数理解析研究所講究録
1150 巻 2000 年 2-102Isometric immersions of homogeneous spaces
Per Tomter
Department of Mathematics,

University of Oslo,

Norway.
Abstract
In the theory of isometric immersions of submanifolds there are fundamental theoremsof John Nash for the$C^{r}$-case and Burstin-Cartan-Janet-Schl\"afly for the analytic case(also see Robert Greene (2) for the case of local isometric immersions). These theorems

require, however, a large codimension and are of practically no help in considering concrete
questions in low codimensions. An obvious way of producing large varieties of isometrically
immersed homogeneous submanifolds is to take the orbits of Lie group actions. In low
codimensions the following theorem should often be true: Let a compact, connected Lie

group

$G$

act on the connected manifold $N$ with principal orbit type $M=G/H$ . Then,among all the$G$-homogeneous metrics on $G/H$ the only ones which allow an isometricimmersion into $N$ are those which are already realized as the orbit metrics of this action.
Obviously this is true for the spheres $S^{n-1}(r)$ of$\mathbb{R}^{n}$under the standard $SO(n)$ -action.

With a little work it is also easy to prove for the larger classes of metrics invariant under
the unitary or symplectic groups. A slightly more challenging example is to prove this3
result for the second Stiefel manifold of 2-frames: $SO(n)/SO(n-2)$ of

diagonal embedding $SO(n)arrow SO(n)\cross SO(n)$ acting on$\mathbb{R}^{2n}$$M$$\hat{R}$$\langle\hat{R}(X\wedge Y)Z, W\rangle=$is the curvature operator of the submanifold

and $B$ is the second fundamental form. In the case of a non-Euclidean surrounding

space $N$ , however, the Gauss equation reads:$B(X, W)B(Y, Z)$ , where$N$

under the.

Such theorems follow from a careful study of the Gauss equation:$B(X, W)B(Y, Z)-B(X, Z)B(Y, W)$ , where$\mathbb{R}^{2n}$$R^{t}=\overline{R}^{t}-\hat{R}$and$\overline{R}^{t}$$\langle R^{t}(X\wedge Y)Z, W\rangle=B(X, Z)B(Y, W)$

is the part of the curvature operator$\overline{R}$-oftangential to $M$ . This varies with $M’ s$ position in $N$ , so the left hand side is also not

given. There has not been much study of this, but we will report on the solution of theprobably most basic question in isometric immersions into non-classical geometries.

Let $N=\mathbb{C}P(n)$ with metric normalized such that sectional curvatures are in $[1, 4]$ .
Each geodesic sphere $S^{2n-1}(r)=U(n)/U(n-1)$ determines a different homothety class,$\alpha^{2}\gamma_{r}$of metrics

Theorem:$S^{2n-1}$$(r \in(0, \frac{\pi}{2}),$ $\alpha^{2}\in(0, \infty))$

These Berger metrics.

$\gamma_{r}(r\in(0, \frac{\pi}{2}))$

which allow an isometric immersion into

Remark:are the only $U(n)$ -invariant metrics on$\mathbb{C}P(n)$.

This is a 1-dimensional set in the 2-dimensional variety of all

metrics. Thus no non-trivial homothetic image$\alpha^{2}\gamma_{r}$of$\gamma_{r}(\alpha^{2}\neq 1)$$U$(n)-imvariant

or no Berger metricfrom a geodesic sphere in complex hyperbolic space allows such an isometric immersion.
We give the flavor of the argument for one of the easier cases:
Let $\{e_{i}, Je_{i};i=0, \ldots, n-1\}$ be a adapted orthonormal basis for$T_{p}\mathbb{C}P(n)\cong \mathbb{C}^{n}\cong \mathbb{R}^{2n}$. Then$\overline{R}(e_{i}\wedge e_{j})=-e_{i}$A $e_{j}-Je_{i}\wedge Je_{j}$$\overline{R}(e_{i}\wedge Je_{i})=-2e_{0}\wedge Je_{0}-\cdots-4e_{i}$A $Je_{i}-\cdots-2e_{n-1}$ A $Je_{n-1}$

