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Let $f:M\to\overline{M}$ an isometric immersion and assume $\dim(M)=\dim(\overline{M})-1$.

I'm asked to show that the Ricci equation offers no information. I guess what I have to show is that the Ricci equation gives something like $0=0$.

What I have as Ricci equation is that for all $X,Y$ vector fields on $M$ and for all $\eta,\xi$ normal vector fields on the image of $f$ we have

$\langle \overline{R}(\overline{X},\overline{Y})\overline{\eta},\overline{\xi}\rangle=\langle R^\perp (X,Y)\eta,\xi\rangle -\langle [A_\eta,A_\xi](X),Y\rangle $

where $R$ and $\overline{R}$ are the curvature tensors on $M$ and $\overline{M}$, the bar represents an extension to $\overline{M}$ and $A_\eta$ is the Weingarten endomorphism.

I've been manipulating this equation using facts like $A_\eta(X)=-\nabla_X\eta$ and that $T_p^\perp M$ has dimension one (and therefore all normal vector fields are proportional). However, I haven't been able to reach any conclusion comparable to "the Ricci equation offers no information".

For instance, on the RHS, on the one hand I get

$ \langle [A_\eta,A_\xi](X),Y\rangle = \langle A_\eta(A_\xi(X))-A_\xi(A_\eta(X)),Y\rangle =\langle A_\eta(A_\xi(X)),Y\rangle- \langle A_\xi(A_\eta(X)),Y\rangle = $

$ \langle A_\xi(X),A_\eta(Y)\rangle-\langle A_\eta(X),A_\xi(Y)\rangle= \langle \nabla_X\xi,\nabla_Y\eta\rangle -\langle \nabla_X\eta, \nabla_Y\xi\rangle=\langle \nabla_X(\xi-\eta),\nabla_Y(\eta-\xi)\rangle $

On the other hand

$ R^\perp(X,Y)\eta=\nabla_Y^\perp \nabla_X^\perp\eta-\nabla_X^\perp \nabla_Y^\perp\eta+\nabla_{[X,Y]}^\perp\eta $

So

$ \langle R^\perp (X,Y)\eta,\xi\rangle= \langle\nabla_Y^\perp \nabla_X^\perp\eta,\xi\rangle -\langle\nabla_X^\perp \nabla_Y^\perp\eta,\xi\rangle+\langle\nabla_{[X,Y]}^\perp\eta,\xi\rangle $

which doesn't seem to be equal to the previous expression.

On the LHS it is essentially the same but without $\perp$ and writing bars, so I don't see why that should vanish.

What can be then deduced from the Ricci equation in the codimension $1$ case?

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    $\begingroup$ I think the fact that $\eta$ and $\xi$ are proportional should fairly quickly tell you that both sides are zero. $\endgroup$ – Anthony Carapetis Feb 17 at 2:12
  • $\begingroup$ @AnthonyCarapetis How? I've added my calculations to the question, maybe I'm going in the wrong direction or I'm missing something. $\endgroup$ – Javi Feb 17 at 11:14
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    $\begingroup$ It's easy to see that $\nabla^\perp_X \eta = 0$ for all $X \in \mathfrak{X}(M)$, $\eta \in \mathfrak{X}^\perp(f)$, using the fact that $A_\tilde{\eta} \tilde{X} = -\overline{\nabla}_\tilde{X} \tilde{\eta}$, so your $\langle R^\perp(X,Y)\eta$ must vanish. $\endgroup$ – Javier González Feb 17 at 11:38
  • $\begingroup$ @JavierGonzález thanks, that simplifies the story, but know I have to show that $\langle \nabla_X(\xi-\eta),\nabla_Y(\eta-\xi)\rangle =0$. If $\xi=\eta$ that's obvious, but in general each term of the product is of the form $\nabla_X(g\eta)=g\nabla_X\eta+X(g\eta)$ $\endgroup$ – Javi Feb 17 at 11:44
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We have the equation $$\langle \overline{R}(\overline{X},\overline{Y})\overline{\eta},\overline{\xi}\rangle=\langle R^\perp (X,Y)\eta,\xi\rangle -\langle [A_\eta,A_\xi](X),Y\rangle.$$In codimension $1$, we have $\nabla^\perp = 0$. So $R^\perp = 0$ and the first term on the right vanishes. Now, since $\overline{\eta} = f\overline{\xi}$ and $\eta = f\xi$, we have that $\overline{R}(\overline{X},\overline{Y}, \overline{\eta},\overline{\xi}) = f\overline{R}(\overline{X},\overline{Y}, \overline{\xi},\overline{\xi}) = 0$ since $\overline{R}$ is skew in the last two entries. So the left side vanishes. Lastly, we have that $[A_\eta,A_\xi] = f[A_\xi,A_\xi] = 0$, since $[\cdot,\cdot]$ is also skew, and thus the last remaining term also vanishes, and it all reduces to $0=0$.

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