most recent 30 from http://mathoverflow.net 2013-05-23T03:28:54Z http://mathoverflow.net/feeds/user/3261 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16888/when-do-divisors-pull-back Randy Reddick 2010-03-02T18:18:48Z 2011-12-27T04:37:45Z

<p>If we have a nonconstant map of nonsingular curves $\varphi:X\rightarrow Y$, then Hartshorne defines a map $\varphi^* Div(Y)\rightarrow Div(X)$ using the fact that codimension one irreducibles are just points, and looking at <code>$\mathcal{O}_{Y,f(p)} \rightarrow \mathcal{O}_{X,p}$</code>. My question is if we don't have a nice map of curves, what conditions can we put on the morphism so that we may pull divisors back? Clearly it's not true in general, since we can take a constant map and then topologically the inverse image doesn't even have the right codim.</p> <p>Thinking about this in terms of Cartier divisors (and assuming the schemes are integral), it seems like we just need a way to transport functions in $K(Y)$ to functions in $K(X)$. If $\varphi$ is dominant, then we'll get such a map. Is this sufficient? Also is there something we can say when $\varphi$ is not dominant? Something like we have a way to map divisors with support on $\overline{\varphi(X)}$ to divisors on $X$?</p>

http://mathoverflow.net/questions/79998/weils-descent-criteria-for-covers-from-the-critereon-for-varieties Randy Reddick 2011-11-04T02:00:58Z 2011-11-04T08:00:46Z

<p>I have read several articles which use a version of the Weil decent criterion for covers, but the reference is always to Weil's original paper (1956 - The field of definition of a variety). I would like to know how one makes the transition. Note that when I say covers I mean a morphism between covers $(f:V\rightarrow W), (f':V'\rightarrow W)$ is a $g:V\rightarrow V'$ commuting with the covering maps.</p> <p>For reference, here is the theorem from Weil's paper, which I simplified assuming Galois:</p> <p>Let $k/k_0$ be a finite, Galois extension, $H=Gal(k/k_0)$. Let $V$ be a projective variety defined over $k$. Elements of $H$ have a natural action on varieties and morphisms defined over $k$ (for example by acting on the coefficients if we embed into projective space). Suppose for each $\sigma, \tau \in H$, we have an isomorphism $f_{\tau,\sigma}:\sigma(V)\rightarrow \tau(V)$. Then we have a model $V_0$ over $k_0$ for $V$ if the following are satisfied:</p> <p>(i) $f_{\tau,\rho}=f_{\tau,\sigma}\circ f_{\sigma,\rho}$ for all $\sigma,\tau,\rho\in H$.</p> <p>(ii) $f_{\tau \omega, \sigma \omega}=\omega(f_{\tau,\sigma})$ for all $\sigma,\tau\in H$, $\omega \in Gal(k_0^{sep}/k_0)$.</p> <p>Now I suspect that to make the translation, one takes $f:V\rightarrow W$ and looks at the graph $\Gamma_f\subseteq V\times W$. Let's suppose we can satisfy the criteria above (with the $f_{\tau,\sigma}$ morphisms of covers) for $\Gamma_f$. Then we get some $\Gamma_0$ defined over $k_0$ and an isomorphism $\varphi:\Gamma_0\times k \rightarrow \Gamma_f$. There are two points I can't resolve:</p> <p>(1) How do we know we have $\Gamma_0\subseteq V_0\times W_0$ for some models $V_0,W_0$ of $V,W$ (we can get the models of $V$ and $W$ over $k_0$ from the covering data). And further that it is the graph of a morphism $V_0\rightarrow W_0$.</p> <p>(2) How do we know that the map $\varphi$ corresponds to a morphism of covers of $W$?</p> <p>Thanks</p>

http://mathoverflow.net/questions/79764/reference-request-deformations-of-a-map-bijective-to-global-sections-of-the-pull Randy Reddick 2011-11-01T20:40:15Z 2011-11-02T11:27:44Z

