Definition 1 : Let $B$ be a topological space. A complex vector bundle of rank $n$ over $B$ consists of the data $(E,B,\pi,\{U_i\}_i,\{\phi_i\}_i)$ where $E$ is a topological space, $\pi:E\to B$ is a continuous map called the projection map, $\{U_i\}_i$ is an open cover of $B$ and $\phi_i:\pi^{-1}(U_i)\to U_i\times \mathbb C^n$ is a homeomorphism such that

  1. $\pi=p_{U_i}\circ\phi_i$ (where $p_{U_i}:U_i\times\mathbb C^n\to U_i$ is the projection onto first factor)

  2. $\phi_j\circ\phi_i^{-1}:(U_i\cap U_j)\times\mathbb C^n\to(U_i\cap U_j)\times\mathbb C^n$ takes $(u,x)$ to $(u,\theta_{ij}(x))$ where $\theta_{ij}\in GL_n(\mathbb C)$.

Projectivizing this vector bundle is essentially replacing the $\mathbb C^n$ by $\mathbb C^N\setminus \{0\}/\mathbb C^* =\mathbb CP^{n-1}$. So I tried to write this definition as follows -

Definition 2 : Let $B$ be a topological space. A complex projective vector bundle of rank $n-1$ over $B$ consists of the data $(E',B,\pi',\{U_i\}_i,\{\psi_i\}_i)$ where $E'$ is a topological space, $\pi':E'\to B$ is a continuous map called the projection map, $\{U_i\}_i$ is the same open cover of $B$ as in definition 1 and $\psi_i:(\pi')^{-1}(U_i)\to U_i\times \mathbb CP^{n-1}$ is a homeomorphism such that

  1. $\pi'=p_{U_i}\circ\phi_i$ (where $p_{U_i}:U_i\times\mathbb CP^{n-1}\to U_i$ is the projection onto first factor)

  2. $\psi_j\circ\psi_i^{-1}:(U_i\cap U_j)\times\mathbb CP^{n-1}\to(U_i\cap U_j)\times\mathbb CP^{n-1}$ takes $(u,[x])$ to $(u,[\theta_{ij}(x)])$ where $\theta_{ij}\in GL_n(\mathbb C)$ are the same as in definition 1.

Question 1: Is this definition correct? (I ask because all google results for projectivizing a vector bundle define it using the language of sheaves and I want to know if I have done it correctly without sheaves)

Given a vector bundle we can always define the zero section as follows - $s_0:B\to E$ is the map $s_0(b)=\phi_i^{-1}(b,0)$ where $b\in U_i$. This is well defined by condition 2 of definition 1.

Now $0$ obviously doesn't belong to $\mathbb CP^{n-1}$. But instead we some special points. Let $x_l=(0,\cdots,1,\cdots,0)$ where $1$ is in the $l^{th}$ position. Suppose I want to define sections $s_l:B\to E'$ by $s_l(b)=\psi_i^{-1}(b,[x_l])$ where $b\in U_i$

Question 2: Why is this $s_l$ a well defined map (independent of the choice of $U_i$)? Is there something missing in definition 2 that hinders the proof of well definedness here?

For question 2, I need to show that $\psi_i^{-1}(b,[x_l])=\psi_j^{-1}(b,[x_l])$ if $b\in U_i\cap U_j$. This will certainly happen if $[ x_l]=[\theta_{ij}(x_l)]$. But why should this be true?

Thank you.

