Moment map of $\mathbb{S}^1$ acting on $\mathbb{C}$

Question

I am trying to work through this set of lecture notes and am having trouble understanding the example of $\mathbb{S}^1$ acting on $\mathbb{C}$.

Here is a screens shot of the example in question

Firstly, I don't see how to show that the $\mathbb{S}^1$ action is generated by the vector field $X$.

Secondly, I don't understand why it follows that the moment map is given as $H(z)=-\frac{1}{2}|z|^2$.

Answer 1

You can think of a vector field $X$ on a smooth manifold $M$ as a derivation on the algebra $C^{\infty}(M)$ of smooth functions on $M$, and you can think of the flow it generates as an exponential

$$\exp(tX) : C^{\infty}(M) \to C^{\infty}(M)$$

from which you can recover $X$ itself, as a derivation, by taking the derivative with respect to $t$ and then substituting $t = 0$.

Two examples

The simplest case of this to understand is $M = \mathbb{R}$ and the translation flow $\phi_t(x) = x + t$. This acts on smooth functions $f : M \to \mathbb{R}$ via pullback as $f(x) \mapsto f(\phi_{t}(x))$, hence $f(x) \mapsto f(x + t)$. Differentiating this action with respect to $t$ at $t = 0$ gives

$$\frac{d}{dt}_{t=0} f(x + t) = f'(x)$$

which gives that the derivation exponentiating to this flow is $X = \frac{d}{dx}$ (I guess the author's notation for this would be $\partial x$, which personally I think is not a good convention). The exponential formula

$$\exp \left( t \frac{d}{dx} \right) f(x) = \sum_{k=0}^{\infty} \frac{t^k}{k!} f^{(n)}(x) = f(x + t)$$

then describes the Taylor series of $f$ centered at $x$.

Let's do another example on $\mathbb{R}$: instead of translation let's consider scaling $\phi_t(x) = e^t x$, which is more similar to this case. We have

$$\frac{d}{dt}_{t=0} f(e^t x) = x f'(x)$$

which gives that the derivation exponentiating to this flow is $X = x \frac{d}{dx}$.

The vector field generating a rotation

The $S^1$-action $\phi_t$ in the given problem acts on $\mathbb{R}^2$ via $x + iy \mapsto (\cos t + i \sin t)(x + iy)$, hence on smooth functions via

$$f(x, y) \mapsto f \left( (\cos t) x - (\sin t) y, (\sin t) x + (\cos t) y \right).$$

Differentiating with respect to $t$ at $t = 0$ then gives

$$\frac{d}{dt}_{t=0} f \left( (\cos t) x - (\sin t) y, (\sin t) x + (\cos t) y \right) = -y \frac{\partial f}{\partial x} + x \frac{\partial f}{\partial y}$$

so the derivation exponentiating to this flow is $X = -y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}$ as desired.

Computing the moment map

Next, the moment map $H$, if it exists, is defined by the condition $\iota(X) \omega_0 = dH$, so we need to compute the contraction $\iota(X) \omega_0$. The symplectic form $\omega_0$ is never defined but I assume it's the standard symplectic form $dx \wedge dy$. We have

$$\iota \left( -y \frac{\partial}{\partial x} \right) (dx \wedge dy) = - y \, dy$$

e.g. using the Leibniz rule (contraction is a derivation, and the second term in the Leibniz rule vanishes), and

$$\iota \left( x \frac{\partial}{\partial y} \right) (dx \wedge dy) = -x \, dx$$

(using the Leibniz rule again, this time the first term vanishes), hence

$$\iota(X) \omega_0 = dH = - y \, dy - x \, dx.$$

Finally we integrate to find that

$$H = - \frac{y^2 + x^2}{2} = - \frac{1}{2} \| z \|^2$$

(plus a constant). You can check that this is the right answer by checking that Poisson bracket with $H$ reproduces the derivation $X$: that is, another way of stating the moment map condition is that we should have $\{ H, f \} = Xf$.

Another sanity check is that the fibers $H^{-1}(\lambda)$ are precisely the orbits of the $S^1$-action, which they would have to be.