Define a Riemannian metric on $\mathbb C^{n+1} - \{0\}$ in the following way: If $Z \in \mathbb C^{n+1}-\{0\}$ and $V,W \in T_Z(\mathbb C^{n+1}-\{0\})$, $$\langle V,W \rangle_Z = \frac{\text{Real}(V,W)}{(Z,Z)}.$$ Observe that hte mteric $\langle \,,\,\rangle$ restricted to $S^{2n+1} \subset \mathbb C^{n+1}-\{0\}$ coincides with the metric induced from $\mathbb R^{2n+2}$.

(b) Show that, in this metric, the sectional curvature of $\mathbb P^n(\mathbb C)$ is given by $$K(\sigma)=1+3\cos^2 \varphi,$$ where $\sigma$ is generated by the orthonormal pair $X,Y, \cos \varphi=\left\langle \overline X,i\overline Y\right\rangle$, and $\overline X,\overline Y$ are the horizontal lifts of $X$ and $Y$, respectively. In particular, $1 \le K(\sigma) \le 4$.

This is Exercise 8.12 of Riemannian Geometry by do Carmo. Yes, there is a part (a) but I did not print it here because my question here concerns only part (b).

Also, I realize that $P^n(\mathbb C)$ is the complex projective space of complex dimension $n$ and is denoted in other textbooks as $\mathbb CP^n$.

Hint for (b): Let $Z$ be the position vector describing $S^{2n+1}$. Since $(\frac d{d\theta}e^{i\theta}Z)_{\theta=0}=iZ$, $iZ \in T_Z(S^{2n+1})$ and is vertical. Let $\overline \nabla$ be the Riemannian connection of $\mathbb R^{2n+2} \approx \mathbb C^{n+1}$ and $X,Y \in \mathcal X(P^n(\mathbb C))$. Take $\alpha : (-\epsilon,\epsilon) \to S^{2n+1}$ with $\alpha(0)=Z$, $\alpha'(0)-\overline X$. Then \begin{align} (\overline \nabla_{\overline X}iZ)_Z &= \frac d{dt} iZ \circ \alpha(t))\Big\vert_{t=0} \\ &= \frac d{dt} i\alpha(t) \Big\vert_{t=0} = i\alpha'(0)=i\overline X. \end{align} Therefore, \begin{align} \left\langle [\overline X,\overline Y],iZ \right\rangle &= \left\langle \overline \nabla_{\overline X}\overline Y-\overline \nabla_{\overline Y}\overline X,iZ \right\rangle \\ &= -\left\langle i\overline X,\overline Y \right\rangle + \left\langle i\overline Y,\overline X \right\rangle = 2\cos \varphi. \end{align} Now use Exercise 10(b).

I am trying to provide the details of the last line: "Now use Exercise 10(b)". Exercise 8.10(b) merely asserts O'Neill's formula:

$$ K(\sigma)=\overline K(\overline \sigma) + \frac 34 \left|[\overline X,\overline Y]^v \right|^2 $$

So my approach is as follows: $$ K(\sigma)=\overline K(\overline \sigma) + \frac 34 \left|[\overline X,\overline Y]^v \right|^2 = 1 + \frac 34 \left|[\overline X,\overline Y]^v \right|^2 $$ and $$ 1+3\cos^2 \varphi = 1+\frac 34 |2 \cos \varphi|^2 = 1+\frac 34 \left| \left\langle [\overline X,\overline Y],iZ \right\rangle \right|^2. $$ But I am not sure how to link these together by establishing $$ \left|[\overline X,\overline Y]^v \right| = \left| \left\langle [\overline X,\overline Y],iZ \right\rangle \right|. $$ Clearly, $[\overline X,\overline Y]^v$ is a vertical field, so $\left\langle [\overline X,\overline Y],iZ \right\rangle$ has to be vertical as well?


1 Answer 1


No, $\langle [\overline{X},\overline{Y}], iZ\rangle$ is not vertical - it is not even a vector field!

The claim (which I leave to you) that you need to establish to link the two is the following:

Claim: $iZ$ is a unit vector which spans the vertical space.

Believeing this claim, the point is that $\left(\left\langle [\overline{X},\overline{Y}], iZ\right\rangle\right) i Z$ is the projection of $[\overline{X},\overline{Y}]$ to the vertical subspace. That is, $\left(\left\langle [\overline{X},\overline{Y}], iZ\right\rangle\right) i Z = [\overline{X},\overline{Y}]^v.$ Taking lengths of both sides of this equality gives the equality you need.

  • $\begingroup$ Let me know if you have trouble proving the claim, and then I can provide more details $\endgroup$ Jul 8, 2017 at 5:29
  • $\begingroup$ The first line of the textbook's given hint says "Let $Z$ be the position vector describing $S^{2n+1}$." Because $S^{2n+1}$ is a unit sphere, $|Z|=1$. And obviously $|i|=1$. So $|iZ|=|i||Z|=1\cdot 1=1$. I am still unsure about the part why $iZ \in T_Z(S^{2n+1})$ is vertical, mostly because my fundamental understanding of the "fiber"--let alone a vector being orthogonal to said fiber so that said vector is horizontal--is shaky. $\endgroup$ Jul 8, 2017 at 6:24
  • $\begingroup$ I don't have Do Carmo with me right now, can you remind me his definition of being vertical? As well as his definition of $\mathbb{C}P^n$? For me, $\mathbb{C}P^n$ is the quotient of $S^{2n+1}$ by the natural $S^1$ action coming from complex multiplication. I define "vertical" by "kernel of the differential $\pi_\ast:T_ p S^{2n+1}\rightarrow T_{\pi(p)}\mathbb{C}P^n$. Then, the curve $e^{it} Z$ is on $S^{2n+1}$ and has derivative $iZ$. Then, almost by definition, $\pi(e^{it} Z)$ is constant, so $iZ$ is in the kernel of the differential map. $\endgroup$ Jul 8, 2017 at 15:34
  • $\begingroup$ Here are the original problems (8.8-8.12) of do Carmo in this PDF: docdro.id/hA13UgA. 8.8 defines the vertical vector, 8.9 mentions horizontal lift, and 8.11 introduces the notion of the compex projective space $P^n(\mathbb C)$. $\endgroup$ Jul 8, 2017 at 17:47
  • 1
    $\begingroup$ Well, $\langle \overline{Y}, iZ\rangle = 0$ since $iZ$ is vertical and $\overline{Y}$ is horizontal. So, $X\langle \overline{Y}, iZ\rangle = 0$, so $\langle\overline{\nabla}_X \overline{Y}, iZ\rangle = - \langle \overline{Y}, \nabla_X iZ\rangle.$ Now use the computation above in the book. For that computation, the first equality is because you have the usual Euclidean connection. Then next equality $Z\circ \alpha(t) = \alpha(t)$ is because $Z$ is literally the position vector. $\endgroup$ Jul 9, 2017 at 18:11

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