User mikhail borovoi - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:37:04Z http://mathoverflow.net/feeds/user/4149 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids Connected groupoids and action groupoids Mikhail Borovoi 2013-04-16T18:17:18Z 2013-05-19T21:32:55Z <p>It is written in Wikipedia <a href="http://en.wikipedia.org/wiki/Groupoid" rel="nofollow">http://en.wikipedia.org/wiki/Groupoid</a>, that any <em>connected groupoid</em> $A\rightrightarrows X$ is isomorphic to an <em>action groupoid</em> $G\ltimes X$ coming from a transitive action of some group $G$ on $X$. I do not understand how to construct such a group $G$, and would be grateful for an explanation or a reference.</p> <p>I think that in general one cannot recover $G$ from the action groupoid $G\ltimes X$. Indeed, if $G$ acts <em>simply transitively</em> on $X$, then the action groupoid is given by the equivalence relation $X\times X$ on $X$, hence does not depend on $G$, provided that ${\rm Card}(G)={\rm Card}(X)$. Is this correct?</p> <p>This question is a version of <a href="http://math.stackexchange.com/questions/361254/connected-groupoids-and-action-groupoids" rel="nofollow">my question at Math Stack Exchange</a> to which I got no answers.</p> http://mathoverflow.net/questions/130950/groupoids-vs-action-groupoids Groupoids vs. action groupoids Mikhail Borovoi 2013-05-17T12:54:48Z 2013-05-17T15:34:03Z <p>Let $A\rightrightarrows X$ be a groupoid, where $X$ is the set of objects and $A$ is the set of arrows. My favorite example of a groupoid is an action groupoid. If a group $G$ acts on the left on a set $X$, we set $$A=\{(x,g,y)\mid x,y\in X, g\in G, y=g*x\},$$ then $A\rightrightarrows X$ with the evident maps is called the action groupoid corresponding to the action of $G$ on $X$. It is often denoted by $G\ltimes X$. If $G$ acts on $X$ transitively, then $G\ltimes X$ is a connected groupoid. Conversely, any connected groupoid is isomorphic to an action groupoid, see the answers to my question <a href="http://mathoverflow.net/questions/127729" rel="nofollow">http://mathoverflow.net/questions/127729</a>.</p> <p>Now let $\Gamma$ be a group, and assume that $\Gamma$ acts compatibly on $G$ and on $X$ (see my question <a href="http://mathoverflow.net/questions/130712" rel="nofollow">http://mathoverflow.net/questions/130712</a> for a natural example of such action). We say that the $\Gamma$-groupoid $G\ltimes X$ is an <em>action $\Gamma$-groupoid</em>.</p> <blockquote> <p><strong>Question 1.</strong> Is it true that any connected $\Gamma$-groupoid is isomorphic to an action $\Gamma$-groupoid?</p> <p><strong>Question 2.</strong> Is it true that any connected $\Gamma$-groupoid is weakly $\Gamma$-equivalent to an action $\Gamma$-groupoid?</p> </blockquote> <p>See <a href="http://mathoverflow.net/questions/130712" rel="nofollow">http://mathoverflow.net/questions/130712</a> for the definition of a quasi-isomorphism (weak equivalence). We say that two $\Gamma$-groupoids are weakly $\Gamma$-equivalent if they can be connected by a chain of quasi-isomorphisms of $\Gamma$-groupoids.</p> <p>I expect the answer "No" to Question 1, and therefore I ask Question 2, to which I expect the answer "Yes".</p> http://mathoverflow.net/questions/130712/equivalence-and-weak-equivalence-of-groupoids Equivalence and weak equivalence of groupoids Mikhail Borovoi 2013-05-15T12:58:08Z 2013-05-15T13:54:12Z <p>Let $A\rightrightarrows X$ be a groupoid, where $X$ is the set of objects and $A$ is the set of arrows. My favorite example of a groupoid is an action groupoid. If a group $G$ acts on the left on a set $X$, we set $$A=\{(x,g,y)\mid x,y\in X, g\in G,\ y=g*x\},$$ then $A\rightrightarrows X$ with the evident maps is called the action groupoid corresponding to the action of $G$ on $X$. It is often denoted by $G\ltimes X$.</p> <p>Let $F\colon (A\rightrightarrows X)\to (B\rightrightarrows Y)$ be a morphism of groupoids (a functor). We say that $F$ is an <em>equivalence of groupoids</em> if it is an equivalence of categories.</p> <p>Let $x\in X$. We denote by $A(x)$ the set of arrows $a\colon x\to x$. We denote by $\pi_0(X)$ the set of connected components of $X$ (i.e., the set of equivalence classes in $X$ with respect to the equivalence relation induced by $A$). We say that a morphism $F$ as above is a <em>weak equivalence of groupoids</em> (or a <em>quasi-isomorphism</em>) if $\pi_0(F)\colon \pi_0(X)\to \pi_0(Y)$ is a bijection and, for any $x\in X$, the induced homomorphism $F_x\colon A(x)\to B(y)$ is an isomorphism, where $y=F(x)$.</p> <blockquote> <p><strong>Question 1.</strong> Is it true that any weak equivalence of groupoids is an equivalence?</p> </blockquote> <p>Now assume that a group $\Gamma$ acts on our groupoid $A\rightrightarrows X$. We say that $A\rightrightarrows X$ is a $\Gamma$-groupoid. My favorite example of a $\Gamma$-groupoid comes from an action of an algebraic group $\mathcal{G}$, defined over a field $k$, on a $k$-variety $\mathcal{X}$. Let $k_s$ denote a separable closure of $k$, then we set $\Gamma:={\rm Gal}(k_s/k)$, and we consider the action groupoid $\mathcal{G}(k_s)\ltimes\mathcal{X}(k_s)$, on which $\Gamma$ acts.</p> <p>By a <em>weak equivalence of $\Gamma$-groupoids</em> we mean a $\Gamma$-functor $F\colon (A\rightrightarrows X)\to (B\rightrightarrows Y)$ that is a weak equivalence of groupoids. By an <em>equivalence of $\Gamma$-groupoids</em> we mean a $\Gamma$-functor $F\colon (A\rightrightarrows X)\to (B\rightrightarrows Y)$ such that there exists a a $\Gamma$-functor $F'$ in the opposite direction and each of the composite functors $F\circ F'$ and $F'\circ F$ is $\Gamma$-naturally-isomorphic to the corresponding identity functor.</p> <blockquote> <p><strong>Question 2.</strong> Is it true that any weak equivalence of $\Gamma$-groupoids is an equivalence of $\Gamma$-groupoids?</p> </blockquote> <p>I expect the answer "No" to Question 2, but I cannot construct a counter-example.