User dave anderson - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:46:21Z http://mathoverflow.net/feeds/user/5081 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114414/is-there-any-holomorphic-version-of-the-tubular-neighborhood-theorem/114462#114462 Answer by Dave Anderson for Is there any holomorphic version of the tubular neighborhood theorem? Dave Anderson 2012-11-25T23:02:21Z 2012-11-25T23:02:21Z <p>The existence of tubular neighborhoods is a remarkably strong constraint on submanifolds of projective space. A theorem of Morrow and Rossi (Math. Ann., 1978, if I can find a free link I'll attach it) says the following: Suppose $X$ is a connected holomorphic submanifold of ${\Bbb P}^n$ that has a holomorphic tubular neighborhood. Then $X$ is a linear subspace. (They also prove a similar statement for submanifolds of complex tori.)</p> http://mathoverflow.net/questions/106585/when-is-an-orbit-spherical/106587#106587 Answer by Dave Anderson for When is an orbit spherical? Dave Anderson 2012-09-07T10:16:40Z 2012-09-07T10:16:40Z <p>This was an active problem 20 years ago. I don't know if it's completely resolved, but more can be said for <em>affine</em> spherical homogeneous spaces. (Also---in general, a spherical variety need not be homogeneous, so searching for "spherical homogeneous variety might yield better results.) Try Brion, "Classification des espaces homogènes sphériques" (Compositio, 1987) and more recent work of Knop et al, e.g., <a href="http://arxiv.org/abs/math/0505102" rel="nofollow">http://arxiv.org/abs/math/0505102</a> . The latter has convenient tables at the end.</p> http://mathoverflow.net/questions/99661/pull-backs-and-tensor-product-in-algebraic-k-theory/99753#99753 Answer by Dave Anderson for pull backs (and tensor product) in algebraic K-theory Dave Anderson 2012-06-15T22:49:16Z 2012-06-15T22:49:16Z <p>There are a couple issues here. I think the main confusion is between K-theory of <em>vector bundles</em> ($K^\circ X$) and K-theory of <em>coherent sheaves</em> ($K_\circ X$). In Chriss-Ginzburg, I think they often assume $X$ is smooth, in which case the two are isomorphic. The isomorphism comes from resolving a coherent sheaf by vector bundles, and taking the alternating sum.</p> <p>The former is always a ring under tensor product, and pullback is indeed the naive one. As you say, pullback is not exact in general, but since tensor product with a locally free sheaf <em>is</em> exact, pullback on K-theory of vector bundles is well-defined, for any morphism and any spaces.</p> <p>For coherent sheaves, there are pullbacks for some classes of morphisms. The easiest case is when $X$ and $Y$ are smooth, so that there's an isomorphism with K-theory of vector bundles, and there's a pullback for any morphism $f: X \to Y$. More generally, the intuition is that a pullback $f^*: K_\circ Y \to K_\circ X$ should be defined by $$f^*[\mathcal{F}] = \sum (-1)^i [Tor^Y_i(\mathcal{O}_X,\mathcal{F})].$$ This makes sense whenever $f$ is a <em>perfect</em> morphism, i.e., $\mathcal{O}_X$ has a finite resolution by flat $f^{-1}\mathcal{O}_Y$-modules, because in that case the Tor sheaves are zero all but finitely many $i$. In particular, it does work when $f$ is flat, or a regular embedding.</p> <p>(A word of warning: these Tor sheaves are not computed using the pushforward $f_*\mathcal{O}_X$ in general, although that does work when $f$ is a closed embedding. The correct definition is to cover $Y$ and $X$ by affines, construct the Tor locally, and glue --- see EGA III.6.)</p> <p>The answer to your last question is yes: if $Y$ and $Z$ are transversally intersecting subvarieties of a smooth variety $X$, with $Y\cap Z = W$, then $[\mathcal{O}_Y]\cdot [\mathcal{O}_Z] = [\mathcal{O}_W]$, because transversality implies vanishing of the higher Tor's.</p> http://mathoverflow.net/questions/79943/how-gr2-7-and-gr3-6-are-related/79971#79971 Answer by Dave Anderson for How Gr(2,7) and Gr(3,6) are related? Dave Anderson 2011-11-03T20:17:36Z 2011-11-03T20:48:01Z <p>EDIT: The following idea doesn't work, for kind of obvious reasons (see the comments).</p> <hr> <p>As other answers have pointed out, the answer to the original question is no. However, taking up the idea from David Speyer's comment, both $Gr(3,6)$ and the Schubert divisor $\Delta$ in $Gr(2,7)$ are linear sections of $Gr(3,8)$, <em>by linear subspaces of the same dimension</em>. Specifically, as Schubert varieties inside $Gr(3,8)$, $$Gr(3,6) = \Omega_{(2,2,2)}$$ and $$\Delta = \Omega_{(5,1)}.