User grp - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:16:48Z http://mathoverflow.net/feeds/user/26145 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107250/mostows-theorem-on-algebraic-groups/107342#107342 Answer by grp for Mostow's theorem on algebraic groups grp 2012-09-16T19:39:20Z 2012-09-16T22:15:55Z <p>I prefer the argument as I wrote it in the comments, since the brevity there conveys the structure of the proof most clearly. But since the OP requests it, below is a very detailed version (which might obscure the simplicity of the main idea due to its length).</p> <p>Before explaining the "modern" cohomological proof more fully, note that the result is false in positive characteristic, even over algebraically closed fields. For example, following Tits, for any $n > 1$ and any algebraically closed field $K$ of characteristic $p > 0$, the $K$-group $G = {\rm{SL}}_n(W_2(K))$ (rigorous meaning is clear, I hope) has the evident maximal reductive quotient ${\rm{SL}}_n$ but admits <em>no</em> "Levi factor". This is proved by a root group analysis resting on the fact that the natural quotient map <code>$W_2 \rightarrow {\mathbf{G}}_a$</code> admits no homomorphic section; the same argument works with <code>${\rm{SL}}_n$</code> replaced by any nontrivial connected semisimple $\mathbf{Z}/(p^2)$-group.</p> <p>Now back to (possibly disconnected) linear algebraic groups $G$ over a field $K$ of characteristic 0, with unipotent radical $U = R_u(G) = R_u(G^0)$ (defined over $K$, since $K$ is perfect). We shall prove:</p> <p>${\mathbf{Theorem}.}$ There exist closed $K$-subgroups $M \subset G$ such that $M \ltimes U \rightarrow G$ is an isomorphism, and all such $M$ are $U(K)$-conjugate to each other. (Obviously it is the same to assert $G(K)$-conjugacy rather than $U(K)$-conjugacy.) Equivalently, homomorphic sections to the quotient map $G \rightarrow G/U$ exist and are all $U(K)$-conjugate to each other. </p> <p>In the argument below, we'll highlight exactly where the characteristic 0 hypothesis (rather than just perfectness) is used.</p> <p>Step 1: The case $U = 1$ is trivial, so we may assume $U \ne 1$ and proceed by induction on $\dim U$. As for any smooth connected unipotent group over a perfect field, $U$ admits a "characteristic" composition series $1 = U_0 \subset U_1 \subset \dots \subset U_m = U$ with each $U_i/U_{i-1}$ a nonzero vector group (i.e., a power of $\mathbf{G}_a$). Here, by "characteristic" I mean that each $U_i$ is stable under the automorphism functor of $U$, and so each is normal in $G$. [In characteristic 0 we can use the derived series of $U$, since a commutative unipotent group in characteristic 0 is a vector group. Over perfect fields of positive characteristic we can use the derived series followed by $p$-power filtrations within those steps since a $p$-torsion commutative smooth connected unipotent group over a perfect field of characteristic $p > 0$ is a vector group.]</p> <p>Step 2: Suppose for a moment that the case when $U$ is a vector group is settled, so in particular the cases with $m = 1$ are settled and we can suppose $m > 1$. We'll deduce the general case. The unipotent radical of $G' = G/U_{m-1}$ is the vector group $U/U_{m-1}$, so (by our temporary hypothesis that the case "$U$ is a vector group" is settled) the $K$-group $G'$ admits a Levi factor $M'$. The preimage $H$ of $M'$ under $G \twoheadrightarrow G'$ has unipotent radical $U_{m-1}$ of smaller dimension than $U$ and has maximal reductive quotient $M'$ (that maps isomorphically onto $G/U$). The induction on the dimension of the unipotent radical provides a Levi factor $M$ of $H$ unique up to $U_{m-1}(K)$-conjugacy, and this is clearly a Levi factor of $G$. So we get the existence for $G$, and uniqueness up to $U_{m-1}(K)$-conjugacy for those Levi factors which map isomorphically onto a common Levi factor $M'$ of $G'$ (as all such are Levi factors of the preimage $H$ in $G$ of such an $M'$).