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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. $\Theta$.

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    $\begingroup$ 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

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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.

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