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There's an exercise at the end of Ch. 2 of Mosher & Tangora's "Cohomology Operations and Applications in Homotopy Theory", which says:

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

i. Show that $u \cup_0 u + u\cup_1 u$ 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 $Z_2 \otimes Z_2\rightarrow Z_4$.

First of all, I'm confused by the first sentence. Shouldn't it just say $u$ is a cochain, not a cocycle? Probably more importantly, I don't really see what's supposed to be going on in part (i), since those two addends aren't in the same cohomological dimensions. I'm getting that

$\delta (u\cup_0 u+u\cup_1 u) = 4(a\cup u)+2(a\cup_1 u+u\cup_1 a)$,

and I'm not sure how to cancel the last terms. The addends in the RHS came straight from the addends on the LHS, i.e. there's no interaction between the two terms as far as I can tell, which makes me doubt myself. This could be just a lot of silliness on my part, but I'd appreciate it if someone could clear this up for me. And of course I'd love to hear about fun or unexpected applications of this particular operation...

EDIT

Following Tyler's suggestion, I've found that changing the formula to $u\cup_0 u+u\cup_1 \delta u$ does wonders.

Now that I've reached it, I'm having difficulty with part (iv). I have that

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

So somehow those last four terms are supposed to collapse into $u\cup v$ as computed using the non-trivial pairing. How exactly should this work? I have a vague sense that a representing cochain should only be spitting out the values 0(mod 4) and 2(mod 4), the latter only when the cup guys both spat out 1(mod 2)'s. So I'm slightly inclined to believe that the first two loose terms, which are equal since $\deg(u)=\deg(v)=2p$ and $\cup_0=\cup$ (the usual cup product), might sum to the desired "$u\cup v$" thing. Assuming we represent $u$ and $v$ by the same cochains throughout the equation, which I'm pretty sure is a valid thing to do(???), then if they evaluate to an even number then that sum will be 0(mod 4) while if they evaluate to an odd number then that sum will be 2(mod 4). So I'm thinking that somehow those last two terms should vanish. But why? And is there a concise way of writing everything I've just said, e.g. using notation $\rho:Z_2\otimes Z_2\rightarrow Z_4$ for the pairing, etc.? And probably more importantly, is there any significance to the fact that the cup product measures the failure of the Pontrjagin square to be a group homomorphism?

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  • $\begingroup$ Maybe this is standard notation, but what are $\cup_0$ and $\cup_1$? $\endgroup$
    – Jason DeVito - on hiatus
    Commented May 6, 2010 at 1:58
  • $\begingroup$ At this point I don't know enough to know whether they're standard or what, but M&T construct these "cup-i products" first on the level of cochains, $C^p(X) \otimes C^q(X) --> C^{p+q-i}(X)$, through some sort of craziness involving S^{\infty}, which I'm guessing is for some reason because (I think) it's a homogeneous space for Z_2, i.e. it is contractible and covers B(Z_2)=RP^{\infty}. Eventually these cup-i products descend to cohomology, and for u\in H^k we define Sq^i(u) = u (\cup_{k-i}) u. $\endgroup$
    – Aaron Mazel-Gee
    Commented May 6, 2010 at 8:56
  • $\begingroup$ @Aaron: I think that you should accept Tyler's answer, since he answered your question before you changed it. $\endgroup$
    – Harry Gindi
    Commented May 6, 2010 at 9:57
  • $\begingroup$ @Harry: Yeah, in retrospect that may have been bad form. Luckily for me, Tyler was kind enough to answer my revised question, so no problem! $\endgroup$
    – Aaron Mazel-Gee
    Commented May 6, 2010 at 17:42
  • $\begingroup$ There is now a math.SE question asking for more details on this one: math.stackexchange.com/questions/54578/… $\endgroup$
    – Jonas Meyer
    Commented Jul 30, 2011 at 8:35

1 Answer 1

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I believe the formula shoud be $$ u \cup_0 u + u \cup_1 \delta u $$ instead.

EDIT: In order to answer your second question about part (iv), consider the expression $$ u \cup_0 v + v \cup_0 u + u \cup_1 \delta v + v \cup_1 \delta u. $$ Upon looking at this, you might say that the two cup-0 products should be "basically" the same, because they're multiplied in the opposite order. However, because the cup-product is not actually commutative, you can't just switch $v \cup_0 u$ with $u \cup_0 v$ - 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.

So far as the significance of the additivity formula: The Pontrjagin square constructs a lift of the "squaring" operation from mod-2 cohomology to mod-4 cohomology. However, taking an element to its square isn't additive mod 4 like it is mod 2, because $(x+y)^2 = x^2 + 2xy + y^2$. The formula tells you the lift of the squaring operation to mod-4 cohomology satisfies this formula.

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  • $\begingroup$ Thanks a bunch. That helps enormously. I've expanded on my original question, which is why I haven't accepted your answer... $\endgroup$
    – Aaron Mazel-Gee
    Commented May 6, 2010 at 9:54
  • $\begingroup$ This is much clearer now, thank you. $\endgroup$
    – Aaron Mazel-Gee
    Commented May 6, 2010 at 17:43
  • $\begingroup$ Dear Tyler, do you know if the behaviour of the Pontrjagin square with respect to reduction mod 2, together with the additivity formula, characterises the square uniquely? This is claimed in the online encylopedia entry eom.springer.de/p/p073810.htm (written by Postnikov no less) but my doubtfulness stems from the fact that the standard reference on axiomatics of these operations, by Browder and Thomas, has a different axiom replacing additivity (that it suspends to the Postnikov square). Are there any more modern references? Thanks. $\endgroup$
    – Mark Grant
    Commented Jul 25, 2011 at 11:09
  • $\begingroup$ Mark, I would basically suggest a brute-force computation - namely, to use the Bockstein sequence and the bottom of the Serre spectral sequence to compute enough about cohomology operations that you can show this characterizes the operation uniquely. I don't know a more modern reference off-hand. $\endgroup$
    – Tyler Lawson
    Commented Jul 25, 2011 at 13:50
  • $\begingroup$ Upon closer investigation, it seems to me that on degree-10 classes, if one writes $i$ for the multiplication-by-2 map from mod-2 cohomology to mod-4, then a back-of-the-envelope calculation indicates that the cohomology operation $P_2 + i Sq^6 Sq^3 Sq^1$ satisfies properties (1) and (2) of your link and is nontrivial on the universal example. $\endgroup$
    – Tyler Lawson
    Commented Jul 25, 2011 at 14:07

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