</p> <p>If $M$ is any Levi factor of $G$ (so $M \cap U = 1$) then $M \cap U_{m-1} = 1$ and clearly $M$ maps isomorphically onto a Levi factor of $G'$. But any two Levi factors of $G'$ are related through conjugacy by $(U/U_{m-1})(K) = U(K)/U_{m-1}(K)$ (equality since ${\rm{H}}^1(K,U_{m-1}) = 1$, as $U_{m-1}$ is filtered by vector groups over $K$). Thus, to establish the desired $U(K)$-conjugacy result in $G$ we may apply a preliminary $U(K)$-conjugation to reduce to considering those Levi factors of $G$ which have the same image in $G'$, so these are Levi factors in an $H$ as considered above. We have seen that such $K$-subgroups of $G$ are $U_{m-1}(K)$-conjugate, so we'd be done. </p> <p>Step 3: Now we may assume that $U = V = \mathbf{G}_a^n$ is a vector group over $K$. Let $G' = G/V$, so $(G')^0$ is reductive. Since $V$ is commutative, the conjugation action of $G$ on its normal subgroup $V$ factors through an action by $G'$ on $V$. We will reduce our problem to the vanishing of some higher Hochschild cohomology groups for $G'$ with coefficients in the commutative group functor $V$ equipped with its $G'$-action.</p> <p>Consider the quotient map $G \twoheadrightarrow G'$. This is a $V$-torsor for the etale topology on $G'$ (since $V$ is smooth), so the isomorphism class of this torsor corresponds to a class in the etale cohomology group ${\rm{H}}^1(G',V)$. But since $V$ is a vector group, this is the etale cohomology of the quasi-coherent sheaf on $G'$ associated to a vector bundle. On any scheme, the etale cohomology of a quasi-coherent sheaf coincides with the Zariski cohomology (due to descent theory arguments). But on an <em>affine</em> scheme the higher Zariski cohomology of quasi-coherent sheaves vanishes, so the etale cohomology group ${\rm{H}}^1(G',V)$ vanishes. In other words, for the short exact sequence $$1 \rightarrow V \rightarrow G \stackrel{f}{\rightarrow} G' \rightarrow 1$$ there exists a section $s$ to $f$ (as a map of $K$-schemes, not necessarily a homomorphic section). We can arrange $s(1)=1$ via $G(K)$-translation, so $G$ as a pointed $K$-scheme equipped with an inclusion from $V$ and a surjection onto $G'$ can be identified with $V \times G'$ equipped with a $K$-group structure making it fit into an exact sequence as above and making the resulting $G'$-action on $V$ be the one that we introduced above.</p> <p>Arguing exactly as for extensions in ordinary group theory (for which set-theoretic splittings always exist), the set of such $K$-group structures on $V \times G'$ is identified with the set ${\rm{Z}}^2(G',V)$ of "algebraic" 2-cocycles on $G'$ valued in $V$ (equipped with its $G'$-action!). This cocycle is a 2-coboundary (i.e., in the image ${\rm{B}}^2(G',V)$ of the group of "algebraic" 1-cochains ${\rm{C}}^1(G',V)$ under the usual differential) if and only if the resulting short exact sequence of $K$-groups splits as a semidirect product, which is to say that the obstruction to the existence of a Levi subgroup is a class in the Hochschild cohomology group ${\rm{H}}^2(G',V)$. Moreover, if the obstruction vanishes then the set of such splittings (i.e., the set of Levi $K$-subgroups) up to $V(K)$-conjugation is a principal homogeneous space under the Hochschild cohomology group ${\rm{H}}^1(G',V)$.</p> <p>We have shown that the existence of a Levi $K$-subgroup is reduced to the vanishing of the Hochschild cohomology group ${\rm{H}}^2(G',V)$, and the uniqueness up to $V(K)$-conjugacy is reduced to the vanishing of the Hochschild cohomology group ${\rm{H}}^1(G',V)$. Note that we <em>still</em> haven't used that char($K$)=0, only that $K$ is perfect. (We also haven't yet done anything serious, just basic formalism.)