<p>I have been trying to learn some deformation theory, and came across the following in a paper:</p> <p>The first order deformations of a morphism of smooth curves $f:X\rightarrow Y$ is in bijection with $H^0(X,f^*(\mathcal{T}_Y))$.</p> <p>I would like to understand a proof of this. I understand some simple facts, like $H^1(X,\mathcal{T}_X)$ being bijective with the first order deformations of $X$. The reference given was to a paper of Ravi Vakil's, but I am unfamiliar with anything but the very basics of stacks. Does anyone know of a reference for this fact that doesn't use stacks?</p>

http://mathoverflow.net/questions/67978/additive-form-of-hilbert-90-for-schemes Randy Reddick 2011-06-16T17:46:15Z 2011-06-16T17:56:31Z

<p>First, I am by no means well-versed on cohomology so I apologize if this is too elementary.</p> <p>I have been going through some basics of etale cohomology, with my ultimate goal being an understanding of some basic applications. I have gone through the Kummer and Artin-Schreier sequences, and wanted to get an idea for how these sequences can help us classify $\mathbb{Z}/n$ and $\mathbb{Z}/p$ torsors.</p> <p>I found a short exposition by Artin that said $H^1_{et}(X,\mathbb{G}_m)=Pic(X)$, and this was was labelled as Hilbert 90. This presumably has something to do with $\mathbb{G}_m$ being related to $\mathcal{O}_X^*$. Can someone tell me what the additive version of this is, ie what $H^1_{et}(X,\mathbb{G}_a)$ is equal to? Also if anyone has a reference for this being worked out that would be great as well.</p>

http://mathoverflow.net/questions/60743/tameness-for-the-galois-closure-of-a-map-of-curves Randy Reddick 2011-04-05T21:05:05Z 2011-04-07T17:41:22Z

<p>Say we are working over some $K=\overline{K}$, of characteristic $p>0$. Let $\phi: Y\rightarrow X$ be a nonconstant map of smooth projective curves. To this map we can associate a map $\psi: Z\rightarrow X$, where on the level of fields this is the Galois closure of $k(X)\subseteq k(Y)$. I would like to know about the tameness of this map.</p> <p>Let $e_P$ denote the ramification indices (with the maps understood to be either $\psi$ or $\phi$ depending on where $P$ lives). Now obviously if $p|e_P$ and if $Q$ lies above $P$, $p|e_Q$ as well, so $\psi$ has wild ramification at $Q$. I am wondering when we can ensure this map is (everywhere) tamely ramified. For instance if $d=deg(\phi) &lt; p$, then the degree of the Galois closure of $k(Y)$ over $k(X)$ has degree dividing $d!$, and hence $\psi$ remains tame.</p> <p>My question is this: Suppose we can show for each $P\in Y$ such that $e_P \geq p$ that each point above $P$ is tamely ramified. Can we conclude that $\psi$ is (everywhere) tamely ramified? It seems to me that this isn't true but I cannot produce a counterexample. It would be fortuitous if it were true, however. Any help is greatly appreciated.</p>

http://mathoverflow.net/questions/21781/oriention-reversing-diffeomorphisms-of-a-manifold Randy Reddick 2010-04-18T22:06:10Z 2010-04-20T19:28:43Z

<p>I am trying to figure out when a closed, oriented manifold admits an orientation reversing diffeomorphism. My naive argument that the orientation cover should allow you to switch orientations is apparently wrong, since not every manifold admits such a diffeomorphism.</p> <p>Can anyone give me some criteria for when such a morphism should exist, or why some of the standard counterexamples (such as $\mathbb{P}^{2n}$) fail to admit one?</p> <p>Thanks</p>

http://mathoverflow.net/questions/13410/degree-of-divisors-and-degrees-of-the-corresponding-maps-to-projective-space Randy Reddick 2010-01-29T20:10:33Z 2010-01-30T17:53:31Z

