Let $(M,\omega)$ be a symplectic manifold. There are a priori two ways of evaluating $\omega$ on an element $A \in \pi_2(M)$:

  1. we can integrate $\omega$ on any representative $u : S^2 \to M$ of the class $A$: $$ \omega(A) := \int_{S^2} u^* \omega $$ Indeed, if $u$ and $u'$ represent the same class in $\pi_2(M)$, then the integrals $\int_{S^2} u^* \omega$ and $\int_{S^2} u^{'*} \omega$ differ by that of a $\omega$ on the boundary of a three dimensional manifold, which vanishes since $d \omega = 0$.
  2. Let $\pi_2(M) \overset{\rho_1}{\longrightarrow} H_2(M;\mathbb{Z}) \overset{\rho_2}{\longrightarrow} H_2(M; \mathbb{R})$ be the composition of the Hurewicz map $\rho_1$ and the natural homomorphism $\rho_2 : H_2(M; \mathbb{Z}) \to H_2(M; \mathbb{R})$. Then one can define $$ \omega(A) := \langle [\omega], \rho_2 \rho_1 (A) \rangle, \quad A \in \pi_2(M), $$ where $\langle . , . \rangle$ denotes the natural pairing $H^2(M; \mathbb{R}) \times H_2(M; \mathbb{R}) \to \mathbb{R}$.

Suppose that $\rho_1$ is an isomorphism $\pi_2(M) \simeq H_2(M; \mathbb{Z})$. Then are these two definitions equivalent ? In other words, do we have: $$ \langle [\omega], \rho_2 \rho_1 (A) \rangle = \int_{S^2} u^* \omega, $$ for any $A \in \pi_2(M)$ and representative $u : S^2 \to M$ of $A$ ?


1 Answer 1


Recall that the Hurewicz homomorphism $\rho_1 : \pi_2(M) \to H_2(M; \mathbb{Z})$ is given by $A \mapsto u_*[S^2]$ where $A = [u]$; here $[S^2] \in H_2(S^2; \mathbb{Z})$. So

$$\int_{S^2}u^*\omega = \langle [u^*\omega], \rho_2[S^2]\rangle = \langle u^*[\omega], \rho_2[S^2]\rangle = \langle [\omega], u_*\rho_2[S^2]\rangle.$$

Now note that we have the following commutative diagram

$$\require{AMScd} \begin{CD} H_2(S^2; \mathbb{Z}) @>{u_*}>> H(M;\mathbb{Z})\\ @V{\rho_2}VV @VV{\rho_2}V \\ H_2(S^2; \mathbb{R}) @>{u_*}>> H(M;\mathbb{R}) \end{CD}$$

i.e. $u_*\rho_2 = \rho_2 u_*$. Using singular homology, this follows because pushforwards are defined on singular simplicies (by composition) and extended linearly.

So we see that

$$\int_{S^2}u^*\omega = \langle [\omega], u_*\rho_2[S^2]\rangle = \langle[\omega], \rho_2 u_*[S^2]\rangle = \langle[\omega], \rho_2\rho_1(A)\rangle.$$

  • $\begingroup$ Thank you very much for the answer @Michael. $\endgroup$
    – BrianT
    Commented May 1, 2019 at 15:40

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