</p> <p>Step 4: Now it suffices to prove that the Hochschild cohomology groups ${\rm{H}}^i(G',V)$ vanish for all $i > 0$ and any action on the $K$-group $V = \mathbf{G}_a^n$ by a smooth linear algebraic group $G'$ whose identity component is reductive. For this we will finally use that $K$ has characteristic 0. </p> <p>The first key point is to prove that the $G'$-action on $V$ is necessarily <em>linear</em> (which can certainly fail in positive characteristic). To see this, we just have to check that the endomorphism functor of $\mathbf{G}_a^n$ is represented by the functor ${\rm{Mat}}_n$ of $n \times n$ matrices (so its automorphism functor is represented by ${\rm{GL}}_n$, recovering also the uniqueness of the linear structure on vector groups in characteristic 0). This immediately reduces to the case $n = 1$, which is the obvious statement that an "additive polynomial" over a $\mathbf{Q}$-algebra is precisely a scalar multiple of $x$. Note in particular that $V^{G'}$ is a linear subspace of $V$.</p> <p>Our problem is now one in the category of algebraic linear representations of a smooth linear algebraic group $G'$ with reductive identity component. Ignoring the reductivity property for a moment, let's consider the category of all algebraic linear representations of $G'$ (by which I mean algebraic linear actions on vector spaces of possibly infinite dimension, exhausted by $G'$-stable algebraic finite-dimensional subrepresentations). This category has enough injectives, and by a suitable "induction" construction inspired by the case of ordinary groups we see that the Hochschild cohomology $\delta$-functor ${\rm{H}}^{\bullet}(G',\cdot)$ on this category is erasable and thus is the derived functor of the functor $W \rightsquigarrow W^{G'}$ of $G'$-invariants. (This is valid without restriction on the characteristic.) Thus, our vanishing assertion when $(G')^0$ is reductive (and ${\rm{char}}(K)=0$) is <em>equivalent</em> to the (right-)exactness of the formation of $G'$-invariants, which in turn suffices to be checked for finite-dimensional representations.</p> <p>The formation of the Hochschild cohomology groups commutes with extension of the ground field, so we can assume $K$ is algebraically closed. Thus, $G'$ has a finite composition series whose successive quotients are of 3 types: finite constant, split torus, or connected semisimple. To check that the functor $W \rightsquigarrow W^{G'}$ on finite-dimensional algebraic representations is right-exact, we thereby reduce to each of those 3 basic cases separately. The case of split tori is obvious by consideration of gradings, and the case of finite constant groups is obvious by averaging (since we're in characteristic 0). There remains the case of connected semisimple groups over a field $K$ of characteristic 0. </p> <p>Since ${\rm{char}}(K)=0$, so surjective homomorphisms between linear algebraic $K$-groups induce surjections between Lie algebras, the <em>connected</em> smooth closed image of a $K$-homomorphism $G' \rightarrow {\rm{GL}}(W)$ is contained inside the smooth closed $K$-subgroup $P$ of automorphisms that restrict to the identity on a specific subspace of $W$ if and only if the induced map of Lie algebras ${\rm{Lie}}(G') \rightarrow {\rm{End}}(W)$ factors through the Lie subalgebra ${\rm{Lie}}(P)$ of endomorphisms that kill that specified subspace. Thus, the subspace of $G'$-invariants on a finite-dimensional algebraic representation $W$ of $G'$ is the same as the subspace of ${\rm{Lie}}(G')$-invariants under the corresponding representation of ${\rm{Lie}}(G')$ on $W$. </p> <p>Step 5: We're reduced to proving the semisimplicity of the finite-dimensional representation theory of the Lie algebra $\mathfrak{h}$ of a connected semisimple group $H$ over a field $K$ of characteristic 0. By the theory of Lie algebras in characteristic 0, it suffices to show that $\mathfrak{h}$ is semisimple in the sense of Lie algebras. (This is not a tautology: it requires some input from the theory of algebraic groups, since solvable normal Lie subalgebras do not necessarily "integrate" to Zariski-closed subgroups. But I don't know a reference, so I give a proof below.) </p> <p>It is harmless to extend the ground field so that $K$ is algebraically closed and hence $H$ admits a (split) maximal torus $T$ and the associated root system formalism. In particular, the weight-space decomposition of $\mathfrak{h}$ under $T$ has ${\rm{Lie}}(T)$ as the weight space for 0, and the nonzero weight spaces are 1-dimensional and come in pairs <code>$\mathfrak{h}_a, \mathfrak{h}_{-a}$</code> generating the non-solvable $\mathfrak{sl}_2$.