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 $x\mapsto x^2$. 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.)


2 Answers 2


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.

  • $\begingroup$ 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). $\endgroup$
    – John Klein
    Mar 25, 2011 at 20:30
  • $\begingroup$ 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. $\endgroup$ Mar 25, 2011 at 20:58

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.

  • $\begingroup$ You're right, that's not quite what I had in mind, but this formula is also worth knowing. $\endgroup$
    – John Klein
    Mar 25, 2011 at 20:01
  • 3
    $\begingroup$ Wow. I had never seen such a statement. Are there other similar results involving other operations? $\endgroup$ Mar 25, 2011 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.