The submanifold $S^2$ in the symplectic product $S^2\times S^2$

Question

Let $S^2$ come equipped with the usual symplectic form and $S^2\times S^2$ come equipped with the product symplectic form and coordinates $(x,y)$ with $x\in S^2$. Consider the "diagonal sphere" $(x,x)$. Is the sphere symplectic? I feel like it is as the tangent space seems to be the same as the tangent space of one of the factors and so the form should be symplectic on it, but I would like to make sure...

Answer 1

Let $\omega$ be the symplectic form on $S^2$, then the product symplectic form on $S^2\times S^2$ is $\pi_1^*\omega + \pi_2^*\omega$ where $\pi_i : S^2\times S^2 \to S^2$ denotes the projection onto the $i^{\text{th}}$ factor. Let $\Delta : S^2 \to S^2\times S^2$ denote the diagonal map, $\Delta(x) = (x, x)$. Then we have

\begin{align*} \Delta^*(\pi_1^*\omega + \pi_2^*\omega) &= \Delta^*\pi_1^*\omega + \Delta^*\pi_2^*\omega\\ &= (\pi_1\circ\Delta)^*\omega + (\pi_2\circ\Delta)^*\omega\\ &= \operatorname{id}_{S^2}^*\omega + \operatorname{id}_{S^2}^*\omega\\ &= \omega + \omega\\ &= 2\omega. \end{align*}

Therefore, the diagonal $S^2$ is a symplectic submanifold, but the restriction of the product symplectic form is not $\omega$, but $2\omega$. Note, there was nothing special about $S^2$ here. The same calculation shows that if $(M, \omega)$ is a symplectic manifold, then the diagonal submanifold of $M\times M$ with its product symplectic form is symplectic, but the restriction of the symplectic form is $2\omega$, not $\omega$.