$$ (I'm using notation where $\Omega_\lambda$ has codimension $|\lambda|$, for $\lambda$ a partition inside the $3 \times (8-3)$ rectangle.) One checks that these two Schubert varieties are both defined by the vanishing of 36 Plücker coordinates (on $Gr(3,8)$). </p> <p>Taking any curve in the Grassmannian of codimension-36 subspaces inside ${\Bbb P}^N$, where $N = \binom{8}{3}-1$, connecting the two linear spaces cutting out $Gr(3,6)$ and $\Delta$, you should get a flat family having these two as fibers, explaining why they have the same Hilbert polynomial.</p> http://mathoverflow.net/questions/78483/reference-request-representation-of-type-g2-lie-algebras/78737#78737 Answer by Dave Anderson for Reference request: representation of type G2 Lie algebras. Dave Anderson 2011-10-21T02:43:12Z 2011-10-21T02:43:12Z <p>Another direction, that works (only?) for groups of rank at most 2, is Kuperberg's "spider" model: <a href="http://arxiv.org/abs/q-alg/9712003" rel="nofollow">http://arxiv.org/abs/q-alg/9712003</a> . The web diagrams he constructs generalize tableaux, in the sense that the $A_1$ and $A_2$ versions are in bijection with tableaux (at least of rectangular shape). This is not quite the same as constructing representations via Schur functors, but it has a combinatorial appeal.</p> http://mathoverflow.net/questions/75888/line-bundles-on-ind-schemes/75898#75898 Answer by Dave Anderson for Line bundles on Ind Schemes Dave Anderson 2011-09-20T00:38:35Z 2011-09-20T00:38:35Z <p>I don't know the general story, but I think the correspondence between Weil divisors and line bundles breaks down already for $\mathbf{A}^\infty$, which is pretty much the smoothest ind-variety out there. Using the standard ind-structure on $\mathbf{A}^\infty$, the example I have in mind is the union of all hyperplanes $x_n-x_1=0$. This is a codimension one closed ind-subscheme, but it's not the zero locus of any polynomial in $\mathcal{O}_{\mathbf{A}^\infty}$.</p> http://mathoverflow.net/questions/68568/is-there-always-a-toric-isomorphism-between-isomorphic-toric-varieties/68572#68572 Answer by Dave Anderson for Is there always a toric isomorphism between isomorphic toric varieties? Dave Anderson 2011-06-23T01:10:31Z 2011-06-23T01:10:31Z <p>This is a partial answer. Let $X$ be the given abstract variety. I think the question is equivalent to asking whether all maximal tori in the group $\mathrm{Aut}(X)$ are conjugate. When $X$ is complete, this is a linear algebraic group, so all maximal tori are conjugate, and the answer is affirmative. (See Cox's famous paper on the homogeneous coordinate ring.)</p> <p>If $X$ is not complete, the automorphism group may of course be infinite-dimensional, but perhaps you can argue by compactifying.</p> http://mathoverflow.net/questions/67959/cone-of-effective-divisors/67987#67987 Answer by Dave Anderson for Cone of effective divisors! Dave Anderson 2011-06-16T18:53:55Z 2011-06-16T18:53:55Z <p>As mentioned in the comments, the <em>(pseudo)effective cone</em> $\overline{\mathrm{Eff}}(X)$, defined as the closure of the cone of all effective divisors on $X$, is certainly an object of study, and Lazarsfeld's book is a good reference. Your complaint that he doesn't say much about its structure is surely related to the fact that so little is known! Here are a few general things I'm aware of:</p> <ul> <li><p>The interior of the effective cone is the <em>big cone</em>, i.e., the cone of line bundles with positive volume.</p></li> <li><p>The dual of the effective cone is the <em>cone of moveable curves</em>, see Boucksom-Demailly-Paun-Peternell.</p></li> <li><p>As part of their work on the minimal model program, Birkar-Cascini-Hacon-McKernan prove that log Fano varieties have finitely generated effective cones.</p></li> </ul> <p>And here are a couple specific instances where one knows more:</p> <ul> <li><p>When $X$ admits an action by a solvable group with a dense orbit, the effective cone is generated by the components of the complement of the orbit. (This works when $X$ is, e.g., a toric variety or a Schubert variety.)</p></li> <li><p>There's been a lot of recent work on the case $X=\overline{M}_{0,n}$, see e.g., Hu-Keel, Hassett-Tschinkel, Castravet-Tevelev.</p></li> </ul> http://mathoverflow.net/questions/64793/bounding-the-number-of-critical-values-of-a-map-between-varieties/65182#65182 Answer by Dave Anderson for Bounding the number of critical values of a map between varieties Dave Anderson 2011-05-16T21:44:00Z 2011-05-16T21:44:00Z <p>Hi Matt,</p> <p>I'm not sure this helps, but you could write your $X$ as the zero locus of a single degree $4$ polynomial. Of course, if your coordinate projection is sufficiently general, it's a Morse function and the Milnor-Thom bounds on betti numbers should work for you. For a special projection, it seems like critical points would collide (and so decrease in number) but I could be completely wrong.