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Consider the projective space $\mathbb{P}^n = \mathbb{P}(\mathbb{C}^{n+1})$ and the open sets $$U_i = \{ [Z_0:Z_1:\ldots:Z_n] \in \mathbb{P}^n: \ Z_i \neq 0\}$$ for $i = 0 \ldots n$. These sets cover $\mathbb{P}^n$ and we can use each $U_i$ to define a coordinate system (affine coordinates) $(U_i, \varphi_i)$ such that $$\varphi_i([Z_0:\ldots:Z_n]) = \left( \frac{Z_0}{Z_i}, \ldots, \frac{Z_{i-1}}{Z_i}, \frac{Z_{i+1}}{Z_i}, \ldots, \frac{Z_n}{Z_i}\right) = (z_1, \ldots, z_n).$$

With respect to this coordinate system the Fubini-Study metric is given by the (1,1)-form $$\omega = \frac{i}{2\pi}\sum_{i,j=1}^n h_{ij} dz_i \wedge d\overline{z}_j,$$ where $$h_{ij} = \frac{(1+\|z\|^2)\delta_{ij} - \overline{z}_i z_j}{(1+\|z\|^2)^2}$$ and $z = (z_1, \ldots, z_n)$.

Now consider another possible coordinate system. We have the open sets $\mathbb{A}_Z = \{ [Y+Z] \in \mathbb{P}^n: \ Y \in Z^\perp\}$ for each $Z \in \mathbb{C}^{n+1}$ (we are considering $\mathbb{C}^{n+1}$ with the usual Hermitian inner product to define $Z^\perp$). These sets also cover $\mathbb{P}^n$ and we can use each $\mathbb{A}_Z$ to define the new coordinate system $(\mathbb{A}_Z, \psi_Z)$ such that $$\psi_Z([W_0:\ldots:W_n]) = \frac{\|Z\|^2}{\langle W, Z \rangle} W - Z,$$ where $W = (W_0, \ldots, W_n) \in \mathbb{C}^{n+1}$. Although $\text{Im}(\psi_Z) \subset \mathbb{C}^{n+1}$, it is a subspace of $\mathbb{C}^{n+1}$ of dimension $n$.

This coordinate system has a very nice geometric interpretation and is as natural as the affine coordinate system. But in all texts I've seen everybody only talks about Fubini-Study metric in affine coordinates. How is the computation of the Fubini-Study metric is this particular coordinate system?

Any recommendation of reading is welcome. Specially if this specific question is addressed! I already read this topic on Griffiths-Harris (Principles of Algebraic Geometry) and Daniel Huybrechts (Complex Geometry). Both texts make the same thing: they introduce the Fubini-Study metric in affine coordinates. I also gave a look at the Wikipedia article but it didn't help me.

PS: In Griffiths-Harris they say the form $\omega$ is invariant by unitary action. This means $\omega_{Z}(u,v) = \omega_{UZ}(Uu, Uv)$ for all $Z \in \mathbb{C}^{n+1}\backslash\{0\}$, all $U \in \mathbb{C}^{(n+1) \times (n+1)}$ unitary and all tangent vectors $u,v \in z^\perp \cong T_{[Z]}\mathbb{P}^n$ ?

Thank you for your help.

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    $\begingroup$ These are the standard charts up to base change, aren't they? Have you tried something like $ \omega_Z = \frac{i}{2\pi} \partial \bar{\partial}\log\left(\left\|\frac{\|Z\|^2}{\langle W, Z \rangle} W \right\|^2\right) $? $\endgroup$ – Alan Muniz Dec 8 '17 at 14:12
  • $\begingroup$ @AlanMuniz In fact I tried this today and it seems to be the correct way to go. There is just one issue to make everything work. When I try to expand this form I should get something like $$\frac{i}{2\pi} \sum_{i,j} h_{ij} \ dz_i \wedge d\overline{z_j}.$$ But do I make $i,j=0 \ldots n$ or $i,j = 1 \ldots n$ ? The way $W$ and $Z$ are represented make it necessary to use all $n+1$ variables (in contrast to affine coordinates where one of the coordinates is fixed equal to 1), but the tangent space only has $n$ coordinates, so I probably I won't consider $dz_0$ and $d\overline{z_0}$. $\endgroup$ – Integral Dec 8 '17 at 15:18
  • $\begingroup$ Representing $\omega_Z$ in affine coordinates works fine because the number of variables in $(1, z_1, \ldots, z_n)$ equals the dimension of the tangente space. I just don't know how to relate the terms $h_{ij}$ in affine coordinates and in this other system of coordinates where we have $n+1$ coordinates, but in a subspace of dimension $n$. $\endgroup$ – Integral Dec 8 '17 at 15:22
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    $\begingroup$ Any one you choose to put coordinates on $Z^{\perp}$ $\endgroup$ – Alan Muniz Dec 8 '17 at 16:37
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    $\begingroup$ @AlanMuniz Don't you want put all these comments together to form an answer? I will accept it. $\endgroup$ – Integral Dec 8 '17 at 17:08
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I got very helpful comments thanks to AlanMuniz, but not an official answer. So I'll try to give mine and I hope this can be helpful to someone. Also, if there is something wrong I hope someone can point it out to me.

Following Griffiths-Harris terminology, the map $W\backslash\{0\} \in \mathbb{C}^{n+1} \mapsto L_Z(W) = \displaystyle \frac{\|Z\|^2}{\langle W,Z \rangle}W$ is not a chart but a lifting, i.e., a holomorphic section of the total space $\mathbb{C}^{n+1}$ over the base $\mathbb{P}^n$. Since $\omega_Z$ is invariant by unitary action, we have that $$\omega_{Z}(u,v) = \omega_{UZ}(Uu, Uv)$$ for every unitary matrix $U \in \mathbb{C}^{(n+1) \times (n+1)}$ and every tangent vectors $u,v \in Z^\perp \cong T_{[Z]}\mathbb{P}^n$. Associated to this change we have the lifting $L_{UZ}(W) = \displaystyle \frac{\|UZ\|^2}{\langle W,UZ \rangle}W$, so $$\omega_{UZ} = \frac{i}{2\pi} \partial \overline{\partial}\log\left( \|L_{UZ}\|^2 \right) = \frac{i}{2\pi} \partial \overline{\partial}\log\left( \left\|\frac{\|UZ\|^2}{\langle W,UZ \rangle}W\right\|^2 \right) = \frac{i}{2\pi} \partial \overline{\partial}\log\left( 1+\|z\|^2 \right),$$ where $z = \left( \frac{(UZ)_1}{(UZ)_0}, \ldots, \frac{(UZ)_n}{(UZ)_0} \right)$ and $(UZ)_i$ denotes the $i$-th coordinate of $UZ$. Now we are working in affine coordinates. In this case we have that $$\omega_Z = \frac{i}{2\pi} \sum_{i,j=1}^n h_{ij} dz_i \wedge d\overline{z}_j,$$ where $$h_{ij} = \frac{(1+\|z\|^2) \delta_{ij} - \overline{z}_i z_j}{(1+\|z\|)^2}.$$

Extra: We can take one step further. In particular, there is a matrix $U$ such that $UZ = e_0 = (1,0,\ldots,0)$. Note that $h_{ij} = \delta_{ij}$. Therefore, $$\omega_{e_0} = \frac{i}{2\pi} \sum_{j=1}^n dz_j \wedge d\overline{z}_j.$$

These results agree with the text of Griffiths-Harris.

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