We prove Darboux's theorem using Moser's deformation method, showing that any symplectic form can be transformed to standard form near a point.

Setup: Let $(M, \omega)$ be a $2n$-dimensional symplectic manifold and $p \in M$. We construct coordinates $(q_i, p_i)$ near $p$ with $\omega = \sum dq_i \wedge dp_i$.

Step 1: Linear algebra at $p$

The tangent space $(T_pM, \omega_p)$ is a symplectic vector space. By the symplectic basis theorem, there exist $e_1, \ldots, e_n, f_1, \ldots, f_n \in T_pM$ with:

$\omega_p(e_i, e_j) = 0, \quad \omega_p(f_i, f_j) = 0, \quad \omega_p(e_i, f_j) = \delta_{ij}$

Choose coordinates $(x_1, \ldots, x_{2n})$ centered at $p$ such that:

$\frac{\partial}{\partial x_i}\bigg|_p = e_i, \quad \frac{\partial}{\partial x_{n+i}}\bigg|_p = f_i$

In these coordinates, $\omega_p = \omega_0|_p$ where $\omega_0 = \sum_{i=1}^n dx_i \wedge dx_{n+i}$.

Step 2: Taylor expansion near $p$

Write $\omega = \omega_0 + \alpha$ where $\alpha$ is a 2-form vanishing at $p$ (i.e., $\alpha_p = 0$).

Since $\omega$ is closed, $d\omega = 0$, and $\omega_0$ is closed, we have $d\alpha = 0$.

By the Poincaré lemma, on a small neighborhood $U$ of $p$, there exists a 1-form $\beta$ with $d\beta = \alpha$.

Moreover, since $\alpha_p = 0$, we can choose $\beta$ with $\beta_p = 0$ (by subtracting an exact form if necessary).

Step 3: Apply Moser's method

Define a path of 2-forms:

$\omega_t = \omega_0 + t\alpha = \omega_0 + t \, d\beta, \quad 0 \leq t \leq 1$

For $t$ sufficiently small, $\omega_t$ is non-degenerate (since $\alpha_p = 0$ means $\alpha$ is small near $p$).

We seek a vector field $X_t$ satisfying:

$\mathcal{L}_{X_t} \omega_t = -\alpha$

By Cartan's formula:

$\mathcal{L}_{X_t} \omega_t = d(\iota_{X_t} \omega_t) + \iota_{X_t}(d\omega_t) = d(\iota_{X_t} \omega_t)$

since $d\omega_t = 0$. Thus we need:

$d(\iota_{X_t} \omega_t) = -d\beta$

So we can take:

$\iota_{X_t} \omega_t = -\beta$

Since $\omega_t$ is non-degenerate and $\beta_p = 0$, this uniquely defines $X_t$ with $X_t(p) = 0$.

Step 4: Integrate the flow

Let $\phi_t$ be the flow of $X_t$: $\frac{d}{dt}\phi_t = X_t \circ \phi_t$ with $\phi_0 = \text{id}$.

By construction:

$\frac{d}{dt}(\phi_t^* \omega_t) = \phi_t^*(\mathcal{L}_{X_t} \omega_t + \dot{\omega}_t) = \phi_t^*(-\alpha + \alpha) = 0$

Therefore $\phi_t^* \omega_t = \phi_0^* \omega_0 = \omega_0$.

At $t = 1$:

$\phi_1^* \omega_1 = \phi_1^* \omega = \omega_0$

So $\phi_1$ is a local diffeomorphism near $p$ taking $\omega$ to $\omega_0$.

Step 5: Define Darboux coordinates

The coordinates $(q_1, \ldots, q_n, p_1, \ldots, p_n) = (x_1 \circ \phi_1, \ldots, x_{2n} \circ \phi_1)$ are Darboux coordinates:

$\omega = (\phi_1^{-1})^* \omega_0 = \sum_{i=1}^n dq_i \wedge dp_i$

This completes the proof.