</p> http://mathoverflow.net/questions/64613/a-bound-on-the-top-homology-of-a-complement-to-a-variety-in-mathbb-cn/64661#64661 Answer by Dave Anderson for A bound on the top homology of a complement to a variety in $\mathbb C^n$ Dave Anderson 2011-05-11T17:19:09Z 2011-05-11T17:31:56Z <p>The sharp bound is this: For any closed algebraic set $V$ of codimension $d$ in ${\Bbb C}^n$, with $U={\Bbb C}^n \setminus V$, one has $\pi_i(U) = 0$ for $0 &lt; i\leq 2d-2$ and $\pi_{2d-1}(U) \neq 0$. Using the Hurewicz isomorphism, you get the same vanishing and non-vanishing for homology.</p> <p>A simple proof is in the appendix to my notes (with Fulton) on equivariant cohomology, <a href="http://www.math.washington.edu/~dandersn/eilenberg/" rel="nofollow">http://www.math.washington.edu/~dandersn/eilenberg/</a> . A slick reason for vanishing was pointed out by David Speyer: given a (nice) map of an $i$-sphere into $U$, the (real) lines between the points in image of the sphere and points in $V$ sweep out a space of dimension at most $(2n-2d)+i+1$. When $i&lt;2d-1$, you can pick a point in $U$ not lying on any such line, and contract your sphere down to that point. The non-vanishing happens because all algebraic sets have nontrivial fundamental classes in Borel-Moore homology. (Vanishing can also be proved using B-M homology.)</p> http://mathoverflow.net/questions/56440/what-does-a-homogeneous-space-of-a-linear-algebraic-group-know-about-the-group/56445#56445 Answer by Dave Anderson for What does a homogeneous space of a linear algebraic group know about the group? Dave Anderson 2011-02-23T20:42:24Z 2011-02-23T20:42:24Z <p>For the special case where $X$ is projective, and reducing (by the "trivial" examples of surjections and normal quotients) to $G$ a semisimple group of adjoint type, with $H=P$ a parabolic, one version of the answer is given by Demazure in "Automorphismes et déformations des variétés de Borel", <em>Invent. Math.</em>, 1977. In a nutshell, the automorphism group of such an $X$ is the subgroup of $Aut(G)$ preserving the conjugacy class of $P$ --- unless the pair $(G,P)$ is one of several "exceptional" cases. (E.g., the $5$-dimensional quadric is both $G_2/P$ and $SO_7/P$.)</p> <p>In other words, for the non-exceptional cases, you can recover $G$ as the connected component of the identity in $Aut(X)$.</p> http://mathoverflow.net/questions/56363/list-of-classifying-spaces-and-covers/56371#56371 Answer by Dave Anderson for List of Classifying Spaces and Covers Dave Anderson 2011-02-23T08:24:40Z 2011-02-23T08:24:40Z <p>As Craig Westerland points out, the space $V_n$ of full-rank $\infty\times n$ complex matrices (aka the infinite Stiefel manifold) works for any subgroup of $GL_n({\Bbb C})$. That applies to all the examples listed in the question, except the free group -- but of course, it's nice to have alternative descriptions, such as the ones given above, where it's easier to get a handle on $BG$. (A compendium of these things would be nice to have.) Here are a few more:</p> <p>(1) For $G$ upper triangular matrices in $GL_n({\Bbb C})$, $EG=V_n$ gives $BG=Fl(1,\ldots,n;{\Bbb C}^\infty)$.</p> <p>(2) For $G=({\Bbb C}^\ast)^n$, you can take $EG=(V_1)^n$ and get $BG=({\Bbb CP}^\infty)^n$. Or you could take $EG=V_n$ and get $BG$ as the space of $n$-dimensional subspaces of ${\Bbb C}^\infty$, together with a splitting into lines. The latter has the advantage of coming with an obvious map to the space from (1), realizing the homotopy equivalence explicitly. (I mention this to point out that it can be useful to have different choices available, even for "decomposable" groups.)</p> <p>(3) For $G=Sp(2n)$ (the compact symplectic group), take $EG$ to be "full-rank" $\infty\times n$ quaternionic matrices, and get $BG = Gr({\Bbb H^n},{\Bbb H}^\infty)$. (It has the same cell structure as the complex Grassmannian, but with cells in dimensions $4k$ instead of $2k$.)</p> <p>(4) For $G=Sp_{2n}({\Bbb C})$, you can take $BG=Sp_{\infty}/(Sp_{2n}\times Sp_{\infty})$ (interpreted as a suitable limit).</p> http://mathoverflow.net/questions/55565/question-on-kahler-ample-cone-cone-of-curves/55643#55643 Answer by Dave Anderson for Question on Kähler/ample cone, cone of curves.... Dave Anderson 2011-02-16T16:57:21Z 2011-02-16T16:57:21Z <p>To augment Sándor's answer, especially for "Question 3.5": On a smooth variety $X$, for cycles of any codimension, homological equivalence always implies numerical equivalence, so you're asking when the two are the same. They certainly differ by torsion, so let's use ${\Bbb Q}$ coefficients and ask when $$Z^k/Hom^k \to Z^k/Num^k$$ is an isomorphism. This question is part of Grothendieck's "standard conjectures" and is unresolved in general; however it is always true for $k=1$, and apparently also for $k=2$, as well as for any $k$ on abelian varieties; see Fulton's <em>Intersection Theory</em>, 19.3. (In case the notation isn't self-explanatory, $Z^k$ is codimension $k$ cycles, $Hom^k$ is the kernel of the map to $H^{2k}$, and $Num^k$ is cycles numerically equivalent to zero.)