User david corwin - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T17:23:26Z http://mathoverflow.net/feeds/user/1355 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51204/is-there-a-crash-course-book-on-abelian-varieties-e-g-an-introduction-for-ph/131561#131561 Answer by David Corwin for Is there a "crash-course" book on Abelian varieties (e.g., an introduction for physicists)? David Corwin 2013-05-23T05:09:49Z 2013-05-23T05:09:49Z <p>There are also some <a href="http://dept.math.lsa.umich.edu/~idolga/phys2.pdf" rel="nofollow">notes</a> on physics by Dolgachev that mention abelian varieties and theta functions.</p> http://mathoverflow.net/questions/128657/finite-flat-group-schemes-for-modular-forms-of-higher-weight Finite Flat Group Schemes for Modular Forms of Higher Weight David Corwin 2013-04-24T20:33:56Z 2013-04-24T21:06:19Z <p>Let $f$ be a newform (normalized, cuspidal) of weight $k \ge 2$. Then for a prime $\ell$ there is an $\ell$-adic Galois representation associated to $f$. If $k=2$, this comes from an abelian variety, hence the local representation at $\ell \nmid N$ arises from an $\ell$-divisible group, or finite flat group scheme of order $\ell^2$ if we look at the mod $\ell$ reduction. My understanding is that this is true in the case $k>2$, but I'm having trouble seeing why or finding a reference. I understand the construction of the representation using $H^1_{\acute{e}t}(X,\mathrm{Sym}^{k-2}(R^1_{\acute{e}t}f_* \underline{\mathbb{Z}_\ell}))$.</p> <p>The reason for this is to use Raynaud's results to get a good local picture of the mod $\ell$ representation at primes dividing $\ell$.</p> <p>I imagine one could just use the crystalline property in p-adic Hodge theory, but I'd like to see it in the more elementary way (and I understand that better).</p> http://mathoverflow.net/questions/33807/angle-maximizing-the-distance-of-a-projectile Angle Maximizing the Distance of a Projectile David Corwin 2010-07-29T16:49:06Z 2013-04-22T14:22:22Z <p>It is well-known that to maximize the horizontal distance traveled by a projectile fired from the ground at a given speed, one should fire it at a $45^\circ$ angle. What's less-known, though not too difficult to prove with some manipulation of derivatives, is that if one is on an incline of angle $\phi$ below the horizontal, then to maximize horizontal distance traveled, one should fire the projectile at an angle of $\frac{\pi}{4}-\frac{\phi}{2}$ above the horizontal.</p> <p>To see this, suppose we fire it at an angle of $\theta$ at speed $v_0$ with gravitational constant $g$. Then the equation of the path is $(v_0 t \cos{\theta}, -gt^2+v_0 t \sin{\theta})$, so we are looking for the $t$ at which $v_0 t \cos{\theta} \sin{\phi}-gt^2 \cos{\phi}+v_0 t \sin{\theta} \cos{\phi} = 0$, or $t = \frac{v_0(\cos{\theta}\sin{\phi}+\sin{\theta}\cos{\phi})}{g} = \frac{v_0}{g} \sin(\theta+\phi)$. Then we wish to maximize $\cos{\theta} t$ at this point, or equivalently $\cos{\theta} \sin{(\theta+\phi)}$. Taking the derivative with respect to $\theta$, we find $-\sin{\theta} \sin{\theta+\phi}+\cos{\theta} \cos(\theta+\phi) = \cos{(2\theta+\phi)} = 0$. Thus $2\theta+\phi = \frac{\pi}{2}$, giving us our answer.</p> <p>Since this formula looks so nice, I'm wondering whether there is a nicer proof of this fact, possibly using symmetry and/or with more physical intuition. In particular, it might involve some kind of rotation (maybe by $\frac{\phi}{2}$?).</p> http://mathoverflow.net/questions/121769/etale-homology-via-etale-cosheaves Etale homology via étale cosheaves David Corwin 2013-02-14T03:26:16Z 2013-04-10T06:54:48Z <p>Can one develop a theory of étale homology via étale cosheaves? The hope is that this would, for example, return the Tate module (and not its dual) for an elliptic curve, and it would return group homology for $\rm Spec$ of a field.</p> <p>The first step in this direction is to notice that you can define an étale cosheaf on $\rm Spec$ of a field, in which the sections are coinvariants, rather than invariants, of the associated Galois module. In this case I believe we would recover group (Galois) homology.</p> <p>(See <a href="http://mathoverflow.net/questions/99969/cosheaf-homology-and-a-theorem-of-beilinson-in-a-paper-on-mixed-tate-motives" rel="nofollow">http://mathoverflow.net/questions/99969/cosheaf-homology-and-a-theorem-of-beilinson-in-a-paper-on-mixed-tate-motives</a> if you don't know what cosheaves are.)</p> <p>I'm not expecting this to prove anything new, but it would make certain formulations nicer (for some definition of "nice"). E.g., I'm hoping we would get theorems that look similar to the relations between singular homology and cohomology, and we would get a comparison with singular homology for finite coefficients.</p> <p>I feel like this should work, though I'm not entirely sure about the ability to cosheafify.</p> http://mathoverflow.net/questions/126123/how-did-takagi-prove-kroneckers-jugendtraum-for-qi How did Takagi prove Kronecker's Jugendtraum for Q(i)? David Corwin 2013-03-31T18:58:26Z 2013-03-31T20:34:25Z <p>In Noah Snyder's historical undergraduate thesis on Artin L-Functions, it mentions that Takagi proved Kronecker's Jugendtraum in the case of Q(i) in his doctoral thesis. Since I don't know how to get access to Takagi's thesis, does anyone have any idea how he did this?</p> <p>I know how to prove complex multiplication (e.g. as in Silverman's book) by first assuming class field theory. But Takagi was working before class field theory had been demonstrated. So how did he do it?</p> <p>I understand how one can directly prove the reciprocity law directly for extensions generated by torsion points on elliptic curve. The nontrivial part is this: how does one prove that an <em>arbitrary</em> abelian extension of an imaginary quadratic field arises in such a manner?</p> <p>Note that this question has everything to do with my <a href="http://mathoverflow.net/questions/100276/can-one-prove-complex-multiplication-without-assuming-cft" rel="nofollow">previous question</a>.</p> http://mathoverflow.net/questions/124754/what-is-the-difference-between-an-automorphic-form-and-a-modular-form What is the difference between an automorphic form and a modular form? David Corwin 2013-03-17T05:42:38Z 2013-03-17T20:37:32Z <p>This is more of a question about terminology than about math.</p> <p>The term "automorphic form" is clearly a generalization of the term "modular form." What is not clear is exactly which generalization it is. Many sources use the term in different ways.</p> <p>Any classical holomorphic modular form for $\mathrm{SL}_2(\mathbb{Z})$ is called a modular form, and usually (but not always) so are modular forms for congruence subgroups. Often, "automorphic form" is used when one considers either other Fuchsian groups, forms on groups other than $\mathrm{GL}_2$, or non-holomorphic forms (such as Maass forms and real analytic Eisenstein series). Alternatively, <a href="http://www.amazon.com/First-Course-Modular-Graduate-Mathematics/dp/038723229X" rel="nofollow">Diamond and Shurman</a> define an "automorphic form" in Section 3.2 as being like a modular form but possibly meromorphic instead of holomorphic.</p> <p>As another example, Miyake's book <a href="http://www.amazon.com/Modular-Forms-Springer-Monographs-Mathematics/dp/3540295925" rel="nofollow">Modular Forms</a> writes on p.114, "Automorphic functions and automorphic forms for modular groups are called modular functions and modular forms, respectively," and his definition of "modular group" seems to coincide with that of congruence subgroup.