Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question follows up this previous question, which has an accepted answer that I am having trouble believing. (Edit: the answer has now been fixed; thanks Georges!)

I have been working on giving myself as concrete an understanding as possible of Serre's twisting sheaf $\mathcal{O}(1)$. My reference is Hartshorne, ch. 2, section 5.

On their face, $\mathcal{O}(1),\mathcal{O}(-1)$ are invertible sheaves; I know it is standard to view them as algebraic line bundles. In the case $X=\mathbb{P}_\mathbb{R}^1$ (viewed as a scheme), I am interested in visualizing these line bundles.

Background: I am using the definition of algebraic vector bundle in Hartshorne ch. 2 sec. 5 exercise 18 on p. 128: a scheme $Y$, a morphism $f:Y\rightarrow X$, and an open cover $\{U_i\}$ of $X$ such that for each $U_i$, $f^{-1}(U_i)$ is isomorphic to $\mathbb{A}_{U_i}^n$ for some fixed $n$, and the isomorphisms $\psi_i:f^{-1}(U_i)\rightarrow\mathbb{A}_{U_i}^n$ satisfy the property that for every pair $i,j$ and for every open affine $V\subset U_i\cap U_j$, the appropriately restricted composition $\psi_j\circ\psi_i^{-1}$ is a linear automorphism of $\mathbb{A}_V^n$.

To fix notation, let $Y,Z$ be algebraic line ($n=1$) bundles over $X=\mathbb{P}_\mathbb{R}^1$ such that the sheaves of sections $\mathscr{S}(Y/X)=\mathcal{O}_X(1)$ and $\mathscr{S}(Z/X)=\mathcal{O}_X(-1)$.

After a fair amount of work, I have come to believe that we can obtain $Y,Z$ explicitly as follows:

Let $U_1=\operatorname{Spec} \mathbb{R}[x,y]$ and let $U_2=\operatorname{Spec} \mathbb{R}[x',y']$, and identify $U_1\setminus \{x=0\} = \operatorname{Spec} \mathbb{R}[x,x^{-1},y]$ with $U_2\setminus \{x'=0\} = \operatorname{Spec}\mathbb{R}[x',x'^{-1},y']$ via the scheme isomorphism induced from the ring isomorphism $x'\mapsto x^{-1}$, $y'\mapsto x^{-1}y$. Let $Y$ be the quotient of the disjoint union $U_1\cup U_2$ by this identification. Let the projection morphism $f:Y\rightarrow X$ be given by projection to the $x$-axis on $U_1$ and to the $x'$-axis on $U_2$. (In scheme-theoretic terms, the projection $U_1\rightarrow \mathbb{A}^1$ is the morphism coming from the ring inclusion $\mathbb{R}[x]\hookrightarrow \mathbb{R}[x,y]$ and similarly for $U_2$.)

We get $Z$ from the exact same construction except the isomorphism of $U_1\setminus\{x=0\}$ with $U_2\setminus \{x'=0\}$ is given by $x'\mapsto x^{-1}, y'\mapsto xy$ this time.

My initial question:

Do you agree with these constructions?

As a plausibility check, we know $\mathcal{O}_X(1)$ has a 2-dimensional vector space of global sections; these should correspond to sections of $Y/X$. I believe we get a basis of sections from the section $u$ that looks like $y=1$ on $U_1$ and $y'=x'$ on $U_2$ and the section $v$ that looks like $y'=1$ on $U_2$ and $y=x$ on $U_1$. On the other hand, we know $\mathcal{O}_X(-1)$ has no nonzero global sections; this results from the fact that if a section looks like $y'=g(x')$ on $U_2$ (for $g$ some nonzero polynomial) it should look like $y = x^{-1}g(x^{-1})$ on $U_1$, but this can not possibly be regular at $x=0$.

My main question:

$Y, Z$ are not manifolds because they're schemes, but we can copy their construction with copies of the smooth manifold $\mathbb{R}^2$ instead of the scheme $\mathbb{A}_\mathbb{R}^2$; it seems to me that this will give us real vector bundles over $S^1$ that are both diffeomorphic to the Möbius band (without boundary) and I'd like to know if you agree. To be explicit about the construction:

Let $U_1=\mathbb{R}^2$ with coordinate functions $x,y$. Let $U_2=\mathbb{R}^2$ with coordinate functions $x',y'$. Identify $U_1\setminus \{x=0\}$ with $U_2\setminus\{x'=0\}$ by $(x',y')\sim (x^{-1},x^{-1}y)$. The result is a smooth manifold because this identification gives smooth transition functions between the two coordinate patches. Restricting the construction to the $x$ and $x'$-axes gives $\mathbb{R}P^1\cong S^1$. Furthermore, for fixed $x$, the transition function $y'=x^{-1}y$ is linear in $y$, so it seems to me this implies the result is a line bundle over $S^1$. Let this be $\mathbf{Y}$.

Create $\mathbf{Z}$ the same way except using $(x',y')\sim (x^{-1},xy)$.

If these really are real line bundles, then they both have to be Möbius bands because that's the only nontrivial real line bundle over $S^1$.

Is my thinking that these constructions yield real line bundles correct?

Thanks in advance.

share|cite|improve this question
I agree. In algebraic geometry $\mathrm{Pic}(\mathbb{P}^1)=\mathbb{Z}$ (in particular $\mathcal{O}(1)$ and $\mathcal{O}(-1)$ are not isomorphic), but for manifolds we have $\mathrm{Pic}(\mathbb{P}^1)=\mathbb{Z}/2$. More generally, if $X$ is a CW complex, then $\mathrm{Pic}(X) \cong H^1(X,\mathbb{Z}/2)$ (see Hatcher's book). In particular $\mathcal{L} \cong \mathcal{L}^{\otimes -1}$ for every line bundle $\mathcal{L}$. – Martin Brandenburg Apr 12 '13 at 18:41
Dear Ben, I have modified my answer to the question you mention, since it was indeed incorrect. You might want to have a look at this modified version, which confirms what you write here. – Georges Elencwajg Apr 12 '13 at 19:41
Thanks! The linked question now answers this one. – Ben Blum-Smith Apr 16 '13 at 22:42

1 Answer 1

up vote 3 down vote accepted

Leaving a community wiki answer to remove this from the unanswered list (once someone upvotes it): Yes, this is true. Georges Elencwacj edited his answer to another question in order to address this one.

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.