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.

Let $\mathbb{Z}$ be the constant sheaf on $\mathbb{S}^1$, $f: \mathbb{S}^1 \to\mathbb{R}P^1$ the double cover and $\mathcal{A} = f\mathbb{Z}$.

Then $\mathcal{A}$ has stalks $\mathbb{Z}\oplus\mathbb{Z}$ but I am having difficulties describing the topology of $\mathcal{A}$. That is, I do not understand why the sheaf is 'twisted' via the automorphism $(n,k)\to (k,n)$.

edit: my idea was flawed so I might as well remove it.

share|improve this question
    
@ZhenLin: Thank you for your reply. But is that even isomorphic to the éspace étalé I've written above? Quoting from Bredon: "this has stalks $\mathbb{Z}\otimes\mathbb{Z}$ and is ''twisted'' via the automorphism $(n,k)\to (k,n)$". Now if it indeed is isomorphic to the one Bredon writes down I do not know how to prove that they in fact are isomorphic. It feels to me that in order to prove that I would have to use the same ideas as I would have had to use in order to find out how Bredon reaches his conclusion. Sadly, I have spent hours trying to figure out just that. –  M.B. Jun 5 '12 at 22:01
    
I got the following idea when riding my bike for home: what if we think of sections over $U$ as pairs $(\gamma_1, \gamma_2)$ where $\gamma_1$ corresponds to the section of the upper half of the circle and $\gamma_2$ of the lower half. Upon taking an open set around the equator (i.e. the identified equator in the projective plane) $\gamma_1$ travels from the upper half to the lower half and $\gamma_2$ travels from lower to upper. So this means that under this reprsentation of sections $(\gamma_1, \gamma_2)$ "twists" to $(\gamma_2, \gamma_1$) as we pass the equator. Does this sound reasonable? –  M.B. Jun 5 '12 at 23:23
    
Sorry, I got confused by some exceptional homeomorphisms happening in this example; I retract my earlier comments. –  Zhen Lin Jun 6 '12 at 9:31

2 Answers 2

up vote 3 down vote accepted

Locally, $\mathcal A$ is a constant sheaf with stalks $\mathbb Z \oplus\mathbb Z$.
More precisely, if $U$is an open of $S^1$, $\underline {\mathbb Z}(U)$ is the product of one copy of $\mathbb Z$ for each connected component of $U$, and if $V$ is an open of $\mathbb P^1$, then $\mathcal A (V) = \underline{\mathbb Z}(f^{-1} (V))$ = the product of as many copies of $\mathbb Z$ as there are connected components of $f^{-1} (V)$, which is usually twice the number of connected components of $V$.

For example, if we take two distinct points $x_1,x_2 \in \mathbb P^1$, and take $V_i = \mathbb P^1 - \{x_i\}$, then $\mathcal A |_{V_i}$ is a constant sheaf with stalks $\mathbb Z \times\mathbb Z$. (the connected components of $f^{-1}(U)$ are always twice the connected components of $U$ when $U$ is an open subset of $V_i$)

But, as a whole, $\mathcal A$ is not the constant sheaf on $\mathbb P^1$, because $\mathcal A(\mathbb P^1) = \underline {\mathbb Z} (S^1) = \mathbb{Z}$, which is not $\mathbb Z \oplus \mathbb Z$.

We can describe it instead as a locally constant sheaf, which is constant on $V_1$ and $V_2$, and we need to describe how it glues.
Let $W = V_1 \cap V_2$.$W$ has two components, call them $W = W_a \cup W_b$. The fact that $\mathcal A$ is constant on $V_i$ means that it gives two different isomorphisms from $\mathcal A(W_a) \to \mathcal A(W_b)$, one going through $V_1$ and the other going through $V_2$.

Denote $W_a^1, W_a^2$ the connected components of $f^{-1}(W_a)$, do the same for $W_b^1, W_b^2$, and put the exponents such that $W_a^1$ and $W_b^1$ are in the same connected component of $f^{-1}(V_1)$, so that the connected components of $f^{-1}(V_1)$ correspond to $W_a^1 \cup W_b^1$ and $W_a^2 \cup W_b^2$, and those of $f^{-1}(V_2)$ correspond to $W_a^1 \cup W_b^2$ and $W_a^2 \cup W_b^1$

Then, we have restriction isomorphisms $\rho_i^a : \mathcal A(V_i) \to \mathcal A(W_a)$, which look like this : $\rho_1^a(x,y) = (x,y)$ and $\rho_2^a(x,y) = (x,y)$ ; and $\rho_i^b : \mathcal A(V_i) \to \mathcal A(W_b)$ : $\rho_1^a(x,y) = (x,y)$ and $\rho_2^a(x,y) = (y,x)$.