  • 2
    $\begingroup$ Your structure group should be PGL, not GL; note that scalar multiplication acts as the identity on projective space. Of course you can always lift a map from PGL to GL if say $U_i \cap U_j$ is always contractible. $\endgroup$
    – user98602
    Jul 14, 2016 at 16:15
  • $\begingroup$ @MikeMiller, So if I take an element $\theta_{ij}\in$PGL will I have $\theta_{ij}[x_l]=[x_l]$? $\endgroup$
    – R_D
    Jul 15, 2016 at 5:30
  • 1
    $\begingroup$ You should realize that $(0,\ldots,1,0,\ldots,0)\in E$, with $E$ a vector space, does not really make sense, you need a basis for it. The same with the equivalence class in $P(E)$. This obstructs you to write a section of the projectivized bundle. However, if $E=L_1\oplus L_2$, with $L_1$ and $L_2$ one dimensional vector spaces, the point $[1,0]\in P(L_1\oplus L_2)$ does make sense (think about it!). Therefore a bundle of this form also admits this section. You can look at Atiyahs book on K theory. I believe it is in the second chapter. $\endgroup$
    – Thomas Rot
    Jul 15, 2016 at 8:32
  • 1
    $\begingroup$ (Note that $E=L_1\oplus L_2$ is saying that $E$ has a basis. $\endgroup$
    – Thomas Rot
    Jul 15, 2016 at 9:34
  • $\begingroup$ @ThomasRot, I understand now. However, assuming the standard basis for $\mathbb C^n$ shouldn't what I have done above make sense? $\endgroup$
    – R_D
    Jul 15, 2016 at 13:15

1 Answer 1


For exact $\theta_{ij}:x\in U_{ij}=U_i\cap U_j\to\theta_{ij}(x)\in GL_n(\mathbb{C})$; but this is the minus.

Let $\pi:E\to B$ a (topological) rank $n$ vector bundle over $B$, the projectivization $\mathbb{P}(E)$ of $E$ is a principal bundle over $B$ with structure group $PGL_n(\mathbb{C})$!

Let $\varpi:GL_n(\mathbb{C})\to PGL_n(\mathbb{C})$ the canonical projection, we can consider the maps $\eta_{ij}=\varpi\circ\theta_{ij}:U_{ij}\to PGL_n(\mathbb{C})$ and get \begin{equation*} \mathbb{P}(E)=\coprod_{i\in I}U_i\times PGL_n(\mathbb{C})_{\displaystyle/\sim} \end{equation*} where \begin{gather*} (P_i,g)\in U_i\times PGL_n(\mathbb{C}),(P_j,h)\in U_j\times PGL_n(\mathbb{C}),\\(P_i,g)\sim(P_j,h)\iff P_i=P_j=P,h=\eta_{ij}(P)g. \end{gather*} Easily, one can prove that $\pi^{\prime\prime}:[P,g]_{\sim}\in\mathbb{P}(E)\to P\in B$ is a well defined map; by construction, $(\pi^{\prime\prime})^{-1}(U_i)$ is in bijection with $U_i\times PGL_n(\mathbb{C})$, then we can require that this bijection is a homeomorphism, we introduce a topology on $\mathbb{P}(E)$; in particular, $\pi^{\prime\prime}$ is a continuous map and in this way $\pi^{\prime\prime}:\mathbb{P}(E)\to B$ is a principal $PGL_n(\mathbb{C})$-bundle!

Vice versa: one can associate canonically a fibre bundle $E^{\prime}$ over $B$ to $\mathbb{P}(E)$ with fibre $\mathbb{CP}^{n-1}$!

We get \begin{equation*} E^{\prime}=\coprod_{i\in I}U_i\times\mathbb{CP}^{n-1}_{\displaystyle/\sim} \end{equation*} where \begin{gather*} (P_i,A)\in U_i\times\mathbb{CP}^{n-1},(P_j,B)\in U_j\times\mathbb{CP}^{n-1},\\(P_i,A)\sim(P_j,B)\iff P_i=P_j=P,B=\eta_{ij}(P)(A); \end{gather*} and in the same way, one can that $\pi^{\prime}:E^{\prime}\to B$ is a fibre bundle over $B$ with fibre $\mathbb{CP}^{n-1}$.

All this answers to your first question; as a consequence, we can't define a zero global section of $\mathbb{P}(E)$ (or equivalently of $E^{\prime}$).

About your second question: the equation $[\theta_{ij}(b)(x_l)]=[x_l]$ doesn't hold in general!

  • $\begingroup$ Does $[\eta_{ij}(b)(x_l)]=[x_l]$ hold? $\endgroup$
    – R_D
    Jul 12, 2017 at 5:45
  • $\begingroup$ Yes, it does! Do you understand because it is hold? $\endgroup$ Jul 13, 2017 at 12:01
  • $\begingroup$ I do not understand how it holds. Would you mind explaining? I know that once its holds my problem is solved. $\endgroup$
    – R_D
    Jul 18, 2017 at 8:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .