User ronggang - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:17:22Z http://mathoverflow.net/feeds/user/11056 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128516/arithmetic-group-over-function-fields-and-its-fundamental-domain arithmetic group over function fields and its fundamental domain ronggang 2013-04-23T18:21:29Z 2013-04-23T21:19:42Z <p>Let $G$ be a semi-simple algebraic group defined over a global function field $K$. Let $S$ be a finite set of places of $K$. For a place $v$ of $K$ let $K_v$ be the completion under $v$. We take $K_S=\prod_{v\in S}K_v$ and $O={x\in K: x \mbox{ is integral in }K_v \mbox{ if } v\not \in S }$. It is well known that $G(O)$ is a lattice in $G(K_S)$. Are there any fundamental domain of $G(O)$ expressed in terms of Ziegel sets?</p> <p>More precisely we want the following type result. If $H$ is semi-simple $\mathbb Q$-group, then it is proved in Theorem 15.5 of Borel's book "Introduction aux groupes arithmétic" that there exists a Ziegel set $F$ and a finite subset $C\subset G(Q)$ such that $FCG(\mathbb Z)=G(\mathbb R)$. To my understanding this result is not contain in the paper of Borel and Harish-Chandra 1962 on arithmetic groups.</p> <p>It will be great if one can suggest some references (in English) about it. </p> http://mathoverflow.net/questions/125782/local-field-and-number-field local field and number field ronggang 2013-03-28T03:07:46Z 2013-03-28T07:29:00Z <p>Let $K$ be a local field (locally compact topological field) of characteristic zero. Is it true that $K$ is isomorphic to the completion of a number field under some valuations? If yes, then how to prove it?</p> <p>I ask this question since in a paper it is said that $k_v^*/(k_v^*)^2$ where $k_v$ is the completion of a number field at some place $v$ has order 1, 2, 4 or 8. From the structure theory of local fields this order argument is incorrect for general local fields of characteristic zero. </p> <p>If not every local field of characteristic zero comes from the completion of some number field and the above claim about the order of $k_v^*/(k_v^*)^2$ is correct. Can some one prove it or give a reference for it?</p> http://mathoverflow.net/questions/106259/a-question-on-tits-note-reductive-groups-over-local-fields a question on TITS' note "Reductive groups over local fields" ronggang 2012-09-03T16:50:05Z 2013-03-23T03:29:09Z <p>This note appears in "Proceedings of Symposia in pure mathematics" vol.33 1979 part 1 pp. 26-69. The question will be about materials on page 31-32. </p> <p>Let $G$ be a reductive algebraic group (not necessarily connected) defined over a local field $K$. We fix a maximal $K$-split torus $S$ of $G$ and take N(resp. Z) to be normalizer (resp. the centralizer) of $S$ in $G$. Let $X_*=Hom_K (Mult, S)$ (Resp. $X^*=Hom_K (S, Mult)$) be the group of cocharacters (resp. characters) of $S$. Let $V=X_*\otimes_{\mathbb Z}\mathbb R$. We fix a discrete valuation $\omega: K\to (-\infty, \infty]$. Let $\nu: Z(K)\to V$ be the homomorphism defined by $$\chi (\nu (z))=-\omega (\chi (z)) \quad \mbox{for}\ z\in Z(K) \mbox{ and } \chi \in X^*(\mathbb Z).$$ Let $Z_c$ be the kernel of $\nu$. Then we have a short exact sequence of gorups $$0\to Z(K)/Z_c\to N(K)/Z_c\to N(K)/Z(K)\to 0$$ where $N(K)/Z(K)$ is a finite group.</p> <p>Then it is claimed that the map $\nu$ induces a group homomorphism $\phi$ from $N$ to the group of affine transformations of $V$ such that for $z\in Z(K)$ and $x\in V$ one has $\phi(z)x=x+\nu (z)$. </p> <p>I do not understand in which way this function $\phi$ is defined. </p> http://mathoverflow.net/questions/89777/how-many-q-forms-of-sl-nr-are-there-for-a-given-q-rank how many Q-forms of SL_n(R) are there for a given Q-rank ronggang 2012-02-28T16:42:11Z 2013-01-12T11:17:30Z <p>Let $G$ be a linear algebraic group defined over $\mathbb Q$. Suppose that $G$ is isomorphic to $SL_n$ over $\mathbb R$. Suppose the $\mathbb Q$-rank of $G$ is fixed, say $m$. How many types are there for $G$ up to $\mathbb Q$-isomorphism? Are they finite especially for $m>2$?