</p> <p>The Princeton Companion to Mathematics writes in section III.21, "automorphic forms, which are generalized versions of the classical analytic functions called modular forms [III.61]," but it does not specify what the generalization is. The book contains other, similar statements (in III.61, "And indeed, automorphic forms, which are generalizations of modular forms").</p> <p>So what exactly is the* definition of modular form as opposed to automorphic form? Since there is likely no "right" answer, what I really want to know is what is the history and what are the different conventions and the relations between them.</p> http://mathoverflow.net/questions/120749/what-is-an-algebro-geometric-family-of-modular-forms What is a(n algebro-geometric) family of modular forms? David Corwin 2013-02-04T10:00:42Z 2013-02-06T09:28:03Z <p>We know that a family of elliptic curves is a morphism of schemes $f:X \to Y$ such that the fiber of every point of $Y$ is an elliptic curve (and we usually require the morphism to be smooth, proper, etc).</p> <p>Given such a family, we can take the $\ell$-adic representation associated to any given fiber, and in this sense we also have a "family" of Galois representations. (Alternatively, by the proper base change theorem in étale cohomology, we can take $R^1f_*(\mathbb{Z}_{\ell})$, which is a sheaf on $Y$, and the stalks of this sheaf are the duals of the above Galois representations).</p> <p>Now consider different question. Can we have a "family" of cuspidal eigenforms whose associated Galois representations fit into a family in the above sense?</p> <p>I'll consider the case of weight 2 (though I'm most interested in higher weight). Then such family should lead to a family of RM abelian varieties, i.e. those associated to the weight 2 cusp forms.</p> <p>Let's go back to the elliptic curve (or abelian variety) side a bit, and think about what this would mean. The level of a modular form corresponds to the conductor of the associated elliptic curve, so the level of the modular forms in such a family should be just as bizarre a function of the base as is the conductor of a family of curves.</p> <p>To try to engineer such a family, suppose we had a family of elliptic curves that were all known to be modular. I'm most interested in rational families, i.e. with open subsets of projective space as bases. Then if the family is defined over $\mathbb{Q}$, we at least know that the fibers of rational points are modular, and we get a "family" of modular forms over the rational points. What would this "family" look like? I have a feeling it would be pretty strange from the point of view of modular forms.</p> <p>A different approach is to try to to construct a family of modular curves, then view a family of modular forms as a section of the relative cotangent sheaf or some power thereof. Maybe one could try to make it an "eigensection" of some sort of relative Hecke operators. Of course, the very idea of a family of modular curves seems strange, as there are countably-many modular curves!</p> <p>In fact, this points to a general problem with this attempt: modular forms are based on discrete data, a discrete set of levels, and a discrete set of eigenforms within each level.</p> <p>I have a feeling that it's impossible to make this notion work, but please let me know if you have good ideas. In particular, it's possible that experts in modular forms and curves would have more ideas.</p> http://mathoverflow.net/questions/120612/trichotomies-in-mathematics/120750#120750 Answer by David Corwin for Trichotomies in mathematics David Corwin 2013-02-04T10:13:03Z 2013-02-04T10:13:03Z <p>The endomorphism ring of an elliptic curve is either $\mathbb{Z}$, an order in a quadratic field, or an order in a quaternion algebra (ranks $1,2$, and $4$, respectively).</p> http://mathoverflow.net/questions/119015/what-precisely-does-kleins-erlangen-program-state What, precisely, does Klein's Erlangen Program state? David Corwin 2013-01-15T18:56:21Z 2013-01-28T18:23:51Z <p>People write that the <a href="http://en.wikipedia.org/wiki/Erlangen_program" rel="nofollow">Erlangen Program</a> is a "program" (like the "Langlands Program"), i.e. a series of related conjectures, which in this case were all solved. There are various intuitive accounts, describing that this program is about relating algebra and geometry, about relating transformation groups of spaces (Lie groups) and different geometries, invariants, etc.</p> <p>What I haven't been able to find is a precise statement in modern language (e.g. not that of his original paper) of what Klein's conjectures were. What precisely were his conjectures, or equivalently, what results constitute their resolution? Or was there never truly a precise statement?</p> http://mathoverflow.net/questions/99969/cosheaf-homology-and-a-theorem-of-beilinson-in-a-paper-on-mixed-tate-motives Cosheaf homology and a theorem of Beilinson (in a paper on Mixed Tate Motives) David Corwin 2012-06-19T04:07:26Z 2013-01-20T00:40:07Z <p>I'm trying to understand the proof of Theorem 4.1 in the paper Multiple Polylogarithms and Mixed Tate Motives by AB Goncharov (http://arxiv.org/pdf/math/0103059v4.pdf). In it, the author uses cosheaf homology.</p> <p>As far as I can tell, the global sections functor for cosheaves is right exact, so homology should be given by the <em>left</em> derived functors of the global sections functor. Similarly, the higher direct image functors should be the <em>left</em> derived functors of the standard direct image. The Grothendieck spectral sequence should be a homological spectral sequence. </p> <p>I have four questions:</p> <ol> <li><p>Is this correct?</p></li> <li><p>I understand that (see e.g. Hartshorne, Chapter III, Proposition 8.1) the cosheaf sending an open set $U$ to $H_q(p^{-1}(U))$ should be the $q$th direct image of the constant sheaf $\mathbb Z$. However, when on p.46 we are defining the cosheaf $\mathcal{R}_c$ and we are taking relative homology, what cosheaf are we taking the direct image of?</p></li> <li><p>Where does the exact sequence (110) come from? Do we always get an exact sequence from the joining of two complexes?</p></li> <li><p>Between (109) and (110), I assume that $R_i$ means left derived functor, since, as I mentioned, higher direct images of cosheaves are left, not right, derived functors. But what on Earth does he mean by the higher direct image of a subvariety (or complex of subvarieties)?</p></li> </ol> <p>I'm guessing that the $q$th direct image of the complex of varieties should be interpreted as the cosheaf corresponding to the homology relative to the union of the varieties associated to that complex? Assuming that's the case, I'm a little unsure how to deal with the higher direct images of the truncation (maybe it corresponds to the hyperhomology of the complex on the inverse image of $U$? Then the special case of the whole complex makes sense since the hyperhomology of that complex is the relative homology. But if that's so, I don't know what to make of the hyperhomology of the truncation...)</p> http://mathoverflow.net/questions/108860/anabelian-geometry-study-materials/108920#108920 Answer by David Corwin for Anabelian geometry study materials? David Corwin 2012-10-05T14:07:11Z 2013-01-07T04:07:21Z <p><a href="http://www.amazon.com/Fundamental-Mathematical-Sciences-Institute-Publications/dp/0521808316" rel="nofollow">This volume</a>, <em>Galois Groups and Fundamental Groups</em>, edited by Leila Schneps has a great collection of articles, as does <a href="http://www.amazon.com/Geometric-Galois-Actions-Grothendiecks-Mathematical/dp/0521596424" rel="nofollow">this volume</a>, <em>Geometric Galois Actions</em>, including a nice article by Florian Pop on "Glimpses of Grothendieck's anabelian geometry."