</p> <p>Let $\mathfrak{s}$ be the solvable radical of $\mathfrak{h}$, so we want to show that $\mathfrak{s} = 0$. This radical is stable under all automorphisms of $\mathfrak{h}$, and so under the adjoint action of $H$. Thus, it admits a weight space decomposition under the $T$-action. If the $T$-action is nontrivial then $\mathfrak{s}$ would have to contain one of the lines $\mathfrak{h}_a$, and then applying the adjoint action of a suitable element of $N_H(T)(K)$ (representing the reflection $r_a$ that swaps $a$ and $-a$) we see that $\mathfrak{h}_{-a}$ is also contained in $\mathfrak{s}$. This would imply that the solvable $\mathfrak{s}$ contains the Lie subalgebra generated by <code>$\mathfrak{h}_a$</code> and <code>$\mathfrak{h}_{-a}$</code>, a contradiction since this subalgebra is the non-solvable $\mathfrak{sl}_2$.</p> <p>We conclude that the $T$-action on $\mathfrak{s}$ is trivial. But $T$ was arbitrary and (as for any connected reductive group over an algebraically closed field) $H$ is generated as an algebraic group by its maximal tori, so the adjoint action of $H$ on $\mathfrak{s}$ is trivial. The subspace of invariants in $\mathfrak{h}$ under the adjoint action of $H$ is the Lie algebra of the scheme-theoretic center $Z$ of $H$ (as in any characteristic), so since $Z$ is finite etale (as $H$ is connected semisimple and ${\rm{char}}(K) = 0$) we conclude that ${\rm{Lie}}(Z) = 0$, so $\mathfrak{s} = 0$. This proves that $\mathfrak{h}$ is semisimple.</p> http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/106603#106603 Answer by grp for Philosophy behind Mochizuki's work on the ABC conjecture grp 2012-09-07T13:28:40Z 2012-09-08T23:27:15Z <p>[The answer below is a response to an earlier version of the question that was rather different in certain respects. Minhyong Kim's answer gives excellent insight into ideas that Mochizuki had back in 2000 and that provide essential building blocks for the more recent work. But I still believe that it is too premature for a non-expert to seek insight into the <em>new</em> work, for reasons explained below, given that many top experts are presently trying to absorb the ideas Mochizuki developed back in 2000.]</p> <p>This question appears to be inspired by an historical fallacy: the only "vision" of a proof of the Weil Conjectures that Grothendieck had when he began developing ideas related to his work on the problem (i.e., etale cohomology) was the one laid out in Weil's original paper. The yoga around the standard conjectures came much later. </p> <p>That being said, although the new ABC developments are potentially very exciting, and it is understandable to want to "share in the excitement", for reasons specific to this situation it seems to be much too premature to ask for a sketch on MO or in a blog of Mochizuki's vision/proof with <em>an expectation of insight</em> into the new work. Let me try to indicate why this is the case.</p> <p>As has been explained clearly by JSE elsewhere, there are plenty of top experts in arithmetic geometry who are presently struggling to get <em>even a small handle</em> on what is really going on in Mochizuki's papers (due entirely to the experts' lack of prior study of these ideas; Mochizuki's writing is extremely precise, detailed, thorough, and full of intuitive asides!). So the situation seems to be rather different from that of other tremendous advances in recent decades (by Perelman, Faltings, Wiles, etc.), for which the deep new work took place within a context that was already somewhat familiar to a good-sized community of experts in the field (who could then use their experience and expertise to quickly disseminate a "bird's eye view" to others of some of the key new ideas). </p> <p>Because of the rather unique circumstances of this