Impact
~103k
people reached
- 0 posts edited
- 0 helpful flags
- 2 votes cast
Oct
14 |
comment |
Are extensions of linear algebraic groups (over a field) themselves linear algebraic?
Dear Michael: I agree with your final comment, but I think that the notion of "reduced group scheme" (of finite type, say) over a field is uninteresting/useless except over a perfect field, for which it implies smoothness (and so is preserved under base change). The point is that if $G$ is a possibly non-reduced group scheme (of finite type) over an imperfect field then $G_{\rm{red}}$ is typically not a subgroup scheme. Results like the one you cite from SGA3 seem to be examples of largely useless generality. |
Oct
13 |
comment |
Are extensions of linear algebraic groups (over a field) themselves linear algebraic?
Dear Michael: The point of my Frobenius example was just that the notion of "reduced group" is very delicate when the ground field isn't perfect (and as I guessed, OB meant to say geometric reducedness). Anyway, bringing in stacks for group quotients over a field seems like overkill: by SGA3, for any lft group $G$ over an artin local ring and flat closed subgroup scheme $H$, there is an fppf scheme map $G \rightarrow X$ identifying $X$ with the quotient sheaf $G/H$ (so a group when $H$ is functorially normal) and having all properties one could desire, such as compatibility with base change. |
Oct
12 |
comment |
Are extensions of linear algebraic groups (over a field) themselves linear algebraic?
@Olivier: Since Michael seems to be working over a general field, one should say "smooth" rather than "reduced". As you know, over any imperfect field there are reduced linear algebraic groups that are not smooth, and their relative Frobenius morphism is a finite surjective homomorphism which is not flat. Also, I see your intent by saying "flatness should somehow be part of the definition of being surjective", but this seems a bit risky since surjective has its own useful (ordinary) meaning for scheme maps. However,"fppf" requires fewer letters than "surjective", so using French solves it. :) |
Oct
12 |
comment |
Are extensions of linear algebraic groups (over a field) themselves linear algebraic?
@Michael: I assumed (and probably Angelo did too) that you know the equivalence of several equivalent definitions of "group extension" (without which it is hard to work with this concept in a nice way). What definition are you using (especially if you aren't assuming smoothness of the groups)? |
Oct
12 |
comment |
Are extensions of linear algebraic groups (over a field) themselves linear algebraic?
Sure. If $1 \rightarrow G' \rightarrow G \rightarrow G'' \rightarrow 1$ is a short exact sequence of fppf group sheaves over a scheme $S$ with $G''$ representable and $G'$ is $S$-affine and fppf over $S$ then $G$ is representable and $G \rightarrow G''$ is affine and fppf (so $G$ is $S$-affine if $G''$ is, same for fppf). This is proved by identifying $G$ as a $G'$-torsor sheaf over $G''$ for the fppf topology (sheaf quotient maps have "local" sections!) and using effectivity of fppf descent for affine morphisms. It is explained in Oort's LNM book on commutative (!) group schemes. |
Oct
8 |
comment |
complex reductive Lie groups which are not defined over the real numbers
The way you seem to be defining "complex reductive group" is not the standard procedure, and almost defines the answer to be negative (once you clarify what you mean by "defined over the real numbers"; e.g., the complexification structure you give makes the answer negative tautologically). But even if you use the "right" definition (in terms of the theory of linear algebraic groups) there are no examples because any connected reductive group over an alg. closed field of char. 0 is defined over $\mathbf{Q}$ (e.g., by inspecting the classification of simply connected cases and their centers). |
Oct
5 |
comment |
Nonsingular zeroes are algebraic?
This is a fact of algebraic geometry for any field $K$: for $J = (f_1,\dots,f_n) \subset K[x] := K[x_1,\dots,x_n]$ and $d = \det(\partial f_i/\partial x_j) \in K[x]$, if we let $A= (K[x]/J)[1/d]$ (so $K$-algebra maps from $A$ into an extension field $L/K$ correspond to $Q$ you care about) then $A$ is finite-dimensional as a $K$-vector space (and even is an etale $K$-algebra: a finite product of finite separable extensions of $K$). Intuitively, $f:\{d\ne 0\}\rightarrow \mathbf{A}^n$ with components $f_i$ satisfies an "algebraic inverse function theorem", so it has finite geometric fibers. |
Oct
5 |
comment |
splitting field for a division algebra
Or flip ahead several pages to read Theorem 4.12 in BA II... |
Oct
5 |
comment |
Jacobians defined over smaller fields
@Piotr: It sounds like you ask just that the p.p. does not descend to $K$ respecting the given $K$-structure on the abelian surface. In principle, it might happen that the p.p. abelian surface over $L$ admits another $K$-descent as a p.p. abelian surface (i.e., with a $K$-structure different from the given one on the abelian surface), so the associated curve over $L$ would then be defined over $K$. So to make an example in this way one needs a stronger "does not descend to $K$" property. Perhaps you were already aware of this, in which case all I'm doing is clarifying your comment. |
Oct
4 |
comment |
Jacobians defined over smaller fields
The genus-1 curve $ax^3 + by^3 + cz^3 = 0$ has Jacobian (away from characteristics 2 and 3) depending only on $abc$, so you can probably make some explicit examples based on that. |