</p> <p>If you'd like videos, <a href="http://www.newton.ac.uk/programmes/NAG/seminars/index.html" rel="nofollow">here</a> is a series of lectures on related topics, including a long series by Pop on anabelian geometry. At MSRI, you can find some <a href="http://www.msri.org/realvideo/index1.html" rel="nofollow">lectures from Fall 1999</a>, including one specifically about anabelian geometry.</p> http://mathoverflow.net/questions/54430/video-lectures-of-mathematics-courses-available-online-for-free/118241#118241 Answer by David Corwin for Video lectures of mathematics courses available online for free David Corwin 2013-01-07T03:00:21Z 2013-01-07T03:00:21Z <p>This collection has a mixture of French and English, but <a href="http://video.upmc.fr/differe.php?collec=S_C_galois_2011" rel="nofollow">here</a> you can find videos given at the Bicentennial of the Birth of Evariste Galois (Bicentennaire de la naissance d'Evariste Galois) at the Institut Henri Poincaré in Paris.</p> http://mathoverflow.net/questions/1714/best-online-mathematics-videos/118234#118234 Answer by David Corwin for Best online mathematics videos? David Corwin 2013-01-06T23:12:01Z 2013-01-06T23:12:01Z <p>The first <a href="http://www.math.princeton.edu/events/seminars/minerva-lectures/inaugural-minerva-lectures-i-equidistribution" rel="nofollow">Minerva Lecture</a> by Jean-Pierre Serre at Princeton in Fall 2012 is online. There were two other lectures, and they did videotape them, but I can't find them online.</p> http://mathoverflow.net/questions/117595/points-in-sites-etale-fppf/117605#117605 Answer by David Corwin for Points in sites (etale, fppf, ... ) David Corwin 2012-12-30T09:33:50Z 2013-01-06T22:14:33Z <p>See <a href="http://library.msri.org/books/sga/sga/ps/sga4-2.ps" rel="nofollow">SGA 4</a>, Exposé VIII, 7.8, which defines an abstract point of a site as a functor from the topos of that site (i.e. the category of sheaves of sets on that site) to the category of sets that commutes with arbitrary inductive (direct) limits and finite projective (inverse) limits.*</p> <p>One should think of a functor as being the functor that takes a sheaf to its stalk at that point (the "fiber functor" of the point). Since every fiber functor preserves such limits, these are reasonable axioms to postulate.</p> <p>Grothendieck then proves in this exposé that in the étale site, every functor satisfying the axiom above comes from a (unique) geometric point (up to isomorphism in a suitably-defined sense).</p> <p>Furthermore, <strike>and I can only find this referenced in Brian Conrad's unavailable draft book on the Ramanujan conjecture,</strike> in a (classical) topological space where every irreducible subset has a unique generic point (also known as a <em>sober</em> space), such as a normal Hausdorff space or the underlying space of a scheme, every abstract point of the category of sheaves on the topological space corresponds to a unique classical point. This is Proposition 2, Section 3, Chapter IX of <a href="http://www.springer.com/mathematics/algebra/book/978-0-387-97710-2" rel="nofollow">Sheaves in Geometry and Logic</a>.</p> <p>Why characterize points in terms of their functors? This fits in with the general Tannakian formalism, which emphasizes fiber functors, and in particular, defines the fundamental group to be the group of automorphisms of the fiber functor. I.e., we should think of a point as the functor that assigns to a sheaf its stalk at that point.</p> <p>This also fits into a more general philosophy that we should ignore the site and focus on the topos entirely. One can in fact put a Grothendieck topology on the topos so that it is equal to the category of sheaves on itself.</p> <p>*<strong>Edit</strong>: One can also find this in <a href="http://www.springer.com/mathematics/algebra/book/978-0-387-97710-2" rel="nofollow">Sheaves in Geometry and Logic</a>, by Maclane and Moerdijk, Chapter VII, Section 5. </p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology Intuition for Group Cohomology David Corwin 2010-01-06T04:19:01Z 2013-01-06T05:54:12Z <p>I'm beginning to learn cohomology for cyclic groups in preparation for use in the proofs of global class field theory (using ideal-theoretic arguments). I've seen the proof of the long exact sequence and of basic properties of the Herbrand quotient, and I've started to look through how these are used in the proofs of class field theory.</p> <p>So far, all I can tell is that the cohomology groups are given by some ad hoc modding out process, then we derive some random properties (like the long exact sequence), and then we compute things like $H^2(\mathrm{Gal}(L/K),I_{L})$, where $I_L$ denotes the group of fractional ideals of a number field $L$, and it just happens to be something interesting for the study of class field theory such as $I_K/\mathrm{N}(I_L)$, where $L/K$ is cyclic and $\mathrm{N}$ denotes the ideal norm. We then find that the cohomology groups are useful for streamlining the computations with various orders of indexes of groups.</p> <p>What I don't get is what the intuition is behind the definitions of these cohomology groups. I do know what cohomology is in a geometric setting (so I know examples where taking the kernel modulo the image is interesting), but I don't know why we take these particular kernels modulo these particular images. What is the intuition for why they are defined the way they are? Why should we expect that these cohomology groups so-defined have nice properties and help us with algebraic number theory? Right now, I just see theorem after theorem, I see the algebraic manipulation and diagram chasing that proves it, but I don't see a bigger picture.</p> <p>For context, if $A$ is a $G$-module where $G$ is cyclic and $\sigma$ is a generator of $G$, then we define the endomorphisms $D=1+\sigma+\sigma^2+\cdots+\sigma^{|G|-1}$ and $N=1-\sigma$ of $A$, and then $H^0(G,A)=\mathrm{ker}(N)/\mathrm{im}(D)$ and $H^1(G,A)=\mathrm{ker}(D)/\mathrm{im}(N)$. Note that this is a slight modification of group cohomology, i.e. Tate cohomology, which the cohomology theory primarily used for Class Field Theory. Group cohomology is the same but with $H^0(G,A) = \mathrm{ker}(N)$. The advantage of Tate cohomology is that it is $2$-periodic for $G$ cyclic.</p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/33649#33649 Answer by David Corwin for Intuition for Group Cohomology David Corwin 2010-07-28T12:46:00Z 2013-01-06T05:54:12Z <p>See also this Math.SE post I wrote for some more motivation: <a href="http://math.stackexchange.com/a/270266/873" rel="nofollow">http://math.stackexchange.com/a/270266/873</a>. Recall that <code>$\mathrm H^*(G,M)=\mathrm{Ext}^*(\mathbb{Z},M)$</code>.</p> <p>After learning some more math, I've come across the following example of a use of group cohomology which sheds some light on its geometric meaning. (If you want to see a somewhat more concrete explanation of how group cohomology naturally arises, skip the next paragraph.)