</p> http://mathoverflow.net/questions/54929/multiplication-tables-for-hg-p/54973#54973 Answer by Dave Anderson for Multiplication tables for H*(G/P)? Dave Anderson 2011-02-10T01:44:26Z 2011-02-10T01:44:26Z <p>Some people have written code to do this, but not necessarily the most efficient possible way. Alex Yong wrote one that works for general type, and it's linked in the references for the paper by H. Thomas and A. Yong, "A combinatorial rule for (co)minuscule Schubert calculus" (in <em>Adv. Math.</em>, or on the arXiv). As for published tables, for small rank some are given in Griffeth-Ram, "Afﬁne Hecke algebras and the Schubert calculus". (I also wrote down the table for $G_2$ in my thesis.) There are probably many other sources, these are just the ones that come to mind.</p> http://mathoverflow.net/questions/54502/affine-bundles-over-varieties/54511#54511 Answer by Dave Anderson for Affine bundles over varieties Dave Anderson 2011-02-06T09:23:18Z 2011-02-06T09:23:18Z <p>I think the term "affine bundle" is used for at least two things: (1) A map $p:Y\to X$ such that for some open cover (in your choice of topology) there are isomorphisms $p^{-1}(U)={\Bbb A}^n \times U$ --- just like you said. (2) A torsor for a vector bundle, i.e., like (1) but with the added condition that the transition functions are affine-linear.</p> <p>In the situation of (2), it's a vector bundle exactly when there's a section (like any torsor). A simple non-vector-bundle-example is the complement of the diagonal in ${\Bbb P}^1 \times {\Bbb P}^1$, projecting onto one of the factors.</p> <p>For (1), I don't know any general (non-trivial) criterion for such a thing to be a vector bundle. (Maybe because the group $Aut({\Bbb A}^n)$ is so complicated...) A simple non-example is the 2nd-order jet scheme $\mathrm{Hom}(\mathrm{Spec}(k[t]/(t^3)),{\Bbb P}^1) \to {\Bbb P}^1$. The fibers are ${\Bbb A}^2$, and there's a section, but it's not linear. I suppose one test is whether the sheaf of $O_X$-algebras $p_*O_Y$ admits a grading generated in degree one. (This fails for the jet schemes, though there is a natural grading by scaling $t$.)</p> http://mathoverflow.net/questions/53306/what-can-we-learn-from-the-tropicalization-of-an-algebraic-variety/53310#53310 Answer by Dave Anderson for What can we learn from the tropicalization of an algebraic variety? Dave Anderson 2011-01-26T01:48:28Z 2011-01-26T17:07:05Z <p>I hope others will have lots more to say, but one nice property is $$g(X) \geq b_1(\mathrm{Trop}(X))$$ when $X$ is a (plane) curve, where $g$ is the genus and $b_1$ is the betti number (=number of cycles) of the graph. (Thanks to quim for the correcting the inequality.)</p> http://mathoverflow.net/questions/53000/cohomology-of-structure-sheaves-algebraic-constructible-and-more/53020#53020 Answer by Dave Anderson for Cohomology of Structure Sheaves: Algebraic, Constructible and more Dave Anderson 2011-01-24T01:51:28Z 2011-01-24T01:51:28Z <p>A couple little things to try to complement the previous answers:</p> <p>(1) Partially addressing your first question: If you are interested in complex projective varieties, Serre's GAGA theorem says the cohomology groups of the structure sheaf of analytic functions (or other coherent analytic sheaf) are the same as those of the algebraic structure sheaf. So in this case, nothing changes when you move from regular algebraic functions to analytic ones.</p> <p>(2) For real (semi)algebraic varieties, I think the natural sheaf would be that of semi-algebraic functions, though I know fairly little about it. The place to look would be the book by Bochnak, Coste, and Roy.</p> http://mathoverflow.net/questions/49827/chern-classes-of-pushforwards/49829#49829 Answer by Dave Anderson for Chern classes of pushforwards Dave Anderson 2010-12-18T22:34:59Z 2010-12-18T23:49:09Z <p>Do you know the (generic) degree of your map $f$? As you probably know, standard intersection theory says $f_*[D] = n[f(D)]$ as classes in $A_{d-1}Y$, where $n$ is the degree of $f$ (restricted to $D$) and $d=\dim X = \dim Y$. No flatness or smoothness hypotheses on $f$ are needed for this; the sticky point is in identifying these divisors with line bundles. But since you're dealing with smooth DM stacks, that should be ok (over ${\Bbb Q}$ at least).