# Why the Skyscraper Sheaf is Named as Such, Intuition and Interpretation

I have reviewed all the other questions on SE regarding the skyscraper sheaf and I believe my question is not a duplicate. I am also aware that the questions I pose below may be considered ill defined since there is not 'one right answer' - if my question gets closed, so be it - but I just started learning sheaves and they are complicated. With a complicated object like a sheaf, I like to discuss these sorts of things early on with people who have more maturity on the topic to make sure I am interpreting the object in a good way. Please at least skim my high lighted questions to avoid just answering the title, which does not encompass all of my related questions.

I have mostly been working with sheaves of abelian groups, but answers involving sheaves that take values in any concrete category should be understandable to me. Probably best to avoid examples and answers involving abstract sheaves that take values in complicated categories like Cat.

First of all, there are two definitions of a sheaf. The more modern/common is the functorial definition given on wikipedia, and then there is the étalé space definition given by Serre on the first page of FAC. I give this definition in the questions I asked here, and here. The definition Serre gives is the one I have been using, but the discussion in the comments on the second link briefly discusses why they are 'equivalent'.

Some of the explicit sheaves I have encountered that have names seem to make more sense using one definition than the other. For example, the constant sheaf makes more sense to me using the étalé definition, and the skyscraper sheaf, so far, seems to make more sense using the functorial definition. The skyscraper sheaf is the one I would like to discuss.

The standard definition given for the skyscraper sheaf (using a sheaf of sets for simplicity), such as in Vakil's notes, is as follows.

For a topological space $X$, and a fixed point $x \in X$, and some set $S$, for each open set $U \subset X$, we assign the set of sections

$$F(U) := \begin{cases} S, &\text{ if } x \in U,\\ \{e\}, &\text{ if } x \notin U. \end{cases}$$

For sheaves of algebraic objects replace $\{e\}$ by the appropriate terminal object in the category of choice.

One can check that the stalks of this sheaf are $$\mathscr{F}_y := \begin{cases} S, &\text{ if } y \in \overline{\{x\}},\\ \{e\}, &\text{ if } y \notin \overline{ \{x\}}. \end{cases}$$

Now we are prepared for questions.

1. Do some skyscraper sheaves not live up to the name?

How I think of a sheaf intuitively, or decide what a sheaf should look like, is pretty much determined by its stalks. This makes sense since I use the étalé definition where the actual object that is called the sheaf is the disjoint union of stalks, as opposed to being the functor that determines such data. Now if $X$ is T1, or more generally $x$ is a closed point of $X$, I see that if $S \neq \{b\}$ we get one non-zero stalk $\mathscr{F}_x$ and all other stalks are $\{e\}$. However, if $S = \{b\}$, or if we are in the algebro-geometric case where the space $X$ is usually not T1, $x$ could be a generic point so that its closure was the whole space. Then no matter what $S$ was, every stalk would be $S$. These are two different ways that a, by definition, skyscraper sheaf could turn out to be a constant sheaf. Is there a way I am missing to still interpret these sheaves as skyscrapers, or are they just perverse examples that don't live up to the name?

In my next question, for simplicity, I will speak as though $X$ is T1 so that the skyscraper sheaf has exactly one non zero stalk (so $S \neq \{b\}$ also).

2. Is a skyscraper sheaf actually 'tall', or is it just tall compared to everything else around it?

When I first heard the name 'skyscraper sheaf' I automatically thought I was about to encounter a massive object like the monster group. However, since I 'picture' a sheaf as a collection of stalks, it doesn't seem that it's one non-zero stalk at $x$ needs to be necessarily 'tall' or 'big'. Since the stalk is a direct limit, and direct limits tend to make bigger things out of smaller things, I first thought maybe I would be getting lots of copies of $S$ for the stalk at $x$, perhaps one for each open neighborhood, which will (maybe) make that stalk very tall. But I quickly thought through this and realized the direct system determining that direct limit is just the identity maps and the same object over and over, yielding the same object for the direct limit, so this can't be why the one non-zero stalk would be a 'skyscraper'. I tend to think of the 'size' or 'height' of a stalk as its size as a set, or group, etc. So if I chose $S = \{a,b\}$ for my one non-zero stalk, by definition I have a sky scraper sheaf, but really to me this is more like a mole hill sheaf because the stalk corresponding to $S$ would not be very tall, even compared to all the 0 stalks around it. But maybe if I quit thinking about a sheafs physical interpretations as only being determined by stalks, I could also consider stalks and sections. Then, even if I choose $S = \{a,b\}$, I have a copy of $S$ at $x$, then as I transition from local to global, taking nested open subsets containing $x$, I get a copy of $S$ for each one. Maybe this is how the sheaf is 'tall' around one point? Then if I do this process anywhere else, my sequence of sections would be 0 until the open subsets were big enough that they also included $x$ (yes, I am not abstractly thinking of the topology, and am thinking of it as being 'like' a euclidean topology or a Hausdorff space, but oh well).

