Let $M$ be a $3$-Riemannian manifold and $\Sigma$ an embedded surface in $M$.

Using the Gauß equation, I want to show the following identity on $\Sigma$: $$R-2\mathrm{Ric}(\nu,\nu)-|A|^2=2K-H^2,$$ where $R$ is the scalar curvature, $\mathrm{Ric}$ is the Ricci tensor, $\nu$ is a unitary vector normal to $\Sigma$, $A$ is the second fundamental form, $K$ and $H$ are the Gauß and mean curvatures of $\Sigma$, respectively.

This appears for example at the beginning of the proof of Proposition 2, here.

Gauß equation: $\langle \overline{\boldsymbol{R}}(X,Y)Z,W\rangle=\langle \boldsymbol{R}(X,Y)Z,W\rangle-\langle A(Y,W),A(X,Z)\rangle+\langle A(X,W),A(Y,Z)\rangle$, where the bold letters $\overline{\boldsymbol{R}},\boldsymbol{R}$ indicates the curvature tensors of $M$ and $\Sigma$, respectively.

What I have tried:

  1. First of all, I was doubtful about the normal vector $\nu$. Since we don't know if $\Sigma$ is orientable, it might be that a (unitary differentiable) normal vector field does not exist on $\Sigma$. However, looking at the definitions, we have (is this really right?) $R(\nu)=R(-\nu)$, $\mathrm{Ric}(\nu,\nu)=\mathrm{Ric}(-\nu,-\nu)$, etc., in such a way that the identity makes sense, regardless the (non)orientability of $\Sigma$.
  2. Then, with the definitions of $R$ and $\mathrm{Ric}(\nu,\nu)$, I tried to work on the term $R-2\mathrm{Ric}(\nu,\nu)$, getting an expression which does not seem to clarify very much...
  3. Then came the next doubt: what is the usual norm for $A$? There are many equivalent ones, right? Generally, this is not important when we talk about continuity, but in the present case, it seems important because we deal with an equation/identity.

I was not able to do much more...

Other useful definitions:

Given an orthonormal basis $\{x_1,x_2,x_3=\nu\}$ of $T_pM$, $p\in \Sigma$, $$\mathrm{Ric}(\nu,\nu)=\frac{1}{2}\big(\langle \overline{\boldsymbol{R}}(\nu,x_1)\nu,x_1\rangle+\langle \overline{\boldsymbol{R}}(\nu,x_2)\nu,x_2\rangle\big)$$ $$R=\frac{1}{3}\sum_{j=1}^3\mathrm{Ric}(x_j,x_j)$$ $$A(X,Y)=\overline{\nabla}_{\overline{X}}\overline{Y}-\nabla_XY.$$

  • $\begingroup$ Hello, how did you embed $\Sigma$ in $M$, please? Is $\Sigma$ itself a subset of $M$? Or is $\Sigma$ embedded in $M$ via an embedding map $F:\Sigma\to M$? Thank you. $\endgroup$
    – Boar
    Nov 7, 2023 at 7:21

2 Answers 2


Let's do calculation following your idea (some different notations). $$\operatorname{Ric}(\nu,\nu)=\sum_{i=1}^3<\tilde{R}(e_i,e_3)e_3,e_i>=\sum_{i=1}^2<\tilde{R}(e_i,e_3)e_3,e_i>=\tilde{R}_{1331}+\tilde{R}_{2332}$$ and $$S=\sum_{i,j}^3<\tilde{R}(e_i,e_j)e_j,e_i>=2\sum_{i<j}^3<\tilde{R}(e_i,e_j)e_j,e_i>=2\tilde{R}_{1221}+2\tilde{R}_{1331}+2\tilde{R}_{2332}.$$ Thus \begin{eqnarray*} S-2\operatorname{Ric}(\nu,\nu)&=&2\tilde{R}_{1221}\\&=&2<\tilde{R}(e_1,e_2)e_2,e_1>\\ &=& 2<R(e_1,e_2)e_2,e_1>-2h(e_1,e_1)h(e_2,e_2)+2h^2(e_1,e_2)\\ &=& 2K-2k_1k_2\\ &=& 2K+(k_1^2+k_2^2)-(k_1+k_2)^2\\ &=& 2K+|A|^2-H^2 \end{eqnarray*} where we used the Gussian euqation in the third equation; in the fourth equation we choose $e_1,e_2$ to be the principal direction, i.e. for shape operator $Se_i=k_ie_i$; in the last equation I guess in your context the mean curvature is the sum of principal curvatures instead of half of them.