4
For $S^{2n-1}=U(n)/U(n-1)$ we choose an adapted orthonormal basis$J’Y_{n-1}$

on. $T_{p}J’Y_{n-1}T_{p},$$\mathbb{R}^{2n-2}$$S^{2n-1}\cong \mathbb{R}^{2n-1}=\mathbb{R}Y_{0}\oplus \mathbb{R}^{2n-2}$

is by the standard representation andby this action on$\mathbb{R}^{2n-2}$. Now, let$\alpha^{2}\gamma_{r}$$J’$$Y_{0},$ $Y_{1},$$J’Y_{1},$$\ldots,$$Y_{n-1}$,

where the isotropy action of $U(n-1)$

is the almost complex structure definedbe the metric of $S^{2n-1}(r)$ multiplied by$\alpha^{2}$. Then$\hat{R}(Y_{0}\wedge Y_{i})=-\frac{\cot^{2}r}{\alpha^{4}}Y_{0}\wedge Y_{i}$$\hat{R}(Y_{i}\wedge Y_{j})=-\frac{1+\cot^{2}r}{\alpha^{4}}Y_{i}\wedge Y_{j}-\frac{1}{\alpha^{4}}J’Y_{i}\wedge J’Y_{j}$$\hat{R}$($Y_{i}$A$J’Y_{i}$-.

Now, choose the basis$Je_{1}\sin\varphi$..

)$=- \frac{2}{\alpha^{4}}Y_{1}\wedge J’Y_{1}$$- \frac{4+\cot^{2}r}{\alpha^{4}}Y_{i}$$\{e_{i}, Je_{i}\}$such thatA$J’Y_{i}$$Je_{0}$–.

..

$- \frac{2}{\alpha^{4}}Y_{n-1}\wedge J’Y_{n-1}$.

is normal to $M$ at . Then $Y_{0}=e_{0}\cos\varphi+$$p$, and let $Y_{1}=e_{1}$ .

Lemma 1 $R^{t}(X\wedge Y)\wedge R^{t}(X\wedge Z)=0$ .

Proof.

The Gauss equation may be written: $\langle R^{t}(X\wedge Y)W_{1}, W_{2}\rangle=-(B\wedge B)(X, Y, W_{1}, W_{2})$$=-\overline{B\wedge B}$( $X$ A$Y$

) $(W_{1}, W_{2})$ where $B\wedge B$ is defined by: $(B\wedge B)(X, Y, Z, W)=B(X, Z)$$B(Y, W)-B(X, W)B(Y, Z)$ . Let$\hat{B}$be the shape operator, i.e.

Then: $R^{t}(X\wedge Y)=\hat{B}(X)\wedge\hat{B}(Y)$ . Hence $R^{t}(X\wedge Y)$ A$\hat{B}(X)\wedge\hat{B}(Z)=0$$R^{t}$( $X$ A$\langle\hat{B}(X), Y\rangle=B(X, Y)$$Z$.

) $=\hat{B}(X)\wedge\hat{B}(Y)\wedge$.

q.e.d.

Proposition 2 We have $\sin\varphi=0$ , hence we may assume $Y_{0}=e_{0}$ .

Proof.
. +2$..$$R^{t}$($Y_{0}$

A$Y_{1}$$e_{n-1}\wedge Je_{n-1}$$e_{3}\wedge Je_{3}+\cdots=0$)$= \cos\varphi(-1+\frac{\cot^{2}r}{\alpha^{4}})e_{0}$]. For $n\geq 4$ it is easy:

A $e_{1}+ \sin\varphi[(4-\frac{\cot^{2}r}{\alpha^{4}})e_{1}\wedge Je_{1}+2e_{2}\wedge Je_{2}+$$R^{t}(Y_{0}\wedge Y_{1})\wedge R^{t}(Y_{0}\wedge Y_{1})=\cdots 8\sin^{2}\varphi e_{2}\wedge Je_{2}\wedge$, i.e. $\sin^{2}\varphi=0$ . For $n=3$ it follows by more delicate choices.

Proposition 3 We have$\alpha^{4}=1,$$i.e$. $S^{2n-1}=S^{2n-1}(r)$ .

q.e.d.

