What can be the analogue of Frobenius in complex geometry? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:25:51Z http://mathoverflow.net/feeds/question/31608 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31608/what-can-be-the-analogue-of-frobenius-in-complex-geometry What can be the analogue of Frobenius in complex geometry? Bo Peng 2010-07-12T19:48:33Z 2010-07-14T19:49:33Z <p>The Frobenius map is usually only defined in char p, or when we are working with $\mathbb{Q}_p$, etc.</p> <p>What can be the analogue of Frobenius in complex geometry?</p> <p>As a vague guess, about which I probably heard some years ago somewhere, is it somehow "related" to the Laplacian or Dirac operators? Do we need some very heavy machineries for a better analogue? Will $\mathbb{F}_1$ say something about it?</p> <p>Thank you very much.</p> http://mathoverflow.net/questions/31608/what-can-be-the-analogue-of-frobenius-in-complex-geometry/31614#31614 Answer by Jack Huizenga for What can be the analogue of Frobenius in complex geometry? Jack Huizenga 2010-07-12T20:36:15Z 2010-07-12T20:36:15Z <p>While this is really more of a non-answer to your question (I'm not really fond of such philosophical generalities), I think part of what sets complex geometry apart is the <em>lack</em> of a Frobenius.</p> <p>If one wants to make use of the Frobenius in studying complex geometry, it is still possible to do so by reducing to characteristic $p$. For instance, the simplest proof of the next result uses this method (the following discussion is a summary from the beautiful book "Higher-Dimensional Algebraic Geometry" by Debarre).</p> <p><strong>Theorem.</strong> If $X$ is a smooth Fano variety (i.e. $-K_X$ is ample) over $\mathbb{C}$, then through any point of $X$ there is a rational curve.</p> <p>It is relatively easy to prove this result in characteristic $p$ by making use of the Frobenius. For an outline, first fix a point $x\in X$. If you can produce a map $f:C\to X$ from a pointed curve $(C,c)$ taking $c$ to $x$ in such a way that this map can be deformed while holding $f(c)=x$ constant, then by Mori's bend-and-break lemma it is possible to find a rational curve through $c$. Now we can estimate the dimension of the space of such deformations, and it follows that if $$-K_X \cdot f_\ast C - n g(C) \geq 1 \qquad (n=\dim X)$$ that we will be done.</p> <p>Now the question becomes: how can we produce a curve $C$ (or rather a map $f$) such that this inequality holds? If you play around in complex geometry, you will find that standard constructions, such as taking branched covers of a given curve, won't get you anywhere: if you manage to increase the $(-K_X)$-degree $-K_X\cdot f_\ast C$, then this will lead to an increase in the genus $g(C)$ as well, and you won't have made any progress towards proving the inequality.</p> <p>However, in characteristic $p$, it is possible to increase $-K_X\cdot f_\ast C$ while keeping the genus of $C$ fixed! If we start from any random map $f:C\to X$ with $f(c)=x$, we can precompose it as many times as we wish by the Frobenius $Frob:C\to C.$ This has the effect of multiplying $-K_X\cdot f_\ast C$ by $p$, while the genus $g(C)$ remains constant since the domain curve is the same (although different as a $k$-scheme). Then since $-K_X$ was ample, it follows that if we compose with the Frobenius enough times our inequality will be satisfied.</p> <p>Proving that the characteristic $p$ case implies the characteristic $0$ case is a relatively formal argument; I'll refer the reader to Debarre to see it.</p> http://mathoverflow.net/questions/31608/what-can-be-the-analogue-of-frobenius-in-complex-geometry/31616#31616 Answer by Stopple for What can be the analogue of Frobenius in complex geometry? Stopple 2010-07-12T21:07:06Z 2010-07-12T21:07:06Z <p>An observation: the existence of the Frobenius automorphism in characteristic p was crucial to proving the Riemann Hypothesis in that case. The whole program of Alain Connes to prove RH in general is to come up with an appropriate definition of Frobenius in characteristic 0; then the characteristic p case provides a roadmap to follow. He now thinks he has such a definition - but it's way over my head to say what it is...</p> <p>My point is that coming up with the correct definition is thought to be a significant step towards proving the Riemann Hypothesis - it's going to be hard!