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. $\Theta$.

  • 1
    $\begingroup$ These metrics are called Berger metrics, and they are even $U(k+1)$-invariant. Indeed, with the correct biinvariant metric on $U(k+1)$, some of them are even normally homogeneous. The others are naturally reductive (this is similar to normally homogeneous, but you allow degenerate or indefinite biinvariant metrics on $U(k+1)$. And then of course you can restrict the action to $SU(k+1)$ if you prefer. You should find enough literature on that topic. $\endgroup$ Commented Apr 7, 2016 at 16:40
  • $\begingroup$ It helps to write out explicitly the contact 1-form, and then you can just add a suitable multiple of its square to the metric, I think. $\endgroup$
    – Ben McKay
    Commented Apr 8, 2016 at 6:17

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.