case, as just indicated, I believe that quid's initial urging of patience (if one isn't going to be directly engaged with the struggle to read the actual papers and the prior work upon which they depend) is appropriate. </p> <p>But to end on a semi-positive note, let me explain why quid's mention of Mochizuki's survey papers is very apt. Some of those surveys are relatively short (e.g., less than 20 pages), and if you find them difficult to grok then you will get a real sense of the difficulties that a lot of top experts are current facing in their efforts to try to understand what Mochizuki has achieved. Please be patient! As quid has noted, in due time, as experts eventually come to acquire a genuine understanding of the overall structure of the arguments in these papers, plenty of expositions for wide dissemination of the ideas will emerge. Mochizuki has put a lot of effort into providing indications of his motivation and insights throughout his papers (which are a serious challenge even for top experts to absorb), and to respect his remarkable effort it seems best to engage with it directly (whether through reading the surveys or the main papers). </p> http://mathoverflow.net/questions/106259/a-question-on-tits-note-reductive-groups-over-local-fields/106264#106264 Answer by grp for a question on TITS' note "Reductive groups over local fields" grp 2012-09-03T19:06:55Z 2012-09-03T21:44:34Z <p>Are you sure that $G$ isn't required to be connected? I think this is needed in order to construct the "valued root datum" structure which underlies Bruhat-Tits structure theory. Anyway, the key point is that there is the concept of "valuation" on the root datum, which is really a collection of "valuations" on the possibly non-commutative groups $U_a(K)$ subject to axioms defined in the first big Bruhat-Tits paper in IHES, which I'll call BTI. The existence of this kind of structure on $G(K)$ requires the full power of the theory of the 2nd Bruhat-Tits IHES paper (developed in more "modern" terms in later work of others, such as J-K. Yu), and it requires connectedness of $G$. On the set of such "valuations" there is a natural free action of $V$ and elements of the same equivalence class are called "parallel". The equivalence classes are naturally affine spaces for $V$, and the group $N(K)$ acts naturally on the entire set preserving each equivalence classes through an action by affine transformations (with $Z(K)$ acting through the translation formulas as you have written down). This is all pure group theory formalism (but far from trivial to set up), the definitions of which have nothing to do with any topological structure on $K$. The specification of a valuation on $K$ selects out a preferred equivalence class, and that is the one used to define $\phi$. </p> <p>(For example, if $G = {\rm{SL}}_2$ and $S$ is the diagonal split maximal torus of $G$ over a field $F$ then the parallelism classes of "valuations" on $G(F)$ in accordance with the root datum for $(G,S)$ correspond exactly to choices of nontrivial non-archimedean $\mathbf{R}$-valued valuations on the abstract field $F$.)</p> <p>So what you're missing is the (highly non-trivial to develop!) definition of the principal homogeneous space for $V$ which supports the action of $N(K)$. In other words, although one can say in concrete terms that the target of $\phi$ is the group of affine transformations of $V$, this is conceptually misleading: it is really the group of automorphisms of a more intrinsic affine space for $V$ in which there is absolutely no canonical base point (intrinsic to $(G,S,\Phi^+,\omega)$). I suppose there could be a way to make the definition of $\phi$ by bare hands (or at least give formulas, without proving things are well-defined), but my understanding (which could be incomplete) is that using a specific parallelism class of of "valuations" as indicated above provides the only natural way to make the definition. Take a look at section 6 of BTI to learn what a valued root datum is, and