</p> http://mathoverflow.net/questions/31608/what-can-be-the-analogue-of-frobenius-in-complex-geometry/31653#31653 Answer by Victor Protsak for What can be the analogue of Frobenius in complex geometry? Victor Protsak 2010-07-13T03:22:51Z 2010-07-13T03:22:51Z <p>From a purely formal point of view, the analogues of "Frobenius symmetries" given by the action of <code>$Gal(\mathbb{\overline{F}}_q/\mathbb{F}_q)$</code> on various cohomology spaces are "Hodge symmetries" given by the action of $Gal(\mathbb{C}/\mathbb{R})$, and more precisely, (mixed) Hodge structure on the cohomology of a (possibly singular) complex variety. This remark is contained in Gelfand and Manin's "Homological algebra" volume of the Russian (Yellow Springer) Mathematics Encyclopaedia written more than 20 years ago.</p> http://mathoverflow.net/questions/31608/what-can-be-the-analogue-of-frobenius-in-complex-geometry/31698#31698 Answer by Heinrich Hartmann for What can be the analogue of Frobenius in complex geometry? Heinrich Hartmann 2010-07-13T12:55:10Z 2010-07-13T12:55:10Z <p>I think of the Frobenius as an anlogue of a small loop around a point $p$ in the curve $Spec(\mathbb{Z})$.</p> <p>This is for the following reason. As $\mathbb{Z}$ is a regular ring of Krull dimension one, the complex analogue is an (affine) curve $C/\mathbb{C}$.</p> <p>A prime $p$ corresponds to a point $P \in C$. The Forbenius automorphism arises as generator of the Galois group of finite extension of the quotient field $k(p)=\mathbb{Z}/p=\mathbb{F}_p$. In the complex case the qoutient field $k(P)=\mathbb{C}$ is algebraically closed, there is no direct analogue of the Frobenius as action on $k(P)$.</p> <p>Nevertheless we can do something different. Lets consider the completion $\mathbb{Z}_p$ of $\mathbb{Z}$ at $p$. It corresponds to a small disc round $P$ in the complex setting. The qoutient field $\mathbb{Q}_p$ corresponds to the pointed disc. Let $K/\mathbb{Q}_p$ be a finite unramified field-extension of $K/\mathbb{Q}_p$ then $Gal(K/\mathbb{Q}_p)$ is cyclic an generated by an automorphism $F$.</p> <p>In the complex setting this is the familiar statement that a (connected) unramified cover of a pointed disc has cyclic Galois group. Moreover the generator lifts the action of a small loop on a nearby fiber to an automorphism of the cover.</p> <p>The relation with the Frobenius automorphism arises as follows. There is a canonical valuation ring $R \subset K$ with maximal ideal $M \subset R$ such that $R/M = \mathbb{F}_q$ for $q=p^d$ where $d=[K:\mathbb{Q}_p]$. The automorphism $F$ on $K$ induces the Frobenius on $\mathbb{F}_q$.</p> <p>Disclaimer: Let me appologize for not being an arithmetic geometer. This philosophy is probably well known to those who work in this field. As I cannot remember where I learned this from, I would also appreciate references to more thorough treatments of this ideas.</p> http://mathoverflow.net/questions/31608/what-can-be-the-analogue-of-frobenius-in-complex-geometry/31896#31896 Answer by Donu Arapura for What can be the analogue of Frobenius in complex geometry? Donu Arapura 2010-07-14T19:29:25Z 2010-07-14T19:49:33Z <p>I've been hesitating about whether or not to answer this question. Not because it's uninteresting, but because I think it's <strong>too</strong> interesting. I'm glad you didn't ask what the influence of Frobenius is on characteristic $0$ geometry, because it's enormous and would be too much to get into. Since some of these interactions have been discussed in other people's answers and elsewhere, I'll just make a list. These include: Mori's "bend and break", the theory of weights, the decomposition theorem for perverse sheaves, Frobenius splitting, Deligne-Illusie splitting, tight closure... </p> <p>I'd rather get off the beaten track a bit, and discuss some answers which are more obscure, but hopefully interesting. Actually, these are closer to answers to the question you did ask.</p> <ol> <li><p>Some special classes of complex varieties (e.g. toric varieties, modular curves, CM elliptic curves) carry endomorphisms or self correspondences which are in fact lifts of the Frobenius. For $\mathbb{P}^n$, one can take $[x_0,\ldots x_n]\mapsto [x_0^p,\ldots x_n^p]$. Unfortunately, such lifts usually don't exist. (I'd be happy with more examples, if people know of any.)</p></li> <li><p>On K-theory the Adams operations $\psi^p:K^0(X)\to K^0(X)$ behave like the action of Frobenius. Defining these is a long story, suffice to say that for a line bundle $\psi^p(L) = L^p$, exactly like the Frobenius.</p></li> <li><p>This is a bit of a crazy thing to do, but you can take the ultraproduct of the mod $p$ reductions of your complex variety and you will get a nonnoetherian (!) scheme dominating $X$ with a Frobenius like endomorphism.</p></li> </ol> <p>As I said in my comment above, there are many answers. The best one really depends on what you want to do.</p>