</p> http://mathoverflow.net/questions/105293/proofs-of-ergodicity-of-sl2-z-action-on-r2-without-using-duality proofs of ergodicity of Sl(2, Z) action on R^2 without using duality ronggang 2012-08-23T03:18:09Z 2012-10-03T19:50:42Z <p>The group $G=SL(2, R)$ acts linearly on $\mathbb R^2$. The Lebesgue measure of $\mathbb R^2$ is invariant and ergodic for $G$. There is a proof using duality theorem: Let $U$ be the upper triangular unipotent subgroup of $G$ and let $\Gamma=SL(2, \mathbb Z)$. Then $U$ acts ergodically on $G/\Gamma$. Therefore duality tells us that $\Gamma$ acts ergodically on $G/U\cong \mathbb R^2\backslash 0$.</p> <p>This method is good but it can not be extended to the group action on non homogeneous manifolds. Are there any other proofs? It is better that I can see different proofs to learn methods.</p> http://mathoverflow.net/questions/101157/the-group-g-of-algebraic-groups-over-local-fields The group G^+ of algebraic groups over local fields ronggang 2012-07-02T16:24:16Z 2012-07-03T08:59:42Z <p>Let $G$ be an algebraic group defined over a char 0 local field $k$. Following Borel and Tits (73) we define the group $G^+(k)$ or $G^+$ by the subgroup of $G(k)$ generated by the unipotent elements of $G(k)$. </p> <p>Suppose $G$ is generated by a finite set of unipotent $k$-subgroups, say $U_1,\cdots, U_n$. Is it true that the group generated by $U_1(k), \cdots, U_n(k)$ is $G^+$? </p> <p>I feel the answer is positive but do not know how to prove it. It seems that the ideas of the original paper of Borel and Tits can help, but I still do not read French (which I always plan to learn) yet. </p> http://mathoverflow.net/questions/100847/surjectivity-of-rational-points-induced-by-surjective-map-from-affine-space surjectivity of rational points induced by surjective map from affine space ronggang 2012-06-28T07:07:04Z 2012-06-28T08:22:23Z <p>Let $k$ be a local field of char $0$ (which is the case I concern). Let $V$ be a variety defined over $k$ and let $f: \mathbb A^n\to V$ be a surjective map (over the algebraic closure of $k$) defined over $k$. Is it true that the restriction of $f$ to $k$ rational points $k^n\to V(k)$ surjective?</p> <p>After it is answered I realized that I simplified what I want to know too much. Please see the comments for the answer for more information.</p> http://mathoverflow.net/questions/97840/is-the-image-of-the-representation-of-the-fundamental-group-associated-to-a-local Is the image of the representation of the fundamental group associated to a local system discrete? ronggang 2012-05-24T14:41:28Z 2012-05-25T11:23:50Z <p>If $f: X \to S$ is a projective smooth morphism between complex algebraic varieties. Does the $\pi_1(S)$-representation corresponding to the local system $R^i f_* (C_X)$ on $S$ maps $\pi_1(S)$ onto a discrete subgroup of $GL(r, C)$?</p> http://mathoverflow.net/questions/49810/unitary-representation-of-semisimple-lie-groups-in-view-of-moores-ergodicity-thm unitary representation of semisimple lie groups in view of Moore's ergodicity thm ronggang 2010-12-18T15:39:33Z 2011-08-27T06:31:52Z <p>Let $G=G_1\times\ldots\times G_n$ be a product of (connected) simple Lie groups and $(H, \pi)$ be a unitary representation of $G$. In a proof of Moore's ergodicity thm it uses the following fact $$\pi=\oplus_{I\subset [1,\ldots, n]} \pi_I$$ where $(\pi_I, H_I)$ be the subrepresentation of $\pi$ whose kernel contains $G_i$ for $i\in I$ but no $G_j$ for $j\not \in I$.</p> <p>How do we prove this fact?</p> http://mathoverflow.net/questions/63765/unipotent-group-and-translation-invariant-metric unipotent group and translation invariant metric ronggang 2011-05-03T02:44:07Z 2011-05-03T15:13:13Z <p>Let $U$ be a unipotent upper triangluar group over a local field $K$ of characteristic zero. Can we guarantee that there is a right translation invariant metric on $U$ such that any ball of finite radius is relatively compact?