</p> <p>We define an elliptic curve to be $E=\mathbb{C}/L$ for a two-dimensional lattice $L$. Note that the first homology group of this elliptic curve is isomorphic to $L$ precisely because it is a quotient of the universal cover $\mathbb{C}$ by $L$. A theta function is a section of a line bundle on an elliptic curve. Since any line bundle can be lifted to $\mathbb{C}$, the universal cover, and any line bundle over a contractible space is trivial, the line bundle is a quotient of the trivial line bundle over $\mathbb{C}$. We can define a function $j(\omega,z):L \times \mathbb{C} \to \mathbb{C} \setminus {0}$. Then we identify $(z,w) \in \mathbb{C}^2$ (i.e. the line bundle over $\mathbb{C}$) with $(z+\omega,j(\omega,z)w)$. For this equivalence relation to give a well-defined bundle over $\mathbb{C}/L$, we need the following: Suppose $\omega_1,\omega_2 \in L$. Then $(z,w)$ is identified with $(z+\omega_1+\omega_2,j(\omega_1+\omega_2,z)w$. But $(z,w)$ is identified with $(z+\omega_1,j(\omega_1,z)w)$, which is identified with $(z+\omega_1+\omega_2,j(\omega_2,z+\omega_1)j(\omega_1,z)w)$. In other words, this forces $j(\omega_1+\omega_2,z) = j(\omega_2,z+\omega_1)j(\omega_1,z)$. This means that, if we view $j$ as a function from $L$ to the set of non-vanishing holomorphic functions $\mathbb{C} \to \mathbb{C}$, with (right) L-action on this set defined by $(\omega f)(z) \mapsto f(z+\omega)$, then $j$ is in fact a $1$-cocyle in the language of group cohomology. Thus $H^1(L,\mathcal{O}(\mathbb{C}))$, where $\mathcal{O}(\mathbb{C})$ denotes the (additive) $L$-module of holomorphic functions on $\mathbb{C}$, classifies line bundles over $\mathbb{C}/L$. What's more is that this set is also classified by the sheaf cohomology $H^1(E,\mathcal{O}(E)^{\times})$ (where $\mathcal{O}(E)$ is the sheaf of holomorphic functions on $E$, and the $\times$ indicates the group of units of the ring of holomorphic functions). That is, we can compute the sheaf cohomology of a space by considering the group cohomology of the action of the homology group on the universal cover! In addition, the $0$th group cohomology (this time of the meromorphic functions, not just the holomorphic ones) is the invariant elements under $L$, i.e. the elliptic functions, and similarly the $0$th sheaf cohomology is the global sections, again the elliptic functions.</p> <p>More concretely, a theta function is a meromorphic function such that $\theta(z+\omega)=j(\omega,z)\theta(z)$ for all $z \in \mathbb{C}$, $\omega \in L$. (It is easy to see that $\theta$ then gives a well-defined section of the line bundle on $E$ given by $j(\omega,z)$ described above.) Then, note that $\theta(z+\omega_1+\omega_1)=j(\omega_1+\omega_2,z)\theta(z) = j(\omega_2,z+\omega_1)j(\omega_1,z) \theta(z)$, meaning that $j$ must satisfy the cocycle condition! More generally, if $X$ is a contractible Riemann surface, and $\Gamma$ is a group which acts on $X$ under sufficiently nice conditions, consider meromorphic functions $f$ on $X$ such that $f(\gamma z)=j(\gamma,z)f(z)$ for $z \in X$, $\gamma \in \Gamma$, where $j: \Gamma \times X \to \mathbb{C}$ is holomorphic for fixed $\gamma$. Then one can similarly check that for $f$ to be well-defined, $j$ must be a $1$-cocyle in $H^1(\Gamma,\mathcal{O}(X)^\times)$! (I.e. with $\Gamma$ acting by precomposition on $\mathcal{O}(X)^\times$, the group of units of the ring of holomorphic functions on $X$.) Thus <em>the cocycle condition arises from a very simple and natural definition</em> (that of a function which transforms according to a function $j$ under the action of a group). A basic example is a modular form such as $G_{2k}(z)$, which satisfies $G_{2k}(\gamma z) = (cz+d)^{2k} G_{2k}(z)$, where $\gamma = \left(\begin{array}{cc} a &amp; b \ c &amp; d \end{array}\right) \in SL_2(\mathbb{Z})$ acts as a fractional linear transformation. It follows automatically that something as simple as $(cz+d)^{2k}$ is a cocycle in group cohomology, since $G_{2k}$ is, for example, nonzero.</p> http://mathoverflow.net/questions/118044/an-isomorphism-between-different-exts-coming-from-group-cohomology An isomorphism between different Ext's coming from group cohomology David Corwin 2013-01-04T11:41:31Z 2013-01-04T12:25:49Z <p>Let $G$ be an abelian group and $M$ a $G$-module with trivial action. It is well-known that $H^2(G,M)$ classifies extensions of $G$ by $M$, which is $\mathrm{Ext}^1_{Ab}(G,M)$.</p> <p>On the other hand $H^2$ is $\mathrm{Ext}^2_{G-mod}(\mathbb Z,M)$ (by the definition of group cohomology), where $\mathbb Z$ is given the trivial $G$-action.</p> <p>Is there a nice way to see this isomorphism $$\mathrm{Ext}^1_{Ab}(G,M) \cong \mathrm{Ext}^2_{G-mod}(\mathbb Z,M)?$$</p> <p>Or is it just an accident that both happen to classify the same type of object?</p> <p>Is there a high-brow reason? Is there a generalization? Why should $\mathrm{Ext}^1$ and $\mathrm{Ext}^2$ be related in such a way? Might this result from a spectral sequence? I wonder if this might be a special case of something well-known.</p> http://mathoverflow.net/questions/117668/new-grand-projects-in-contemporary-math/117675#117675 Answer by David Corwin for New grand projects in contemporary math David Corwin 2012-12-30T21:57:58Z 2013-01-02T22:30:07Z <p>Manjul Bhargava's new field of arithmetic invariant theory is a perfect example of a new grand project. It began with Manjul's <a href="http://www.jstor.org/discover/10.2307/3597249?uid=3739696&amp;uid=2&amp;uid=4&amp;uid=3739256&amp;sid=21101483414553" rel="nofollow">doctoral thesis</a>, in which he presented a completely new view of Gauss's composition law for binary quadratic forms in a way that led to generalizations. This led to a <a href="http://annals.math.princeton.edu/?s=Manjul+Bhargava" rel="nofollow">series of papers</a> on generalizations to cubic forms and beyond, with a general framework coming from representation theory.</p> <p>In addition to being intrinsically interesting, this has led to new results on counting quadratic rings, cubic rings, etc, whose crowning achievement is an important new result on the Birch and Swinnerton-Dyer conjecture (c.f. <a href="http://arxiv.org/abs/1007.0052" rel="nofollow">Bhargava and Shankar</a>). There is much more to research in the theory, and numerous people (including some of his students) have found manifold connections with other areas of math, such as knot theory and algebraic geometry.</p> <p>See <a href="http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_13.pdf" rel="nofollow">here</a> for an overview of Manjul's work.</p> <p>See <a href="https://web.math.princeton.edu/~arabindr/ait.html" rel="nofollow">here</a> for some notes from a seminar at Princeton that shows the vast reach of this theory.</p> http://mathoverflow.net/questions/117415/old-books-still-used/117591#117591 Answer by David Corwin for Old books still used David Corwin 2012-12-30T05:35:24Z 2013-01-02T22:06:45Z <p>Tate's thesis, <em>Fourier analysis in number fields, and Hecke's zeta-functions</em>, is from 1950 and is certainly still considered a primary on the subject (in addition to being the original resource).</p> http://mathoverflow.net/questions/117715/extending-systems-of-l-adic-representations-to-other-l Extending systems of l-adic representations to other l David Corwin 2012-12-31T11:18:56Z 2012-12-31T13:33:43Z <p>I'm asking this not because I have an idea how one might approach it, but because it seems natural and inherently interesting.</p> <p>Let $K$ be a number field, $G_K$ its absolute Galois group, and $\ell\neq\ell'$ prime numbers. An $\ell$-adic representation of $K$, i.e. a continuous homomorphism $\rho_\ell$ from $G_K$ to $\mathrm{GL}(n,\mathbb Q_{\ell})$ is said to be compatible with an $\ell'$-adic representation $\rho_{\ell'}$ of $K$ if for almost all (i.e. all but finitely many) places $v$ of $K$, the representations $\rho_\ell,\rho_{\ell'}$ are unramified, and the characteristic polynomials of $\rho_\ell(F_v)$ and $\rho_{\ell'}(F_v)$ are equal and have rational coefficients, where $F_v \in D_v/I_v \subseteq G_K$ denotes the Frobenius at $v$.