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Consider $\Bbb C^n$ with the usual $\newcommand{\d}{{\rm d}}$symplectic form $$\omega =\frac{i}{2}\sum_{k=1}^n \d z^k \wedge \d \overline{z}^k$$and the action $\Bbb S^1 \circlearrowright \Bbb C^n$ given by $$e^{i\theta} \cdot (z^1,\ldots, z^n) \doteq (e^{i\theta}z^1,\ldots, e^{i\theta}z^n).$$It is easy to see that this action is symplectic.

In my lecture notes it is stated that the infinitesimal action is given by $1\mapsto X_{(z^1,\ldots, z^n)} = (iz^1,\ldots, iz^n)$, and that the action is Hamiltonian with moment map $\mu(z) = \frac{1}{2}\sum_{k=1}^n |z^k|^2$.

I'd like to check these assertions, using the bare minimum of identifications possible. The Lie algebra of $\Bbb S^1$ is the vertical line tangent to the circle at $1$, so it is $i\Bbb R \cong \Bbb R$. With this, given $a \in \Bbb R$, we consider the curve $\alpha\colon \Bbb R \to \Bbb S^1$ given by $\alpha(t) = e^{iat}$. Since $\alpha(0) = 1$ and $\alpha'(0) = ia$, we can use that to compute the infinitesimal generator associated to $a$ as $$(a^\#)_{(z^1,\ldots, z^n)} = \frac{\d}{\d{t}}\bigg|_{t=0} e^{iat} \cdot(z^1,\ldots, z^n) = (iaz^1,\ldots, iaz^n),$$good. Since the action is symplectic, the vector field $a^\#$ is symplectic, right? But $$a^\# = ia\sum_{k=1}^n z^k \frac{\partial}{\partial z^k} \implies \iota_{a^\#}\omega = -\frac{a}{2}\sum_{k=1}^n z^k \d \overline{z}^k.$$The form $\iota_{a^\#}\omega$ is not even closed. What is going on? I was planning to find a primitive, then finding the comoment map, and finally computing the moment map from the comoment map.

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  • $\begingroup$ The identification $\mathbb{C} \cong \mathbb{R}^2$ is given by $z = x+iy \mapsto (x,y)$ or, through identifications, $z \mapsto x \partial_x + y \partial_y$. In terms of $x, y$, you'll see that $z \partial_z \neq x \partial_x + y \partial_y$. If you unravel your proposed expression for $a^{\sharp}$, you'll observe that it's equal to $(\nabla \mu + i X_{\mu})/2$, which expresses that since the Euclidean metric, the symplectic form and the complex structure $i$ form a Kaehler structure, the gradient $\nabla \mu$ and the symplectic gradient $X_{\mu}$ together form a 'holomorphic gradient'. $\endgroup$ Jun 21, 2018 at 10:56
  • $\begingroup$ Typo: Your proposed $a^{\sharp}$ is $(i/2)(\nabla \mu \pm i X_{\mu})$ (where the choice for $\pm$ is a matter of convention). The bottom line of my comment is : $(iz^1, \dots, iz^n)$ is to be identified with $X_{\mu}(z^1, \dots, z^n)$ as it is (relatively clearly) equal to $i \nabla \mu$, hence not to be identified with your proposed expression for $a^{\sharp}$. Side remark: your computation is incidentally really close to proving that $\nabla \mu$ is a Liouville vector field for $\omega$. $\endgroup$ Jun 21, 2018 at 11:36

