# Ricci equation is trivial in codimension $1$

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?

• I think the fact that $\eta$ and $\xi$ are proportional should fairly quickly tell you that both sides are zero. – Anthony Carapetis Feb 17 at 2:12
• @AnthonyCarapetis How? I've added my calculations to the question, maybe I'm going in the wrong direction or I'm missing something. – Javi Feb 17 at 11:14
• 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. – Javier González Feb 17 at 11:38
• @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 $\nabla_X(g\eta)=g\nabla_X\eta+X(g\eta)$ – Javi Feb 17 at 11:44

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 $$0=0$$.