</p> <p>Note that by the Chebotarev density, this uniquely determines the trace of $\rho_{\ell'}$, which, in turn, uniquely determines it up to semi-simplification.</p> <p>If $\mathcal{L}$ is a collection of prime numbers, then a strictly compatible system of $\ell$-adic representations consists of an $\ell$-adic representation $\rho_\ell$ of $K$ for each $\ell \in \mathcal{L}$ such that there is a finite set $S$ of places of $K$ (called the <em>exceptional set</em>) such that for $\ell\neq\ell'$, $v \notin S$, and $v \nmid \ell\ell'$, the representations $\rho_\ell$ and $\rho_{\ell'}$ are unramified at $v$, and the characteristic polynomials of $\rho_\ell(F_v)$ and $\rho_{\ell'}(F_v)$ are equal and have rational coefficients.</p> <p>Can we find an $\ell$-adic representation unramified almost everywhere with integral traces of Frobenius and a prime $\ell'$ such that there is no $\ell'$-adic representation compatible with it? The Fontaine-Mazur conjecture would predict that the representation cannot be geometric.</p> <p>But what's more, I'm wondering the following: suppose we have a strictly compatible system of $\ell$-adic representations for almost all $\ell$. Then can we necessarily extend this to a strictly compatible system for all $\ell$?</p> http://mathoverflow.net/questions/14574/your-favorite-surprising-connections-in-mathematics/32023#32023 Answer by David Corwin for Your favorite surprising connections in Mathematics David Corwin 2010-07-15T15:59:29Z 2012-12-31T00:26:13Z <p>The beautiful analogy between number fields and function fields (and in general, algebra and geometry) that one learns about in arithmetic algebraic geometry.</p> <p>Some specific examples:</p> <p>The idea that a Galois group and a fundamental group (one algebro-number theoretic, the other geometric/topological) are two instances of the same thing.</p> <p>The use of the term ramification in both number theory and geometry. Describing $\mathbb{Z}$ as simply connected because $\mathbb{Q}$ has no unramified extensions.</p> <p>The appearance of integral closure in both algebraic geometry and algebraic number theory. The integral closure, in the former case, actually corresponds to a distinct geometric idea: non-singularity.</p> <p>The idea of considering a prime number to be a point; then viewing localization at that prime, -adic completion at that prime, and the residue field of that prime as if they were the corresponding geometric objects. In particular, using the term "local" in number theory, as if we were talking about geometry! This idea is built into scheme theory.</p> <p>There are many more examples.</p> <p><a href="http://www.renyi.hu/~szamuely/fg.pdf" rel="nofollow">This book</a> looks deeply into the relationships between Galois groups and fundamental groups and eventually develops a theory which covers both.</p> <p><a href="http://books.google.co.il/books?id=c8vnGts4994C&amp;printsec=frontcover&amp;dq=arithmetic+geometry+lorenzini&amp;source=bl&amp;ots=w2hYlzwjbL&amp;sig=qf1I_2wkJeaycxkh8_T4yJ-azl8&amp;hl=en&amp;ei=uC8_TM3cMMmNONjQ6J8H&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CBYQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow">This book</a> explores the beautiful relation between algebraic curves and algebraic number theory.</p> http://mathoverflow.net/questions/117401/is-every-functor-inducing-a-homotopy-equivalence-a-composition-of-adjoint-functor Is every functor inducing a homotopy equivalence a composition of adjoint functors? David Corwin 2012-12-28T11:36:14Z 2012-12-28T13:01:22Z <p>It was asked <a href="http://mathoverflow.net/questions/335/is-every-functor-a-composition-of-adjoint-functors/336#336" rel="nofollow">here</a> whether every functor is a composition of adjoint functors. The answer is no, because all adjoint functors induce homotopy equivalences on the nerve, and we can construct functors that do not induce homotopy equivalences.</p> <p>My question is the following: can all functors inducing homotopy equivalences on the nerve be expressed as compositions of adjoint functors?</p> http://mathoverflow.net/questions/25054/different-ways-of-thinking-about-the-derivative/117387#117387 Answer by David Corwin for Different ways of thinking about the derivative David Corwin 2012-12-28T05:19:14Z 2012-12-28T05:19:14Z <p>The intuitive idea is that the derivative is zero if there is a local minimum or maximum.</p> <p>The derivative is then the (tangent of) the angle by which we have to rotate the graph around that point in order to get a local minimum or maximum.</p> <p>You could probably cook up some sort of way to think about what happens when the function vanishes to odd order (first of all, no matter how you rotate it, there will not be a local minimum or maximum).</p> http://mathoverflow.net/questions/117374/why-is-the-leibniz-rule-a-definition-for-derivations/117376#117376 Answer by David Corwin for Why is the Leibniz rule a definition for derivations? David Corwin 2012-12-28T01:39:41Z 2012-12-28T05:09:34Z <p>So I hope you're convinced that a directional derivative <em>should</em> satisfy the product rule. In that case, it's fine to postulate that a directional derivative must satisfy the product rule. The question then is: Why do we assume the Leibniz rule (+linearity) and nothing else?</p> <p>I would suggest the answer is simply that it gives you the correct result. I.e., you might think that there are too many functionals* that satisfy the Leibniz rule, and we have to impose some other condition to ensure that our functionals are directional derivative operators. But one can prove that the set of functionals that satisfy the Leibniz rule has dimension equal to the dimension of the manifold, so you can be confident that you don't have too many.</p> <p>Here's a different definition of the tangent space that I find more intuitive. If you have a point $P$ on your manifold $M$ and a smooth curve $\gamma:\mathbb{R} \to M$ such that $\gamma(0)=P$, then this curve defines a functional on functions around $P$. I.e., if $f$ is a smooth real-valued function defined in a neighborhood of $P$, then we get $f \circ \gamma:\mathbb{R} \to \mathbb{R}$. We can then differentiate this function at $0$ to get a number, which we denote $r_\gamma(f)$.</p> <p>We can then define the tangent space to be the set of linear functionals (on the space of smooth functions on $M$ in a neighborhood of $P$) that arise from smooth paths, as above (i.e. the set of functionals of the form $r_\gamma$ for some $\gamma$). Notice that this purposely is not the set of paths; two paths give rise to the same functional iff they're tangent to each other. This is an equally valid way of defining the tangent space.</p> <p>The reason the Leibniz rule suffices is then the fact that every functional that satisfies the Leibniz rule actually arises as the directional derivative with respect to some smooth path.</p> <p>In fact, you can ignore functionals entirely. You can define the tangent space to be the set of smooth paths through $P$ modulo the following equivalence relation. Two paths are equivalent if in some coordinate patch around $P$, they have the same tangent vector (in the classical sense). Note that this is independent of coordinate path, so it's well-defined. The only reason we cannot just take tangent vectors in some coordinate patch is that this is not coordinate-independent, whereas these equivalence classes of paths are.