Expression for integral of a particular 1-form doesn't convince me

Question

Hofer-Zehnder, Symplectic invariants and Hamiltonian dynamics, defines $\omega_0$ as the standard symplectic form, $\sum_1^ndy_j\wedge dx_j$, where $x_1,\dotsc,x_n,y_1,\dotsc,y_n$ are the coordinates of $\mathbb{R}^{2n}$. It also says:

$$\omega_0(u,v)=\langle Ju,v\rangle,$$

where $J$ is a block matrix having 0 blocks on the diagonal, a $-I_n$ block in the BL corner and a $I_n$ block in the TR corner, blocks being all $n\times n$. Then it notes:

$$\omega_0=\sum_{j=1}^ndy_j\wedge dx_j=d\left(\sum_{j=1}^ny_jdx_j\right)=d\lambda,$$

thus defining $\lambda=\sum_{j=1}^ny_jdx_j$. And I'm perfectly OK with all that. If I integrate this form over a curve $\gamma$ parametrized by $x(t)$ with $t\in[0,1]$, I expect to get:

$$\int_\gamma\lambda=\int_0^2\left(\sum_{j=1}^nx_{n+j}\dot x_j\right)dt.$$

Then the book says the integral equates to:

$$\int_\gamma\lambda=\frac12\int_0^1\langle-J\dot x,x\rangle dt.$$

If I spell that out, I would get:

$$\int_\gamma\lambda=\frac12\int_0^1\left(\sum_{j=1}^n\dot x_{n+j}x_j-\sum_{j=1}^nx_{n+j}\dot x_j\right)dt.$$

So there is a $\frac12$ that shouldn't be there, the first term is out of the blue, and the second one has the wrong sign. Am I missing something? How are these two forms equal?

Answer 1

There are two points to be made here:

1. I misexpanded the scalar product, that is, I gave it the wrong sign. I probably was thinking of $\dot x^TJx$, but we have $\dot x^TJ^Tx=-\dot x^TJx$.
2. The term with the dot in the wrong place can be integrated by parts to move the dot to the right place. This gives it an extra minus sign, making it sum with the other one to cancel the $\frac12$, and the endpoint (boundary) terms that appear cancel each other out because $x(1)=x(0)$, i.e. $\gamma$ is a closed curve.

More explicitly:

$$\langle -J\dot x,x\rangle=\sum_{j=1}^n(-\dot x_{n+j}x_j)+\sum_{j=1}^nx_{n+j}\dot x_j \tag{1}$$

$$-\int_0^1\left(\sum_{j=1}^n\dot x_{n+j}x_j\right)dt=-\sum_{j=1}^nx_{n+j}x_j\Big|_0^1+\int_0^1\left(\sum_{j=1}^nx_{n+j}\dot x_n\right)dt \tag{2}$$

So the integral term in (2) sums with the other one in (1) to give twice my original expression for $\int_\gamma\lambda$ and the $\frac12$ cancels the "twice", whereas the boundary terms give you a sum over $j$ of $x_{n+j}x_j|_0^1$, but these terms are all 0 since $x_{n+j}(0)=x_{n+j}(1)$ and $x_j(0)=x_j(1)$ for all $j$ because $\gamma$ is closed.