Let $S^n$ be the $n$-sphere. If $n=2k+1$ is odd, then we can identify $S^n$ as a subset of $\mathbb{C}^{k+1}$. We define the $S^1$ action on $S^n$ by multiplication, namely 
$$ \Psi \colon S^1 \times S^n \to S^n, \ (c, (z\_0, \dots , z\_k)) \mapsto (cz\_0, \dots cz\_k).$$


If we endow $S^n$ with the standard-metric $g$, then this action is isometric. Now I want to define a new family $g^t$ of metrics on $S^n$. We multiply the standard-metric with $t^2$ in the directions tangent to the $S^1$-orbits.


Furthermore we can define the action


$$ \Theta \colon SU(k+1) \times S^n \to S^n, \ (A, (z\_0, \dots, z\_k)) \mapsto (Az\_0, \dots, Az\_k).$$


Since $SU(k+1)$ acts by isometries on $\mathbb{C}^{k+1}$ and $S^n$ is an invariant submanifold of $\mathbb{C}^{k+1}$, we have that $\Theta$ defines also an isometric action on $(S^n, g)$.


My question is now:


1) How can I prove, that $SU(k+1)$ acts isometrically on $(S^n, g^t)$?


For that my idea was to use, that $(S^n,g)$ is a homogeneous space with $SU(k+1)/SU(k) = S^n$. That means we choose a point $q \in S^n$ and we have the projection $\pi \colon SU(k+1) \to S^n, \quad A \mapsto Aq$. 


Now this induces a map $\overline{\pi} \colon SU(k+1) \to S^n, \quad A \mapsto A.SU(k)$


So we have a left-invariant metric $\langle \cdot, \cdot \rangle$ on $\mathfrak{su}(n+1)$ and a corresponding orthogonal decomposition $\mathfrak{su}(n+1) = \mathfrak{su}(n) \oplus \mathfrak{p}$ such that the standard metric restricted to $g(q)$ can be identified with $\langle \cdot, \cdot \rangle |\_{\mathfrak{p}}$.


Now we find left-invariant vectorfields $X\_1, \dots, X\_{n^2} \in \mathfrak{su}(n)$ and $Y\_0, \dots, Y\_l \in \mathfrak{p}$ that are orthonormal. With $\sigma\_0, \dots, \sigma\_l$ we denote the dual elements of $Y\_0, \dots, Y\_l$. Furthermore we can choose $Y\_0$ to be the tangentvector in the direction of the above defined $S^1$-action. Then $g(q) = (\sigma\_0)^2+ (\sigma\_1)^2+ \cdots + (\sigma\_l)^2$ and we get $g^t(q) = t^2(\sigma\_0)^2 + (\sigma\_1)^2 + \cdots + (\sigma\_l)^2$.


How could I use this description of $g^t$ to show that it is $SU(k+1)$-invariant?


Edit: Maybe that way? For $x \in S^n$ and $v,w \in T\_xS^n$ we have $\lambda\_j, \mu\_j \in \mathbb{R}$ such that $g(x)(v,w)= \langle (\lambda\_0 Y\_0 +\sum\_{j=1}^l \lambda\_j Y\_j), (\mu\_0 Y\_0 + \sum\_{j=1}^l \mu\_j Y\_j) \rangle$. 
Then $g^t(x)(v,w) = \langle (t\lambda\_0 Y\_0 +\sum\_{j=1}^l \lambda\_j Y\_j), (t\mu\_0 Y\_0 + \sum\_{j=1}^l \mu\_j Y\_j) \rangle$ and so
$(\Theta\_A)^\*g^t(x)(v,w) = \langle \Theta\_A(t\lambda\_0 Y\_0 +\sum\_{j=1}^l \lambda\_j Y\_j), \Theta\_A(t\mu\_0 Y\_0 + \sum\_{j=1}^l \mu\_j Y\_j) \rangle = \langle (t\lambda\_0 Y\_0 +\sum\_{j=1}^l \lambda\_j Y\_j), (t\mu\_0 Y\_0 + \sum\_{j=1}^l \mu\_j Y\_j) \rangle=g^t(x)(v,w)$ 


since $\langle \cdot, \cdot \rangle$ is left-invariant w.r.t. $\Theta$.



isometric action on the $n$-sphere
The actions $\Psi$ and $\Theta$ commute and hence $\Theta$ maps $\Psi$ orbits into $\Psi$ orbits. Since $\Theta$ acts by $g$-isometries, it follows that it also preserves the splitting of the tangent space into $\Psi$-direction and its ortho-complement. You modify $g$ only in the $\Psi$-direction so it's clear that $\Theta$ will also act by isometries for this modified metric.


