Geometric interpretation of the Pontryagin square

The following geometric interpretation of $\mathfrak{P}_2$, due to Morita, is not exactly what you are looking for but maybe it can be interesting as well. Anyway, it was too long to be a comment.

Assume $q=2k$, so that the Pontrjagin square is a map $$\mathfrak{P}_2 \colon H^{2k}(X, \mathbb{Z}_2) \longrightarrow \mathbb{Z}_4.$$ Set $$z:= \sum_{t \in H^{2k}(X, \mathbb{Z}_2)} e^{2 \pi i \mathfrak{P}_2(t)}.$$ Then $$\textrm{Arg}(z)= \frac{\sigma(M)}{8} \in \mathbb{Q}/\mathbb{Z},$$ where $\sigma(X)$ denotes the signature of $M$.

For a proof, see Gauss Sums in Algebra and Topology, by Laurence R. Taylor.


This is just a guess, based on your comment about the case when $Q$ and $M$ are both oriented.

Claim: If $M$ is oriented and $Q$ is unoriented (and perhaps nonorientable), then the self-intersection of $Q$ is well defined in $\mathbb Z$ if $q$ is odd and well-defined in $\mathbb Z_4$ is $q$ is even.

Proof: Let $Q'$ be a parallel copy of $Q$ which intersects $Q$ transversely. Choose local orientations of $Q$ and $Q'$ near their intersection points so that the local orientations of corresponding points of $Q$ qnd $Q'$ agree. Define an intersection number (dependent on these choices) by summing the signs of the intersections. There are two types of intersections of $Q$ and $Q'$: (a) those coming from the non-triviality of the normal bundle of $Q$, and (b) those coming from intersections of $Q$ with itself (if $Q$ if immersed but not embedded). Changing a local orientation does not affect the contribution of type (a) intersection points to the total intersection, since both the $Q$ and $Q'$ local orientations change together. Type $b$ intersections come in pairs. If $q$ is odd then the signs of these pairs always cancel, so changing local orientation has no effect. If $q$ is even then changing a local orientation flips the signs of both points, so the total intersection changes by $\pm 4$.

So perhaps in the case when $M$ is oriented the Pontryagin square is given by the above mod 4 intersection number, just as in the case when $Q$ is oriented.