</p> <p>It just so happens two paths are equivalent in the sense I just described iff the define the same functional on the space of smooth functions on $M$, which is why we can use functionals to define the tangent space, and this is the same as the definition above. But I find the definition in terms of equivalence classes of paths to be the most natural.</p> <p>*By a functional, I mean a linear map from the vector space of smooth real-valued functions defined in a neighborhood of $P$ to $\mathbb{R}$ that depends only on the values of the function in a neighborhood of $P$</p> http://mathoverflow.net/questions/100276/can-one-prove-complex-multiplication-without-assuming-cft Can one prove complex multiplication without assuming CFT? David Corwin 2012-06-21T19:25:21Z 2012-12-27T05:50:41Z <p>The Kronecker-Weber Theorem, stating that any abelian extension of $\mathbb Q$ is contained in a cyclotomic extension, is a fairly easy consequence of Artin reciprocity in class field theory (one just identifies the ray class groups and shows that each corresponds to a cyclotomic extension). However, one can produce a more direct and elementary proof of this fact that avoids appealing to the full generality of class field theory (see, for example, the exercises in the fourth chapter of <em>Number Fields</em> by Daniel Marcus). In other words, one can prove class field theory for $\mathbb Q$ using much simpler methods than for the general case.</p> <p>The theory of complex multiplication is similar to the theory of cyclotomic fields (and hence the Kronecker-Weber Theorem) in that it shows that any abelian extension of a quadratic imaginary field is contained in an extension generated by the torsion points of an elliptic curve with complex multiplication by our field. To prove this, one normally assumes class field theory and then shows that the field generated by the $m$-torsion (or, more specifically, the Weber function of the $m$-torsion) is the ray class field of conductor $m$.</p> <p>My question is: Can one prove that any abelian extension of an imaginary quadratic field $K$ is contained in a field generated by the torsion of an elliptic curve with complex multiplication by $K$ without resorting to the general theory of class field theory? I.e. where one directly proves class field theory for $K$ by referring to the elliptic curve. Is there a proof in the style of the exercises in Marcus's book?</p> <p>Note: Obviously there is no formal formulation of what I'm asking. One way or another, you <em>can</em> prove complex multiplication. But the question is whether you can give a proof of complex multiplication in a certain style.</p> http://mathoverflow.net/questions/37021/is-fourier-analysis-a-special-case-of-representation-theory-or-an-analogue Is Fourier analysis a special case of representation theory or an analogue? David Corwin 2010-08-29T03:47:43Z 2012-12-06T00:36:58Z <p>I'm asking this question because I've been told by some people that Fourier analysis is "just representation theory of $S^1$."</p> <p>I've been introduced to the idea that Fourier analysis is related to representation theory. Specifically, when considering the representations of a finite abelian group $A$, these representations are all $1$-dimensional, hence correspond to characters $A \to \mathbb{R}/\mathbb{Z} \cong S^1 \subseteq \mathbb{C}$. On the other side, finite Fourier analysis is, in a simplistic sense, the study of characters of finite abelian groups. Classical Fourier analysis is, then, the study of continuous characters of locally compact abelian groups like $\mathbb{R}$ (classical Fourier transform) or $S^1$ (Fourier series). However, in the case of Fourier analysis, we have something beyond characters/representations: We have the Fourier series / transform. In the finite case, this is a sum which looks like $\frac{1}{n} \sum_{0 \le r &lt; n} \omega^r \rho(r)$ for some character $\rho$, and in the infinite case, we have the standard Fourier series and integrals (or, more generally, the abstract Fourier transform). So it seems like there is something more you're studying in Fourier analysis, beyond the representation theory of abelian groups. To phrase this as a question (or two):</p> <p>(1) What is the general Fourier transform which applies to abelian and non-abelian groups?</p> <p>(2) What is the category of group representations we consider (and attempt to classify) in Fourier analysis? That is, it seems like Fourier analysis is more than just the special case of representation theory for abelian groups. It seems like Fourier analysis is trying to do more than classify the category of representations of a locally compact abelian group $G$ on vector spaces over some fixed field. Am I right? Or can everything we do in Fourier analysis (including the Fourier transform) be seen as one piece in the general goal of classifying representations?</p> <p>Let me illustrate this in another way. The basic result of Fourier series is that every function in $L^2(S^1)$ has a Fourier series, or in other words that $L^2$ decomposes as a (Hilbert space) direct sum of one dimensional subspaces corresponding to $e^{2 \pi i n x}$ for $n \in \mathbb{Z}$. If we encode this in a purely representation-theoretic fact, this says that $L^2(S^1)$ decomposes into a direct sum of the representations corresponding to the unitary characters of $S^1$ (which correspond to $\mathbb{Z}$). But this fact is not why Fourier analysis is interesting (at least in the sense of $L^2$-convergence; I'm not even worrying about pointwise convergence). Fourier analysis states furthermore an <em>explicit</em> formula for the function in $L^2$ giving this representation. Though I guess by knowing the character corresponding to the representation would tell you what the function is.</p> <p>So is Fourier analysis merely similar to representation theory, or is it none other than the abelian case of representation theory?</p> <p>(Aside: This leads into a more general question of mine about the use of representation theory as a generalization of modular forms. My question is the following: I understand that a classical Hecke eigenform (of some level $N$) can be viewed as an element of $L^2(GL_2(\mathbb{Q})\ GL_2(\mathbb{A}_{\mathbb{Q}})$ which corresponds to a subrepresentation. But what I don't get is why the representation tells you everything you would have wanted to know about the classical modular form. A representation is nothing more than a vector space with an action of a group! So how does this encode the information about the modular form?)</p> http://mathoverflow.net/questions/115462/on-the-derived-category-of-constructible-etale-sheaves On the derived category of constructible étale sheaves David Corwin 2012-12-05T02:44:54Z 2012-12-05T02:59:10Z <p>The derived category <code>$D^{\flat}_{c}(X,R)$</code> of constructible sheaves of $R$-modules on $X_{et}$ is defined as the full subcategory of $D^b(X,R)$ whose cohomology sheaves are all constructible.</p> <p>Clearly, given a sequence of constructible sheaves indexed by $\mathbb{Z}$ can be realized as the cohomology of a complex of sheaves consisting entirely of constructible sheaves (just put the desired cohomology sheaf in each degree and make the differential zero). But one might imagine that there is a complex of sheaves, not all of which are constructible, whose cohomology sheaves are constructible, and which is not quasi-isomorphic to a complex consisting entirely of constructible sheaves.