the many nontrivial properties of this kind of structure. I think that BTI is more illuminating in certain conceptual respects than the Corvallis paper (though of course it doesn't have the rich supply of interesting examples as in the Corvallis paper, and is a rather challenging paper to read). </p> http://mathoverflow.net/questions/106150/commutator-of-algebraic-subgroups-is-connected/106160#106160 Answer by grp for Commutator of algebraic subgroups is connected grp 2012-09-02T05:14:07Z 2012-09-02T09:51:14Z <p>Using classical varieties (and classical points only), since $G^n$ does not have the product topology (in the algebraic group setting) it isn't clear what can be useful for $G$ concerning "topological" statements (as in Todd's answer) concerning the product topology on the subset $T \times T$ inside $G \times G$ when $T$ is just some random subset of $G$ (not yet known to be constructible). So although topological groups provide valuable intuition that can sometimes be transported to the case of algebraic groups (which are of course not themselves topological groups in general), in this case the central issue is not addressed by thinking about topological groups. </p> <p>The purpose of the longer delicate arguments one finds in the basic textbooks on algebraic groups is that the commutator subgroup is reached in "finitely many steps" (even <em>without</em> connected hypotheses on $H$ or $K$, which is very important for applications) and so is constructible. It is for <em>constructible</em> $T$ that $T \times T$ with the "right" topology (inherited from $G \times G$) is connected when $T$ is connected, etc. The hard part therefore involves a problem which doesn't arise in the topological group setting (unless one poses finer topological question, such as closedness of $(H,K)$ under some reasonable hypotheses, which is a deeper problem than mere connectedness). </p> <p>For an arbitrary (not necessarily constructible) connected subset $T$ of $G$ is the subset $T \times T$ inside $G \times G$ (the latter given the Zariski topology) connected?</p> http://mathoverflow.net/questions/109456/are-extensions-of-linear-algebraic-groups-over-a-field-themselves-linear-algebr/109462#109462 Comment by grp grp 2012-10-14T06:29:50Z 2012-10-14T06:29:50Z Dear Michael: I agree with your final comment, but I think that the notion of &quot;reduced group scheme&quot; (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. http://mathoverflow.net/questions/109456/are-extensions-of-linear-algebraic-groups-over-a-field-themselves-linear-algebr/109462#109462 Comment by grp grp 2012-10-13T18:46:52Z 2012-10-13T18:46:52Z Dear Michael: The point of my Frobenius example was just that the notion of &quot;reduced group&quot; 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. http://mathoverflow.net/questions/109456/are-extensions-of-linear-algebraic-groups-over-a-field-themselves-linear-algebr/109462#109462 Comment by grp grp 2012-10-12T17:36:51Z 2012-10-12T17:36:51Z @Olivier: Since Michael seems to be working over a general field, one should say &quot;smooth&quot; rather than &quot;reduced&quot;. 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 &quot;flatness should somehow be part of the definition of being surjective&quot;, but this seems a bit risky since surjective has its own useful (ordinary) meaning for scheme maps. However,&quot;fppf&quot; requires fewer letters than &quot;surjective&quot;, so using French solves it. :) http://mathoverflow.net/questions/109456/are-extensions-of-linear-algebraic-groups-over-a-field-themselves-linear-algebr/109462#109462 Comment by grp grp 2012-10-12T16:08:21Z 2012-10-12T16:08:21Z @Michael: I assumed (and probably Angelo did too) that you know the equivalence of several equivalent definitions of &quot;group extension&quot; (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)? http://mathoverflow.net/questions/109456/are-extensions-of-linear-algebraic-groups-over-a-field-themselves-linear-algebr Comment by grp grp 2012-10-12T13:50:53Z 