Let $T_{X}\rightarrow X$ be the tangent bundle over a complex manifold $X.$ Let $\pi:PT_{X}\rightarrow X$ be a projectivization of that bundle. Let $L$ be the tautological line bundle of $PT_{X}.$

  1. How do we construct the isomorphism between the relative tangent bundle $T_{\pi}$ and $R=L^{*}\otimes \pi^{*}T_X/L$? For example, for every vector $v\in T_{\pi}$ we know that $d\pi(v)=0,$ but how we can see that vector as an element of $R$?

  2. What are the trivialization maps for both of these bundles $T_{\pi}$ and $R$?


1 Answer 1


There is nothing special about the tangent bundle: for any vector bundle $E \rightarrow X$ with projectivisation $\pi : \mathbb{P}E \rightarrow X$, and tautological bundle $\mathcal{L} \rightarrow \mathbb{P}E$, we have $T\mathbb{P}E \cong (\pi^*E / \mathcal{L}) \otimes \mathcal{L}^*$. In fact the base $X$ is not really involved, so you may as well work with the case where $X$ is a point, and $E$ is just a vector space $V$. Then the result follows from the standard Euler sequence

$$ 0 \rightarrow \mathcal{O}_{\mathbb{P}V} \rightarrow V \otimes \mathcal{O}_{\mathbb{P}V}(1) \rightarrow T\mathbb{P}V \rightarrow 0, $$

where we regard $V$ as a constant sheaf on $\mathbb{P}V$.

Added in response to comment: Explicitly, given a point $l \in \mathbb{P}V$, viewed as a line in $V$, and a vector $v \in T_l \mathbb{P}V$, we can lift $v$ (under the projection $V \setminus \{0\}\rightarrow \mathbb{P}V$) to a vector in $V/l$ at each point of $l \setminus \{0\}$. Moreover, this lift is equivariant under the action of $k^\times$ on $l\setminus \{0\}$, where $k$ is our ground field (e.g. $\mathbb{C}$). In other words, it defines a linear $V/l$-valued function on $l$.

Linear functions on $l$ are precisely elements of $\mathcal{L}^*_l$, and $V/l$ is precisely $(\pi^*V/\mathcal{L})_l$, so $v$ defines an element of $((\pi^*V/\mathcal{L}) \otimes \mathcal{L}^*)_l$.

Added in response to further comments: Even more explicitly, if $e_0, e_1, \dots, e_n$ is a basis for $V$, with corresponding coordinates $w_0, w_1, \dots, w_n$, then the $w_i$ form homogeneous coordinates on $\mathbb{P}V$. On the affine patch $w_0 \neq 0$, we have local coordinates given by $z_j = w_j/w_0$, for $j=1, \dots, n$. Given a point $l = [1: l_1: \dots: l_n] \in \mathbb{P}V$, the vector

$$ \sum_{j=1}^n v_j \frac{\partial}{\partial z_j} $$

is tangent to the curve $z_j(t) = l_j + tv_j$. This lifts to the curve in $V$ given by $w_0(t)=f(t)$, $w_j(t)=f(t)(l_j+tv_j)$ for $j\geq 1$, where $f$ is any smooth $\mathbb{C}^\times$-valued function. Its tangent at $t=0$, i.e. a lift of $v$ to $V$, is

$$ f'(0) \bigg( e_0 + \sum_{j=1}^n l_je_j \bigg) + \sum_{j=1}^n v_j e_j. $$

The ambiguity coming from $f'(0)$ is exactly what is killed when we quotient by $l$, so the lift to $V/l$ is represented by $\sum v_je_j$.

Explicit computation for $V=\mathbb{C}^2$: Take coordinates $w_0, w_1$ on $\mathbb{C}^2$, and let $U_j = \{w_j \neq 0\} \subset \mathbb{P}^1$. We have

$$(\pi^*V / \mathcal{L})_{[w_0 : w_1]} = \mathbb{C}^2 \Big/ \mathbb{C} \begin{pmatrix} w_0 \\ w_1 \end{pmatrix}.$$

On $U_0$ a complement to $\mathcal{L}$ in $\pi^*V$ is given by $\mathbb{C}e_1$; we can therefore take $[e_1]$ as a trivialising frame for the quotient bundle (where the square brackets denote the image in the quotient). Similarly on $U_1$ a frame is given by $[e_0]$. On $U_0 \cap U_1$ we have for all local holomorphic functions $f$ that

$$f e_1 = \frac{f}{w_1} \begin{pmatrix} w_0 \\ w_1 \end{pmatrix} - \frac{fw_0}{w_1} e_0 $$


$$ f[e_1] = -\frac{fw_0}{w_1}[e_0].$$

The transition function $\psi_{10}$ is therefore $-w_0/w_1$, which (up to the minus sign, which we could have eliminated by choosing $[-e_0]$ as our frame on $U_1$) is the transition function for $\mathcal{O}(1)$.

  • $\begingroup$ @AndRub What do you mean "$v$ has coordinates $(a, b, c)$"? $T\mathbb{P}V$ is $2$-dimensional. $\endgroup$
    – user81684
    May 13, 2016 at 18:26
  • $\begingroup$ @AndRub I'm still a little confused. I'll add a bit more to my answer, which will hopefully explain what I'm saying. $\endgroup$
    – user81684
    May 15, 2016 at 7:28
  • $\begingroup$ @AndRub I'll add another bit to my answer. $\endgroup$
    – user81684
    May 19, 2016 at 19:27
  • $\begingroup$ @ Jezz what is precisely an element of $(\mathcal(L)^{*} \otimes (\pi^{*}V/\mathcal(L)))_{[l]}? $ How it can be seen, as linear function on $l\backslash \{0\}$ that has for result some vector from $\pi^{*}V/\mathcal(L))? $ Am I wrong? How it can be presented? $\endgroup$
    – And Rub
    May 22, 2016 at 13:50
  • $\begingroup$ @AndRub This is a standard fact from linear algebra: if $U$ and $V$ are finite dimensional vector spaces then there is a natural isomorphism between $U^* \otimes V$ and $\mathrm{Hom}(U, V)$. A concrete way to think about this is to pick bases for $U$ and $V$ and compare an element of $U^* \otimes V$ in the corresponding basis with a matrix for a linear map from $U$ to $V$. $\endgroup$
    – user81684
    May 23, 2016 at 7:13

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