</p> http://mathoverflow.net/questions/130256/second-nonabelian-group-cohomology-cocycles-vs-gerbes Second nonabelian group cohomology: cocycles vs. gerbes Mikhail Borovoi 2013-05-10T12:26:15Z 2013-05-11T16:36:41Z <p>In 1965 Jean Giraud published two Comptes Rendus notes titled "Cohomologie non abélienne", and in 1971 he published a book with the same title. In 1966 Tonny A. Springer's paper "Nonabelian $H^2$ in Galois cohomology" appeared, where he, in particular, constructs nonabelian $H^2$ of a group in terms of group extensions and in terms of cocycles. Springer writes that his definition for group cohomology "seems to be essentially equivalent to to that of Dedecker and Giraud". Giraud in his book (page 452) writes that "la définition de $H^2$ en termes de gerbes ... redonne, dans ce cas, la théorie de Springer".</p> <p>I do not understand the latter assertion. Let $\Gamma$ be a group, and let $G$ be a group together with a "$\Gamma$-kernel": a homomorphism $\kappa\colon \Gamma\to {\rm Out}(G)$, where ${\rm Out}(G):= {\rm Aut}(G)/{\rm Inn}(G)$ is the group of outer automorphisms of $G$. Springer defines $H^2(\Gamma, G,\kappa)$ in terms of group extensions $$1\to G\to E\to \Gamma\to 1$$ inducing the "kernel" $\kappa$. He also describes $H^2(\Gamma, G,\kappa)$ in terms of 2-cocycles coming from the group extension. (A 2-cocycle is a pair of maps $(f,g)$ of maps $f\colon \Gamma\to {\rm Aut}(G)$, $g\colon \Gamma\times \Gamma\to G$ satisfying certain conditions.) Giraud defines $H^2$ in terms of gerbes (on the category of $\Gamma$-sets?).</p> <p><strong>Question:</strong> How can I get a gerbe $\mathcal{G}$ (i.e., a stack over the category of $\Gamma$-sets) from a group extension? In other words, for any $\Gamma$-set $S$, I want to get a groupoid $\mathcal{G}_S$ defined in terms of the given group extension. Conversely, I would like get a group extension from a gerbe.</p> http://mathoverflow.net/questions/125829/quotient-of-a-reductive-group-by-a-non-smooth-subgroup Quotient of a reductive group by a non-smooth subgroup Mikhail Borovoi 2013-03-28T14:15:01Z 2013-05-09T16:22:00Z <p>This is a continuation of my question <a href="http://mathoverflow.net/questions/16261/" rel="nofollow">http://mathoverflow.net/questions/16261/</a>.</p> <p>Let $G$ be a smooth, connected, reductive $k$-group over a field $k$ of characteristic $p>0$. Let $H\subset G$ be a $k$-subgroup, not necessarily smooth. <strong>Question 1:</strong> Does the quotient $G/H$ exist as a $k$-variety?</p> <p>I am interested in the following special case. Let $H^{\rm mult}$ denote the largest quotient of $H$ which is a $k$-group ($k$-group scheme) of multiplicative type. Set $H_1=\ker[H\to H^{\rm mult}]$. I assume that $H_1$ is smooth, connected and semisimple. <strong>Question 2:</strong> Does the quotient $G/H$ exist as a $k$-variety under this assumption? (I do not assume that $H^{\rm mult}$ is smooth.)</p> <p>All comments and references are welcome!</p> http://mathoverflow.net/questions/128508/gluing-gerbes-over-a-spectrum-of-a-field/128543#128543 Answer by Mikhail Borovoi for gluing gerbes over a spectrum of a field Mikhail Borovoi 2013-04-23T21:24:31Z 2013-04-23T21:29:33Z <p>I don't think so. I think a gerbe bound by $A$ over the spectrum of a field $k$ gives a cohomology class $\eta\in H^2(k,A)$, and the gerbe trivializes over an extension $k'/k$ if and only if this cohomology class trivializes. Now take $k=\mathbb{R}$, $A=\mathbf{G}_{m,\mathbb{R}}$, then $H^2(k, A)={\rm Br}(\mathbb{R})=\frac{1}{2}\mathbb{Z}/\mathbb{Z}$. Let $\eta\in{\rm Br}(\mathbb{R})$ be the nontrivial element (corresponding to Hamilton's quaternions). Then $\eta$ trivializes over $\mathbb{C}$, but not over $\mathbb{R}$, so $k'=\mathbb{C}$, $G={\rm Gal}(\mathbb{C}/\mathbb{R})$, and $H^2(G,A)=H^2(\mathbb{C}/\mathbb{R},\mathbf{G}_m)={\rm Br}(\mathbb{R})\neq 0$, so $H^2(G,A)$ does not vanish.</p> http://mathoverflow.net/questions/127531/constructing-a-stack-gerbe-from-a-connected-groupoid Constructing a stack (gerbe) from a connected groupoid Mikhail Borovoi 2013-04-14T13:09:33Z 2013-04-16T04:43:50Z <p>Let $\mathcal{G}=(A\rightrightarrows X)$ be a groupoid. Here $X={\rm Ob}(\mathcal{G})$, $A={\rm Ar}(\mathcal{G})$, and we have 5 maps: $s,t\colon A\to X$ (the source and the target, surjective), $m\colon A\times_X A\to A$ (multiplication of composable arrows), ${\rm id}\colon X\to A$ ($x\mapsto{\rm id}_x$, injective), and $i\colon A\to A$ ($a\mapsto a^{-1}$), satisfying the usual axioms. I say that my groupoid is connected if for any two objects $x,y\in X$ there exists an arrow $a\colon x\to y$.</p> <p>Assume that a finite group $\Gamma$ acts on $\mathcal{G}$, i.e., it acts on $X$ and $A$ such all the 5 maps are $\Gamma$-equivariant. We say that $\mathcal{G}$ is a $\Gamma$-groupoid.</p> <p>Now I want to construct a fibered category (gerbe) $\mathbb{G}$ over the category (site) of finite $\Gamma$-sets, starting from a connected $\Gamma$-groupoid $\mathcal{G}$. In other words, for any finite $\Gamma$-set $S$, I want to construct a groupoid $\mathbb{G}(S)$, and for a morphism $S\to T$ of finite $\Gamma$-sets, I want to define a restriction functor $\mathbb{G}(T)\to \mathbb{G}(S)$. How can I do that? I could not find this in Giraud's book.</p> http://mathoverflow.net/questions/121814/on-z-extensions/121833#121833 Answer by Mikhail Borovoi for on z-extensions Mikhail Borovoi 2013-02-14T20:16:22Z 2013-02-14T20:16:22Z <p>No, if the center $Z(G)$ is not connected, we cannot construct $G'$ with connected $Z(G')$.</p> <p>Indeed, let $\pi\colon G'\to G$ be an epimorphism of connected reductive $F$-groups with central kernel. Choose a maximal torus $T\subset G$. We have $$Z(G)=\ker[T\to {\rm Aut}\ {\rm Lie}(G)]=\ker[T\to {\rm Aut}\ {\rm Lie}( G_{der})].$$ Set $T'=\pi^{-1}(T)\subset G'$, then $$Z(G')=\ker[T'\to {\rm Aut}\ {\rm Lie}( G'_{der})].$$ It follows easily that $\pi(Z(G'))=Z(G)$.</p> <p>Now, if $Z(G')$ is connected, then $Z(G)=\pi(Z(G'))$ must be connected. Therefore, if $Z(G)$ is not connected, then we cannot construct $G'$ with connected center $Z(G')$.