Then, the twisting map is the isomorphism $\tau = \rho_2^a \circ (\rho_2^b)^{-1} \circ \rho_1^b \circ (\rho_1^a)^{-1} : \mathcal A(W_a) \to \mathcal A(W_a) $. You should get that $\tau(x,y) = (y,x)$. This isomorphism describes what happens to sections on $W_a$ when you go once through $\mathbb P^1$ : the two connected components of $f^{-1}(W_a)$ get switched around, and you are basically right when you describe it in your comment.

And in fact, this map is enough to recover all the information you need to describe the sheaf $\mathcal A$ completely, so we can describe $\mathcal A$ as a locally constant sheaf whose stalks are $\mathbb Z \oplus \mathbb Z$ twisted by $\tau$ when we do one loop around $\mathbb P ^1$

share|improve this answer
    
Right, thank you! –  M.B. Jun 7 '12 at 12:33

Let's recall some general facts.

Fact 1. If $\mathscr{F}$ is a sheaf on $X$ and $f : X \to Y$ is a continuous map, then the direct image sheaf $f_* \mathscr{F}$ is the presheaf defined by $U \mapsto \mathscr{F}(f^{-1} U)$. In other words, a section of $f_* \mathscr{F}$ over $U$ is the same thing as a section of $\mathscr{F}$ over $f^{-1} U$.

Fact 2. If $\mathscr{F}$ is a sheaf on $X$, then its éspace étalé $E$ is constructed as follows: as a set, $$E = \coprod_{x \in X} \mathscr{F}_x$$ where $\mathscr{F}_x$ denotes the stalk of $\mathscr{F}$ at $x$, and the topology of $E$ is the one generated by the base $$\hat{s} = \{ (x, s_x) : x \in \operatorname{dom}(s) \}$$ for all local sections $s$ of $\mathscr{F}$. Consequently, it is possible to express $E$ as a colimit of open subsets of $X$.


Now, if $X$ is Hausdorff and $f : X \to Y$ is a finite covering map, then it is very easy to describe the stalk of $f_* \mathscr{F}$ at a point $y$: it is simply the product of the stalks $\mathscr{F}_x$ as $x$ varies over $f^{-1} \{ y \}$. (If $f$ does not have finite discrete fibres, something more complicated happens.) The way to think about this is that the germ of a section of $f_* \mathscr{F}$ at $y$ must specify the germs of a section of $\mathscr{F}$ at all the preimages of $y$.

Now, let $g : X \to X$ be an automorphism over $Y$. It is not quite true that this induces an automorphism of every sheaf on $X$. But for constant sheaves it does: if $E$ is the éspace étalé of a constant sheaf $\mathscr{F}$ and $E = X \times F$ for some discrete space $F$, then the induced automorphism on $E$ is the obvious one, i.e. $g \times F : X \times F \to X \times F$. Therefore there is an induced automorphism on the stalks of $f_* \mathscr{F}$, which just permutes the factors of the product $\prod_{x \in f^{-1} \{ y \}} \mathscr{F}_x$. (More generally, this holds when $\mathscr{F} = f^* \mathscr{G}$ for some sheaf $\mathscr{G}$ on $Y$ – then $\mathscr{F}$ inherits an action by the automorphisms of $X$.)

share|improve this answer
    
Dear Zhen, I am not convinced that the fiber $(f_*\mathcal F)_y$ is equal to the product of the fibers $\mathcal F_x$ for $x\in f^{-1}(y)$. Because you could take a collection of germs upstairs defined on small neigbourhoods of the points of the fiber of $y$ and I don't see why they should come from a section defined on the inverse image of some open neighbourhood of $y$. –  Georges Elencwajg Jun 6 '12 at 11:05
    
Ah, that's true. I'll fix that. In the case we care about, $f$ is a covering map, so everything behaves nicely. (We could just work in the topos $\textbf{Sh}(Y)$ and pretend everything is a funny kind of set!) –  Zhen Lin Jun 6 '12 at 11:19
    
I think the result you state is false even for a trivial infinite covering map. Take $Y=$ the unit disk of $\mathbb C$ and $X=Y\times \mathbb N^*$. Then $(f_*(\mathcal O_X))_0\neq \Pi _n\mathcal O_{X,(0,n)}$ because in the second member you may take germs with convergence radius tending to zero but in the first member you may only consider germs which all have a common domain of definition. Of course equality holds for finite covering maps, as in the original question. –  Georges Elencwajg Jun 6 '12 at 13:31
    
Ah, quite right, of course. Thanks! What I was thinking of was the description of the sheaf $\mathscr{G}^\mathscr{F}$ in terms of sections, which is very simple. But I guess translating what happens to the sections to what happens to the stalks is the hard part. –  Zhen Lin Jun 6 '12 at 16:45

Your Answer

 
discard

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.