5
Proof. We haveperpendicular to$Y_{0}=e_{0},$$e_{0},$$e_{1},$$Je_{1},$$J’Y_{1}=\cos\varphi Je_{1}+\sin\varphi e_{2}$$Y_{1}=e_{1},$$e_{2},$(and hence to$Je_{2}$$Y_{0},$$Y_{1},$$J’Y_{1}$. Now choose $Y_{3}=e_{3}$). Then$R^{t}(Y_{1}\wedge Y_{3})=\overline{R}^{t}(e_{1}\wedge$$e_{3})- \hat{R}(Y_{1}\wedge Y_{3})=(-1+\frac{1+\cot^{2}r}{\alpha^{4}})e_{1}\wedge e_{3}-Je_{1}\wedge Je_{3}+\frac{1}{\alpha^{4}}J’e_{1}\wedge J’e_{3}.\cdot R^{t}(Y_{1}\wedge Y_{3}.)\wedge R^{t}(Y_{1}\wedge Y_{3}.)=$$2(-1+ \frac{1+\cot^{2}r}{\alpha^{4}})e_{1}\wedge e_{3}\wedge$$2(-1+ \frac{\cot^{2}r}{\alpha^{4}})e_{0}\wedge e_{1}\wedge$$\alpha^{4}=1$(( $-Je_{1}$ A$Je_{3}+ \frac{1}{\alpha^{4}}J’e_{1}\wedge J’e_{3}$$-Je_{1} \wedge e_{3}+\frac{1}{\alpha^{4}}J’e_{1}$A$J’e_{3}$) $=0$ .

). Hence$R^{t}$($Y_{0}$

A$Y_{1}$) A $R^{t}(Y_{1}\wedge Y_{3})=$$Je_{1} \wedge Je_{3}=\frac{1}{\alpha^{4}}J’e_{1}\wedge J’e_{3}$, andq.e.d.

.

We now deal with the more complicated cases $n=3$ and (especially) $n=2$ , and givea couple of typical arguments. Assume first $n=3$ . Then

$R^{t}(Y_{0}\wedge Y_{1})\wedge R^{t}(Y_{0}\wedge Y_{1})=$$2 \sin\varphi\cos\varphi(\frac{\cot^{2}r}{\alpha^{4}}-1)e_{0}\wedge e_{1}\wedge e_{2}\wedge Je_{2}+2\sin^{2}\varphi(4-\frac{\cot^{2}r}{\alpha^{4}})e_{1}\wedge Je_{1}\wedge e_{2}\wedge Je_{2}+terms$of other type $=0$ . Hence, either $\sin\varphi=0$ or

the term 6

Choose$\sin\varphi\cos\varphi e_{0}\wedge e_{1}\wedge e_{2}\wedge Je_{2}$$Y_{2}=e_{2}$orthonormal to$e_{0},$. In the latter case we have

, hence $\cos\varphi=0$ ; i.e. we let $Y_{0}=Je_{1},$ $Y_{1}=e_{1}$ .

$Je_{1},$ $J’Y_{1}$$e_{1},$$\frac{\cot^{2}r}{\alpha^{4}}=4$(and hence to$Y_{1},$$Y_{0}$). Then$R^{t}(Y_{0}\wedge Y_{2})=$.

$\overline{R}^{t}(Je_{1}\wedge e_{2})-\hat{R}^{t}(Y_{0}\wedge Y_{2})=-Je_{1}\wedge e_{2}+e_{1}\wedge Je_{2}+\frac{\cot^{2}r}{\alpha^{4}}Y_{0}\wedge Y_{2}=(\frac{\cot^{2}r}{\alpha^{4}}-1)Je_{1}\wedge e_{2}+e_{1}\wedge Je_{2}$Hence$R^{t}(Y_{0} \wedge Y_{2})\wedge R^{t}(Y_{0}\wedge Y_{2})=2(\frac{\cot^{2}r}{\alpha^{4}}-1)Je_{1}\wedge e_{2}\wedge e_{1}\wedge Je_{2}=6Je_{1}\wedge e_{2}\wedge e_{1}\wedge Je_{2}\neq 0$,

which is a contradition. Hence $\sin\varphi=0$ .
This proves the following:
Proposition 4 For $n=3$ we have$\sin\varphi=0,\dot{?}.e$.

$Y_{0}=e_{0}$.

We also need:
Proposition 5 For $n=3$ we have$\alpha^{2}=1$.

Proof.

$Y_{0}=e_{0},$ $Y_{1}=e_{1},$ $J’Y_{1}=\cos\psi Je_{1}+\sin\psi e_{2}$thenis orthonormal to$e_{2}$$Y_{2}$, hence:$Y_{0},$$Y_{1},$$J’Y_{1}$(A$R^{t}(Y_{1}\wedge J’Y_{2})=\overline{R}^{t}$$e_{1}$. We define: $Y_{2}=-\sin\psi Je_{1}+\cos\psi e_{2}$ ,. By dimension $J’Y_{2}=\pm Je_{2}$ .

