Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Reading Barth, Peters: Compact complex surfaces, i stumbled across the following:

Let $Y$ be an algebraic surface over $k =\mathbb{C}$, and $\mathcal{L}$ an invertible sheaf on $Y$. Denote by $p: L \rightarrow Y$ the total space of $\mathcal{L}$ and consider the invertible sheaf $p^*(\mathcal{L})$ on $L$.

Now according to the book, this bundle is supposed to have a "tautological section", $$ t \in \Gamma(L, p^*(\mathcal{L})) $$ but i have no idea what is meant. I thought it was the zero section, but in this case it is clear (from the context in the book) that they do not mean that! Could anybody help me out here?

It's on page 54, on the bottom, in section 17 on cyclic coverings.


share|cite|improve this question
up vote 3 down vote accepted

I'll describe the tautological section by telling you what its value $t(l)$ is at $l\in L$ .

Put $p(l)=y\in Y$.
The fiber $p^*(\mathcal{L})[l]$ of $p^*(\mathcal{L})$ at $l$ is $$p^*(\mathcal{L})[l]=\lbrace l\rbrace \times p^{-1}(y)=\lbrace l\rbrace \times \mathcal L[y]$$
This last set contains $(l,l)$ i.e. $(l,l)\in \lbrace l\rbrace \times p^{-1}(y)$.
And the value I promised to give you is $t(l)=(l,l) $

I have used the following notation:
Given an invertible sheaf $\mathcal M$ on a complex manifold $X$, one associates to it a geometric vector bundle of rank one $p:M\to X $.
The fiber of $p$ at $x\in X$ is the one dimensional $\mathbb C$-vector space $$\mathcal L[x]=p^{-1}(x)=\mathcal L_x\otimes_{\mathcal O_{X,x}} \mathcal O_{X,x}/\mathcal m_x=\mathcal L_x\otimes_{\mathcal O_{X,x}} \mathbb C$$ Needless to say, $L=\sqcup_{x\in X} \mathcal L[x]$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.