</p> http://mathoverflow.net/questions/121408/what-are-the-symmetries-of-a-principal-homogeneous-bundle/121425#121425 Answer by Mikhail Borovoi for What are the symmetries of a principal homogeneous bundle? Mikhail Borovoi 2013-02-10T21:13:24Z 2013-02-10T21:13:24Z <p>No, in general $G=G(\mathbf{Q})$ can be strictly smaller that ${\rm Aut}(\mathbf{Q})$.</p> <p>Let $G$ be a Lie group and $H\subset G$ be a Lie subgroup. Set $P=G$, $\ Q=G/H$, and define the maps in the obvious way. We get a principal homogeneous bundle $\mathbf{Q}=(Q,P,\dots)$ in your sense.</p> <p>Now take $G=SL(n,\mathbb{C})$, $\ H=SO(n,\mathbb{C})$. Consider the automorphism $\tau$ of $G$ given by $\tau(g)= (g^T)^{-1}$, there $g^T$ denotes the transpose of $g$. Since $\tau$ takes $H$ to itself, it induces an automorphism of the principal homogeneous bundle $\mathbf{Q}$. It is well known that for $n\ge 3$, the automorphism $\tau$ of $G$ is <em>not inner</em>, i.e. is not of the form $c_u(g)=ugu^{-1}$, which gives the negative answer to your question. </p> http://mathoverflow.net/questions/118988/automorphisms-of-sl-n-as-a-variety Automorphisms of $SL_n$ as a variety Mikhail Borovoi 2013-01-15T16:12:22Z 2013-01-15T19:45:09Z <p>What are the automorphisms of $SL_n$ as an algebraic variety?</p> <p>In other words, let $k$ be an algebraically closed field of characteristic 0 (e.g., $k=\mathbb{C}$). Let $\tau$ be an automorphism of $SL_n$ regarded as an <em>algebraic variety</em> over $k$. Assume that $\tau$ takes the unit element $e$ of $G$ to itself. Is it true that $\tau$ is an automorphism of $SL_n$ as an <em>algebraic group</em> over $k$?</p> http://mathoverflow.net/questions/118554/non-trivial-representation-of-second-smallest-dimension/118621#118621 Answer by Mikhail Borovoi for Non-trivial representation of second-smallest dimension Mikhail Borovoi 2013-01-11T12:41:04Z 2013-01-15T18:57:10Z <p>The irreducible complex representations of the simply connected simple group $G=Sp_{r,{\mathbb C}}$ of type $C_r$, for $r>1$, of dimension $n&lt;{\rm dim}\ G$ are listed in the <a href="http://link.springer.com/content/pdf/10.1007%2FBF01076005" rel="nofollow">paper of Andreev, Vinberg, and Elashvili</a>, Table 1 (see also the <a href="http://www.mathnet.ru/links/419af2a1d9d33839a49ff8898866b056/faa2839.pdf" rel="nofollow">Russian version</a>). They are the fundamental irreducible representations $R(\pi_1)$ of dimension $2r$, $R(\pi_2)$ of dimension $2r^2-r-1$, and, for $r=3$, $R(\pi_3)$ of dimension 14. For all $r\ge 2$, $r\neq 3$, we have ${\rm dim}\ R(\pi_1)=2r&lt;2r^2-r-1={\rm dim}\ R(\pi_2)$, hence $R(\pi_2)$ is the nontrivial irreducible representation of second smallest dimension. For $r=3$, as Jim Humphreys noted, the dimensions are $6,14,14$, so ${\rm dim}\ R(\pi_2)={\rm dim}\ R(\pi_3)>{\rm dim}\ R(\pi_1)$, and $R(\pi_2)$ is <em>a</em> nontrivial irreducible representation of second smallest dimension.</p> http://mathoverflow.net/questions/118472/transitive-action-on-the-sphere/118581#118581 Answer by Mikhail Borovoi for Transitive action on the sphere Mikhail Borovoi 2013-01-10T22:24:55Z 2013-01-11T10:08:21Z <p>(I add details to my comments.) The answer depends on $n=4r$. Write $G=Sp(r)/\mu_2$. If $r=1$, then $G\simeq SO_3$, so $G$ admits a faithful 4-dimensional representation into $SO_4$. Similarly, if $r=2$, then $G\simeq SO_5$, hence $G$ admits a faithful 8-dimensional representation into $SO_8$. (Of course, in these cases $S^{4r-1}$ is not an orbit.) For $r\ge 3$ the group $G$ has no nontrivial representations of dimension $4r$, see below, hence it cannot be embedded into $SO_{4r}$.</p> <p>An irreducible real $n$-dimensional representation of the real algebraic group $G$ induces an irreducible complex $n$-dimensional representation of $G_{\mathbb C}=Sp_{r,{\mathbb C}}/\mu_2$. The irreducible complex representations of the simple group $\widetilde G_{\mathbb C} =Sp_{r,{\mathbb C}}$ of type $C_r$ for $r>1$ of dimension $n&lt;{\rm dim}\ \widetilde{G}_{\mathbb C}$ are listed in the <a href="http://link.springer.com/content/pdf/10.1007%2FBF01076005" rel="nofollow">paper of Andreev, Vinberg, and Elashvili</a>, Table 1. They are the fundamental irreducible representations $R(\pi_1)$ of dimension $2r$, $R(\pi_2)$ of dimension $2r^2-r-1$, and, for $r=3$, $R(\pi_3)$ of dimension 14. Since the representations $R(\pi_1)$ and $R(\pi_3)$ are nontrivial on the center $Z(\widetilde G_{\mathbb C})\simeq\mu_2$, we see that the only nontrivial irreducible representation of $G_{\mathbb C}$ of dimension $n&lt;{\rm dim}\ G_{\mathbb C}$ is the representation $R(\pi_2)$ of dimension $2r^2-r-1$, hence $R(\pi_2)$ is the irreducible representation of $G_{\mathbb C}$ of the smallest dimension. For $r\ge 3$ we have $2r^2-r-1>4r$, hence $G_{\mathbb C}$ has no nontrivial representations of dimension $4r$.</p> http://mathoverflow.net/questions/118356/conjugation-of-homogeneous-spaces Conjugation of homogeneous spaces Mikhail Borovoi 2013-01-08T12:57:46Z 2013-01-09T16:39:55Z <p>Let $X$ be a smooth irreducible algebraic variety over the field of complex numbers ${\mathbb{C}}$. Let $x\in X({\mathbb{C}})$. Let $\tau$ be an automorphism of ${\mathbb{C}}$ (not necessarily continuous), and let $\tau X$ denote the $\tau$-conjugated ${\mathbb{C}}$-variety obtained from $X$ by transport of structure (i.e. by action of $\tau$ on the coefficients of equations defining $X$). We consider the <em>topological</em> fundamental groups $\pi_1(X({\mathbb{C}}),x)$ and $\pi_1((\tau X)({\mathbb{C}}),\tau x)$.</p> <p>In the papers of Serre, Exemples de variétés projectives conjuguées non homéomorphes, C. R. Acad. Sci. Paris <strong>258</strong> (1964), 4194–4196, and of Milne and Suh, <a href="http://arxiv.org/abs/0804.1953" rel="nofollow">Nonhomeomorphic conjugates of connected Shimura varieties</a>, one can find examples of $X$ and $\tau$ such that $\pi_1((\tau X)({\mathbb{C}}),\tau x)$ and $\pi_1(X({\mathbb{C}}),x)$ are <em>not</em> isomorphic. The authors conclude that in these cases the topological spaces $(\tau X)({\mathbb{C}})$ and $X({\mathbb{C}})$ are not homeomorphic.