Talk:Stalk (sheaf)

I saw some mention of ``inverse limit in this page which might be meaning to say ``direct limit instead. But I am just learning this stuff so I won't make the changes.164.67.235.219 (talk) 01:59, 1 April 2009 (UTC)

The skyscraper sheaf associated to a point x and a group or ring G has the stalks 0 off x

Does the definition (Where is it anyway? All I can find is http://en.wikipedia.org/wiki/Skyscraper_sheaf#Examples) of skyscraper sheaves require x to be closed?


 * The point can be non-closed. This is used in algebraic geometry, for example. The statement concerning the stalks of the skyscraper sheaf are correct on Kolmogorov spaces (which includes schemes. Otherwise one has to assume that the point is closed to get the desired behaviour. The place you mention is the time being the only definition. Feel free to create a new article on the subject if there is some more material... Jakob.scholbach (talk) 10:16, 28 February 2008 (UTC)


 * But what if you take the skyscraper sheaf in x over the space X={x,y} with x being the only open point? Then the stalk in y is G although X is T0, or am I making a mistake? 129.69.181.4 (talk) 13:10, 29 February 2008 (UTC)


 * Yes, good point. Originally, I somehow got T0 and T1 confused.  I'm just going to adjust the text of the section again to require that the point be closed in order for the stalk to have the stated property.  Silly rabbit (talk) 13:35, 29 February 2008 (UTC)


 * Thanks for clarifying. When x is closed we don't even need the T0 assumption, do we? 129.69.181.4 (talk) 14:33, 29 February 2008 (UTC)


 * It would seem not. Good point.  Let's see what Jacob says.  Silly rabbit (talk) 15:12, 29 February 2008 (UTC)

Sorry, I messed up T0 and T1, too. If the space is T1 then the stalks in all points have the stated property: lets calculate the stalk in y of a skyscraper sheaf concentrated in x. If y equals x then every open neighbourhood of y contains x, so the stalk is the given group. If x is not equal to y then there is by definition of T1 an open neighbourhood of y which does not contain x. Therefore the limit in this case is 0. (T0 may only give a nbh. of x not meeting y). In closed points the property always holds, of course, because the complement of this point shows up in the limit, hence kills this limit. Jakob.scholbach (talk) 12:11, 2 March 2008 (UTC)

I extended the statement in the text to T1 spaces. (The example above shows is T0, but not T1; for y does not have a nbh. not meeting x). OK? Jakob.scholbach (talk) 12:18, 2 March 2008 (UTC)


 * I've changed the wording again to indicate that every point of a T1 space is closed. I hope you don't mind.  Also, the text seems to give the impression that the skyscraper sheaf construction requires the point to be closed.  This assumption certainly isn't necessary. Silly rabbit (talk) 13:57, 2 March 2008 (UTC)


 * Thank you. In Monopoly, I would probably have to go back to the starting point without getting $4.000 ... Jakob.scholbach (talk) 17:29, 2 March 2008 (UTC)


 * Heh. Nobody's perfect.  Cheers, Silly rabbit (talk) 18:30, 2 March 2008 (UTC)