# nLab tautological line bundle

### Context

#### Bundles

bundles

fiber bundles in physics

# Contents

## Idea

The canonical line bundle over a projective space is sometimes called its “tautological line bundle”. For more see at classifying space.

## Definition

### As a topological line bundle

We discuss the tautological line bundle as a topological vector bundle. Hence let $k$ be the topological field either

• $k = \mathbb{R}$ the real numbers

• or $k = \mathbb{C}$ the complex numbers

equipped with their Euclidean metric topology.

###### Definition

(topological projective space)

Let $n \in \mathbb{N}$. Consider the Euclidean space $k^{n+1}$ equipped with its metric topology, let $k^{n+1} \setminus \{0\} \subset k^{n+1}$ be the topological subspace which is the complement of the origin, and consider on its underlying set the equivalence relation which identifies two points if they differ by multiplication with some $c \in k$ (necessarily non-zero):

$(\vec x_1 \sim \vec x_2) \;\Leftrightarrow\; \left( \underset{c \in k}{\exists} ( \vec x_2 = c \vec x_1 ) \right) \,.$

The equivalence class $[\vec x]$ is traditionally denoted

$[x_1 : x_2 : \cdots : x_{n+1}] \,.$

Then the projective space $k P^n$ is the corresponding quotient topological space

$k P^n \;\coloneqq\; \left(k^{n+1} \setminus \{0\}\right) / \sim \,.$
###### Definition

(standard open cover of topological projective space)

For $n \in \mathbb{N}$ the standard open cover of the projective space $k P^n$ (def. ) is

$\left\{ U_i \subset k P^n \right\}_{i \in \{1, \cdots, n+1\}}$

with

$U_i \coloneqq \left\{ [x_1 : \cdots : x_{n+1}] \in k P^n \;\vert\; x_i \neq 0 \right\} \,.$

To see that this is an open cover:

1. This is a cover because with the orgin removed in $k^n \setminus \{0\}$ at every point $[x_1: \cdots : x_{n+1}]$ at least one of the $x_i$ has to be non-vanishing.

2. These subsets are open in the quotient topology $k P^n = (k^n \setminus \{0\})/\sim$, since their pre-image under the quotient co-projection $k^{n+1} \setminus \{0\} \to k P^n$ coincides with the pre-image $(pr_i\circ\iota)^{-1}( k \setminus \{0\} )$ under the projection onto the $i$th coordinate in the product topological space $k^{n+1} = \underset{i \in \{1,\cdots, n\}}{\prod} k$ (where we write $k^n \setminus \{0\} \overset{\iota}{\hookrightarrow} k^n \overset{pr_i}{\to} k$).

###### Definition

(tautological topological line bundle)

For $k$ a topological field and $n \in \mathbb{N}$, the tautological line bundle over the projective space $k P^n$ is topological $k$-line bundle whose total space is the following subspace of the product space of the projective space $k P^n$ with $k^n$:

$T \coloneqq \left\{ ( [x_1: \cdots : x_{n+1}], \vec v) \in k P^n \times k^{n+1} \;\vert\; \vec v \in \langle \vec x\rangle_k \right\} \,,$

where $\langle \vec x\rangle_k \subset k^{n+1}$ is the $k$-linear span of $\vec x$.

(The space $T$ is the space of pairs consisting of the “name” of a $k$-line in $k^{n+1}$ together with an element of that $k$-line)

This is a bundle over projective space by the projection function

$\array{ T &\overset{\pi}{\longrightarrow}& k P^n \\ ([x_1: \cdots : x_{n+1}], \vec v) &\mapsto& [x_1: \cdots : x_{n+1}] } \,.$
###### Proposition

(tautological topological line bundle is well defined)

The tautological line bundle in def. is well defined in that it indeed admits a local trivialization.

###### Proof

We claim that there is a local trivialization over the canonical cover of def. . This is given for $i \in \{1, \cdots, n\}$ by

$\array{ U_i \times k &\overset{}{\longrightarrow}& T\vert_{U_i} \\ ( [x_1 : \cdots x_{i-1}: 1 : x_{i+1} : \cdots : x_{n+1}] , c ) &\mapsto& ( [x_1 : \cdots x_{i-1} : 1 : x_{i+1} : \cdots : x_{n+1} ], (c x_1, c x_2, \cdots , c x_{n+1}) ) } \,.$

This is clearly a bijection of underlying sets.

To see that this function and its inverse function are continuous, hence that this is a homeomorphism notice that this map is the extension to the quotient topological space of the analogous map

$\array{ ( (x_1, \cdots, x_{i-1}, x_{i+1}, \cdots, x_{n+1}) , c) &\mapsto& ( (x_1, \cdots, x_{i-1}, x_{i+1}, \cdots, x_{n+1}) , (c x_1, \cdots c x_{i-1}, c, c x_{i+1}, \cdots, c x_{n+1}) ) } \,.$

This is a polynomial function on Euclidean space and since polynomials are continuous, this is continuous. Similarly the inverse function lifts to a rational function on a subspace of Euclidean space, and since rational functions are continuous on their domain of definition, also this lift is continuous.

Therefore by the universal property of the quotient topology, also the original functions are continuous.

## References

Lecture notes include

See also

Last revised on June 11, 2017 at 09:58:43. See the history of this page for a list of all contributions to it.