$, there is an $A\in SU(n+1)$ with $A[p] = [q]$.  This follows from the fact that the action of $SU(n+1)$ on $\mathbb{C}^{n+1}$ acts transitively on the unit sphere and that any equivalence class $[p]$ contains points of unit length.
That's enough for this description, let's go to the $SU(n+1)/U(n+1)$ description.  I'm going to call these Lie groups $G$ and $H$ to save on typing.  On the Lie Algebra level, we get $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{p}$ (as vector spaces) where $\mathfrak{p}$ is orthogonal to $\mathfrak{h}$ (with respect to the bi-invariant metric).  It turns out, there is a canonical way to identify $\mathfrak{p}$ with $T_{eH} G/H$ by sending $X\in \mathfrak{p}$ to $\frac{d}{dt}|_{t=0} exp(tX)H$.
Further, the adjoint action of $H$ on $\mathfrak{g}$ preserves $\mathfrak{h}$, and thus, also preserves $\mathfrak{p}$.  Likewise, $H$ acts on $G/H$ by multiplication, with $h\ast gH = hgH (= hgh^{-1} H)$ and one can check that the identification between $\mathfrak{p}$ and $T_{eH}G/H$ is $H$-equivariant.
Now, via this identification, one gets the following fact:

$G$ invariant metrics on $G/H$ are in natural bijective correspondance with $Ad_H$ invariant inner products on $T_{eH}G/H$ which are in natural bijective correspondance with $Ad_H$ invariant inner products on $\mathfrak{p}$.

The proof is not too difficult:  A $G$ invariant metric on $G/H$ must automatically be $Ad_H$ invariant at $T_{eH}G/H$, and conversely, given an $Ad_H$ invariant metric, use left translation to move it around to all of $G/H$.  $Ad$ invariance guarantees this is well defined.
The action of $H$ on $\mathfrak{p}$ is linear, i.e., is a representation of $H$.  Since $H$ is compact, this decomposes into irreducible representations.  We have the following fact:

If no two irreducibles appearing are isomorphic to each other, then the dimension of $Ad_H$ invariant inner products on $\mathfrak{p}$ is given by the number of irreducible summands.

The proof proceeds by induction.  For the base case, if the action is already irreducible, Schur's lemma implies any invariant inner product must be a multiple of the Killing form.
In our particular setting, the action of $U(n)$ on $\mathfrak{p}$ is irreducible.  In fact, it's even better:  The $U(n)$ action on $\mathfrak{p}$ is nothing but the standard action of $U(n)$ on $\mathbb{C}^n$.  This action is transitive on the unit sphere, and this easily implies the actions is irreducible.
The upshot is that any two $Ad_{U(n)}$-invariant metrics on $\mathfrak{p}$ are the same up to scaling, and hence, any two metrics on $SU(n+1)/U(n)$ which are $SU(n+1)$-invariant are the same up to scaling.  Since the Fubini-Study metric is $SU(n+1)$-invariant, and (so is the one induced from the bi-invariant metric), it follows that they are the same up to scaling.
In order to figure out the scaling constant (if you care), simply take the canonical diffeo which maps a matrix in $SU(n+1)$ to its first row (or column), and see what it does to lengths of a single tangent vector.