</p> <p>My question: Can such a complex of sheaves exist? If so, under what conditions might it not exist?</p> <p>I asked this question in a course of N. Katz, who answered that he did not know the answer but that he knew of some results in the affirmative when asking this question for quasi-coherent or coherent sheaves in place of constructible sheaves.</p> http://mathoverflow.net/questions/112245/rigid-uniformization-vs-grothendiecks-local-monodromy-theory Rigid Uniformization vs Grothendieck's Local Monodromy Theory David Corwin 2012-11-13T02:42:42Z 2012-11-29T06:22:00Z <p>I've noticed that some interesting results about abelian varieties can each be proven using one of two ways: the theory of rigid uniformization of abelian varieties or Grothendieck's local monodromy theory of SGA 7 1. I'm therefore wondering about possible interconnections between the two.</p> <p><strong>Detailed Motivation</strong>:</p> <p>The first example of a result referred to above is the fact that CM abelian varieties have (potentially) good reduction everywhere. The proof using rigid uniformization is discussed in Silverman, <em>Advanced Topics in the Arithmetic of Elliptic Curves</em> (ATAoEC) in Chapter V. ATAoEC also gives a proof in Chapter II Section 6 using local class field theory, Neron-Ogg-Shafarevich, and the fact that a pro-$p$ group can only map trivially into a pro-$\ell$ group. I consider this latter proof to be part of Grothendieck's local monodromy theory, as one uses a similar method to prove the local monodromy theorem (at least, as demonstrated to me in Nicholas Katz's course at Princeton this fall; the original should be in the elusive SGA 7 1 Exposé III).</p> <p>The next example is the following. SGA 7 1 Exposé IX proves that if $A/K$ has semistable reduction over a local field $K$ with inertia group $I$ with dimension $g$ and toric dimension $\mu$, then $T_\ell(A)^f := T_\ell(A)^I$ has rank $2g-\mu$, and $I$ acts trivially on the quotient as well. Furthermore, it has a complement under the Weil pairing (for a fixed polarisation), denoted $T_\ell(A)^t$, of rank $\mu$. See 2.2.5, 2.4, 2.5.4, and 3.5 of the Exposé notes.</p> <p>Of course, one can prove the same result using rigid uniformization, where $T_\ell(A)^t$ corresponds to the $\ell^n$th roots of unity in $\bar{K}^*$. See Ribet, <em>Galois Action on Division Points of Abelian Varieties with Real Multiplications</em>, Section III, or <a href="http://math.stanford.edu/~vakil/snowbird/mihranjun21.pdf" rel="nofollow">these notes</a> by Mihran Papikian.</p> <p><strong>Specific Question</strong>: Why do these two theories seem to prove the same results? </p> <p>This would make sense if I saw similar arguments being used to develop both theories. But I don't see how analyzing the inertia using local class field theory and then looking at profinite groups is the same as writing down $p$-adic power series. While they both have a $p$-adic and $\ell$-adic "flavor," they seem to be very different proofs.</p> <p>However, please tell me if I'm wrong - could it be that one can trace the arguments developing each theory to find a common thread?</p> <p>More specifically, can one prove in general that if one can prove a result with one theory, then one can do it with the other? Is one theory strictly stronger than the other? Is there a common generalization?</p> http://mathoverflow.net/questions/114395/result-that-follows-from-zfc-and-not-zf-but-are-strictly-weaker-than-choice Result that follows from ZFC and not ZF but are strictly weaker than choice David Corwin 2012-11-25T03:58:50Z 2012-11-26T12:23:32Z <p>A number of results that people use that require the axiom of choice (i.e. do not follow from ZF alone) are known to actually imply the axiom of choice. Therefore, one might naturally wonder whether there are results that require choice to prove yet which, on the contrary, do not imply choice.</p> <p>The most natural answer to this question is countable choice, or $\kappa$-choice for cardinals $\kappa$.</p> <p>Therefore, I wonder, are there other (i.e. not equivalent to any of the above) examples of results with this property? What about which are strictly stronger or weaker than all the cardinal-choice axioms (but not equivalent to ZF or ZFC)? In particular, I am interested in results occurring naturally in algebra, geometry, etc. And whether or not there are, is there a deeper reason why results that use choice tend to imply it?</p> http://mathoverflow.net/questions/54932/how-were-moduli-spaces-defined-before-functors/112487#112487 Answer by David Corwin for How were moduli spaces defined before functors? David Corwin 2012-11-15T15:09:21Z 2012-11-17T22:46:46Z <p>I will give an answer, but first I would like to clarify the question. It seems to me that most commenters have misinterpreted the question. The question is not how people managed to construct different examples of moduli spaces before they had the tool of the language of functors. The question is the following: A moduli space is supposed to be a space whose points are in bijection with the isomorphism classes of some type of object. But it's not just that; for, if we only needed to find a bijection, then the sets would merely need to have the same cardinality. But we need more: namely, we need the geometry of the space to somehow reflect the nature of the objects in a "natural way." Now, we may have an intuitive idea of what this means, and in many cases we might be able to recognize when some space is not just in bijection with a class of objects but actually has geometry that reflects those objects. But the question is: how can we precisely state what this means?</p> <p>Nowadays, we have the language of functors and functor of points, and we look not just at a single set, but at isomorphism classes of the given type of object over arbitrary bases, given a functor. We then say that a space is a moduli space for those objects if it represents that functor (or represents it up to isomorphisms - the coarse vs fine distinction is not too relevant for this discussion); note that now, this determines not only the set of points of our space, but it actually (by Yoneda's lemma) determines the geometry of our space.</p> <p>So the question is the following: before the notion of functors and functor of points, how did people rigorously define what it meant for the geometry of the moduli space to reflect the geometry nature of the set of isomorphism classes of objects to be parametrized. I should add that this is a question that I have been curious about myself for a long time.</p> <p>Now, according to Newstead's text <a href="http://www.amazon.com/Introduction-Moduli-Problems-Orbit-Spaces/dp/8184871627" rel="nofollow">Introduction to Moduli Problems and Orbit Spaces</a>,</p> <blockquote> <p>The word "moduli" is due to Riemann, who showed in his celebrated paper of 1857 on abelian functions that an isomorphism class of Riemann surfaces of genus $p$... However, it is only very recently that one has been able to formulate moduli problems in precise terms and in some cases to obtain solutions to them.</p> </blockquote> <p>This book (at least the edition I'm looking at) was written in 1977, which gives some perspective on this statement.</p> <p>In the ensuing chapter, Newstead goes on to define a family of objects parametrized by a variety. The definition is quite simple: it is a morphism of varieties $X \to S$ such that the fiber of any point $s$ of $S$ (i.e. its pre-image in $X$) is an object of the type in question. Even if one does not have the language of functors, my guess is that this idea could motivate a more precise notion of what a moduli space is.