</p> <p>In my very recent <a href="http://arxiv.org/abs/1301.1046" rel="nofollow">preprint with Cyril Demarche</a> (excuse me for advertising my own work!) we consider the following situation. Let $X=G/H$, where $G$ is a connected linear algebraic group over ${\mathbb{C}}$, and $H\subset G$ any algebraic subgroup, <em>not necessarily connected</em>. Set $x:=eH\in X({\mathbb{C}})$. We prove that in this case $\pi_1((\tau X)({\mathbb{C}}),\tau x)$ and $\pi_1(X({\mathbb{C}}),x)$ are canonically isomorphic. I am trying to understand, what this really means.</p> <blockquote> <p><strong>Question.</strong> For a homogeneous space $X=G/H$ over ${\mathbb{C}}$ as above, and for $\tau\in {\rm Aut}({\mathbb{C}})$, is it always true that<br> (1) $(\tau X)({\mathbb{C}})$ and $X(\mathbb{C})$ are homotopically equivalent, or even<br> (2) $(\tau X)({\mathbb{C}})$ and $X(\mathbb{C})$ are homeomorphic, or even<br> (3) $\tau X$ and $X$ are isomorphic ${\mathbb{C}}$-varieties?</p> </blockquote> http://mathoverflow.net/questions/118356/conjugation-of-homogeneous-spaces/118463#118463 Answer by Mikhail Borovoi for Conjugation of homogeneous spaces Mikhail Borovoi 2013-01-09T16:32:45Z 2013-01-09T16:39:55Z <p>I answer the question in the comment of Tom Goodwillie: What is known when $H=1$?</p> <blockquote> <p><strong>Theorem.</strong> Let $G$ be a connected linear algebraic group over ${\mathbb{C}}$. Let $\tau$ be an automorphism of ${\mathbb{C}}$. Then the complex varieties $G$ and $\tau G$ are isomorphic.</p> </blockquote> <p>Note that I do not claim that the <em>algebraic groups</em> $G$ and $\tau G$ are always isomorphic, see the comment of Yves Cornulier.</p> <p><em>Proof.</em> It suffices to show that <em>as a variety</em> $G$ can be defined over $\mathbb Q$.</p> <p>Write $G^{\rm u}$ for the unipotent radical of $G$, and set $G^{\rm red}:=G/G^{\rm u}$, then $G^{\rm red}$ is a connected reductive ${\mathbb{C}}$-group. By Mostow's theorem $G\simeq G^{\rm u}\rtimes G^{\rm red}$, hence $G\simeq G^{\rm u}\times G^{\rm red}$ as a ${\mathbb{C}}$-variety. Using the exponential map, one sees easily that $G^{\rm u}$ is isomorphic to an affine space (the Lie algebra of $G^{\rm u}$) as a variety, hence <em>as a variety</em> it can be defined over ${\mathbb{Q}}$. Now it suffices to show that the reductive ${\mathbb{C}}$-group $G^{\rm red}$ admits a ${\mathbb{Q}}$-form (as an algebraic group).</p> <p>Set $G^{\rm ss}=[G,G]$, it is a connected semisimple group. Let $G^{\rm sc}$ denote the universal covering of $G^{\rm ss}$, it is a simply connected semisimple $\mathbb{C}$-group. Let $Z^0$ denote the identity component of the center $Z$ of $G^{\rm red}$, it is a $\mathbb{C}$-torus. We have a canonical epimorphism $\phi\colon G^{\rm sc}\times_{\mathbb{C}} Z^0\to G$ with finite central kernel $\mu$.</p> <p>Let $G_{1,{\mathbb{Q}}}$ be the direct product over ${\mathbb{Q}}$ of a split ${\mathbb{Q}}$-form (Chevalley's form) of $G^{\rm sc}$ and a split ${\mathbb{Q}}$-form of the torus $Z^0$. We have an epimorphism $\phi\colon G_{1,{\mathbb{C}}}\to G^{\rm red}$. Since $\mu\subset T_{1,{\mathbb{C}}}$ for some <em>split</em> maximal torus $T_{1,{\mathbb{Q}}}\subset G_{1,{\mathbb{Q}}}$, we see that $\mu$ is defined over ${\mathbb{Q}}$ in $T_{1,{\mathbb{C}}}$, i.e. $\mu=\mu_{\mathbb{Q}}\times_{\mathbb{Q}} {\mathbb{C}}$ for some central ${\mathbb{Q}}$-subgroup $\mu_{\mathbb{Q}}\subset G_{1,{\mathbb{Q}}}$.</p> <p>Now we set $G_{\mathbb Q}^{\rm red}$ to be $G_{1,{\mathbb{Q}}}/\mu_{\mathbb{Q}}$, it is a ${\mathbb{Q}}$-form of $G^{\rm red}$.</p> http://mathoverflow.net/questions/107250/mostows-theorem-on-algebraic-groups Mostow's theorem on algebraic groups Mikhail Borovoi 2012-09-15T12:09:38Z 2012-09-19T23:15:58Z <p>In his classical 1956 paper <a href="http://www.jstor.org/stable/2372490?origin=crossref" rel="nofollow">Fully reducible subgroups of algebraic groups</a> Mostow proves the following theorem:</p> <blockquote> <p><strong>Theorem 7.1.</strong> Let $G$ be an algebraic group over a field $K$ of characteristic 0, $\mathfrak{N}$ the set of nilpotent elements in the radical of its Lie algebra, and $N$ the connected algebraic group with Lie algebra $\mathfrak{N}$. Then $G=M\cdot N$ (semi-direct) with $M$ a maximal fully reducible group. Furthermore any two maximal fully reducible subgroups of $G$ are conjugate under an inner automorphism from $\mathfrak{N}$.</p> </blockquote> <p>This paper was written long before the Three Books on linear algebraic groups were published, and I do not fully understand Mostow's terminology. Is "a fully reducible group" the same as "an algebraic group with reductive identity component" (in characteristic 0)? Does the following result follow from Mostow's theorem?</p> <blockquote> <p><strong>Theorem?</strong> Let $G$ be a linear algebraic group, not necessarily connected, over a field $K$ of characteristic 0. Let $G^0$ denote the identity component of $G$, and let $N$ denote the unipotent radical of $G^0$. Then the extension $$1\to N\to G\to G/N\to 1$$ splits, i.e. $G=M\cdot N$ (a semidirect product), where $M\subset G$ is a $K$-subgroup of $G$ isomorphic to $G/N$.</p> </blockquote> <p>Are there modern proofs of Mostow's theorem? References will be appreciated.</p> http://mathoverflow.net/questions/106838/isomorphic-general-linear-groups-implies-isomorphic-fields/106916#106916 Answer by Mikhail Borovoi for Isomorphic general linear groups implies isomorphic fields? Mikhail Borovoi 2012-09-11T14:22:06Z 2012-09-11T15:30:23Z <p>The answer is "yes", see below.</p> <p>Dieudonné in his book "La géométrie des groupes classiques" considers the abstract group $SL_n(K)$ for a field $K$, not necessarily commutative, and writes $PSL_n(K)$ for $SL_n(K)$ modulo the center. In Ch. IV, Section 9, he considers the question whether $PSL_n(K)$ can be isomorphic to $PSL_m(K')$ for $n\ge 2,\ m\ge 2$. He writes that they can be isomorphic only for $n=m$, except for $PSL_2(\mathbb{F}_7)$ and $PSL_3(\mathbb{F}_2)$. If $n=m>2$, then the isomorphism is possible only if $K$ and $K'$ are isomorphic or anti-isomorphic. The same is true for $m=n=2$ if both $K$ and $K'$ are commutative, except for the case $K=\mathbb{F}_4$, $K'=\mathbb{F}_5$. Dieudonné gives ideas of proof and references to Schreier and van der Waerden (1928), to his paper "On the automorphisms of classical groups" in Mem. AMS No. 2 (1951) and to the paper of Hua L.