To show that $S^4$ has no symplectic structure, we can use the following theorem:

Theorem: Any closed oriented manifold $M$ with $H^2(M;\mathbb{Z})=0$ has no symplectic structure.

Proof: Suppose $M$ has a symplectic structure. Then by the non-degeneracy of the symplectic form, we have $H^2(M;\mathbb{R})\cong H^2_{dR}(M;\mathbb{R})\neq 0$. But by Poincaré duality and the Universal Coefficient Theorem, we have $H^2(M;\mathbb{R})\cong H^2(M;\mathbb{Z})\otimes\mathbb{R}\cong \operatorname{Hom}(H_2(M;\mathbb{Z}),\mathbb{R})\cong H^2(M;\mathbb{Z})$ since $M$ is closed and oriented. Therefore, $H^2(M;\mathbb{Z})\neq 0$, which is a contradiction. Hence, $M$ has no symplectic structure.

Now, $H^2(S^4;\mathbb{Z})=0$ since $S^4$ is simply connected and $H^2(S^4;\mathbb{Z})$ is the abelianization of $\pi_1(S^4)$, which is trivial. Therefore, by the above theorem, $S^4$ has no symplectic structure.

To show that $S^2 \times S^4$ has no symplectic structure, we can use the following lemma:

Lemma: If $M$ and $N$ are symplectic manifolds of dimensions $2m$ and $2n$, respectively, then $M\times N$ has a symplectic structure of dimension $2(m+n)$.

Proof: Let $\omega_M$ and $\omega_N$ be symplectic forms on $M$ and $N$, respectively. Then the product form $\omega_M\oplus \omega_N$ is a symplectic form on $M\times N$ of dimension $2(m+n)$.

Since $S^4$ has no symplectic structure, it follows that $S^2 \times S^4$ has no symplectic structure.

In the symplectic manifold $T^*M$, a Hamiltonian flow is a flow generated by a Hamiltonian function $H:T^*M \rightarrow \mathbb{R}$ via the Hamiltonian vector field $X_H$ defined as:

$X_H=\sum_{i=1}^n\left(\frac{\partial H}{\partial p_i} \frac{\partial}{\partial q_i}-\frac{\partial H}{\partial q_i} \frac{\partial}{\partial p_i}\right)$

where $(q_1,\ldots,q_n,p_1,\ldots,p_n)$ are local coordinates on $T^*M$.

Given a function $f \in C^{\infty}(M)$, we can define a Hamiltonian function on $T^*M$ by pulling back $f$ via the natural projection $\pi:T^*M\rightarrow M$, i.e., $H = \pi^*f$. The Hamiltonian vector field generated by $H$ is given by:

$X_H=\sum_{i=1}^n\left(\frac{\partial f}{\partial q_i} \frac{\partial}{\partial p_i}-\frac{\partial f}{\partial p_i} \frac{\partial}{\partial q_i}\right)$

which generates the flow:

$\frac{d q_i}{d t}=\frac{\partial H}{\partial p_i}=\frac{\partial f}{\partial p_i} \quad \frac{d p_i}{d t}=-\frac{\partial H}{\partial q_i}=-\frac{\partial f}{\partial q_i}$

This is Hamilton’s equations of motion.

Now, consider a coordinate chart $U \subset M$. Let $(q_1,\ldots,q_n)$ be local coordinates on $U$. We can then define a local coordinate chart on $T^*U$ by taking $(q_1,\ldots,q_n,p_1,\ldots,p_n)$ as coordinates on $T^*U$. Since $\pi|{T^*U}:T^*U\rightarrow U$ is a diffeomorphism, we can pull back functions from $U$ to $T^*U$ via the inverse of this map. In particular, we can define Hamiltonian functions on $T^U$ by pulling back functions from $U$, i.e., for $f\in C^{\infty}(U)$, we can define $H_f=\pi^|{T^*U} f:T^*U\rightarrow \mathbb{R}$.

The Hamiltonian vector field generated by $H_f$ is given by:

$X_{H_f}=\sum_{i=1}^n\left(\frac{\partial f}{\partial q_i} \frac{\partial}{\partial p_i}-\frac{\partial f}{\partial p_i} \frac{\partial}{\partial q_i}\right)$

which is the same as the Hamiltonian vector field generated by $\pi^*f$ on $T^*M$, but restricted to $T^*U$. Note that the restriction of a Hamiltonian vector field to a submanifold is still a Hamiltonian vector field with the same Hamiltonian.

To show that $d \rho = 0$ for a left-invariant and right-invariant form $\rho$, we will use the Cartan’s magic formula:

$\mathcal{L}_X=d \circ i_X+i_X \circ d$,

where $\mathcal{L}_X$ is the Lie derivative with respect to the vector field $X$, $i_X$ is the interior product with $X$, and $d$ is the exterior derivative. Since $\rho$ is left-invariant, we have that $\mathcal{L}_X \rho = 0$ for all left-invariant vector fields $X$. Similarly, since $\rho$ is right-invariant, we have that $\mathcal{L}_Y \rho = 0$ for all right-invariant vector fields $Y$.

Let $X$ be a left-invariant vector field, and $Y$ be a right-invariant vector field. Then, we have:

\begin{align*} d \rho(X,Y) &= \mathcal{L}_X \rho(Y) – \rho(\mathcal{L}_X Y) &&\text{(Cartan’s magic formula)}\ &= \mathcal{L}_X \rho(Y) &&\text{(since $Y$ is right-invariant)}\ &= 0 &&\text{(since $\rho$ is left-invariant)} \end{align*}

Similarly, we can show that $d \rho(X,Y) = 0$ when $X$ is right-invariant and $Y$ is left-invariant. Since any vector field on $G$ can be written as a linear combination of left-invariant and right-invariant vector fields, we conclude that $d \rho = 0$ for any left-invariant and right-invariant form $\rho$ on $G$.