</p> <p>János Kollár's <a href="https://web.math.princeton.edu/~kollar/book/chap1.pdf" rel="nofollow">draft book on moduli spaces</a> also gives some hints:</p> <blockquote> <p>The classical literature never diﬀerentiates between the linear system as a set and the linear system as a projective space. There are, indeed, few reasons to distinguish them as long as we work over a ﬁxed base ﬁeld $k$. If, however, we pass to a ﬁeld extension $K/k$, the advantages of viewing $|L|$ as a $k$-variety appear.</p> </blockquote> <p>The first sentence suggests that there was not a precise definition of moduli space in classical literature. It also suggests a natural idea leading to the functor of points, i.e. that over a field k, we might want to look and objects parametrized over different field extensions of k.</p> <p>In general, as you can see from the history mentioned in his text, people didn't necessarily have precise definitions of moduli spaces, but they did understand that the geometry (well, the parameters) of the moduli space should correspond to the coefficients of the defining equations of the objects in question to be parametrized.</p> <p>Finally, in Dieudonné's <a href="http://www.math.brown.edu/~wgillam/math1040/HistoryOfAG.pdf" rel="nofollow">Historical Development of Algebraic Geometry</a>, the author states on p.837</p> <blockquote> <p>the precise meaning of this result that Riemann surfaces of genus $g$ are parametrized by $3g−3$ complex parametrized was to remain until very recently among the least clarified concepts of the theory.</p> </blockquote> <p>While this doesn't answer the question, it might be of interest to note that Dieudonné later notes in his section on Grothendieck's functor of points</p> <blockquote> <p>in particular, one transfers in that way to the theory of schemes many classical constructions such as projective spaces..., and one is able to give a general meaning to the concept of "moduli" introduced by Riemann for curves</p> </blockquote> <p>suggesting slightly that there was no general meaning before this point.</p> <p>That's the best I can do for now. An expert in the history of algebraic geometry might have more to say, but the sources I have seem to point in the direction of saying that there was not a precise definition until much later</p> http://mathoverflow.net/questions/131255/objects-which-cant-be-defined-without-making-choices-but-which-end-up-independen/131262#131262 Comment by David Corwin David Corwin 2013-05-21T02:38:22Z 2013-05-21T02:38:22Z Then I don't feel this example fits the thread. This thread is about objects that can't be defined without making choices. The existence of singular cohomology shows that it can be defined without making choices. http://mathoverflow.net/questions/44045/naturally-occuring-k-pi-n-spaces-for-n-geq-2 Comment by David Corwin David Corwin 2013-05-20T05:26:02Z 2013-05-20T05:26:02Z Shouldn't this be CW? http://mathoverflow.net/questions/130540/an-example-of-a-tensor-product-consisting-of-only-simple-tensors Comment by David Corwin David Corwin 2013-05-14T03:21:04Z 2013-05-14T03:21:04Z Take $A'$ a field, $M$ a vector space of dimension m, $A$ a vector space of dimension n given some ring structure (e.g. product ring). Then we can easily find m,n (I think if both are &gt;1?) where there are non-simple tensors. http://mathoverflow.net/questions/54932/how-were-moduli-spaces-defined-before-functors/112487#112487 Comment by David Corwin David Corwin 2013-05-14T01:50:19Z 2013-05-14T01:50:19Z I should add what Brian Conrad writes in an unpublished draft book: &quot;Grothendieck's reformulation of moduli spaces in terms of representable functors gave the theory of moduli more strength and flexibility.&quot; http://mathoverflow.net/questions/99923/is-the-ell-adic-cohomology-of-a-non-proper-variety-unramified-at-good-primes Comment by David Corwin David Corwin 2013-05-09T12:14:35Z 2013-05-09T12:14:35Z See the argument in Section 4 of Deligne's &quot;Formes modulaires et representations l-adiques&quot; for an example where one can make such conclusions for a non-proper variety (i.e. the bottom of p.13 of <a href="http://math.bu.edu/people/potthars/writings/deligne-l-adic.pdf" rel="nofollow">math.bu.edu/people/potthars/writings/&hellip;</a>). It uses a version of Abhyankar's lemma discussed in SGA 1 Expose XIII. http://mathoverflow.net/questions/129364/why-is-it-hard-to-prove-that-the-euler-mascheroni-constant-is-irrational Comment by David Corwin David Corwin 2013-05-05T05:13:11Z 2013-05-05T05:13:11Z +1 because the question produced a great answer. I think it might have been a perfectly good question if it asked what is the philosophy behind proving transcendence, why have we been able to prove what we proved, and why the limits we have exist, etc. http://mathoverflow.net/questions/30611/relation-between-sheaf-and-group-cohomology Comment by David Corwin David Corwin 2013-05-02T20:50:36Z 2013-05-02T20:50:36Z To expand on Brian's comment: The pullback of the sheaf $G_m$ to the universal cover has vanishing higher cohomology, and the results in Mumford then imply that group cohomology agrees with sheaf cohomology. http://mathoverflow.net/questions/128874/tangent-space-in-algebraic-geometry-and-differential-geometry Comment by David Corwin David Corwin 2013-05-02T03:06:56Z 2013-05-02T03:06:56Z Another nicer way to think about this is local in the etale topology. This is just as good since etale morphisms preserve the tangent space (of course, then you'd have to define etale without the tangent space, but I still think it's a nice way of thinking about it). The use is that you can just use A^1 rather than arbitrary curves. http://mathoverflow.net/questions/2791/understanding-gal-bar-q-q/2812#2812 Comment by David Corwin David Corwin 2013-04-29T13:16:58Z 2013-04-29T13:16:58Z I feel like this is true of most groups. E.g. the symmetric group - you need to choose an initial ordering of the letters you're permuting. http://mathoverflow.net/questions/7509/reference-for-delignes-construction-of-galois-representations-attached-to-modula/14117#14117 Comment by David Corwin David Corwin 2013-04-14T20:39:09Z 2013-04-14T20:39:09Z <a href="https://www.dpmms.cam.ac.uk/~ajs1005/preprints/mf.pdf" rel="nofollow">dpmms.cam.ac.uk/~ajs1005/preprints/mf.pdf</a> http://mathoverflow.net/questions/127157/status-of-global-langlands-conjecture-for-gl-2-over-mathbbq Comment by David Corwin David Corwin 2013-04-11T00:49:45Z 2013-04-11T00:49:45Z +1. And I had been thinking of asking the exact same question! http://mathoverflow.net/questions/127041/surreals-and-nsa-some-foundational-issues Comment by David Corwin David Corwin 2013-04-11T00:47:41Z 2013-04-11T00:47:41Z Aww, and I was hoping that the NSA was using surreal numbers! http://mathoverflow.net/questions/59071/what-elementary-problems-can-you-solve-with-schemes/59129#59129 Comment by David Corwin David Corwin 2013-04-04T01:05:25Z 2013-04-04T01:05:25Z +1 for the comment about Hartshorne http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/119006#119006 Comment by David Corwin David Corwin 2013-03-20T18:37:38Z 2013-03-20T18:37:38Z +1: &quot;Poof&quot; is a great new term for an &quot;incorrect proof,&quot; whether you intended it or not ;) http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/97416#97416 Comment by David Corwin David Corwin 2013-03-20T18:02:38Z 2013-03-20T18:02:38Z But this proves the result if there is at least one equivalence?