-K. and Wan in J. Chinese Math. Soc. 2 (1953), 1-32.</p> <p>This answers affirmatively the question for $SL_n$, because if $SL_n(K)\cong SL_n(K')$, then $PSL_n(K)\cong PSL_n(K')$. In the case $n=2$, $K=\mathbb{F}_4$, $K'=\mathbb{F}_5$, the orders $|SL_2(\mathbb{F}_4)|=60$ and $|SL_2(\mathbb{F}_5)|=120$ are different, and therefore these groups are not isomorphic.</p> <p>This also answers affirmatively the question for $GL_n$, because $SL_n(K)$ is the commutator subgroup of $GL_n(K)$, except for $GL_2(\mathbb{F_2})$, see Dieudonné, Ch. II, Section 1. In the case $n=2$, $K=\mathbb{F}_2$, we have $|GL_2(\mathbb{F}_2)|=6$ , which is less than $|GL_2(\mathbb{F}_q)|=q(q-1)(q^2-1)$ for any $q=p^r>2$, hence $GL_2(\mathbb{F}_2)\not\cong GL_2(\mathbb{F}_q)$ for $q>2$.</p> http://mathoverflow.net/questions/106436/crossed-modules-quasi-isomorphic-to-a-quasi-abelian-crossed-module Crossed modules quasi-isomorphic to a quasi-abelian crossed module Mikhail Borovoi 2012-09-05T13:47:25Z 2012-09-05T13:47:25Z <p>This is a follow-up of <a href="http://mathoverflow.net/questions/106236" rel="nofollow">this question</a>, where the definition of a quasi-abelian crossed module was given. Namely, a crossed module $\partial\colon F\to G$ is quasi-abelian if the embedding $\partial_Z\hookrightarrow \partial$ of its center into it is a quasi-isomorphism.</p> <p><strong>Question 1.</strong> Assume that $\partial\colon F\to G$ is a crossed module which can be related by a chain of quasi-isomorphisms to a quasi-abelian crossed module. Then $\pi_1(\partial):={\rm coker}\ \partial$ is abelian and acts trivially on $\pi_2(\partial):=\ker\partial$ (is the assertion about the action true?). What else can be said about $\partial$? </p> <p><strong>Question 2.</strong> Assume that $\partial$ and $\partial'$ are two quasi-abelian crossed modules which can be related by a chain of quasi-isomorphisms. Is it true that their centers $\partial_Z$ and $\partial'_Z$ are isomorphic in the derived category of complexes of abelian groups, i.e. can be related by a chain of quasi-isomorphisms of complexes of abelian groups?</p> <p>Both questions are motivated by cohomology with coefficients in crossed modules.</p> http://mathoverflow.net/questions/106236/non-quasi-abelian-braided-crossed-modules Non-quasi-abelian braided crossed modules Mikhail Borovoi 2012-09-03T11:47:00Z 2012-09-04T14:43:28Z <p>A right crossed module is a homomorphism of groups $\partial\colon F\to G$ together with a right action of $G$ on $F$, written $(g,f)\mapsto f^g$, satisfying certain conditions.</p> <p>The question is, whether a crossed module $F\to G$ is in some sense abelian.</p> <p>Following <a href="http://archive.numdam.org/article/BSMF_1990__118_2_129_0.pdf" rel="nofollow">Norrie</a>, we set $$Z_F=\{f\in F\mid f^g=f \ \ \forall g\in G\},\ \text{ then }\ Z_F\subset Z(F);$$ $$Z_G=\{g\in Z(G)\mid f^g=f\ \ \forall f\in F\}.$$ Here $Z(G)$ denotes the center of $G$. One easily checks that $\partial(Z_F)\subset Z_G$. We write $\partial_Z$ for the induced homomorphism $\partial_Z\colon Z_F\to Z_G.$ We say that the abelian crossed module $\partial_Z\colon Z_F\to Z_G$ is the <em>center</em> of the crossed module $\partial\colon F\to G$.</p> <p>The following definition is a version of a definition of <a href="http://arxiv.org/abs/1110.4542" rel="nofollow">González-Avilés</a>.</p> <p><em>Definition.</em> A crossed module $F\to G$ is called <em>quasi-abelian</em>, if the morphism of crossed modules $$(Z_F\to Z_G)\to (F\to G)$$ is a quasi-isomorphism. This means that the induced homomorphisms $$\ker\partial_Z\to\ker\partial\ \ \text{ and }\ \ {\rm coker}\ \partial_Z\to{\rm coker}\ \partial$$ are isomorphisms. In other words,</p> <p>(i) $\partial(F)\cdot Z_G=G$ and</p> <p>(ii) $\partial(Z_F)=Z_G\cap \partial(F)$.</p> <p>We obtain an example of a quasi-abelian crossed module $G^{\rm sc}\to G$ from a connected reductive group $G$ over a field $k$ of characteristic 0. Here $G^{\rm sc}$ is the universal covering of the commutator subgroup $G^{\rm ss}:=[G,G]$ of our reductive group $G$. We have a differential $$\partial\colon G^{\rm sc}\twoheadrightarrow G^{\rm ss}\hookrightarrow G.$$ By functoriality, $G$ acts on $G^{\rm sc}$ on the right. Let $\bar k$ denote an algebraic closure of $k$, then we obtain a quasi-abelian crossed module $G^{\rm sc}(\bar k)\to G(\bar k)$.</p> <p>A <em>braiding</em> of a crossed module $F\to G$ is a map $$\{\ \} \colon G\times G \to F,\ g_1,g_2\mapsto \{g_1,g_2\}$$ satisfying certain conditions, in particular, $$\partial\{g_1,g_2\}=g_1^{-1} g_2^{-1} g_1 g_2.$$ A braiding is called symmetric if $\{g_1,g_2\}\{g_2,g_1\}=1$. A braiding is called Picard if it is symmetric and also $\{g,g\}=1$.</p> <p>We define a canonical braiding of a quasi-abelian crossed module as follows. Let $g_1,g_2\in G$. By (i) we can write $$g_1=z_1\cdot\partial(f_1),\ g_2=z_1\cdot\partial(f_2), \ \text{ where } z_1,z_2\in Z_G.$$ Then we set $$\{g_1,g_2\}=f_1^{-1}f_2^{-1}f_1 f_2.$$ Using (ii), one can prove that this braiding is well defined. It is symmetric and even Picard. We see that any quasi-abelian crossed module admits a Picard braiding.</p> <p><strong>Question.</strong> What are examples of a non-quasi-abelian crossed module admitting a braiding? Admitting a symmetric braiding?</p> http://mathoverflow.net/questions/102746/structure-of-abelian-connected-complex-linear-algebraic-groups/102752#102752 Answer by Mikhail Borovoi for Structure of abelian connected complex linear algebraic groups? Mikhail Borovoi 2012-07-20T17:28:08Z 2012-07-20T17:28:08Z <p>A reference: Lie Groups and Algebraic Groups (Springer Series in Soviet Mathematics) by Arkadij L. Onishchik, Ernest B. Vinberg and Dimitry A. Leites (new printing will appear on Amazon.com on Jul 31, 2012). See Ch. 3, Section 2.5, page 116, Corollary of Theorem 8: Any irreducible commutative (linear) algebraic group is the direct product of a torus and a vector group.