Proving $R_{XY}Z=K[\langle Z, X \rangle Y-\langle Z,Y\rangle X]$

Question

For a semi-Riemannian surface:

a) $R_{XY}Z=K[\langle Z, X \rangle Y-\langle Z,Y\rangle X]$

b)$\text{Ric}=Kg$

There is a corollary whose proof would be useful for a), but I do not understand it.

Corollary: If $M$ has constant curvature $C$, then:

$$R_{xy}z=C\left(\langle z, x \rangle y-\langle z,y\rangle x\right)$$.

Proof: A routine computation shows that the formula:

$$F(x,y,v,w)=C\left( \langle v,x \rangle\langle y,w\rangle-\langle v,y \rangle\langle x,w\rangle \right)$$

defines a curvaturelike function at each point and $F(x,y,x,y)=CQ(x,y)$. So if $x$ and $y$ span a nondegenerate plane,

$$K(x,y)=C=\frac{F(x,y,x,y)}{Q(x,y)}$$

$\blacksquare$

Edit:

I am going to try to prove part a) of the exercise: Consider point $p$ and its tangent space $T_pM$ Then: $$R_{XYXY}=F(X,Y,X,Y)$$

By previous corollary it is proven:

$$R_{XYZW}=F(X,Y,Z,W)$$

for $Z,W$ on the neighborhood of $X,Y$.

Then $Q(X,Y,Z,W)=\langle Z, X \rangle \langle Y,W\rangle-\langle Z,Y\rangle \langle X,W\rangle$.

By definition:

$K(X,Y,Z,W)=\frac{R_{X,Y,Z,W}}{Q(X,Y,Z,W)}\implies K(X,Y,Z,W) Q(X,Y,Z,W)=R_{X,Y,Z,W}$

$R_{X,Y,Z,W}=K(X,Y,Z,W)(\langle Z, X \rangle \langle Y,W\rangle-\langle Z,Y\rangle \langle X,W\rangle)$

Question: I do not know how to prove b), that is $Ric=Kg$, where $K$ is the sectional curvature and it is not constant. After some research, I believe that Ricci=Kg would indicate the surface is an Einstein manifold.

How do I prove b)

Thanks in advance.

Answer 1

It seems that this is for your homework and because of this, no one can give you a full answer. I show you some hints:

1. Note that $R_{x,y}z$ is a tensor. So if you could prove (a) for a orthonormal base (ONB) $\{v_1,v_2,\dots, v_n\}$ that span the tangent space then you are done! (here the underlying manifold is a surface, so its tangent space is of dim = ?)
2. consider an ONB for surface at $p$ and apply that ONB to LEMMA 39 of the book then obtain $\langle R_{vw}v,w \rangle$.
3. In this step, there is a common trick to obtain $A(x,y)$ from $A(x,x)$. that is replace $x$ with $x+y$ then simply the result. can you apply this to $\langle R_{vw}v,w \rangle$ to obtain $\langle R_{xy}z,w \rangle$?
4. Next, you should arrive to an equation like $\langle R_{xy}z,w \rangle = C\langle A+B+\dots,w \rangle$ that implies $ R_{xy}z = C( A+B+\dots)$ by comparing term by term and using properties of metric.

For $Ric$ use LEMMA 52. Again consider an ONB (note that we are dealing with surface so dim = 2 and we need just two vectors say $E_1\perp E_2$. also note that Ric is also a tensor.) It is just a little simplifying issue. just do it with help of (a). Finally, note that $Ric(x,y) = A \langle x,y \rangle \implies Ric = A g$.