While reading Chow, Lu, and Ni's book on Hamilton's Ricci flow I found the following about curvature tensor. They say that the Riemannian curvature is in the kernel of the map $$ b : \Lambda^{2}T^{*}M \otimes_{S} \Lambda^{2}T^{*}M \to \Lambda^{3}T^{*}M \otimes_{S}T^{*}M $$ where $$ b(\Omega)(X,Y,Z,W) = \frac{1}{3} (\Omega(X,Y,Z,W) + \Omega(Y,Z,X,W) + \Omega(Z,X,Y,W)). $$ I can see that the curvature lies in the kernel of this map. My question is what does it mean to take the symmetric product of two different vector spaces.
I thought it should be that $b(\Omega)(X,Y,Z,W) = b(\Omega)(W,X,Y,Z)$, but the computation shows that $b(\Omega)(X,Y,Z,W) = - b(\Omega)(W,X,Y,Z)$.
So, how should I think about $\Lambda^{3}T^{*}M \otimes_{S}T^{*}M$?
I think this is a typo. It probably should just be the ordinary tensor product. In fact the image of $b$ is exactly $\Lambda^4 T^*M\subseteq \Lambda^3 T^*M \otimes T^*M$, so if you apply any kind of symmetrization operator to something in the image of $b$, you get zero. See the proof of Proposition 7.21 in my Introduction to Riemannian Manifolds (2nd ed.).