</p> <p>I highly recommend this book for beginners, and not only for beginners. It is mainly a collection of problems (with hints of proofs at the end of each section). </p> http://mathoverflow.net/questions/101322/cayley-transform-for-all-reductive-groups-a-k-a-an-algebraic-logarithm/101343#101343 Answer by Mikhail Borovoi for Cayley Transform for all reductive groups a.k.a an algebraic logarithm Mikhail Borovoi 2012-07-04T21:14:15Z 2012-07-05T06:15:08Z <p>A <em>Cayley map</em> is a $G$-equivariant birational isomorphism $\lambda: G\to \mathfrak{g}$ (which does not have to exist). A connected linear algebraic group $G$ over $\mathbb{C}$ is called a <em>Cayley group</em> if it admits a Cayley map, and it is called a <em>stably Cayley group</em> if $G\times (\mathbb{G}_m)^n$ is a Cayley group for some $n=0,1,2,\dots$. These notions were introduced in the paper <a href="http://arxiv.org/abs/math/0409004" rel="nofollow">Cayley groups</a> by Lemire, Popov and Reichstein. As usual, the "stable" question is easier than the original one.</p> <p>The authors classified Cayley and stably Cayley simple groups. They proved the following result:</p> <blockquote> <p><strong>Theorem.</strong> The stably Cayley simple groups over an algebraically closed field $k$ of characteristic 0 are the following: $SL_2$, $SL_3$, $SO_n$, $Sp_{2n}$, $PGL_n$, and $G_2$. All these groups are Cayley, except $G_2$. The group $G_2$ is not Cayley (V. A. Iskovskikh), but $G_2\times (\mathbb{G}_m)^2$ is Cayley.</p> </blockquote> <p>Note that the question whether $G_2\times \mathbb{G}_m$ is Cayley is open. Note also that all the groups of types $E_6$, $E_7$, $E_8$ and $F_4$ are not stably Cayley, hence they are not Cayley. In addition, the groups $SL_2$ and $SL_3$ are Cayley, while $SL_4$, $SL_5$, $SL_6$ and so on are not stably Cayley, hence they are not Cayley.</p> http://mathoverflow.net/questions/95833/density-in-su2-1/95901#95901 Answer by Mikhail Borovoi for density in SU(2,1) Mikhail Borovoi 2012-05-03T17:52:56Z 2012-05-03T19:48:04Z <p>(EDITED taking in account the comment of Yves.) The group denoted in the question by $SU(2,1)(K)$ is the group of $\mathbf{Q}$-points $G(\mathbf{Q})$ for a suitable $\mathbf{Q}$-group $G$, and the group denoted by $SU(2,1)(\mathbf{C})$ is $G(\mathbf{R})$. For <em>any</em> connected linear algebraic group $G$ over $\mathbf{Q}$, the group $G(\mathbf{Q})$ is dense in $G(\mathbf{R})$ for the real topology. This is the real approximation theorem for connected linear algebraic groups, see the reference to Sansuc's paper in my comments to <a href="http://mathoverflow.net/questions/87665/a-question-on-algebraic-torus" rel="nofollow">http://mathoverflow.net/questions/87665/a-question-on-algebraic-torus</a>.</p> <p>Note also that if $G$ is a <em>simply connected</em> semisimple group over $\mathbf{Q}$ (as the group in the question), then $G(\mathbf{Q})$ is dense in $G(\mathbf{Q}_p)$ for any prime $p$. This follows from the weak approximation theorem for simply connected groups (due to Kneser, Harder, Platonov). This is not true for semisimple groups that are not simply connected, see example 5.8 in Sansuc's paper, where a certain semisimple $\mathbf{Q}$-group $G$ is constructed for which $G(\mathbf{Q})$ is not dense in $G(\mathbf{Q}_2)$. </p> http://mathoverflow.net/questions/93771/proper-compact-connected-subgroup-of-spinn/93780#93780 Answer by Mikhail Borovoi for Proper compact connected subgroup of $Spin(n)$ Mikhail Borovoi 2012-04-11T16:16:26Z 2012-04-11T17:45:37Z <p>A subgroup of maximal rank of maximal dimension is certainly a maximal subgroup of maximal rank. Maximal connected subgroups of maximal rank in $Spin(n)$ correspond to maximal reductive Lie subalgebras of maximal rank in $so(n)_{\mathbf{C}}$. Such subalgebras in semisimple Lie algebras were classified by Dynkin in 1952, see Onishchik and Vinberg (Eds.), Lie Groups and Lie Algebras III, Encyclopaedia of Mathematical Sciences, vol. 41, Tables 5 and 6. For $so(n)$ all such subalgebras are $so(2k)\oplus so(n-2k)$, and also $gl(n/2)$ for $n$ even. The subalgebras of highest dimension are probably $so(n-1)$ for $n$ odd and $gl(n/2)$ for $n$ even.</p> <p>EDIT: For $n=2l\ge 10$, the subalgebra of highest dimension and of maximal rank in $so(n)$ is $so(n-2)\oplus so(2)$ of dimension $2l^2-5l+4=l^2+l(l-5)+4$, and NOT $gl(n/2)$ of dimension $l^2$. For example, for $n=10$ we have ${\rm dim} (so(8)\oplus so(2))=29$, while ${\rm dim}\ gl(5)=25$.</p> http://mathoverflow.net/questions/90342/a-kind-of-orthogonal-subtorus/93245#93245 Answer by Mikhail Borovoi for A kind of orthogonal subtorus Mikhail Borovoi 2012-04-05T19:36:13Z 2012-04-05T19:36:13Z <p>Consider the subgroup $N:=\langle k\rangle\subset \mathbb{Z}^n$. There exists a basis $f_1,\dots,f_n$ of $\mathbb{Z}^n$ such that $uf_1$ is a basis of $N$, where $u\in \mathbb{Z}$, $u>0$, see Vinberg, A Course in Algebra, Thm. 9.1.5, or Lang, Algebra, 3d ed., Thm. III.7.8. Changing, if necessary, $f_1$ to $-f_1$, we may think that $k=uf_1$, $u>0$. If you assume that your vector $k$ is not divisible by any positive integer different from 1, you obtain that $k=f_1$. Let $e_1,\dots,e_n$ be the standard basis of $\mathbb{Z}^n$. If follows easily that $S:=k^\perp=f_1^\perp$ is isomorphic to $e_1^\perp=\mathbb{T}^{n-1}$.</p> http://mathoverflow.net/questions/92794/exact-sequence-of-weyl-groups/92822#92822 Answer by Mikhail Borovoi for Exact sequence of Weyl groups Mikhail Borovoi 2012-04-01T16:40:54Z 2012-04-01T16:40:54Z <p>Yes, it is exact.</p> <p>For an algebraically closed field $k$ and a connected reductive $k$-group $G$, one can construct a canonical based root datum of $G$, so we obtain a canonically defined Weyl group $W(G)$. I am not sure that $G\mapsto W(G)$ is a functor on the category of connected reductive $k$-groups and homomorphisms of $k$-groups. However, it is certainly a functor on the category of connected reductive $k$-groups and <em>normal</em> homomorphisms. A homomorphism of connected reductive $k$-groups is called normal if its image is a normal subgroup.