User tobias hartnick - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:46:50Z http://mathoverflow.net/feeds/user/9927 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46358/arithmetic-groups-vs-zariski-dense-discrete-subgroups/46468#46468 Answer by Tobias Hartnick for arithmetic groups VS. Zariski dense discrete subgroups? Tobias Hartnick 2010-11-18T10:39:02Z 2010-11-18T10:39:02Z <p>Arbitrary Zariski-dense subgroups in a semisimple group can be very small from a real-analytic point of view. It seems that algebra cannot distinguish between "small" and "large" Zariski-dense subgroups, so most criteria to distinguish between the two have a strong non-algebraic flavour. (Of course one can also characterize arithmetic groups algebraically, but this has even less to do with the line of argument you seem to suggest.) From a dynamical point of view, the key difference between lattices and arbitrary Zariski-dense subgroups is that the former act transitively on the product of the Furstenberg boundary of the ambient Lie group with itself ("double ergodicity"). This is a sort of "largeness" property. There are various ways to capture this property, the most systematic way seems to me the concept of a generalized Weyl group due to Bader and Furman. </p> http://mathoverflow.net/questions/2144/a-single-paper-everyone-should-read/46345#46345 Answer by Tobias Hartnick for A single paper everyone should read? Tobias Hartnick 2010-11-17T11:31:44Z 2010-11-17T11:31:44Z <p>I am surprised to see that so many people suggest meta-mathematical articles, which try to explain how one should do good mathematics in one or the other form. Personally, I usually find it a waste of time to read these, and there a few statements to which I agree so wholeheartedly as the one of Borel: </p> <p>"I feel that what mathematics needs least are pundits who issue prescriptions or guidelines for presumably less enlightened mortals."</p> <p>The mere idea that you can learn how to do mathematics (or in fact anything useful) from reading a HowTo seems extremely weird to me. I would rather read any classical math article, and there are plenty of them. The subject does not really matter, you can learn good mathematical thinking from each of them, and in my opinion much easier than from any of the above guideline articles. Just to be constructive, take for example (in alphabetical order)</p> <ul> <li>Atiyah&amp;Bott, The Yang-Mills equations over Riemann surfaces.</li> <li>Borel, Sur la cohomologie des espaces fibrés principaux et des espaces homogènes de groupes de Lie compacts.</li> <li>Furstenberg, A Poisson formula for semi-simple Lie groups.</li> <li>Gromov,Groups of polynomial growth and expanding maps.</li> <li>Tate, Fourier analysis in number fields and Hecke's zeta-functions.</li> </ul> <p>I am not suggesting that any mathematician should read all of them, but any one of them will do. In fact, the actual content of these papers does not matter so much. It is rather, that they give an insight how a new idea is born. So, if you want to give birth to new ideas yourself, look at them, not at some guideline.</p> http://mathoverflow.net/questions/45730/transitive-semigroups-of-2-times-2-matrices/45757#45757 Answer by Tobias Hartnick for Transitive Semigroups of $2\times 2$ matrices Tobias Hartnick 2010-11-11T22:27:22Z 2010-11-12T15:58:51Z <p>Here is a complete answer:</p> <p><em>Every semigroup</em> $S$ <em>of invertible</em> $2\times 2$<em>-matrices which is transitive on</em> $\mathbb R^2$ <em>is either conjugate to</em> $SO_2(\mathbb R) \times \mathbb R^+$ <em>or</em> $SO_2(\mathbb R) \times \mathbb R$ <em>or it is a product of</em> $SL_2(\mathbb R)$ <em>and a multiplicative subgroup of</em> $\mathbb R$<em>.</em></p> <p><strong>Proof:</strong> Let S be such a semigroup. Then the intersection $S_0$ with $SL_2(\mathbb R)$ is a subsemigroup of $SL_2(\mathbb R)$. By a theorem of Hilgert and Hofmann (see their beautiful paper on "Old and new on $SL_2$") there are only three choices for $S_0$: Either $S_0$ is all of $SL_2(\mathbb R)$, a circle group or contained in a conjugate of the elements of $SL_2(\mathbb R)$ with only positive entries. If $S_0$ is a circle group, then $S$ will be conjugate to $SO_2(\mathbb R) \times \mathbb R^+$ or $SO_2(\mathbb R) \times \mathbb R$. If $S_0$ happens to be all of $SL_2(\mathbb R)$, then we have $SL_2(\mathbb R)\subset S \subset GL_2(\mathbb R)$, so $S$ is a product of $SL_2(\mathbb R)$ and a multiplicative subgroup of $\mathbb R$. In the third case, we may assume that $S_0$ is actually contained in the semigroup described above. Then $S_0$ maps every vector with two positive entries to a vector with two posiitve entries, hence $S$ maps the upper right quadrant to a subset of itself and the lower left quadrant. In particular, $S$ cannot be transitive.