Oct
1 |
comment |
Decomposition of Matrices in Semisimple and Nilpotent Parts
The example in #3 (which readily adapts to any imperfect field $k$ using the $k$-linear multiplication by $a^{1/p}$ on $V = k(a^{1/p})$) isn't an entirely satisfactory counterexample because it is semisimple over $k$ (though not "geometrically semisimple"; i.e., not diagonalizable over $\overline{k}$). The Wikipedia entry has now been updated to give an example over any imperfect field $k$ in which the operator isn't a sum of two commuting $k$-linear operators that are respectively semisimple (just over $k$!) and nilpotent. |
Sep
28 |
comment |
Closed subgroups of a $p$-adic algebraic group
Concerning your final paragraph: things do make good sense over extensions $F$ of $\mathbf{Q}_p$ provided one replaces the purely topological viewpoint of "closed subgroups" with the more analytic viewpoint of "closed $F$-analytic subgroups" taken up to clopen subgroups (and Lie $F$-subalgebras of the ambient Lie algebra). This is discussed nicely in both Serre's book and Bourbaki. It is analogous to the fact that one has a good Lie correspondence over $\mathbf{C}$ but it requires going beyond the purely topological formulation that works well over $\mathbf{R}$. |
Sep
27 |
comment |
Closed subgroups of a $p$-adic algebraic group
Your statement about exp's is false: p-adic exp has severe convergence problems even for GL$_n$. Read Serre's book "Lie groups and Lie algebras", in which he develops a good Lie correspondence over any non-archimedean field of characteristic 0 from scratch (and carries along the archimedean case, clarifying the special role of $\mathbf{Q}_p$ much as $\mathbf{R}$ has "better" features than $\mathbf{C}$ for a Lie correspondence, due to the density of $\mathbf{Q}$ in $\mathbf{R}$). Also see Bourbaki Lie Ch. III. You cannot expect to nail down exactly subgroups that are exp of their Lie algebra. |
Sep
23 |
comment |
Finite Subgroups of $SL_2(R)$
Why do you want to do this? More specifically, the proofs that any compact (e.g., finite) subgroup of a connected Lie group lies in a maximal compact subgroup and the maximal compact subgroups are pairwise conjugate rest on the idea of invariant forms (through the perspective of a fixed-point theorem, which in turn inspired the fundamental Borel fixed-point theorem in the algebraic theory), so it is both fruitful and natural to use invariant forms. And very efficient too. |
Sep
21 |
comment |
Non-isomorphic finite simple groups
#2 is a "shadow" of the purely inseparable isogeny ${\rm{Spin}}_{2n+1} \rightarrow {\rm{Sp}}_{2n}$ in char. 2 that induces an isomorphism on $\mathbf{F}_{2^m}$-points for all $m>0$. Indeed, by Steinberg (or cohomological arguments over Spec($\mathbf{Z}$)), for a simply connected Chevalley group $G$ and a finite field $k$, $\#G(k)$ is a polynomial in $|k|$ depending only on the "type" of $G$ and not on char($k$), so equality for different types and all $q$ follows from equality as $q$ varies through powers of one prime (such as 2). In this sense, #1 seems more mysterious. |
Sep
20 |
comment |
Non-isomorphic finite simple groups
@Geoff: Here's a "geometric" proof. For $H = \{\sum x_i = 0\}$ in affine 8-space over $\mathbf{F}_2$, and $q = \sum_{i<j} x_i x_j$, $q|_H$ has defect line $L = \{x_1=\dots=x_8\}$. The quadratic space $(H/L,q)$ identifies $S_8$ with ${\rm{O}}_6(\mathbf{F}_2)$ through the $S_8$-action on affine 8-space preserving $H$, $L$, and $q$, so $A_8 = {\rm{SO}}_6(\mathbf{F}_2)$ as the unique index-2 subgroups. The isogeny ${\rm{SL}}_4 \simeq {\rm{Spin}}_6 \rightarrow {\rm{SO}}_6$ induces an isomorphism on $\mathbf{F}_2$-points, and ${\rm{SL}}_4(\mathbf{F}_2)={\rm{GL}}(4,2)$, so ${\rm{GL}}(4,2)=A_8$ |
Sep
19 |
comment |
Solving z^n=a+ib using only radicals of positive real numbers
@Hugo: Yes, you're right, that is what I meant to say. |
Sep
18 |
comment |
Solving z^n=a+ib using only radicals of positive real numbers
@Hugo: If you unravel the proof of my comment, you'll find that for the application to cyclotomic polynomials it is no different from the usual Galois theory argument for analyzing constructibility of polygons. So that particular example isn't as compelling an illustration as it may seem to be at first glance. |
Sep
18 |
comment |
Solving z^n=a+ib using only radicals of positive real numbers
The restriction to degree 3 can be relaxed (and Cardano bypassed): if $f$ is an irreducible polynomial of any odd degree $> 1$ over a field $F$ of characteristic 0 then its splitting field over $F$ contains a root of unity of odd prime order and so cannot be embedded into $\mathbf{R}$. In particular, when $F$ is given as a subfield of $\mathbf{R}$ and the Galois group of $f$ is abelian over $F$ (e.g., cyclic), so the splitting field is generated over $F$ by a single root of $f$, then it is impossible to find any root of $f$ inside a radical tower over $F$ inside $\mathbf{R}$. |
Sep
16 |
comment |
Can we control the size of the intersection of two abelian subfactors of an abelian variety ?
Take $B$ and $B'$ to be non-CM non-isogenous elliptic curves, pick a common $K$-subgroup $G$, and define $A = (B \times B')/G$. The only elliptic curves that are abelian subvarieties of $A$ are the evident copies of $B$ and $B'$, so there's nothing you can do to "fix" the problem in such cases. One way to avoid such a situation is to assume that $A$ admits a polarization of degree prime to $p$, but that's a rather severe assumption. |