</p> http://mathoverflow.net/questions/47202/is-the-subgroup-generated-by-one-parameter-unipotent-subgroups-a-lie-subgroup is the subgroup generated by one-parameter unipotent subgroups a Lie subgroup? ronggang 2010-11-24T06:04:21Z 2011-03-31T12:35:15Z <p>Let $G$ be a Lie group and $H$ be a subgroup generated by some one parameter unipotent subgroups (in group sense). Is it true that $H$ has a Lie group structure which makes it a Lie subgroup of $G$? Is $H$ closed? </p> <p>How about for general subgroups?</p> http://mathoverflow.net/questions/56713/is-restriction-of-scalars-of-simply-connected-algebraic-groups-still-sc Is restriction of scalars of simply connected algebraic groups still SC? ronggang 2011-02-26T06:47:13Z 2011-02-26T07:15:27Z <p>Let $G$ be a simply connected semisimple algebraic $K$-group and $K$ be a finite extension of $k$. Is $R_{K/k}G$ still a simply connected algebraic group?</p> <p>We say $G$ is simply connected if for any central isogeny $G'\to G$ is in fact an isomorphism of algebraic groups. </p> http://mathoverflow.net/questions/50541/quotient-sigma-algebra-on-quotient-space-of-locally-compact-groups quotient sigma-algebra on quotient space of locally compact groups ronggang 2010-12-28T05:40:51Z 2010-12-28T12:39:26Z <p>Let $G$ be a locally compact second countable topological group and $H$ be a closed subgroup of $G$. Let $\pi: G\to G/H$ be the quotient map. For the Borel $\sigma$-algebra $\mathcal A$ on $G/H$. Is it true that $\pi^{-1}(\mathcal A)$ is the Borel subsets of $G$ consisting of right $H$ orbits?</p> http://mathoverflow.net/questions/49702/is-a-subset-that-contains-no-positive-measurable-subsets-contained-in-a-null-meas Is a subset that contains no positive measurable subsets contained in a null measurable set? ronggang 2010-12-17T03:27:59Z 2010-12-18T19:22:16Z <p>Let $(X, \mathcal B, \mu)$ be a "good" measure space, e.g. $\mu$ is a positive Radon measure on a locally compact topological space $X$ with Borel $\sigma$-algebra $\mathcal B$. Let $A\subset X$ such that every measurable subset of $A$ has zero measure. Is it true that there is a zero measure set $B$ such that $A\subset B$?</p> http://mathoverflow.net/questions/47800/is-every-ball-in-a-lie-group-relatively-compact Is every ball in a Lie group relatively compact? ronggang 2010-11-30T15:22:34Z 2010-11-30T15:33:09Z <p>Let $G$ be a connected Lie group with a left invariant Riemannian metric which induces a metric $d$ on $G$. Does the ball $B_r$ centered at the identity have compact closure? </p> <p>It is true in easy examples and looks correct. </p> http://mathoverflow.net/questions/125782/local-field-and-number-field/125790#125790 Comment by ronggang ronggang 2013-03-28T08:28:36Z 2013-03-28T08:28:36Z I see. Thank you. http://mathoverflow.net/questions/125782/local-field-and-number-field Comment by ronggang ronggang 2013-03-28T08:24:07Z 2013-03-28T08:24:07Z Krasner's Lemma can be used to prove that every local field of characteristic zero comes from completion of a number field. Therefore it answers the question. http://mathoverflow.net/questions/125782/local-field-and-number-field/125790#125790 Comment by ronggang ronggang 2013-03-28T07:32:09Z 2013-03-28T07:32:09Z It is known that a local field of characteristic zero is a finite extension of $Q_2$. This why I said that $K*/(K^*)^2$ has order 1,2,4 or 8 is incorrect. The question is whether the order argument is true for local fields coming from completions of number field? If every local field comes from the completion of a number field, then of course the order argument is incorrect. http://mathoverflow.net/questions/101157/the-group-g-of-algebraic-groups-over-local-fields/101177#101177 Comment by ronggang ronggang 2012-07-03T02:24:35Z 2012-07-03T02:24:35Z @Jim: The question is motivated by the p-adic vertion of Rater's Theorem (Margulis Tomanov 94 Invariant measures for actions of unipotent groups over local fields on homogeneous spaces.) They proved the theorem for rational points of unipotent groups, then as a consequence said that the result is also true for groups generated by rational points of unipotent groups. The question arises from my attempt