Suppose $M^m$ and $N^n$ are smooth, connected, closed, oriented manifolds, with $m\ge n$.If we have a smooth map $f:M\rightarrow N$, we can define a graph as the submanifold $\Gamma=\{(x, f(x))\mid x\in M\}\subset M\times N$.

Is there an explicit description of the Poincare dual $\eta_\Gamma\in H^n(M\times N)$ of $\Gamma$?

Here, I'm thinking about a Poincare dual in the sense of Bott-Tu: it is the equivalence class $[\eta_\Gamma]\in H^n(M\times N)$ of a closed $n$-form $\eta_\Gamma$ with the property that, if $[\omega]\in H^m(M\times N)$ is any $m$-form, there is an equality $$ \int_\Gamma \omega_{|\Gamma}=\int_{M\times N}\omega\wedge\eta_\Gamma$$

My gut instinct was that $\eta_\Gamma=\pi_2^\ast\alpha$, where $\alpha\in H^n(N)$ is nonzero, and $\pi_2:M\times N\rightarrow N$ is projection. But this seems wrong, since it doesn't even mention $f$.

However, by taking $\omega=\pi_1^\ast\beta$ above (where $\beta\in H^m(M)$ is nonzero), it would seem that $\eta_\Gamma=\pi_2^\ast\alpha+\cdots$.

Is there an explicit representation of $\eta_\Gamma$ in terms of $\alpha$, $\beta$, and the various maps mentioned earlier?

  • $\begingroup$ Have you understood how to do this when $N=M$ and $\Gamma = \Delta$ is the graph of the identity? That's an important example to understand. $\endgroup$ Commented Nov 6, 2017 at 19:05
  • $\begingroup$ @TedShifrin: No, I don't think I understand that example. Do you have a reference? $\endgroup$
    – Steve D
    Commented Nov 6, 2017 at 19:22
  • $\begingroup$ Bott/Tu work it out carefully somewhere! Try p. 126-127. $\endgroup$ Commented Nov 6, 2017 at 19:25
  • $\begingroup$ @TedShifrin: Ok, I'll find it! Are there difficulties in going from this "base" case to the general one? Is it even possible to say generally what $\eta_\Gamma$ looks like? I realize I'm asking this before I even look at your reference, so feel free to not answer right away. :) $\endgroup$
    – Steve D
    Commented Nov 6, 2017 at 20:02
  • $\begingroup$ @TedShifrin: Thanks again for the pointer. I think I've worked out the general (somewhat ugly) case; I'd be grateful if you'd take a look and check for mistakes. $\endgroup$
    – Steve D
    Commented Nov 7, 2017 at 0:54

1 Answer 1


After @TedShifrin gave me the hint in the comments, I followed the exposition in Bott-Tu, and came up with the following:

$$ \eta_\Gamma=\sum_{\deg(\alpha_i)+\deg(\beta_j)=n} c_{ij}\pi_1^\ast\alpha_i\wedge\pi_2^\ast\beta_j$$

where $\alpha_i$ are a basis for $H^\ast(M)$ and $\beta_j$ are a basis for $H^\ast(N)$. If $\overline{\alpha}_i$ denotes a Poincare-dual basis to $\alpha_i$, and $\overline{\beta}_j$ denotes a Poincare-dual basis to $\beta_j$, then we can write the $c_{ij}$ as

$$ c_{ij} = (-1)^{\deg(\overline{\beta}_j)}\int_M\overline{\alpha}_i\wedge f^\ast\overline{\beta}_j$$

This seems to agree with the case that $\Gamma=\Delta$ is the diagonal, and $f$ is the identity map from $M$ to $N=M$.

I'll also mention this agrees with the case I'm more familiar with, where $N=S^2$. Then if $\beta\in H^2(S^2)$ is the volume form, we should have $$ \eta_\Gamma=\pi_1^\ast f^\ast\beta + \pi_2^\ast\beta $$ and this agrees with the formula above.


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