2012-10-12T13:50:53Z 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 &quot;local&quot; sections!) and using effectivity of fppf descent for affine morphisms. It is explained in Oort's LNM book on commutative (!) group schemes. http://mathoverflow.net/questions/109112/complex-reductive-lie-groups-which-are-not-defined-over-the-real-numbers Comment by grp grp 2012-10-08T04:28:23Z 2012-10-08T04:28:23Z The way you seem to be defining &quot;complex reductive group&quot; is not the standard procedure, and almost defines the answer to be negative (once you clarify what you mean by &quot;defined over the real numbers&quot;; e.g., the complexification structure you give makes the answer negative tautologically). But even if you use the &quot;right&quot; 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). http://mathoverflow.net/questions/108918/nonsingular-zeroes-are-algebraic Comment by grp grp 2012-10-05T15:28:16Z 2012-10-05T15:28:16Z 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 &quot;algebraic inverse function theorem&quot;, so it has finite geometric fibers. http://mathoverflow.net/questions/108877/splitting-field-for-a-division-algebra/108878#108878 Comment by grp grp 2012-10-05T03:41:14Z 2012-10-05T03:41:14Z Or flip ahead several pages to read Theorem 4.12 in BA II... http://mathoverflow.net/questions/108863/jacobians-defined-over-smaller-fields Comment by grp grp 2012-10-05T00:06:23Z 2012-10-05T00:06:23Z @Piotr: It sounds like you ask just that the p.p. does not descend to $K$ <i>respecting</i> the given $K$-structure on the abelian surface. In principle, it might happen that the p.p. abelian surface over $L$ admits <i>another</i> $K$-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 &quot;does not descend to $K$&quot; property. Perhaps you were already aware of this, in which case all I'm doing is clarifying your comment. http://mathoverflow.net/questions/108863/jacobians-defined-over-smaller-fields Comment by grp grp 2012-10-04T23:28:03Z 2012-10-04T23:28:03Z 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. http://mathoverflow.net/questions/108402/decomposition-of-matrices-in-semisimple-and-nilpotent-parts/108408#108408 Comment by grp grp 2012-10-01T11:23:43Z 2012-10-01T11:23:43Z 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 &quot;geometrically semisimple&quot;; 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. http://mathoverflow.net/questions/108282/closed-subgroups-of-a-p-adic-algebraic-group/108288#108288 Comment by grp grp 2012-09-28T02:26:41Z 2012-09-28T02:26:41Z Concerning your final paragraph: things do make good sense over extensions $F$ of $\mathbf{Q}_p$ provided one replaces the purely topological viewpoint of &quot;closed subgroups&quot; with the more analytic viewpoint of &quot;closed $F$-analytic subgroups&quot; 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 $\mathbf{R}$. http://mathoverflow.net/questions/108282/closed-subgroups-of-a-p-adic-algebraic-group Comment by grp grp 2012-09-27T21:27:41Z 2012-09-27T21:27:41Z Your statement about exp's is false: p-adic exp has severe convergence problems even for GL$_n$. Read Serre's book &quot;Lie groups and Lie algebras&quot;, 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 &quot;better&quot; 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. http://mathoverflow.net/questions/107902/finite-subgroups-of-sl-2r Comment by grp grp 2012-09-23T18:38:05Z 2012-09-23T18:38:05Z 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. http://mathoverflow.net/questions/107620/non-isomorphic-finite-simple-groups/107660#107660 Comment by grp grp 2012-09-21T10:08:43Z 2012-09-21T10:08:43Z #2 is a &quot;shadow&quot; 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 $m&gt;0$. 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 &quot;type&quot; of $G$ and <i>not</i> on char($k$), so equality for different types and <i>all</i> $q$ follows from equality as $q$ varies through powers of <i>one</i> prime (such as 2). In this sense, #1 seems more mysterious.