$A$ a finite abelian group, and denote $\Gamma(A)$ its universal quadratic group. The Pontryagin square $\mathfrak{P}\in H^4(B^2A,\Gamma(A))\cong \text{Hom}(\Gamma(A),\Gamma(A))$ is the element corresponding to the identity, and for a 4-manifold $X$ one gets a functorial operation $H^2(X,A)\rightarrow H^4(X,\Gamma(A))$ sending $B\in H^2(X,A)$ into $$ \mathfrak{P}(B)=B^* \mathfrak{P}\in H^4(X,\Gamma(A)) \ . $$ In some sense this is a quadratic refinement of the pairing given by the cup product. In some special cases however the two things coincide: for instance if $A=\mathbb{Z}_n$ with $n$ odd, then $\Gamma(A)=A$ and $\mathfrak{P}(B)=B\cup B$.
If we now keep $A$ generic, is there a similar simplification by assuming $X$ to be a spin manifold? There reason why I believe there should be some simplification is that on a spin manifold the pairing of $H^2$ given by the cup product is even, hence it makes sense to divide $B\cup B$ by $2$, and this seems to me to provide a quadratic refinement and thus I suspect this could be equal to $\mathfrak{P}(B)$ on spin manifolds. Is this true?
