FFXIAH Linkshell Ni
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FFXIAH Linkshell Ni
Dang, accidentally left some valuables behind on a mule I was too quick to delete :<
Morning
tomorrow is friday....the 13th
Carbuncle.Lynxblade said: tomorrow is friday....the 13th Cerberus.Kalyna said: Carbuncle.Lynxblade said: tomorrow is friday....the 13th Don't spoil what? that it's friday? /doze moar stuff on my mule sold :o still not really raking in alot of gil but meh, it'll do me for now at least. Leviathan.Catnipthief said: Cerberus.Kalyna said: Carbuncle.Lynxblade said: tomorrow is friday....the 13th Don't spoil what? that it's friday? /doze moar stuff on my mule sold :o still not really raking in alot of gil but meh, it'll do me for now at least. some guy gave me a +1 wand and I decided to sell it and use that money for better stuff like a new damn body piece and a belt. Asurans aren't as bad as I thought =D Cerberus.Kalyna said: Leviathan.Catnipthief said: Cerberus.Kalyna said: Carbuncle.Lynxblade said: tomorrow is friday....the 13th Don't spoil what? that it's friday? /doze moar stuff on my mule sold :o still not really raking in alot of gil but meh, it'll do me for now at least. some guy gave me a +1 wand and I decided to sell it and use that money for better stuff like a new damn body piece and a belt. Asurans aren't as bad as I thought =D I maaaadeeeee 31k lol. Too bad you didn't come to leviathan ... I could've given you a TON of junk to sell >.> Leviathan.Catnipthief said: Cerberus.Kalyna said: Leviathan.Catnipthief said: Cerberus.Kalyna said: Carbuncle.Lynxblade said: tomorrow is friday....the 13th Don't spoil what? that it's friday? /doze moar stuff on my mule sold :o still not really raking in alot of gil but meh, it'll do me for now at least. some guy gave me a +1 wand and I decided to sell it and use that money for better stuff like a new damn body piece and a belt. Asurans aren't as bad as I thought =D I maaaadeeeee 31k lol. Too bad you didn't come to leviathan ... I could've given you a TON of junk to sell >.> Morning everyone. I trust you remembered to shield yourself from the oncoming barrage of attacks while I was gone?
Asura.Dameshi said: Morning everyone. I trust you remembered to shield yourself from the oncoming barrage of attacks while I was gone? Ramuh.Rowland said: Asura.Dameshi said: Morning everyone. I trust you remembered to shield yourself from the oncoming barrage of attacks while I was gone? Asura.Dameshi said: Ramuh.Rowland said: Asura.Dameshi said: Morning everyone. I trust you remembered to shield yourself from the oncoming barrage of attacks while I was gone? H^k(X, \mathbf{C}) = \bigoplus_{p+q=k} H^{p,q}(X),\, where Hp,q(X) is the subgroup of cohomology classes which are represented by harmonic forms of type (p, q). That is, these are the cohomology classes represented by differential forms which, in some choice of local coordinates z_1, \ldots, z_n, can be written as a harmonic function times dz_{i_1} \wedge \cdots \wedge dz_{i_p} \wedge d\bar z_{j_1} \wedge \cdots \wedge d\bar z_{j_q}. (See Hodge theory for more details.) Taking wedge products of these harmonic representatives corresponds to the cup product in cohomology, so the cup product is compatible with the Hodge decomposition: \cup : H^{p,q}(X) \times H^{p',q'}(X) \rightarrow H^{p+p',q+q'}(X).\, Since X is a compact oriented manifold, X has a fundamental class. Let Z be a complex submanifold of X of dimension k, and let i : Z → X be the inclusion map. Choose a differential form α of type (p, q). We can integrate α over Z: \int_Z i^*\alpha.\!\, To evaluate this integral, choose a point of Z and call it 0. Around 0, we can choose local coordinates z_1,\ldots,z_n on X such that Z is just z_{k+1} = \cdots = z_n = 0. If p > k, then α must contain some dzi where zi pulls back to zero on Z. The same is true if q > k. Consequently, this integral is zero if (p, q) ≠(k, k). More abstractly, the integral can be written as the cap product of the homology class of Z and the cohomology class represented by α. By Poincaré duality, the homology class of Z is dual to a cohomology class which we will call [Z], and the cap product can be computed by taking the cup product of [Z] and α and capping with the fundamental class of X. Because [Z] is a cohomology class, it has a Hodge decomposition. By the computation we did above, if we cup this class with any class of type (p, q) ≠(k, k), then we get zero. Because H^{2n}(X, \mathbf{C}) = H^{n,n}(X), we conclude that [Z] must lie in H^{n-k,n-k}(X, \mathbf{C}). Loosely speaking, the Hodge conjecture asks: Which cohomology classes in Hk,k(X) come from complex subvarieties Z? 42
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-.- I just broke a lvl 7 craft. /rage
And mornin' Asura.Alymorel said: -.- I just broke a lvl 7 craft. /rage And mornin' and /wave How does one break animal glue...?
AND AGAIN?! /RAGE! Ramuh.Rowland said: Asura.Dameshi said: Ramuh.Rowland said: Asura.Dameshi said: Morning everyone. I trust you remembered to shield yourself from the oncoming barrage of attacks while I was gone? /yawn morning everyone :)
lol Row
/pokes Angels Asura.Alymorel said: How does one break animal glue...? AND AGAIN?! /RAGE! Asura.Dameshi said: Ramuh.Rowland said: Asura.Dameshi said: Ramuh.Rowland said: Asura.Dameshi said: Morning everyone. I trust you remembered to shield yourself from the oncoming barrage of attacks while I was gone? /sigh no coffee this am i feel strangely sad :(
Day 3 no coffee and I thought I would be raging by now :I Edit: and honey bunches of oats with almonds is the best cereal ever /wave people who entered
I have many fav cereals lol
Cookie Crisp Apple Jacks Fruit Loops Life Trix Fruity Pebbles Chocolate Pebbles Resees' Puffs Pops Cerberus.Eliane said: Morning o.o |
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