Timeline for Components of a Fiber Product
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 7, 2014 at 17:24 | comment | added | Eric Larson | I don't know how to compute the dimension of $X_z$ in the case I care about; but I have some control over the differential. (My $X$ and $Y$ are certain Kontsevich spaces of stable $n$-pointed maps; and $Z$ is the Hilbert scheme of $n$ points. The differentials are partially-computable since they are induced maps on cohomology groups of the normal bundle.) | |
| Aug 7, 2014 at 15:45 | comment | added | user27920 | A weaker condition than tangential surjectivity of a map is the dimension formula $\dim(X_z) = \dim(X) - \dim(Z)$, which in this setting implies flatness (weaker than smoothness of the morphism) and so is sufficient. But presumably that formula does not hold (or may not hold) in your setting or you wouldn't have raised the question. Can you say something more about the motivation? | |
| Aug 7, 2014 at 14:35 | answer | added | Eric Larson | timeline score: 1 | |
| Aug 7, 2014 at 13:12 | history | edited | Eric Larson | CC BY-SA 3.0 |
added 59 characters in body
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| Aug 7, 2014 at 3:32 | comment | added | user27920 | Is this question being posed for schemes of finite type over an (unspecified) algebraically closed field? Characteristic 0? Please clarify the hypotheses so that a useful answer can be given. | |
| Aug 7, 2014 at 3:31 | answer | added | Dmitry Vaintrob | timeline score: 0 | |
| Aug 7, 2014 at 3:01 | comment | added | Question Mark | Smoothness is a property of a morphism, not of a scheme. | |
| Aug 7, 2014 at 1:40 | history | asked | Eric Larson | CC BY-SA 3.0 |