</p> <p>EDIT (incorporating the comments): For a proper map $f$, there is a map defined at the cycle level by $$f_*[D] = n\cdot [f(D)],$$ where $n$ is the degree of $D$ over $f(D)$ (i.e., degree of the induced field extension) when these have equal dimension, and $n=0$ when $\dim f(D)&lt; \dim D$. This passes to rational equivalence, so defines a map $A_{d-1}X \to A_{d-1}Y$. In particular, if $f$ collapses a divisor $D$, then $f_*[D]=0$.</p> <p>All this is in Fulton's <em>Intersection Theory</em>, Section 1.4.</p> http://mathoverflow.net/questions/48477/never-appeared-forthcoming-papers/48516#48516 Answer by Dave Anderson for Never appeared forthcoming papers Dave Anderson 2010-12-06T23:23:30Z 2010-12-06T23:23:30Z <p>The comment about stacks in the paper that first used them in an essential way probably belongs in this list:</p> <p>"Full details on the basic properties and theorems for algebraic stacks will be given elsewhere." (Deligne-Mumford, <em>The irreducibility of the space of curves of given genus</em>, 1969.)</p> <p>They don't quite say <em>they</em> will give the details in a paper, of course, so maybe it doesn't count.</p> http://mathoverflow.net/questions/48386/the-construction-of-push-forward-in-algebraic-equivarient-k-theory/48389#48389 Answer by Dave Anderson for The construction of push-forward in algebraic equivarient K-theory Dave Anderson 2010-12-05T20:09:38Z 2010-12-05T20:09:38Z <p>It's the same as in non-equivariant $K$-theory. For a $G$-equivariant proper morphism $f:X \to Y$ and an equivariant coherent sheaf $F$ on $X$, define $$f^G_*[F] = \sum (-1)^i [R^i f_*F],$$ which makes sense because each higher direct image is equivariant (and coherent, because $f$ is proper).</p> <p>An interesting situation is when $Y$ is a point, in which case this is the "equivariant Euler characteristic," an alternating sum of (virtual) $G$-representations.</p> http://mathoverflow.net/questions/48103/is-intersection-homology-the-usual-homology-of-something-else/48111#48111 Answer by Dave Anderson for Is intersection homology the usual homology of something else? Dave Anderson 2010-12-02T22:28:28Z 2010-12-03T15:32:14Z <p>I'm not sure this is the kind of answer you want, but if $X$ has a <em>small resolution</em> $f:X' \rightarrow X$ (so $X'$ is a manifold and the dimension of fibers is sufficiently small), then there is an induced isomorphism $IH_{\ast}(X) = IH_{\ast}(X')$, and because $X'$ is smooth, the latter group is $H_{\ast}(X')$.</p> <p>More precisely, a proper (birational) map is <em>small</em> if the set $${x \in X | \dim f^{-1}(x) \geq r }$$ has codimension more than $2r$; such maps induce isomorphisms on IH.</p> <p>Of course, there is nothing natural about $X'$, nor do small resolutions necessarily exist (see the comment below by Mike Skirvin for an easy example). Hopefully someone more knowledgeable about IH will have something to say.</p> http://mathoverflow.net/questions/47660/symmetric-sequence-of-blow-ups-for-the-fulton-macpherson-compactification/47692#47692 Answer by Dave Anderson for Symmetric sequence of blow-ups for the Fulton-MacPherson compactification Dave Anderson 2010-11-29T16:02:30Z 2010-11-29T16:02:30Z <p>The interesting question of to what extent "wonderful compactifications" like the Fulton-MacPherson space depend on the order of blowups was studied -- and I think, mostly resolved -- by Li Li in his thesis. The paper <a href="http://arxiv.org/abs/math/0611412" rel="nofollow">http://arxiv.org/abs/math/0611412</a> gives "a condition on the order of blow-ups in the construction....such that each blow-up is along a nonsingular center."</p> http://mathoverflow.net/questions/46895/quasiprojectiveness-of-bundle/46899#46899 Answer by Dave Anderson for Quasiprojectiveness of bundle Dave Anderson 2010-11-22T03:21:00Z 2010-11-22T03:21:00Z <p>Yes. The easiest case is when $X$ is affine, say $X=\mathrm{Spec}(A)$. Then $E$ is associated to some locally free $A$-module $M$, and can be realized as $E = \mathrm{Spec} (Sym_A(M^\vee))$. </p> <p>More generally, if $X$ is quasiprojective, take an ample line bundle $L$ on $X$. Then $P(E) = P(E\otimes L)$, and replacing $E$ by a sufficiently high twist by $L$, the line bundle $O_{P(E)}(1)$ is ample.</p> <p>Note that you can realize $E$ as an open subset of the projective bundle $P(E\oplus 1)$, so (quasi)projectivity of the latter implies that of the former.