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Why is the second Stiefel-Whitney Class of a closed oriented 4-manfifold, $M^{4}$, a characteristic element for its intersection form?

  • Precisely, why must the following identity hold for closed oriented 4-manifolds, $$\langle w_2(M),\alpha\rangle \equiv \langle \alpha,\alpha\rangle (mod\ 2)\text{ for all } \alpha \in H^{2}(M,\mathbb{Z/2})$$ where $\langle,\rangle$ denotes the intersection form.
William
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Monkey.D.Luffy
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1 Answers1

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For a $\mathbb{Z}$-module (aka abelian group) $V$ with a symmetric, unimodular, bilinear form $\mu\colon V\otimes V \to V$, a characteristic element is an element $c\in V$ such that

$$\forall v\in V\ \mu(v,v) \equiv \mu(c,v)\mod 2 $$

This means you don't want an element with $\mathbb{Z}/2$ coefficients, you want an integral class. The correct statement is that

For an oriented closed $4$-manifold $M$, an element in $H^2(M;\mathbb{Z})$ is characteristic iff it is an integral lift of $w_2(M)$.

It's not hard to see this algebraically using Wu classes (for a definition and important properties see for example Manifold Atlas or Milnor-Stasheff). In general, for a closed, oriented $4n$-manifold, a characteristic element of its intersection form is an integral lift of the $2n$-th Wu class $v_{2n}$, essentially by definition.

But Wu's formula $Sq(V) = W$ tells us that for a closed, oriented manifold we have $v_2 = w_2$, so if $M$ if $4$-dimensional then $\alpha \in H^2(M;\mathbb{Z})$ is characteristic iff it is an integral lift of $v_2$ iff it is an integral lift of $w_2$.

William
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