Thank you for your answer, H-H.

I also did it with help of a friend, so I will also post an answer.

The identity will follow from the Gauß equation: \begin{align*} \langle {\boldsymbol{R}}(X,Y)Z,W\rangle=\langle \boldsymbol{r}(X,Y)Z,W\rangle-\langle A(Y,W),A(X,Z)\rangle+\langle A(X,W),A(Y,Z)\rangle, \end{align*} where, for simplicity's sake, I will use $\boldsymbol{r}$ for the curvature tensor of $\Sigma$ (instead of those $\overline{\boldsymbol{R}}$ and $\boldsymbol{R}$). Let $\{e_1,e_2,\nu\}$ be an orthonormal base for $T_pM$, $p\in \Sigma$ and $\nu\perp \Sigma$. Doing $Y=W=e_1$, $Y=W=e_2$ and summing up the equations, we get \begin{align*} &\langle {\boldsymbol{R}}(X,e_1)Z,e_1\rangle+\langle{\boldsymbol{R}}(X,e_2)Z,e_2\rangle=\langle \boldsymbol{r}(X,e_1)Z,e_1\rangle-\langle A(e_1,e_1),A(X,Z)\rangle\\ &+\langle A(X,e_1),A(e_1,Z)\rangle+ \langle \boldsymbol{r}(X,e_2)Z,e_2\rangle-\langle A(e_2,e_2),A(X,Z)\rangle+\langle A(X,e_2),A(e_2,Z)\rangle. \end{align*}

Denote $A(X,Y)=h(X,Y)\nu$, and therefore $H:=\text{trace}(A)=h(e_1,e_1)+h(e_2,e_2)$. Now, summing to both sides of the equations the term $\langle \boldsymbol{R}(X,\nu)Z,\nu\rangle$, then in the left side we get $\text{trace}(Y\mapsto R(X,Y)Z)=\mathrm{Ric}(X,Z)$. Note that $\langle \boldsymbol{r}(X,e_1)Z,e_1\rangle+\langle \boldsymbol{r}(X,e_2)Z,e_2\rangle=\text{trace}(Y\mapsto \boldsymbol{r}(X,Y)Z)=\mathrm{ric}(X,Z)$. Hence: \begin{align*} \mathrm{Ric}(X,Z)&=\mathrm{ric}(X,Z)-(h(e_1,e_1)+h(e_2,e_2))h(X,Z)+h(X,e_1)h(e_1,Z)\\ &+h(X,e_2)h(e_2,Z)+\langle \boldsymbol{R}(X,\nu)Z,\nu\rangle\\ &=\mathrm{ric}(X,Z)-Hh(X,Z)+\sum_{i=1}^2h(X,e_i)h(e_i,Z)+\langle \boldsymbol{R}(X,\nu)Z,\nu\rangle. \end{align*}

Now, doing $X=Z=e_1$, $X=Z=e_2$ and then summing up, we have \begin{align*} \sum_{j=1}^2\mathrm{Ric}(e_j,e_j)&=\sum_{j=1}^2\mathrm{ric}(e_j,e_j)-H\sum_{j=1}^2h(e_j,e_j)+\sum_{i,j=1}^2h(e_i,e_j)^2+\sum_{j=1}^2\langle \boldsymbol{R}(e_j,\nu)e_j,\nu\rangle\\ &=r-H^2+|A|^2+\sum_{j=1}^2\langle \boldsymbol{R}(\nu,e_j)\nu,e_j\rangle\\ \\ &=r-H^2+|A|^2+\mathrm{Ric}(\nu,\nu). \end{align*} Finally, summing the term $\mathrm{Ric}(\nu,\nu)$ to both sides and using that $r=2K$, we get \begin{align*} R=2K-H^2+|A|^2+2\mathrm{Ric}(\nu,\nu). \end{align*}


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