</p> http://mathoverflow.net/questions/46418/relation-between-toric-geometry-and-log-geometry/46459#46459 Answer by Dave Anderson for relation between toric geometry and log geometry Dave Anderson 2010-11-18T04:56:54Z 2010-11-18T06:37:26Z <p>Picking up some of Eric Zaslow's reformulation: Assume $P$ is <em>commutative</em>, <em>saturated</em>, and <em>cancellative</em>, as well as finitely generated. The answer to your question is affirmative if and only if the "groupification" $P^{gp}$ of $P$ is torsion-free. (As mentioned in Dustin's answer, saturated means that for all $p$ in $P^{gp}$, $np \in P$ implies $p\in P$. Cancellative means $p_1+q=p_2+q$ implies $p_1=p_2$.) </p> <p>All this amounts to $P$ being embeddable as a sub-monoid of ${\Bbb Z}^n$ for some $n$. Then take the subgroup of ${\Bbb Z}^n$ spanned by $P$. This is isomorphic to some ${\Bbb Z}^m$; take the dual of the convex hull of $P$ in ${\Bbb R}^m$ and you've got your cone $\sigma$, just as Eric says. When $P$ is saturated, it is equal to $\sigma^\vee \cap M$; otherwise, this gives the <em>saturation</em> of $P$, corresponding to the integral closure of $k[P]$.</p> <p>Depending on what references you use, when $P^{gp}$ is torsion-free, $P$ is called either <em>integral</em> or <em>toric</em>. (See, e.g., the toric variety notes on M. Mustata's <a href="http://www.math.lsa.umich.edu/~mmustata/toric_var.html" rel="nofollow">webpage</a> versus the log geometry notes on Danny Gillam's <a href="http://www.math.brown.edu/~wgillam/" rel="nofollow">webpage</a>; both sources are worth looking at.) It seems the latter terminology is more standard in the log geometry world, where "integral" sometimes just means "cancellative".</p> http://mathoverflow.net/questions/46127/g-bundles-on-affine-spaces/46134#46134 Answer by Dave Anderson for $G$-bundles on affine spaces Dave Anderson 2010-11-15T18:58:02Z 2010-11-15T18:58:02Z <p>Over an algebraically closed field, for $G$ connected and reductive, every principal $G$-bundle on ${\Bbb A}^n$ is trivial, also by a theorem of Raghunathan:</p> <p>"Principal bundles on affine space", in <em>C. P. Ramanujam—a tribute</em>, pp. 187–206, Tata Inst. Fund. Res. Studies in Math. <strong>8</strong> (1978).</p> <p>(Unfortunately I can't find this reference free online.)</p> http://mathoverflow.net/questions/43183/structure-of-iterated-mathbbp1-bundles/43190#43190 Answer by Dave Anderson for Structure of iterated $\mathbb{P}^1$-bundles Dave Anderson 2010-10-22T15:21:49Z 2010-10-22T15:21:49Z <p>This is only a partial answer, but the paper</p> <p>M. Willems, "K-théorie équivariante des tours de Bott", Duke Math. J. 132 (2006)</p> <p>includes a combinatorial description of a degeneration of Bott-Samelson varieties to toric ones. (It's based on a construction of Grossberg and Karshon, and was also done algebraically by Pasquier.)</p> http://mathoverflow.net/questions/42548/poincare-dual-in-equivariant-cohomology/42789#42789 Answer by Dave Anderson for Poincare dual in equivariant (co)homology? Dave Anderson 2010-10-19T14:14:55Z 2010-10-21T01:13:46Z <p>There is a paper by Brion: "Poincaré duality and equivariant (co)homology," <em>Michigan Math. J.</em> 48 (2000). As mentioned in one of the comments, he uses Borel-Moore as the homology theory.</p> <p>That said, one should always be careful about the word "duality" in this context -- often people just mean there's a canonical <em>isomorphism</em> $H_G^k X \rightarrow H^G_{n-k}X$. (Which does exist using equivariant Borel-Moore homology, when $X$ is smooth.) The geometric meaning is the same as in the ordinary case: if $V$ is a $G$-invariant sub(manifold/variety/cycle/etc) of codimension $k$, it defines a fundamental class in $H^G_{n-k} X$, which is identified with $H_G^kX$ by means of the isomorphism.</p> <p>Duality could also refer to a pairing $H_G^k X \otimes H_G^\ell X \rightarrow H_G^{k+\ell-n}(point)$, given by the equivariant pushforward (integral): $$\alpha\otimes\beta \mapsto \int \alpha\cdot\beta .$$ Using this pairing, and assuming that $H_G^*X$ is free as a module over $H_G^*(point)$, any $H_G^*(point)$-module basis for $H_G^*X$ has a dual basis, in the usual sense of linear algebra. (Arguably, this duality pairing is what should really be called "Poincaré duality".) Again, the geometric meaning is similar to the ordinary case: if you're lucky enough to have invariant subvarieties whose classes form a basis for $H_G^*X$, the Poincaré dual basis is given by classes of subvarieties intersecting the original ones transversally (if they exist). However, the existence of geometrically defined dual bases is a stronger statement, because the equivariant integral is generally nonzero on classes of degree greater than $\dim X$.</p> <p>(I should say all this is prejudiced toward the equivariant cohomology rings that usually show up in algebraic geometry, e.g., $H_G^*X$ is a free module over $H_G^*(point)$, so that it makes sense to talk about dual bases.)</p> <p>PS: Defining equivariant Borel-Moore homology requires a little more care, since the spaces $EG\times^G X \rightarrow BG$ are infinite-dimensional fiber bundles. But they have finite-dimensional approximations $EG_m \times^G X \to BG_m$, so it makes sense to define $$H^G_k X = H_{k+\dim BG_m}(EG_m\times^G X)$$ for $m\gg0$. The equivariant Poincaré isomorphism is just the ordinary one for these approximation spaces.