$(\pm Je_{2})$)$-\hat{R}$($Y_{1}$A$J’Y_{2}$$Y_{2}$

A $J’Y_{1}=-Je_{1}\wedge$) $=\mp e_{1}\wedge Je_{2}\pm Je_{1}\wedge e_{2}+$$\frac{1+c\circ t^{2}r}{\alpha^{4}}Y_{1}\wedge J’Y_{1}-\frac{1}{\alpha^{4}}J’Y_{1}\wedge Y_{2}=(\frac{1+\cot^{2}r}{\alpha^{4}}-1)Y_{1}\wedge J’Y_{2}+(\pm 1-\frac{1}{\alpha^{4}})J’Y_{1}\wedge Y_{2}$. By setting

6$R^{t}$(

A$Y_{1}$

have$R^{t}$$J’Y_{2}$($Y_{0}$) $\wedge R^{t}(Y_{1}\wedge J’Y_{2})=0$ we get: a) $1+\cot^{2}r=\alpha^{4}$ or b) $\alpha^{4}=1$ . We alsoA

$Y_{1}$)$=( \frac{\cot^{2}r}{\alpha^{4}}-1)Y_{0}\wedge Y_{1}$$(- \frac{1}{\alpha^{4}})(\pm 1-\frac{1}{\alpha^{4}})Y_{0}\wedge Y_{1}\wedge J’Y_{1}$. In case a) we have$R^{t}$($Y_{0}$

A$Y_{1}$)$\wedge R^{t}$($Y_{1}$A$J’Y_{2}$)$=$A $Y_{2}\neq 0$ , which is a contradition. Hence $\alpha^{2}=1$ .
$q.e.d$.

For $n=2$ this argument breaks down, since $\dim M_{p}=3$ and any wedge product of
4 vectors is zero. Indeed, the Gauss equations do have other solutions. Most of thoseare eliminated by the Codazzi equations: $\langle R(X\wedge Y)Z, N\rangle=YB(X, Z)-B(X, \nabla_{Y}Z)-$$XB(Y, Z)+B(Y, \nabla_{X}Z)+B([X, Y], Z)$ .

We note that this is a considerably more com-plicated case and only outline a few highlights of the constructions.

Let $n=2$ and let $SU(2)\cong S^{3}(r)\subset \mathbb{C}P(2)$ be the geodesic sphere of radius$(0, \frac{\pi}{2})$. An orthogonal basis for $SU(2)$ at a point$E_{2}=J’E_{1}=(_{i0}^{0i})$$\mathbb{C}P(2)$$S^{3}$, with, it follows that

Extend$Y_{i}$

is given by:. An orthonormal basis is given byhomothetic image of this by:
to prove that if$p$$Y_{i}$$E_{0}=(\begin{array}{ll}i 00-i \end{array}),$$\frac{1}{\sin r\cos r}E_{0},$$Y_{0}= \frac{1}{\alpha^{2}\sin r\cos r}E_{0)}Y_{i}=\frac{1}{\alpha^{2}\sin r}E_{i},$$\frac{1}{\sin r}E_{1},$$r,$$E_{1}=(\begin{array}{ll}0 1-10 \end{array})$$\frac{1}{\sin r}E_{2}$,, and a$i=1,2$ . We again need

as an orthonormal basis, admits an isometric immersion into$\alpha^{2}=1$.

to left-invariant vector fields on$S^{3}$. Then, in the Koszul formula:$X\langle Y, Z\rangle+Y\langle X, Z\rangle-Z\langle X, Y\rangle+\langle[X, Y], Z\rangle-\langle[Y, Z], X\rangle+\langle[Z, X], Y\rangle$$\langle\nabla_{X}Y, Z\rangle=$

the three first termsof the right hand side vanish.
We compute:(1)$r\in$$[Y_{0}, Y_{1}]= \frac{2}{\alpha^{2}\sin r\cos r}Y_{2}$,$[Y_{0}, Y_{2}]=- \frac{2}{\alpha^{2}\sin r\cos r}Y_{1}$and by repeated use of the Koszul formula we find:,$[Y_{1}, Y_{2}]= \frac{2\cot r}{\alpha^{2}}Y_{0}$

7

(2)