<p>Suppose I have a divisor $D$ on a curve $X$ (Hartshorne curve - smooth, projective, dimension one over an algebraically closed $k$). If the complete linear system $|D|$ is basepoint free then I get a map $\varphi:X\rightarrow\mathbb{P}^n_k$. My question is, say for simplicity our map ends up being to $\mathbb{P}^1_k$, what if anything is the relationship between the degree of the divisor $D$ and the degree of the morphism $\varphi$?</p> <p>It seems for many cases that we have $deg(\varphi)=deg(K)$, however I can't find anywhere that proves that this is always the case.</p> <p>Thanks</p>

http://mathoverflow.net/questions/79764/reference-request-deformations-of-a-map-bijective-to-global-sections-of-the-pull/79767#79767 Randy Reddick 2011-11-02T15:42:17Z 2011-11-02T15:42:17Z

This had the answer I was looking for, and a lot of other things of use to me. Sorry I can't select more than one answer, since a lot of the information provided was helpful too.

http://mathoverflow.net/questions/67978/additive-form-of-hilbert-90-for-schemes/67980#67980 Randy Reddick 2011-06-16T18:16:27Z 2011-06-16T18:16:27Z

Ah thank you. I did not know about that comparison theorem for coherent sheaves.

http://mathoverflow.net/questions/60743/tameness-for-the-galois-closure-of-a-map-of-curves/60807#60807 Randy Reddick 2011-04-07T01:22:48Z 2011-04-07T01:22:48Z

Oh yes sorry Lubin I misread what you wrote. I thought it read &quot;if the Galois closure of $L\supseteq K$ is tame over $K$. Of course you are right that as you wrote it it is false.

http://mathoverflow.net/questions/60743/tameness-for-the-galois-closure-of-a-map-of-curves/60807#60807 Randy Reddick 2011-04-06T20:26:15Z 2011-04-06T20:26:15Z

I was asking what Holger has posted. The converse seems clear to me. Just a silly error though--you meant iv) rather than v), but yes this is exactly the sort of result I was hoping for.

http://mathoverflow.net/questions/21781/oriention-reversing-diffeomorphisms-of-a-manifold Randy Reddick 2010-04-26T16:46:47Z 2010-04-26T16:46:47Z

I accidentally wrote &quot;orientational&quot; instead of orientation in the title, so I apologize for that. I'm not sure how you deduce from that and that I used the common abbreviation $\mathbb{P}^n$ for complex projective space that I had no idea what I was talking about / that this is a homework question. As I said, I'm not a topologist and since every complex manifold is orientable it seemed natural to ask when you can reverse the orientation. I know we don't want to answer calculus questions here but it seems rather silly that I can't ask questions outside my area of specialty.

http://mathoverflow.net/questions/21781/oriention-reversing-diffeomorphisms-of-a-manifold/21799#21799 Randy Reddick 2010-04-19T02:30:01Z 2010-04-19T02:30:01Z

Yes thank you. My definition of orientation was in terms of homology so I was missing out on the ring structure.

http://mathoverflow.net/questions/21781/oriention-reversing-diffeomorphisms-of-a-manifold Randy Reddick 2010-04-18T22:45:17Z 2010-04-18T22:45:17Z

Yes I mean $\mathbb{C}\mathbb{P}^{2n}$. Sorry if this question is actually easy, but I am not a differential geometer so I'm unsure of how to approach this. I checked google and noticed a few theses on when manifolds admit such a morphism so I assumed it wasn't completely trivial.

http://mathoverflow.net/questions/13410/degree-of-divisors-and-degrees-of-the-corresponding-maps-to-projective-space/13489#13489 Randy Reddick 2010-01-31T22:53:38Z 2010-01-31T22:53:38Z

Thanks for this!

http://mathoverflow.net/questions/13410/degree-of-divisors-and-degrees-of-the-corresponding-maps-to-projective-space Randy Reddick 2010-01-31T22:49:35Z 2010-01-31T22:49:35Z

Yes I did intend to mean the degree of the map rather than the degree of the image.