</p> http://mathoverflow.net/questions/40484/mori-cone-of-homogeneous-varieties/40548#40548 Answer by Dave Anderson for Mori cone of homogeneous varieties Dave Anderson 2010-09-29T22:13:54Z 2010-09-29T22:13:54Z <p>One place this is treated is in Brion's notes, <a href="http://arxiv.org/abs/math/0410240" rel="nofollow">http://arxiv.org/abs/math/0410240</a>, particularly Section 1.4, and the references in the notes at the end of that section.</p> <p>A little more precisely, one shows that a divisor (line bundle) is nef iff it is globally generated, and this cone is generated by the Schubert divisors. Using the fact that the Schubert classes form a self-dual basis for $H^*(X,{\Bbb Z})$ (as well as $A^*X$) under the intersection pairing, it follows that the "dual classes" to the Schubert divisors generate the cone of curves. (As with many such things in $G/P$ world, the arguments for cohomology and Chow groups are exactly the same.)</p> <p>By the way, it's an easy consequence of the existence of a dense open $B$-orbit that the pseudo-effective cone is generated by the complement of the open orbit --- that is, again by the Schubert divisors. So for $G/P$, the nef and pseudo-effective cones are the same.</p> http://mathoverflow.net/questions/38599/ampleness-of-the-universal-subbundle/38602#38602 Answer by Dave Anderson for Ampleness of the universal subbundle Dave Anderson 2010-09-13T17:29:40Z 2010-09-13T17:29:40Z <p>I think this is essentially never true, again by restricting to a fiber over $x\in X$. The problem is that (somewhat counterintuitively) the universal quotient bundle on $Gr(k,n)$ is not ample, and for the same reason, neither is the dual of the universal sub. (Except of course when $k=1$!) See Examples 6.1.5 and 6.1.6 in Lazarsfeld's <em>Positivity in Algebraic Geometry II</em>.</p> http://mathoverflow.net/questions/37602/polynomial-invariants-of-the-exceptional-weyl-groups/38353#38353 Answer by Dave Anderson for Polynomial invariants of the exceptional Weyl groups Dave Anderson 2010-09-10T20:44:10Z 2010-09-10T20:44:10Z <p>For the record, these invariants (or rather, the ideals of positive-degree invariants) also come up in the Borel presentation of the cohomology ring of the flag manifold G/B, so one can find generators whenever people have computed these rings. For instance, the preprint <a href="http://arxiv4.library.cornell.edu/pdf/0911.4793v1" rel="nofollow">http://arxiv4.library.cornell.edu/pdf/0911.4793v1</a> gives completely explicit polynomials for E7 in Proposition 2.1 and for E8 in Lemma 2.3. I can't vouch for their correctness -- a nontrivial computational matter, as Jim remarked -- but the calculations are written out in some detail.</p> http://mathoverflow.net/questions/129239/are-all-equivariant-bundles-of-a-total-flag-manifold-constructable-from-line-bund/129240#129240 Comment by Dave Anderson Dave Anderson 2013-04-30T23:34:47Z 2013-04-30T23:34:47Z Sam Gunningham's answer cleared this up, but another point view is: although vector bundles on $SL_n/B$ may not split <i>algebraically</i>, on the <i>real</i> manifold $F(n) = SU(n)/U(1)^{n-1}$, every extension of vector bundles does split: choose a hermitian metric. (Indeed, identifying $F(n)$ this way is essentially doing just that.) http://mathoverflow.net/questions/114011/how-should-the-degree-of-a-variety-be-defined-in-a-weighted-projective-space/114014#114014 Comment by Dave Anderson Dave Anderson 2012-11-21T09:24:53Z 2012-11-21T09:24:53Z The Chow groups are indeed free, thanks to a computation of Al-Amrani from the 1980's. (The K_0 is also free, also by Al-Amrani.) http://mathoverflow.net/questions/106460/bruhat-decomposition-of-a-quadric-hypersurface/106495#106495 Comment by Dave Anderson Dave Anderson 2012-09-06T16:47:16Z 2012-09-06T16:47:16Z @Christopher: Of course there's also the general Lie theory version, but it sounds like you want something more explicit, along the lines of David's nice answer. For that you might try Chapter 6 of Fulton-Pragacz, <i>Schubert varieties and degeneracy loci</i>. (They focus on complete flags, but give very concrete parametrizations via matrices.) I'm sure there are other references... http://mathoverflow.net/questions/100854/is-there-a-concrete-description-of-relative-jet-spaces-that-invokes-neither-inter Comment by Dave Anderson Dave Anderson 2012-07-02T22:14:28Z 2012-07-02T22:14:28Z (+1) This is something I've been asking about for a while. Here is another thing one could desire: there is a grading on $A^{(n)}$ coming from scaling $\epsilon$. Is there an intrinsic description that makes this grading manifest? Again, the $n=1$ case is clear, it's the