# Proving that the first two and last two indices of the Riemann curvature tensor commute

I’ve always been confused with the combinatorial aspect of proving the properties of the Riemann curvature tensor. I want to record my proof of the fact that $R(X,Y,Z,W)=R(Z,W,X,Y)$. This is different from the standard proof given in books. I have been unable to prove this theorem in the past, and hence am happy to write down my proof finally.

Define the function $f(R(X,Y,Z,W))=R(X,Y,Z,W)-R(Z,W,X,Y)$. We want to prove that this function is $0$.

By simple usage of the facts that $R(X,Y,Z,W)+R(Y,Z,X,W)+(R(Z,X,Y,W)=0$ and that switching the first two or last two vector fields gives us a negative sign, we can see that $R(X,Y,Z,W)-R(Z,W,X,Y)=R(X,W,Y,Z)-R(Y,Z,X,W)$.

Hence, $f(R((X,Y,Z,W))=f(R(X,W,Y,Z))$
Now note that $R(X,Y,Z,W)=R(Y,X,W,Z)$. This is obtained by switching the first two and last two indices. However, $R(Y,X,W,Z)-R(W,Z,Y,X)=R(Y,Z,X,W)-R(X,W,Y,Z)=-f(R(X,W,Y,Z)$.

As $f(R(X,Y,Z,W))=$ both positive and negative $f(R(X,W,Y,Z))$, we can conclude that it is $0$.

Hence, $R(X,Y,Z,W)=R(Z,W,X,Y)$.

It is not easy to prove this theorem because just manipulating the indices mindlessly (or even with some gameplan) can lead you down a rabbithole without ever reaching a conclusion. Meta-observations, like the above, are necessary to prove this assertion.