$\nabla_{Y_{0}}Y_{1}=\frac{1+\sin^{2}r}{\alpha^{2}\sin r\cos r}Y_{2}$,$\nabla_{Y_{0}}Y_{2}=-\frac{1+\sin^{2}r}{\alpha^{2}\sin r\cos r}Y_{1}$$\nabla_{Y_{1}}Y_{2}=\frac{\cot r}{\alpha^{2}}Y_{0}$,$\nabla_{Y_{1}}Y_{0}=-\frac{\cot r}{\alpha^{2}}Y_{2}$,$\nabla_{Y_{2}}Y_{0}=\frac{\cot r}{\alpha^{2}}Y_{1}$,,$\nabla_{Y_{2}}Y_{1}=-\frac{\cot r}{\alpha^{2}}Y_{0}$$\nabla_{Y_{0}}Y_{0}=\nabla_{Y_{1}}Y_{1}=\nabla_{Y_{2}}Y_{2}=0$.

$\hat{R}(Y_{0}\wedge Y_{1})Y_{0}=\nabla_{Y_{0}}\nabla_{Y_{1}}Y_{0}-\nabla_{Y_{1}}\nabla_{Y_{0}}Y_{0}-\nabla_{[Y_{0},Y_{1}]}Y_{0}=-\nabla_{Y_{0}}(\frac{\cot r}{\alpha^{2}}Y_{2})-0-\frac{2}{\alpha^{2}\sin r\cos r}\frac{\cot r}{\alpha^{2}}\nabla_{Y_{2}}Y_{0}=$.

$\frac{\cot r}{\alpha^{2}}\frac{1+\sin^{2}r}{\alpha^{2}\sin r\cos r}Y_{1}-\frac{\cot r}{\alpha^{2}}\frac{2}{\alpha^{2}\sin r\cos r}Y_{1}=-\frac{\cot^{2}r}{\alpha^{4}}Y_{1}$Similarly:$Y_{1}$$\hat{R}(Y_{0}\wedge Y_{1})Y_{1}=\frac{\cot^{2}r}{\alpha^{4}}Y_{0}$. By similar computations$S^{3}$. Choose $Y_{1}=e_{1}$ , then$Y_{0}$

A$Y_{1}$)$=\overline{R}$($Y_{0}$

A$Y_{1}$)

and$\hat{R}(Y_{0}\wedge Y_{1})=-\frac{\cot^{2}r}{\alpha^{4}}Y_{0}\wedge$$\hat{R}(Y_{1}\wedge Y_{2})=-\frac{4+\cot^{2}r}{\alpha^{4}}Y_{1}\wedge Y_{2}$$p$

is normal to$Y_{1}$$\overline{R}(e_{1}\wedge e_{2})=-4e_{1}\wedge e_{2}-2e_{0}\wedge Je_{0}$(. Hence

at , then $J(-Je_{0})=e_{0}$ is in

say $Y_{2}=-\sin\varphi e_{0}+\cos\varphi Je_{1}$ . Now$R$$\hat{R}(Y_{0}\wedge Y_{1})Y_{2}=0$$\hat{R}(Y_{0}\wedge Y_{2})=-\frac{\cot^{2}r}{\alpha^{4}}Y_{0}\wedge Y_{2}$Now, let $-Je_{0}$ be normal to$Je_{1}\sin\varphi$

and$e_{0}$

and$Y_{0}$$T_{p}S^{3}$.

$Y_{0}=e_{0}\cos\varphi+$. $Y_{2}=\mp(\sin\varphi e_{0}-\cos\varphi Je_{1})$ ,$\overline{R}(e_{0}\wedge e_{i})=-e_{0}\wedge e_{i}-Je_{0}$A$Je_{i},$$i=1,2$ .

.

$- \hat{R}(Y_{0}\wedge Y_{1})=\overline{R}((e_{0}\cos\varphi+Je_{1}\sin\varphi)\Lambda e_{1})+\frac{\cot^{2}r}{\alpha^{4}}Y_{0}\wedge Y_{1}$$=( \frac{\cot^{2}r}{\alpha^{4}}-1-3\sin^{2}\varphi)Y_{0}\wedge Y_{1}+3\sin\varphi\cos\varphi Y_{1}\wedge Y_{2}+[3\sin\varphi\cos\varphi Y_{0}-(3\sin^{2}\varphi-1)Y_{2}]\wedge Je_{0}$By similar computations

3.

