I'm currently working through this paper from Duke Math J., Volume 51, Number 2 (1984), 243-260 (unfortunately couldn't find a publicly visible link); more specifically, the part about the obstructions to the diagonalizability of Riemannian metrics in $n \geq 4$ dimensions, but there are a few parts where I just can't wrap my head around what's happening.

On page 258, they start by taking an orthonormal coframe $\{\omega^i = f^i d x^i : i = 1, ..., n\}$ altogether with it's dual frame $\{e_i\}$. Their goal is to show that the Weyl-tensor $W$ vanishes, so that one can apply the Weyl-Schouten theorem to get that the manifold is already conformally flat.

Using some calculations with the connection form $\omega_j^i$ and Cartan's structure equations which I can for the most part follow, they arive at the result that

$$ d(\omega^i \wedge \omega^j \wedge \omega_j^i) = \omega^i \wedge \omega^j \wedge \Omega_j^i = 0 \text{ for } i \neq j$$

where $\Omega_j^i$ is the curvature $2$-form.

Now from this, Deturck and Yang make the jump to the sectional curvature $R$ and then on to the Weyl Tensor $W$. Namely, they argue the following:

Thus $\omega^i \wedge \omega^j \wedge \Omega_j^i = 0$ for all $i \neq j$. This will manifest itself as an integrability condition, as we shal see. Rewriting this in terms of the sectional curvature $R$, we must have

$$R(e_i, e_j, e_k, e_l) = 0 \text{ if $i, j, k, l$ are distinct} $$

where $\{e_i\}$ is the dual frame to $\{\omega^i\}$.

How do they make this jump? How can we come from the equation $\omega^i \wedge \omega^j \wedge \Omega_j^i = 0$ where we have only two distinct components $i, j$ to the equation with four distinct components $i, j, k, l$ for which the curvature $R$ vanishes? What exactly do we do when "rewriting" this equation, as they only call it? It might be due to my limited experience with these things, but I fail to see the exact relation between the first equation and the vanishing of the curvature tensor for these four components.

Right thereafter, they continue to carry over this equation about the Weyl tensor, namely

Since $\{e_i\}$ is an orthonormale frame, none of these components of $R$ enter into the Ricci tensor. We conclude that

$$W(e_i, e_j, e_k, e_l) = 0 \text{ if $i, j, k, l$ are distinct}$$

where $W$ is the Weyl tensor.

This seems less obscure to me, but I don't understand it entirely either. My first question would be, how do we get from the $\{e_i\}$ being orthonormal to not entering into the Ricci tensor? Is this some property of the Ricci-tensor that I'm missing?

And my 2nd question here would be: I know that the Riemannian curvature tensor decomposes into the "trace"-part (Ricci-Tensor) and the "traceless"-part (Weyl-Tensor). Is this the argument Deturck and Yang use here to conclude the $W(...) = 0$-equation? That if the components don't enter into the Ricci-tensor, they can only be in the only other part the curvature tensor consists of, namely the Weyl-tensor, so if one of them is $= 0$, the other one is too? Am I understanding this argument correctly? (Even if I probably didn't express it too well...)

From there on, they continue with some calculations that I can roughly follow to show that $W \equiv 0$, but I just can't wrap my head around about this passage. I haven't worked that much with Weyl- and Curvature tensors before, so maybe some stuff I have trouble with has a very easy explanation that I'm just missing out on. Any help would be greatly appreciated.

  • $\begingroup$ This paper sounds interesting. Can you please say where can I find it on the web? I have trouble locating it. $\endgroup$ Aug 18 '17 at 19:12
  • $\begingroup$ @AsafShachar Unfortunately I couldn't find any publicly available web location of it myself, otherwise I would have happily linked it. I've added a link in the post where the document can be bought from. I hope this is of help to you. $\endgroup$
    – moran
    Aug 18 '17 at 21:13

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I suppose the sectional curvature $R$ actually means Riemann curvature instead, since sectional curvatures are usually defined on planes in tangent spaces intead of on $\ts^4T_xM .$

First note that $\om^i\w\om^j\w \Om^i_j$ is a $4$-form, and this is where the four inputs $i,j,k,l$ come from. By the shuffle definition of wedge product, we have $$\om^i\w\om^j(e_m,e_n)=\om^i(e_m)\om^j(e_n)-\om^i(e_n)\om^j(e_m)=\e^{ij}_{mn},$$ where $\e^{ij}_{mn}=1$ when $(m,n)=(i,j),$ $-1$ when $(m,n)=(j,i),$ and $0$ otherwise.

We know that $\Om^i_j(X,Y)=-R(X,Y,e_i,e_j),$ so for $i,j,k,l$ distinct we have \begin{align*} 0=-&\om^i\w\om^j\w \Om^i_j(e_i,e_j,e_k,e_l)\\=&\sum_{\si\in Sh_{2,2}}(-1)^\si\om^i\w\om^j(e_{\si(i)},e_{\si(j)})R(e_{\si(k)},e_{\si(l)},e_i,e_j)\\ =&R(e_k,e_l,e_i,e_j)\\ =&R(e_i,e_j,e_k,e_l), \end{align*} where $Sh_{2,2}$ means $(2,2)$ shuffles, i.e., $\si(i)<\si(j),\si(k)<\si(l).$ $\newcommand{\np}{\wedge\!\!\!\!\!\!\bigcirc}$ Now, we have \begin{align*} W=R-\frac{1}{n-2}(Ric-\frac{S}{n}g)\wedge\!\!\!\!\!\!\bigcirc g-\frac{S}{2(n-2)}g\np g. \end{align*} By definition of Kulkarni-Nomizu product $\wedge\!\!\!\!\!\!\bigcirc,$ we have $T\np g(e_i,e_j,e_k,e_l)=0$ for any 2-tensor $T$ and distinct $i,j,k,l.$ This yields the desired result $$W(e_i,e_j,e_k,e_l)=0.$$

  • $\begingroup$ Thank you! If I may ask, is the fact that $\Omega^i_j(X,Y)=-R(X,Y,e_i,e_j)$ a well-known property? If so, can you please give a short insight into it, or link or refer to a source where I might read up about that? As I didn't know about this relation between the two. $\endgroup$
    – moran
    Aug 20 '17 at 0:51
  • $\begingroup$ Also, about the part: By definition of Kulkarni-Nomizu product $\wedge\!\!\!\!\!\!\bigcirc,$ we have $T\np g(e_i,e_j,e_k,e_l)=0$ for any 2-tensor $T$ and distinct $i,j,k,l.$ If I'm understanding this right, this is because the $\{e_i\}$ are an orthonormal frame, so all the $g(e_i, e_j)$ in the definition of the Kulkarni-Nomizu product are $= 0$ for $i \neq j$? $\endgroup$
    – moran
    Aug 20 '17 at 0:54
  • $\begingroup$ @moran, for the first question, the identity follows from expanding the covariant derivative definition of Riemann curvature using moving frame forms. Please refer to Chapter 1.8 in Riemannian Geometry by Issac Chavel. For the second question, yes. If we follow the sign convention on wikipedia, then $(h {~\wedge\!\!\!\!\!\!\bigcirc~} k)(X_1,X_2,X_3,X_4) := h(X_1,X_3)k(X_2,X_4) + h(X_2,X_4)k(X_1,X_3) - h(X_1,X_4)k(X_2,X_3) - h(X_2,X_3)k(X_1,X_4) .$ $\endgroup$ Aug 20 '17 at 1:34

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