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In Baues "The homotopy category of simply conected 4-manifolds" there is some algebraic description of $\pi_4(M^4)$ where $M^4$ is simply-connected closed 4-manifold, but this description is pure algebraic.

And with it there is geometric description of $\pi_3(M^4)$: $M^4 = \vee S^2 \cup e^4$, and attaching map of $e^4$ is element of $\pi_3(\vee S^2)$. So, $\pi_3(M^4)$ is a $\pi_3(\vee S^2)$ with additional relation. Each element in $\pi_3(\vee S^2)$ has description in terms of linking number of point preimages (circles in $S^3$) of map $S^3\to \vee S^2$.

Does anyone knows geometric description (I think in terms of Whitehead product and typical preimages) of $\pi_4(M^4)$?

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2 Answers 2

First an important distinction: "Each element in $\pi_3(\vee S^2)$ has description in terms of linking number of point preimages (circles in $S^3$) of map $S^3 \to S^2$" is not a fully correct statement. What has a description in terms of such linking numbers is $Hom(\pi_3(\vee S^2), Z).$ As you say, $\pi_3(\vee S^2)$ itself is described as generated by Whitehead products and $\pi_3(M)$ is a quotient of this.

Dually, $Hom(\pi_3(M), Z)$ will consist of a sub-module of these linking numbers, and if you want you can make it more geometric. For any collection of closed two-dimensional cochains $\{ \alpha_i, \beta_i \}$ such that $\sum \alpha_i \smile \beta_i= d \theta$ one can form the generalized linking number which to some $f : S^3 \to S^2$ evaluates the "integral" $\int_{S^3} [ \sum f^* \alpha_i \smile d^{-1} f^* \beta_i - f^* \theta]$. Here $\int_{S^3}$ is evaluation on the fundamental class of $S^3$ and $d^{-1} f^* \beta_i$ indicates a choice of $1$-cochain which cobounds $f^* \beta_i$. If the $\alpha_i$ and $\beta_i$ are Poincare dual to codimension two submanifolds of $M$ this will be a linking number (with "correction" by the $\theta_i$) of the preimages of those submanifolds.

So far, this is just addressing $\pi_3$. But in recent work Ben Walter and I generalize such "linking numbers" and show that the resulting collection yields a finite-index subgroup of $Hom(\pi_n(X), Z)$ for all $n$ for simply connected $X$. One can take the formulae there - in this case one will need to go to "weight two" - and translate them into geometric terms as I did for $\pi_3$ above. We give a number of examples, and I'd be happy to provide closer analysis for simply connected $M^4$ if you think what we do is relevant and I understood better what you're looking for. The caveats are that we are representing functionals on homotopy groups rather than homotopy groups themselves, and what we can do about torsion is very limited (but is likely to suffice in this case).

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Thanks! I will read your article next week. I'm looking for evaluating of self-intersection of sferoid $x\in \pi_4(M^4)$, in the next sense: $x$ is a element $\tilde x\in\pi_3(\varOmega M^4)$, where $\varOmega M^4$ be a loop space of $M^4$. $\tilde x$ has intersection with discriminant of $\varOmega M^4$ (loops with self-intersection). I want to describe this intersection. –  Nikita Kalinin Sep 5 '10 at 15:06

I don't know exactly how to answer your question, but one observation is that if $M^4 \ncong S^4$, then for any map $\alpha : S^4\to M^4$, $H^4(\alpha):H^4(M^4)\to H^4(S^4)$ is the zero map. This follows because $[M]=a\cup b$ for $a,b\in H^2(M^4)$, by the Poincare conjecture and Poincare duality. Thus, one sees that the Hurewicz map is trivial.

If you have a geometric description of $M^4= \vee S^2 \cup e^4$, then for any point $x\in e^4$, the map $\alpha$ may be homotoped to miss $x$. To see this, make $\alpha$ transverse to $x$. Then the preimage of $x$ is finitely many points, whose orientations cancel since $H_4(\alpha)=0$. Choose arcs in $S^4$ pairing up the points in opposite signs, then since $M$ is simply connected, we may homotope neighborhoods of these arcs rel endpoints to be constant by homotopy extension, and then homotope off of $x$ (this is a version of the Whitney trick). Now by cellular approximation, the map may be homotoped onto $\vee S^2$. So the homotopy groups come from the homotopy groups of the 2-skeletion $\vee S^2$. I'm not certain though how to figure out what relations come from adding the 4-cell $e^4$ though.

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Thanks. It is clear up situation. It seems that relations from 4-cell borns from $\pi_4(S^3)$ and attaching map of 4-cell, but explanation is still vague (cut out one point from 4-cell and crush homotopy between elements of $\pi_4(M^4)$ to 2-skeleton. But we need, for example, that preimages of cutted point are points too... that now is not true) –  Nikita Kalinin Sep 1 '10 at 20:53

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