$R(Y_{0} \wedge Y_{2})=(\frac{\cot^{2}r}{\alpha^{4}}-1)Y_{0}$.

A $Y_{2}-Y_{1}\wedge Je_{0}$ , and $R(Y_{1}\wedge Y_{2})=$.

$\sin\varphi\cos\varphi Y_{0}\wedge Y_{1}+(\frac{\cot^{2}r+4}{\alpha^{4}}-1-3\cos^{2}\varphi)Y_{1}\wedge Y_{2}+[3\sin\varphi\cos\varphi Y_{2}+(1-3\cos^{2}\varphi)Y_{0}]\wedge Je_{0}$Now, consider Gauss’ equations:Gl.

$\langle R(Y_{0}\wedge Y_{1})Y_{0}, Y_{1}\rangle=\frac{\cot^{2}r}{\alpha^{4}}-1-3\sin^{2}\varphi=b_{00}b_{11}-b_{01}^{2}$

G2.

$\langle R(Y_{0}\wedge Y_{1})Y_{0}, Y_{2}\rangle=0=b_{00}b_{12}-b_{01}b_{02}$G3.

$\langle R(Y_{0}\wedge Y_{1})Y_{1}, Y_{2}\rangle=3\sin\varphi\cos\varphi=b_{01}b_{12}-b_{02}b_{11}$G4.

$\langle R(Y_{0}\wedge Y_{2})Y_{0}, Y_{2}\rangle=\frac{\cot^{2}r}{\alpha^{4}}-1=b_{00}b_{22}-b_{02}^{2}$

G5.

$\langle R(Y_{1}\wedge Y_{2})Y_{0}, Y_{2}\rangle=0=b_{01}b_{22}-b_{02}b_{12}$G6.

$\langle R(Y_{1}\wedge Y_{2})Y_{1}, Y_{2}\rangle=\frac{4+\cot^{2}r}{\alpha^{4}}-1-3\cos^{2}\varphi=b_{11}b_{22}-b_{12}^{2}$

where $b_{ij}=B(Y_{i}, Y_{j})$8For example, by expanding the determinant$|_{b_{02}b_{12}b_{22}}^{b_{00}b_{01}b_{02}}b_{00}b_{01}b_{02}|=0$after the first line, I obtain:.

$0 \cdot b_{00}-(\frac{\cot^{2}r}{\alpha^{4}}-1)b_{01}+0\cdot b_{02}=0$

Assume$\alpha^{4}\neq\cot^{2}r$(3), then $b_{01}=0$ . Similarly, under this assumption, we get:$b_{12}=0$.

$-3 \sin\varphi\cos\varphi b_{00}+(1+3\sin\varphi-\frac{\cot^{2}r}{\alpha^{4}})b_{02}=0$$( \frac{4+\cot^{2}r}{\alpha^{4}}-1-3\cos^{2}\varphi)b_{02}+3\sin\varphi\cos\varphi b_{22}=0$

Furthermore:(4)$\det B=|_{b_{02}b_{12}b_{22}}^{b_{00}b_{01}b_{02}}b_{01}b_{11}b_{12}|$$( \frac{4+\cot^{2}r}{\alpha^{4}}-1-3\cos^{2}\varphi)b_{00}+3\sin\varphi\cos\varphi b_{02}$$=( \frac{\cot^{2}r}{\alpha^{4}}-1)b_{11}=3\sin\varphi\cos\varphi b_{02}+(\frac{\cot^{2}r}{\alpha^{4}}-1-3\sin^{2}\varphi)b_{22}$(3) and (4) give 4 equations for the 4 unknowns$b_{00},$ $b_{11},$ $b_{22},$ $b_{02}$.

(in terms of ). But we also$\varphi$have the Codazzi equaions: $\langle R(X\wedge Y)Z, N\rangle=YB(X, Z)-B(X, \nabla_{Y}Z)-XB(Y, Z)+$
$B(Y, \nabla_{X}Z)+B([X, Y],$Cl.

$Z\rangle$, where $N$ is a normal vector of$T_{p}S^{3}$.

$\langle R(Y_{0}\wedge Y_{1})Y_{0}, -Je_{0}\rangle=-3\sin\varphi\cos\varphi$$=Y_{1}(b_{00})+ \frac{\cot r}{\alpha^{2}}b_{02}+\frac{2}{\alpha^{2}\sin r\cos r}b_{02}=Y_{1}(b_{00})+\frac{2+\cos^{2}r}{\alpha^{2}\sin r\cos r}b_{02}$Similarly:

C5.

