Take the 2-minute tour ×
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|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
add comment

Your Answer


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

Browse other questions tagged or ask your own question.