</p> <p>If we have a short exact sequence of connected reductive $k$-groups, then we obtain an induced short exact sequence of semisimple groups of adjoint type, which clearly splits, so we obtain a split short exact sequence of Weyl groups. In other words, if we have a short exact sequence $1\to G_1\to G_2\to G_3\to 1$ of connected reductive $k$-groups, then $W(G_2)=W(G_1)\times W(G_3)$.</p> http://mathoverflow.net/questions/69712/existence-of-a-bounded-finitely-generated-torsion-free-resolution Existence of a bounded finitely generated torsion-free resolution Mikhail Borovoi 2011-07-07T12:33:17Z 2011-07-08T06:01:31Z <p>I am looking for a reference for (or a proof of) the following fact:</p> <p>Let $G$ be a profinite group. Let $X^\bullet$ be a complex of discrete $G$-modules. We assume that the cohomology $G$-modules of $X^\bullet$ are nontrivial only for finitely many degrees, and that they are finitely generated over $\mathbb{Z}$. (We do not assume that the $G$-modules $X^i$ are finitely generated over $\mathbb{Z}$.) Then there exists a quasi-isomorphism $M^\bullet \to X^\bullet$, where $M^\bullet$ is a bounded complex of finitely generated (over $\mathbb{Z}$) torsion-free $G$-modules.</p> http://mathoverflow.net/questions/69712/existence-of-a-bounded-finitely-generated-torsion-free-resolution/69774#69774 Answer by Mikhail Borovoi for Existence of a bounded finitely generated torsion-free resolution Mikhail Borovoi 2011-07-08T06:01:31Z 2011-07-08T06:01:31Z <p>Proof (due to Joseph Bernstein). Assume that $H^i(X^\bullet)=0$ for $i>n$. We choose a $G$-morphism $A^n\to \ker[X^n\to X^{n+1}]$ such that the induced morphism $A^n\to H^n(X^\bullet)$ is surjective, where $A^n$ is a finitely generated (over $\mathbb{Z}$) torsion-free $G$-module. We regard $A^n$ as a complex (with one $G$-module $A^n$ in degree $n$). We have a morphism of complexes $\varphi\colon A^n\to X^\bullet$. We denote by $X_{(1)}^\bullet$ the cone of $\varphi$. It is easy to see that $H^n(X_{(1)}^\bullet)=0$. Then we apply this procedure to $X_{(1)}^\bullet$ for $n-1$ to obtain $X_{(2)}^\bullet$ with $H^{n-1}(X_{(2)}^\bullet)=0$, and so on.</p> <p>Assume that $H^i(X^\bullet)=0$ for $i\le n-m$. Then the complex $X_{(m)}^\bullet$ is acyclic. One can check that $X_{(m)}^\bullet$ is the cone of some morphism of complexes $\psi\colon M^\bullet\to X^\bullet$, where $M^\bullet$ is a bounded complex of finitely generated torsion-free $G$-modules. Since the cone $X_{(m)}^\bullet$ of $\psi$ is acyclic, we see that $\psi$ is a quasi-isomorphism.</p> http://mathoverflow.net/questions/65154/norms-in-galois-extensions Norms in Galois extensions Mikhail Borovoi 2011-05-16T16:54:17Z 2011-05-16T16:54:17Z <p>Let $k$ be a field of characteristic 0, and $\overline k$ be a fixed algebraic closure of $k$. Let $k\subset F\subset E$ be a tower of finite Galois extensions in $\overline k$, where both $\mathrm{Gal}(F/k)$ and $\mathrm{Gal}(E/F)$ are nonabelian simple finite groups. We have a norm map $N\colon E\to F$. Choose $x\in F^*$ which is not a norm, i.e. is not contained in the image of $N$.</p> <p>Now let $K/k$ be a finite solvable field extension in $\overline k$. Then $K\cap E=k$, and we have a norm map of composites $N_K\colon KE\to KF$.</p> <blockquote> <p>Does there always exist a finite solvable field extension $K/k$ such that $x$ is contained in the image of the map $N_K$?</p> </blockquote> <p>If $k$ is a number field, the answer is YES. For an arbitrary field $k$ I expect in general the answer NO, but I cannot construct a counter-example.</p> http://mathoverflow.net/questions/64909/rational-points-over-completions-of-a-number-field Rational points over completions of a number field Mikhail Borovoi 2011-05-13T17:32:54Z 2011-05-13T20:29:01Z <p>Let $X$ be a smooth geometrically irreducible $k$-variety over a number field $k$. I do not assume that $X$ has a $k$-point. Is it true that $X$ has $k_v$-points for almost all places $v$ of $k$?</p> http://mathoverflow.net/questions/62456/in-a-compact-lie-group-can-two-closed-connected-subgroups-generate-a-non-closed/62467#62467 Answer by Mikhail Borovoi for In a compact lie group, can two closed connected subgroups generate a non-closed subgroup? Mikhail Borovoi 2011-04-20T22:17:31Z 2011-04-21T07:40:53Z <p>The abstract subgroup generated by $H$ and $K$ is closed.</p> <p>We may assume that $G$ is connected. The groups $G$, $H$, $K$ are the groups of real points of real algebraic groups $\mathbf{G}$, $\mathbf{H}$, $\mathbf{K}$. Let $g$, $h$ , $k$ be their Lie algebras. The complexifications $g_C$, $h_C$, $k_C$ are algebraic Lie algebras, i.e. Lie algebras of complex algebraic groups. Let $l_C$ be the Lie subalgebra of $g_C$ generated by Lie subalgebras $h_C$ and $k_C$. We refer to the book: Onishchik, A. L.; Vinberg, È. B. Lie groups and algebraic groups. Springer-Verlag, Berlin, 1990. By Theorem 3.3.2 of this book, the Lie subalgebra $l_C$ is algebraic, i.e. it is the Lie algebra of a unique connected complex algebraic subgroup $\mathbf{L}_C\subset \mathbf{G}_C$. Clearly $\mathbf{L}_C$ is defined over $R$, i.e. comes from some real algebraic subgroup $\mathbf{L}\subset \mathbf{G}$. Set $L=\mathbf{L}(R)$. Since $\mathbf{L}$ is connected and compact, the group of real points $L$ is connected, see Onishchik and Vinberg, Corollary 1 of Theorem 5.2.1. The Lie algebra $l$ of $L$ is generated by the subalgebras $h$ and $k$. Since $H$ and $K$ are connected, $\mathbf{L}$ contains $\mathbf{H}$ and $\mathbf{K}$, and $L$ contains $H$ and $K$.</p> <p>Let $L'$ denote the abstract subgroup generated by $H$ and $K$, it is contained in $L$. Since the Lie algebra $l$ is generated by $h$ and $k$, one can easily see that for any element $A\in l$ there exists a smooth map $\varphi$ from an interval $(-\varepsilon, \varepsilon)$ to $L$ with image contained in $L'$ and such that $d\varphi|_0=A$. It follows that $L'$ contains an open neighborhood of $1$ in $L$. Since $L$ is connected, we see that $L'=L$. Thus the abstract subgroup $L'$ generated by $H$ and $K$ is closed.