to understand arguments between Theorem 1 and 2 of that paper. http://mathoverflow.net/questions/100847/surjectivity-of-rational-points-induced-by-surjective-map-from-affine-space/100848#100848 Comment by ronggang ronggang 2012-06-29T01:39:26Z 2012-06-29T01:39:26Z @ Will: It is not surjective even in this case since in general $V^+$ or $V(k)^+$ is not equal to $V(k)$. e.g. PGL_2. http://mathoverflow.net/questions/100847/surjectivity-of-rational-points-induced-by-surjective-map-from-affine-space/100848#100848 Comment by ronggang ronggang 2012-06-28T14:33:33Z 2012-06-28T14:33:33Z I mean the group generated by the set $f(k^n)$ has finite index in the group $V(k)$. Here is the main example in my mind. Suppose $V$ is generated by connected unipotent groups defined over $k$. Then we get such a map $f$ according to I Porp. 2.2 of Borel's algebraic groups. Then the question is whether the group generated by $k$ points of these unipotent groups has finite index in $V(k)$. If $V$ is semisimple then the question is whether the group they generate contains $V^+$ in the sense of 1973 paper of Borel and Tits( Homomorphismes &quot;Abstraits&quot; de Groupes Algebriques Simples). http://mathoverflow.net/questions/100847/surjectivity-of-rational-points-induced-by-surjective-map-from-affine-space/100848#100848 Comment by ronggang ronggang 2012-06-28T08:24:51Z 2012-06-28T08:24:51Z Sorry, finite index in $V(k)$. http://mathoverflow.net/questions/100847/surjectivity-of-rational-points-induced-by-surjective-map-from-affine-space/100848#100848 Comment by ronggang ronggang 2012-06-28T08:20:28Z 2012-06-28T08:20:28Z Thanks. In fact what I want to know is the following: Assume in addition that $V$ is a linear algebraic group. Is it true that the group generated by $f(k^n)$ has finite index in $V$? http://mathoverflow.net/questions/56713/is-restriction-of-scalars-of-simply-connected-algebraic-groups-still-sc Comment by ronggang ronggang 2011-02-26T14:06:21Z 2011-02-26T14:06:21Z I think I would assume $K/k$ is separable. Are there any convenient reference for this full weight lattice characterization? http://mathoverflow.net/questions/49810/unitary-representation-of-semisimple-lie-groups-in-view-of-moores-ergodicity-thm Comment by ronggang ronggang 2010-12-19T01:33:17Z 2010-12-19T01:33:17Z I think I should give a ref. for this. Ergodic theory and topological dynamics of group actions on homogeneous spaces, M Bekka &amp; M Mayer, Cambridge University Press, P91. In fact, I don't quite believe this fact. http://mathoverflow.net/questions/47800/is-every-ball-in-a-lie-group-relatively-compact Comment by ronggang ronggang 2010-12-01T12:21:24Z 2010-12-01T12:21:24Z According to Hopf-Rinow, one needs to show the space is complete. This follows from the fact that the metric is left invariant, and it is complete locally! http://mathoverflow.net/questions/47800/is-every-ball-in-a-lie-group-relatively-compact Comment by ronggang ronggang 2010-12-01T00:22:29Z 2010-12-01T00:22:29Z Thanks, Igor!!!! http://mathoverflow.net/questions/47202/is-the-subgroup-generated-by-one-parameter-unipotent-subgroups-a-lie-subgroup/47263#47263 Comment by ronggang ronggang 2010-11-29T02:12:59Z 2010-11-29T02:12:59Z Another correction: The above is true in case of two parabolic $k$-subgroups containing a common minimal parabolic $k$-subgroup of ...........; http://mathoverflow.net/questions/47202/is-the-subgroup-generated-by-one-parameter-unipotent-subgroups-a-lie-subgroup/47263#47263 Comment by ronggang ronggang 2010-11-29T02:11:18Z 2010-11-29T02:11:18Z A correction: $(M_1M_2)(k)=M_1(k)M_2(k)$ http://mathoverflow.net/questions/47202/is-the-subgroup-generated-by-one-parameter-unipotent-subgroups-a-lie-subgroup/47263#47263 Comment by ronggang ronggang 2010-11-29T02:10:07Z 2010-11-29T02:10:07Z Dear Zroslav: This theorem only holds for the rational points over an algebraically closed field. It seems not true that $(M_1M_2)(k)=(M_1M_2)(k)$ which is true in some special cases, e.g. two parabolic $k$-subgroups of a connected reductive $k$-group.