$\langle R(Y_{0}\wedge Y_{2})Y_{1}, -Je_{0}\rangle=1=\frac{\cot r}{\alpha^{2}}b_{00}-\frac{2}{\alpha^{2}\sin r\cos r}b_{11}+\frac{1+\sin^{2}r}{\alpha^{2}\sin r\cos r}b_{22}$C9.

$\langle R(Y_{1}\wedge Y_{2})Y_{2}, -Je_{0}\rangle=-3\sin\varphi\cos\varphi=-Y_{1}(b_{22})+3\frac{\cot r}{\alpha^{2}}b_{02}$(There are 9 such equations altogether.)Now, from the 4 equations from (3) and (4) we may solve$b_{00},$ $b_{11},$ $b_{22}$(in terms of )$\varphi$and substitute into C5. Although this is quite laborious, it works, and we get a non-trivial
expression in$\sin\varphi$

equal to 1. Hencemust vanish. Cl and C9 gives$b_{02}=0$$\sin\varphi$

must be constant, and the terms$\frac{2+\cos^{2}r}{\alpha^{2}\sin r\cos r}b_{02}=\frac{3\cos^{2}r}{\alpha^{2}\sin r\cos r}b_{02}$, since$Y_{1}(b_{00}),$ $Y_{1}(b_{22})$$r \in(0, \frac{\pi}{2})$this implies. But then, by Cl again, $\sin\varphi\cos\varphi=0$ . One can show that $\cos\varphi=0$ leads to a9contradiction. Consider $\sin\varphi=0$ . Then$b_{11}=b_{22}\neq 0$.

$b_{11}b_{22}=b_{11}^{2}= \frac{4+\cot^{2}r}{\alpha^{4}}-4$$b_{11}=\pm\sqrt{\frac{4+\cot^{2}r}{\alpha^{4}}-4}$

and$b_{00}b_{11}= \frac{\cot^{2}r}{\alpha^{4}}-1=b_{00}b_{22}$. C5 now says:. Since$\frac{\cot r}{\alpha^{2}}(b_{00}-b_{11})=1$$b_{00}= \pm(\frac{\cot^{2}r}{\alpha^{4}}-1)(\frac{4+\cot^{2}r}{\alpha^{4}}-4)^{-1/2}$$\alpha^{4}\neq\cot^{2}r$,. By solving

and substituting this into

C5 we now get an equation for :$\alpha$$\alpha^{12}+(2\cot^{2}r-1)\alpha^{8}-6\cot^{2}r\alpha^{4}+4\cot^{2}r=0$

Obviously

is a solution, and by dividing by$\alpha^{4}=1$$4\cot^{2}r=0$$\alpha^{4}-1$.

we get:$\alpha^{8}+2\cot^{2}r\alpha^{4}-$. $\alpha^{4}=\sqrt{\cot^{4}r+4\cot^{2}r}-\cot^{2}r$ as a possible second solution. To eliminate

this we consider Maeda’s condition for the principal curvatures of a hypersurface.

is the principal curvature along the principal direction,$b_{22}=\Delta b\Omega b\lrcorner L_{\frac{+2}{b_{00}}}2b_{11}-=b_{11}$$\frac{8}{\alpha^{4}}=8,$

$i.e$

.

Remark.

$\alpha^{4}=1$, i.e.

$b_{11}$along the direction$e_{1}$$b_{00}$. Then$b_{00}b_{11}+2=2b_{11}^{2}-b_{00}b_{11}.2b_{11}^{2}-2b_{00}b_{11}= \frac{8+2\cot^{2}r}{\alpha^{4}}-8-2\frac{\cot^{2}r}{\alpha^{4}}+2=2$.

$q.e.d$.

Alternatively, we could require that the curvature tensor of the total curvature

constructed from the metric of$S^{3}$

and$B$

be parallell.

Remark.

This is of course only a small part of the work. One needs to check $\cos\varphi=0$ ,$\alpha^{4}=\cot^{2}r$, and then one must check that none of the $U(n)$ -invariant metrics of the spheresin complex hyperbolic or Euclidean space admit an isometric immersion into$\mathbb{C}P(n)$. Allthese results are true for local isometric immersions, obviously. Corresponding results also
hold for complex hyperbolic space and quaternionic projective and hyperbolic spaces, andwill be published shortly.

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