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Jordan's comment was helpful and I talked with a classmate today. Basically, the screw up was that $(iaz^1,\ldots, iaz^n)$ is identified with $$(a^\#)_{(z^1,\ldots, z^n)} = \sum_{k=1}^n iaz^k\frac{\partial}{\partial z^k} - ia\overline{z}^k\frac{\partial}{\partial\overline{z}^k}.$$One checks this by writing $(a^\#)_{(z^1,\ldots, z^n)}$ in terms of $x^k$ and $y^k$, then passing to $\partial/\partial x^k$ and $\partial/\partial y^k$, to finally go back to $\partial/\partial z^k$ and $\partial/\partial \overline{z}^k$. That said, one checks that $$\iota_{a^\#}\omega = \frac{i}{2} \sum_{k=1}^n iaz^k {\rm d}\overline{z}^k + ia\overline{z}^k{\rm d}z^k = -\frac{a}{2}\sum_{k=1}^n {\rm d}(|z^k|^2) = {\rm d}\left(-\frac{a}{2}\sum_{k=1}^n |z^k|^2\right).$$This way, the comoment map is $\hat{\mu}\colon \Bbb R \to \mathcal{C}^\infty(\Bbb C^n)$ given by $$\mu^a(z^1,\ldots, z^n) = -\frac{a}{2}\sum_{k=1}^n|z^k|^2$$and the moment map is $\mu \colon \Bbb C^n \to \Bbb R$ given by $$\mu(z^1,\ldots, z^n) = -\frac{1}{2}\sum_{k=1}^n|z^k|^2.$$The minus sign is just due to a different convention I'm adopting, and since $\Bbb R^* \cong \Bbb R$, we see $\mu(z^1,\ldots, z^n)\colon \Bbb R \to \Bbb R$ as the linear map given my multiplication by the above expression.


Edit: details on the identification of vector fields. Here you have to remember that $$\frac{\partial}{\partial z^k} = \frac{1}{2}\left(\frac{\partial}{\partial x^k}-i\frac{\partial}{\partial y^k}\right)\quad\mbox{and}\quad\frac{\partial}{\partial \overline{z}^k} = \frac{1}{2}\left(\frac{\partial}{\partial x^k}+i\frac{\partial}{\partial y^k}\right). $$ So: $$\begin{align} (iaz^1,\ldots, iaz^n) &= (iax^1-ay^1, \ldots, iax^n-ay^n) \\ &\equiv (-ay^1,\ldots, -ay^n, ax^1,\ldots, ax^n) \\ &= \sum_{k=1}^n\left(-ay^k\frac{\partial}{\partial x^k}+ax^k\frac{\partial}{\partial y^k}\right) \\ &= \sum_{k=1}^n\left(-ay^k\left(\frac{\partial}{\partial z^k}+\frac{\partial}{\partial \overline{z}^k}\right)+ax^ki\left(\frac{\partial}{\partial z^k}-\frac{\partial}{\partial \overline{z}^k}\right)\right) \\ &= \sum_{k=1}^n\left((-ay^k+ax^ki)\frac{\partial}{\partial z^k}+(-ay^k - ax^ki)\frac{\partial}{\partial\overline{z}^k}\right)\\ &= \sum_{k=1}^n\left(ia(iy^k+x^k)\frac{\partial}{\partial z^k}-ia(-iy^k + x^k)\frac{\partial}{\partial\overline{z}^k}\right) \\ &= \sum_{k=1}^n \left(iaz^k\frac{\partial}{\partial z^k} - ia\overline{z}^k\frac{\partial}{\partial\overline{z}^k}\right),\end{align}$$as wanted.

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  • $\begingroup$ how does one "go back" to $\frac{\partial}{\partial z} = \frac{\partial}{\partial x} - i \frac{\partial}{\partial y}$ and $\frac{\partial}{\partial \bar z} = \frac{\partial}{\partial x} + i \frac{\partial}{\partial y}$. Cause as I see it $(a^\#)_{(x^1+iy^1,...,x^n+iy^n)}$ does have $i$ as coordinate. Could you elaborate a wee more on this "go back" part? $\endgroup$ Oct 5, 2018 at 15:33
  • $\begingroup$ @Aaron Sure, see edit. $\endgroup$
    – Ivo Terek
    Oct 5, 2018 at 17:23
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    $\begingroup$ Thanks! I can't believe I've missed that. $\endgroup$ Oct 5, 2018 at 17:40

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