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In their article Instanton counting and Donaldson invariants the authors define the slant product for $\beta \in H_i(X)$ (where $X$ is a manifold) as following.

Let $P \to X$ and SO(3) bundle and $M(P)$ the moduli space of irreducible anti-self-dual connections. Let $\mathcal{P}$ be the universal bundle $X \times M(P)$, then the slant product is defined as: $$ \mu(\beta) := -\frac{1}{4}p_1(\mathcal{P})/\beta $$ This in the end of page 5. What does it even mean to divide by a homology class?

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The slant product is defined in many basic algebraic topology books, eg Hatcher's. If you are willing to work over a field, then you'd take the Kunneth decomposition of $p_1 = \sum x_j \otimes m_j$ with $x_j \in H^*(X)$ and $m_j \in H^*(M(P))$. Then $p_1/\beta = \sum x_j(\beta) \otimes m_j \in H^*(M(P))$, where the evaluation $x_j(\beta)$ is declared to be $0$ unless $x_j$ is in the same dimension as $\beta$.

If you like to think in terms of de Rham cohomology, you are doing a partial integration of a form representing $p_1$ over the cycle $\beta$ to get a $4-i$ form.

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    $\begingroup$ So the division is in some sense "formal"? Or, at least why is it defined as a "division"? $\endgroup$
    – Marion
    Mar 27, 2017 at 2:05
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    $\begingroup$ @Marion Yes, it's formal. As you can see, / takes away a factor from a product. And / reduces degree in the same way that division would. There is even a relation $(x/\alpha)/\beta=x/(\alpha\times\beta)$ which looks like one was actually dividing ... $\endgroup$ Mar 28, 2017 at 19:20

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