# Pontrjagin square (Mosher and Tangora Question)

This is an elaboration on a question/answer posted on MathOverflow.

As per the MO question, the actual question is (with the typo corrected)

Suppose the cocycle $u\in C^{2p}(X;Z)$ satisfies $\delta u=2a$ for some $a$.

i. Show that $u \cup_0 u + u\cup_1 \delta u$ is a cocycle mod 4.

ii. Define a natural operation, the Pontrjagin square, $P_2:H^{2p}(-;Z_2)\rightarrow H^{4p}(-;Z_4)$.

iii. Show that $\rho P_2(u)=u\cup u$, where $\rho:H^*(-;Z_4)\rightarrow H^*(-;Z_2)$ denotes reduction mod 2.

iv. Show that $P_2(u+v)=P_2(u)+P_2(v)+u\cup v$, where $u\cup v$ is computed with the non-trivial pairing $Z_2 \otimes Z_2\rightarrow Z_4$.

I am quite stuck on part (iv). If you just plug in, expand everything and simplify, you get

$P_2(u+v)=P_2(u)+P_2(v)+u\cup_0 v+v\cup_0 u+u\cup_1 \delta v+v\cup_1 \delta u$.

Now the answer given on MathOverflow is that $u \cup_0 v$ is not quite commutative "you need to subtract off a coboundary (involving the cup-1 product). So essentially you need to take the expression that you already have and introduce correction coboundary terms (which don't change the cohomology class) to reduce it to the form you're interested in."

Now the only real cup-1 product that seems viable is $$\delta(u \cup_1 v) = -\delta u \cup_1 v - u \cup_1 \delta v + u\cup_0v - v \cup_0 u$$

(or swap $u$ and $v$ in the formula). I write -1, but I am thinking of the coefficients modulo 4.

No matter what I do I can't seem to get the expression to simplify nicely. Moreover, I realised I can't work out what type of expression gives $u \cup v$ - I don't quite understand the part about computing it using the non-trivial pairing $Z_2 \otimes Z_2\rightarrow Z_4$. What sort of expression should I be seeking?

Update: I'll post some more working here. I am focusing on the expression $$f=u\cup_0 v+v\cup_0 u+u\cup_1 \delta v+v\cup_1 \delta u$$

From the coboundary formula above we see we can write this as (noting that $\delta(u \cup_1 \delta v)$ does not change the cohomology class, and taking everything modulo 4)

\begin{align} f &= \delta u \cup_1 v + u \cup_1 \delta v + v \cup_0 u + v\cup_0 u+u\cup_1 \delta v+v\cup_1 \delta u \\ &= \delta u \cup_1 v + 2\left(v \cup_0 u\right) +v\cup_1 \delta u \end{align}

Now consider $\delta(\delta u \cup_2 v)$. By the coboundary formula we have (again modulo 4)

$$\delta(\delta u \cup_2 v) = - \delta u \cup_1 v - v \cup_1 \delta u$$ and thus we have $$f = 2\left(v \cup_0 u\right)$$

and so

$$P_2(u+v)=P_2(u)+P_2(v)+2\left(v \cup_0 u\right)$$

Is this right? I'm still interested in how to get to an expression involving $u \cup v$?

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Related meta thread for reference –  Juan S Jul 30 '11 at 8:30
I posted a comment to Tyler's answer in order to let him know about your question. –  t.b. Jul 30 '11 at 11:20
Thanks for that Theo! –  Juan S Jul 30 '11 at 12:14

My guess is that the main difficulty in simplifying is that you're stuck with terms $\delta u \cup_1 v$ and $v \cup_1 \delta u$ (signs ignored, because these are multiples of 2 and we're working mod 4). If the cup-1 product was commutative, you would be done. However, it's not commutative, but only commutative up to chain homotopy. The correction factor is a cup-2 product.
Dear Quirk, you've essentially got the correct formula. I see that I was too casual in my post on MathOverflow about the method to get the solution. If you instead add $\delta(u \cup_1 v)$ rather than subtract it, your final formula should be $f = 2 (u \cup_0 v) = 2 (u \cup v)$. –  Tyler Lawson Jul 31 '11 at 1:33
Dear Tyler - don't we require $u \cup v$ not $2(u \cup v)$ ? –  Juan S Jul 31 '11 at 2:09
Quirk, well... In the question you are computing using the nontrivial pairing $Z/2 \otimes Z/2 \to Z/4$, which sends $1 \otimes 1$ to $2$. The notation then becomes a little misleading, in my opinion! If you've chosen integer-valued cochain representatives for $u$ and $v$ then this pairing has a lift by first taking the integer-level cup product, multiplying by two, and then reducing mod 4. –  Tyler Lawson Jul 31 '11 at 5:00