natural grading on the symmetric algebra. (Also, to second Jason Starr's comment about the name, there's an unfortunate clash of terminology: &quot;jet bundle&quot; typically refers to something else---the infinitesimal neighborhoods of diagonals for smooth schemes.) http://mathoverflow.net/questions/99661/pull-backs-and-tensor-product-in-algebraic-k-theory Comment by Dave Anderson Dave Anderson 2012-06-15T23:00:37Z 2012-06-15T23:00:37Z @Rob, I'm not sure my response completely answers your question, but this is partly because it's a little hard to parse what you're asking. Could you tighten up the question a little to clarify? Thanks! http://mathoverflow.net/questions/81035/coincidences-amongst-classifying-spaces-and-eilenberg-mac-lane-spaces/81045#81045 Comment by Dave Anderson Dave Anderson 2011-11-16T23:25:22Z 2011-11-16T23:25:22Z This is neat: in each of these three cases, you have $K( \pi_i(G), i+1 )$. http://mathoverflow.net/questions/79943/how-gr2-7-and-gr3-6-are-related/79971#79971 Comment by Dave Anderson Dave Anderson 2011-11-04T22:15:25Z 2011-11-04T22:15:25Z Dustin, I agree -- there's no reason to expect the flat family to be over ${\Bbb P}^1$. Nonetheless, there must be some family. (And given that we're talking about Grassmannians, one could <i>hope</i> for a nice family!) David's comment about non-equal cohomology classes in $Gr(3,8)$ still prohibits it from being a flat family (by any chain of curves) inside that Grassmannian, though. http://mathoverflow.net/questions/79943/how-gr2-7-and-gr3-6-are-related/79971#79971 Comment by Dave Anderson Dave Anderson 2011-11-04T17:30:22Z 2011-11-04T17:30:22Z @ M I, in some sense, no, this is the only possible reason. If two subvarieties of ${\Bbb P}^n$ have the same Hilbert polynomial, then they are related by a flat family (by connectivity of the Hilbert scheme). http://mathoverflow.net/questions/79943/how-gr2-7-and-gr3-6-are-related/79971#79971 Comment by Dave Anderson Dave Anderson 2011-11-03T20:42:11Z 2011-11-03T20:42:11Z Whoops! Thanks for the correction... after posting, I got a feeling something was off... http://mathoverflow.net/questions/79292/f-4-flag-variety Comment by Dave Anderson Dave Anderson 2011-10-28T16:23:32Z 2011-10-28T16:23:32Z @Alex, another reason, quantifying &quot;easily defined by matrices&quot;, is that the difficulty of these descriptions increases with the ratio (dimension of minimal representation)/(rank of $G$). So $E_8$ is probably hopeless from this point of view. (Which may explain why Skip Garibaldi's paper mentioned in Steven's answer doesn't treat that case.) http://mathoverflow.net/questions/79292/f-4-flag-variety Comment by Dave Anderson Dave Anderson 2011-10-27T23:28:59Z 2011-10-27T23:28:59Z Asad, I just saw your comment -- so my comment basically says: it's (trivially) the line in ${\Bbb P}(V_1)$ fixed by $P$. Perhaps not so helpful. http://mathoverflow.net/questions/79292/f-4-flag-variety Comment by Dave Anderson Dave Anderson 2011-10-27T23:25:23Z 2011-10-27T23:25:23Z Is the question &quot;how to realize this homogeneous space as parametrizing certain subspaces&quot;? If so, I think it's a very interesting one in general, and only partially known. For that specific $F_4$-variety, though, I think it's a hypersurface in the octonionic projective plane. To get a concrete picture, you'll want to look at Albert algebras. Unraveling the definition of OP^2, this does parametrize certain (1-dimensional) subspaces. The more interesting parts will be for other fundamental representations... (I'm assuming you're dealing with complex forms, btw.) http://mathoverflow.net/questions/78483/reference-request-representation-of-type-g2-lie-algebras/78737#78737 Comment by Dave Anderson Dave Anderson 2011-10-21T19:29:52Z 2011-10-21T19:29:52Z @Steven, thanks, glad to know it was useful! http://mathoverflow.net/questions/76101/segre-classes-vs-chern-classes/76127#76127 Comment by Dave Anderson Dave Anderson 2011-09-22T15:57:30Z 2011-09-22T15:57:30Z Read Fulton's <i>Intersection Theory</i>. Prop. 4.1 says the definition you're asking about is equivalent to Sasha's for vector bundles. (Look at Ex. 4.1.6, too, for general coherent sheaves.) http://mathoverflow.net/questions/75888/line-bundles-on-ind-schemes/75898#75898 Comment by Dave Anderson Dave Anderson 2011-09-21T14:34:02Z 2011-09-21T14:34:02Z @Laurent, I may be making a silly mistake, but I think the intersection of my divisor with $X_n$ is the hyperplane arrangement with $x_i=x_1$ for $i=1,...,n$, plus $x_1=0$. (I'm using the standard ind-structure, where $X_n$ has $x_i=0$ for all $i&gt;n$.) I agree that both are Zariski-dense, but I'm using the ind-topology on $\mathbf{A}^\infty$. (A subset is closed iff its intersection with each $X_n$ is.)