3. Is it possible to define a skyscraper sheaf and end up with something more like an “NYC Sheaf"? As I mentioned above, if $x$ is a non closed point, we could get more non-zero stalks on the closure of $x$. Above I mentioned an extreme case from the Zariski topology where a point may have the entire space as its closure. Are there any naturally arising cases or examples where the closure of a single point is not the whole space? In this case we should have non-zero stalks on some closed neighborhood around $x,$ and then 0 everywhere else. I imagine if I stood back and looked at this sheaf it would look kind of like this arial photograph of NYC where the angle and distance of the photograph make it look like a ton of action packed into one (seemingly) small area, then flat everywhere around it for as far as you can see. This sheaf should be called the NYC Sheaf!

Lastly, it is a perfectly acceptable answer that I will gladly upvote (if it is true) to say "It seems your understanding of a sheaf, and a skyscraper sheaf, is adequate and you are horribly overthinking the name. Get back to proving theorems.

• This question should help you come up with some "NYC Sheaf"-examples, I think. math.stackexchange.com/questions/228685/…
– RKD
May 12 '18 at 22:46
• Thank's Ravi! Are you familiar with Vakil's notes that I mention in my post? May 12 '18 at 22:57
• Yes, I am. The way I think of a skyscraper sheaf is take a one point space $\{x\}$. To give a sheaf on this is just to give the global sections. If you now have any inclusion $\{x\}\hookrightarrow X$ of topological spaces, then your skyscraper sheaves at $x$ (for this inclusion) are all just pushforwards of sheaves on $\{x\}$.
– RKD
May 13 '18 at 0:24

1) You should remember that the nickname "skyscraper sheaf" is meant to be evocative of a certain concept - it's true that there are situations (such as the one you point out with a generic point) where this intuition is a little less clear, but there are ways you can reinterpret this to assuage your concerns. Think about buildings that have more than one address - the analogy here is that the plot of land the skyscraper sits on is the generic point and each individual address of the building is a closed point in the closure of the the plot of land.

It seems you are comfortable and familiar with the definition of a skyscraper sheaf as $(i_x)_*A$ for $i: \{x\} \to X$ the inclusion and $A$ a sheaf, so I would not worry that much about the pathological cases until you have to.

2) I would point out that this question makes perfect sense for actual skyscrapers, too: if you shrunk the earth down to the size of a marble, your fingers would not be able to tell it apart from a perfect sphere. In particular, you definitely couldn't tell apart the places where there are skyscrapers and those places where there aren't any skyscrapers.

I would stay with the stalk interpretation here - skyscraper sheaves are a nice concept because they deal with the correct sheafy way to say "stick a copy of $A$ at $x$". The analogy works in terms of stalks.

3) Certainly, this is possible. Easy exercise: if $\mathcal{A}=(i_x)_*A$ is a skyscraper sheaf for the object $A$ at the point $x\in X$, $\mathcal{A}_y$ is the trivial object for $y\notin \overline{\{x\}}$ and $A$ for $y\in \overline{\{x\}}$.

If you have a topological space where points can have nontrivial closures, then you may find skyscraper sheaves that are supported on closed subsets obtained as the closure of such points. Often these closed subsets are rather large, so if your "NYC" will probably need to be large in the same sense (don't think about things like a closed ball in the usual topology, for instance).

Finally, I think I agree a bit with your bolded remark at the end of your question: everything is more or less fine here, and you've worked yourself up a bit. It's fine and reasonable to do this, because sometimes doing this produces new insights, but here, things are pretty much fine.

• Thanks for your answer :) only thing I’ll mention is your exercise is something I mention in my original post, my question was more about the topological space you describe, a space where points have non trivial closures. May 13 '18 at 4:46
• It just hard for me sometimes because I’m the only grad student at my school interested in algebraic geometry and I have always benefited a lot from informal discussions of mathematics with my peers.. I don’t have anyone to talk to anymore so I have to turn to SE to run my interpretations of things by others May 13 '18 at 4:48