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The Pontryagin square (at the prime 2) is a certain cohomology operation $$ \mathfrak P_2: H^q(X;\Bbb Z_2) \to H^{2q}(X;\Bbb Z_4) $$ which has the property that its reduction mod 2 coincides with $x^2$. Furthermore, If $x\in H^q(X;\Bbb Z_2)$ is the reduction of an integral class $y$, then $\mathfrak P_2(x)$ is the mod 4 reduction of $y^2$.

For a definition of $\mathfrak P_2$, see e.g.:

Thomas, E.: A generalization of the Pontrjagin square cohomology operation. Proc. Nat. Acad. Sci. U.S.A., 42 (1956), 266–269.

Suppose now that $X = M^{2q}$ is a closed smooth manifold of dimension $2q$. Then $x\in H^q(X;\Bbb Z_2)$ is Poincaré dual to a class $x' \in H_q(X;\Bbb Z_2)$. By Thom representability, $x'$ is represented by a map $f: Q^q\to M$ in which $Q$ is a (possibly unorientable) closed $q$-manifold (by taking the image of the fundamental class). By transversality, we can even assume that $f$ is an immersion.

Question: Is there an interpretation of $\mathfrak P_2(x)$ as a geometric operation on $f$?

(Note: By the properties of the Pontryagin square, if $Q$ and $M$ are orientable, then then $\mathfrak P_2(x)$ is represented by the self-intersection of $f$ reduced mod 4.)

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up vote 6 down vote accepted

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.

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I'm confused by, "If q is even then changing a local orientation flips the signs of both points, so the total intersection changes by ±4." Don't you mean by ±2? In which case the contribution of the intersection pair is well-defined mod 4 (since +2 = -2 mod 4). – John Klein Mar 25 '11 at 20:30
I meant the contribution of the two points changes from $+2$ to $-2$, i.e. changes by $-4 = -2 - 2$, or it changes from $-2$ to $+2$, i.e. changes by $+4 = +2 - (-2)$. So to total intersection changes by either +4 or -4. So we agree on the arithmetic but perhaps not on the meaning of "changes by" in this context. – Kevin Walker Mar 25 '11 at 20:58

The following geometric interpretation of $\mathfrak{B}_2$, 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

$\mathfrak{B}_2 \colon H^{2k}(X, \mathbb{Z}_2) \longrightarrow \mathbb{Z}_4$.


$z:= \sum_{t \in H^{2k}(X, \mathbb{Z}_2)} e^{2 \pi i \mathfrak{B}_2(t)}$.


$\textrm{Arg}(z)= \frac{\sigma(M)}{8} \in \mathbb{Q}/\mathbb{Z}$,

where $\sigma(X)$ denotes the signature of $M$.

For a proof, see this paper.

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You're right, that's not quite what I had in mind, but this formula is also worth knowing. – John Klein Mar 25 '11 at 20:01
Wow. I had never seen such a statement. Are there other similar results involving other operations? – Mariano Suárez-Alvarez Mar 25 '11 at 20:02

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