</p> http://mathoverflow.net/questions/60781/diagonalizable-subgroups-of-a-connected-linear-algebraic-group Diagonalizable subgroups of a connected linear algebraic group Mikhail Borovoi 2011-04-06T08:00:44Z 2011-04-11T21:52:20Z <p>Let $G$ be a connected linear algebraic group over an algebraically closed field $k$ of characteristic 0. Let $D\subset G$ be a closed diagonalizable subgroup of $G$ (a subgroup of multiplicative type). Is it true that $D$ is contained in some torus $T\subset G$?</p> <p>This is so for $G=\mathrm{GL}_n$. Is this true for any connected linear $G$ (or any connected reductive $G$)?</p> <p>I am stuck with this simple question...</p> <p><strong>Edit.</strong> The answer to the original question is NO, see Angelo's answer. However, is it true that any <em>cyclic</em> finite diagonalizable subgroup $C$ of $G$ is contained in some torus $T\subset G$?</p> http://mathoverflow.net/questions/130950/groupoids-vs-action-groupoids/130973#130973 Comment by Mikhail Borovoi Mikhail Borovoi 2013-05-18T16:45:32Z 2013-05-18T16:45:32Z @Mkouboi: I do not assume that there is a $\Gamma$-fixed point in $X$, and therefore $\Gamma$ does not act on the fibres $A(x,-)$ or $A(-,x)$ or $A(x,x)$. http://mathoverflow.net/questions/130950/groupoids-vs-action-groupoids/130973#130973 Comment by Mikhail Borovoi Mikhail Borovoi 2013-05-17T17:07:54Z 2013-05-17T17:07:54Z @Mkouboi: I do not understand your answer. You write: &quot;Moreover, any connected Γ-groupoid is in particular connected and hence in the form G⋉X. By construction of the group G, Γ acts compatibly on it and X...&quot; But the construction of $G$ from the groupoid A⇉X is <i>not functorial</i>, so I do not understand how you get an action of Γ on G. Could you please add details? http://mathoverflow.net/questions/130712/equivalence-and-weak-equivalence-of-groupoids Comment by Mikhail Borovoi Mikhail Borovoi 2013-05-15T13:52:58Z 2013-05-15T13:52:58Z @The User: Thank you for the answer and the editing suggestion! http://mathoverflow.net/questions/130712/equivalence-and-weak-equivalence-of-groupoids Comment by Mikhail Borovoi Mikhail Borovoi 2013-05-15T13:05:25Z 2013-05-15T13:05:25Z The braces that I typed in the displayed formula are not visible! Maybe somebody can edit the formula to make the braces visible... Thank you, http://mathoverflow.net/questions/130256/second-nonabelian-group-cohomology-cocycles-vs-gerbes/130272#130272 Comment by Mikhail Borovoi Mikhail Borovoi 2013-05-11T06:05:27Z 2013-05-11T06:05:27Z @Urs: Thank you! I will try to read your preprint. Still, I would appreciate if you could write a more explicit answer... http://mathoverflow.net/questions/130249/solve-the-singularities-of-parabolic-orbits-of-schubert-cells Comment by Mikhail Borovoi Mikhail Borovoi 2013-05-10T12:09:03Z 2013-05-10T12:09:03Z I remark that the OP has never voted up or down any question or answer. http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids/127787#127787 Comment by Mikhail Borovoi Mikhail Borovoi 2013-04-18T05:52:02Z 2013-04-18T05:52:02Z @Omar: Thank you, it is fine now. http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids/127747#127747 Comment by Mikhail Borovoi Mikhail Borovoi 2013-04-18T05:50:11Z 2013-04-18T05:50:11Z @Sam: Thank you for detailed answer! Unfortunately, I cannot accept two answers, otherwise I would accept your answer as well... http://mathoverflow.net/questions/127531/constructing-a-stack-gerbe-from-a-connected-groupoid/127664#127664 Comment by Mikhail Borovoi Mikhail Borovoi 2013-04-17T20:44:02Z 2013-04-17T20:44:02Z @DavidRoberts: Thank you! http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids/127787#127787 Comment by Mikhail Borovoi Mikhail Borovoi 2013-04-17T15:19:31Z 2013-04-17T15:19:31Z @Omar: Thank you, now we have a group $G$ and a action of $G$ on $X$. How can we define an isomorphism between $G\ltimes X$ and $A$? Please kindly add more details when you can. http://mathoverflow.net/questions/127729/connected-groupoids-and-action-groupoids/127787#127787 Comment by Mikhail Borovoi Mikhail Borovoi 2013-04-17T08:29:04Z 2013-04-17T08:29:04Z @Omar: You write: &quot;Any group $G$ with subgroup $H$ of the correct index, the action of $G$ on the set of cosets of $H$ has action groupoid isomorphic to $A$&quot;. How can one construct such an isomorphism? http://mathoverflow.net/questions/127531/constructing-a-stack-gerbe-from-a-connected-groupoid/127664#127664 Comment by Mikhail Borovoi Mikhail Borovoi 2013-04-16T17:37:08Z 2013-04-16T17:37:08Z @DavidCarchedi: Thank you! However, I am not quite familiar with your notations $Set^{B\Gamma}$, $Gpd(Set^{B\Gamma})$ and $Gpd^{B\Gamma}$. Your links did not help me. Could you please explain me these notations? Also, what is a weak functor? Where can I read about all this stuff? http://mathoverflow.net/questions/127531/constructing-a-stack-gerbe-from-a-connected-groupoid/127595#127595 Comment by Mikhail Borovoi Mikhail Borovoi 2013-04-16T17:24:03Z 2013-04-16T17:24:03Z @DavidRoberts: Thank you for your detailed answer. This is exactly the kind of answer that I wanted to get! http://mathoverflow.net/questions/127531/constructing-a-stack-gerbe-from-a-connected-groupoid/127595#127595 Comment by Mikhail Borovoi Mikhail Borovoi 2013-04-15T11:49:33Z 2013-04-15T11:49:33Z Probably when you write &quot;such that $\pi(a(p,f))=\pi(a)$&quot;, you mean $\pi(a(p,f))=\pi(p)$. When you write &quot;in addition, we demand that the map $\dots$ is an isomorphism&quot;, you mean a bijection. Is this correct? http://mathoverflow.net/questions/127531/constructing-a-stack-gerbe-from-a-connected-groupoid/127595#127595 Comment by Mikhail Borovoi Mikhail Borovoi 2013-04-15T10:31:45Z 2013-04-15T10:31:45Z I do not understand the line: &quot;isomorphic to $T\times_X S$ for some map $T\to X$&quot;. What is the map $S\to X$ in the fibered product? Is it a typo?