A Concrete Computation of Moment Map

/ Alexander

Let \mathbb{T}^2=\mathbb{S}^1\times\mathbb{S}^1 acts on \mathbb{P}^1\times \mathbb{P}^2 in the following way:

(u,v)\cdot([a,b],[x,y,z]):=([ua,b],[u^kx,y,vz])

then it is a Hamiltonian action, the question is, why? That entails the computation of the moment map \mu:\mathbb{P}^1\times\mathbb{P}^2\longrightarrow\mathfrak{t}^*\cong\mathbb{R}^2.

We know that the \mathbb{P}^1\times\mathbb{P}^2 is endowed with the canonical symplectic structure, so called the Fubini-Study form, but it is still hard to handle with, and computing the fundamental vector field \underline{X} associates to this action, for some X\in\mathfrak{t}\cong\mathbb{R}^2 is also stubborn, so how would we do?

We note that this torus action is actually induced from the \mathbb{T}^2 action on \left(\mathbb{C}^2\setminus0\right)\times\left(\mathbb{C}^3\setminus0\right) via:

(u,v)\cdot((a,b),(x,y,z))=((ua,b),(u^kx,y,vz))

the manifold \left(\mathbb{C}^2\setminus0\right)\times\left(\mathbb{C}^3\setminus0\right) is endowed the standard symplectic structure, and the computation of the fundamental vector field associates with this action is much more easier, hence we will try to compute the moment map \nu associates with this action, and then descending it to the \mathbb{P}^1\times\mathbb{P}^2 level.

First, let’s compute the fundamental vector field, for X=(\alpha,\beta)\in\mathbb{R}^2\cong\mathfrak{t}, the exponential map \exp tX onto the torus group is \exp tX=\left(e^{it\alpha},e^{it\beta}\right), this is because if we take \mathbb{S}^1\times\mathbb{S}^1=\mathrm{U}(1)\times\mathrm{U}(1), the (\alpha,\beta) corresponds to (i\alpha,i\beta)\in\mathfrak{u}(1)\times\mathfrak{u}(1), then the exponential map is the usual exponential in this sense.

Now, we have the fundamental vector field associates to X:

\begin{aligned}\underline{X}((a,b),(x,y,z))&=\left.\frac{d}{dt}\right|_{t=0}\exp tX((a,b),(x,y,z))\\&=((i\alpha,0),(i\alpha kx,0,i\beta y))\end{aligned}

Now, let’s denote

\begin{cases}a&:=x_1+iy_1\\b&:=x_2+iy_2\\x&:=x_3+iy_3\\y&:=x_4+iy_4\\z&:=x_5+iy_5\end{cases}

So that we can use the standard symplectic form on \mathbb{C}^5, namely \omega=\sum_{i=1}^5 dx_i\wedge dy_i, let’s re-write the \underline{X} in these new coordinates:

\underline{X}=(-\alpha y_1)\frac{\partial}{\partial x_1}+\alpha x_1\frac{\partial}{\partial y_1}+(-\alpha ky_3)\frac{\partial}{\partial x_3}+\alpha k x_3\frac{\partial}{\partial y_3}+(-\beta y_5)\frac{\partial}{\partial x_5}+\beta x_5\frac{\partial}{\partial y_5}

We can use a quick formula for interior product: if \omega=\sum dx_i\wedge dy_i, X=\sum \left(X_i\frac{\partial}{\partial x_i}+Y_i\frac{\partial}{\partial y_i}\right), then

\iota_{X}\omega=\sum\left(-Y_idx_i+X_idy_i\right)

So, in our case we have:

\begin{aligned}\iota_{\underline{X}}\omega&=-\left(\alpha x_1 dx_1+k\alpha x_3dx_3+\beta x_5dx_5+\alpha y_1dy_1+k\alpha y_3dy_3+\beta y_5dy_5\right)\\&=-d\left(\frac{\alpha}{2}x_1^2+\frac{k\alpha}{2}x_3^2+\frac{\beta}{2}x_5^2+\frac{\alpha}{2}y_1^2+\frac{k\alpha}{2}y_3^2+\frac{\beta}{2}y_5^2\right)\\&=-d\left(\frac{\alpha}{2}|a|^2+\frac{k\alpha}{2}|x|^2+\frac{\beta}{2}|z|^2\right)\end{aligned}

Hence the moment map \nu is:

\begin{aligned}\nu((a,b),(x,y,z))&=\left(X\mapsto\frac{\alpha}{2}|a|^2+\frac{k\alpha}{2}|x|^2+\frac{\beta}{2}|z|^2\right)\\&=\left(\frac{1}{2}|a|^2+\frac{k}{2}|x|^2,\frac{1}{2}|z|^2\right)\end{aligned}

Then descending it to the level of \mathbb{P}^1\times\mathbb{P}^2 hence we have:

\mu([a,b],[x,y,z])=\left(\frac{1}{2}\cdot\frac{|a|^2}{|a|^2+|b|^2}+\frac{k}{2}\cdot\frac{|x|^2}{|x|^2+|y|^2+|z|^2},\frac{1}{2}\cdot\frac{|z|^2}{|x|^2+|y|^2+|z|^2}\right)

留下评论