</p> http://mathoverflow.net/questions/43286/how-and-how-much-do-the-notations-and-diagrams-influence-our-understanding-of-mat/45766#45766 Answer by Tobias Hartnick for How and how much do the notations and diagrams influence our understanding of mathematical concepts? Tobias Hartnick 2010-11-11T23:41:42Z 2010-11-12T09:01:09Z <p>To support the last remark of Donu Arapura, the following anecdote might be helpful: The late Beno Eckmann, one of the key players of the early developments in algebraic topology in the 40ies and 50ies, was asked to explain, why the revolution in algebraic topology happened in the 50ies. You can find his answer in his "<a href="http://www.fim.math.ethz.ch/preprints/2005/eckmann.pdf" rel="nofollow">Mathematical Miniatures</a>". In short, he explains that the idea to represent a function by an arrow, and a composition of functions by a diagram was completely unknown until the late 1940ies (!!!), when Leray introduced this notation. There seems to be no doubt that even the formulation of modern algebraic topology would have been impossible without the idea of an arrow and/or a diagram!</p> http://mathoverflow.net/questions/16468/topologists-loops-versus-algebraists-loops/45763#45763 Answer by Tobias Hartnick for Topologists loops versus algebraists loops Tobias Hartnick 2010-11-11T23:26:28Z 2010-11-11T23:26:28Z <p>While the answer to your question is negative in general, as pointed out before, the answer is positive for certain type of loop <strong>groups</strong>. This can be proved using topological twin buildings. See Linus Kramer, Loop Groups and Twin Building. (It is not stated very explicitly, but the topology used on the twin building and hence the algebraic loop group is meant to be the ind-topology coming from the Bruhat cell decomposition.)</p> http://mathoverflow.net/questions/45748/what-are-the-compact-symmetric-kahler-algebraic-varieties/45753#45753 Answer by Tobias Hartnick for What are the Compact Symmetric Kahler Algebraic Varieties? Tobias Hartnick 2010-11-11T22:02:12Z 2010-11-11T22:02:12Z <p>If we ignore the trivial case of the affine line, then irreducible symmetric spaces come in pairs compact - non-compact. The compact ones are naturally projective varieties, while the non-compact ones are affine varieties. Thus question (4) is problematic, unless you mean "locally symmetric" or a more general notion of symmetric space than I understand here (i.e. "globally symmetric Riemannian symmetric space"). As far as non-compact symmetric spaces are concerned, they are Kähler if and only if they are biholomorphic to a bounded symmetric domain. Equivalently, there exists a compact quotient with non-trivial H^2 or, equivalently, the point stabilizer of the automorphism group has infinite center... I could give many more characterizations, but I do not quite see what you are after, so maybe you can provide more detailed information?</p> http://mathoverflow.net/questions/44501/what-are-the-canonical-and-earliest-references-to-trivial-symmetries-in-gauge-sys/44535#44535 Answer by Tobias Hartnick for What are the canonical and earliest references to trivial symmetries in gauge systems? Tobias Hartnick 2010-11-02T10:00:22Z 2010-11-02T14:12:40Z <p><strong>Remark:</strong> I think my answer should be ignored. (Apparently I did not understand the problem properly. Probably $S$ should be a linear functional on $C^\infty(M)$ (time-dependent?) etc. I still believe that the question ultimately boils down to something elementary once it is formulated in the right way. I just don't understand how the objects are defined.)</p> <p>I think the question would be easier to answer for mathematicians if formulated in standard math language. I am not sure I am able to translate it, but let me try: You have some configuration space (probably a manifold, maybe infinite-dimensional) $M$, and a one-parameter group ${g_t}$ of $M$ (probably a diffeomorphism) and I guess that you assume that this is contained in some nice Lie group $G$ so that $g_t = \exp(tX)$ with $X \in \mathfrak g$, the Lie algebra of $G$. Now you have a (smooth?) function S on M and you want it to be invariant under ${g_t}$. This just means $XS = 0$. Now what does OS mean? Maybe you could explain this. It seems to me that you want to deduce something about $X$ and the derivative of $S$, but I am not quite sure I understand your notation there. I believe if you reformulate your question along these lines, more people can help.</p> http://mathoverflow.net/questions/43805/when-is-g-isomorphic-to-g-times-g/44027#44027 Answer by Tobias Hartnick for When is $G$ isomorphic to $G \times G$? Tobias Hartnick 2010-10-28T22:20:52Z 2010-10-28T22:20:52Z <p>As a geometric group theorist, one would of course relax the question by allowing passage to finite index subgroup, i.e. one would ask for groups such that G and GxG have finite index subgroups, which are isomorphic. One then calls G an GxG commensurable, and for this weaker property there are a lot of interesting examples. My favourite one right now is the Grigorchuk group. But even commensurability to GxG is a very restrictive property: It implies, for instance, that if G is infinite, then it has infinite asymptotic dimension. I just stumbled over this result in the thesis of J. Smith. The proof is almost trivial: Since G is coarsely equivalent to GxG, it is coarsely equivalent to G^n for all n. Now Z embeds quasi-isometrically into G (since G is infinite), and hence Z^n embeds coarsely into G^n (hence G), so asdim G is at least asdim Z^n for all n, and we conclude. This is in particular the case if G and GxG are isomorphic. The upshot is, that for a group of finite asymptotic dimension one cannot have G=GxG, not even up to finite index.</p> http://mathoverflow.net/questions/41895/conjugacy-classes-with-elliptic-limit-points Conjugacy classes with elliptic limit points Tobias Hartnick 2010-10-12T10:36:55Z 2010-10-14T19:09:55Z <p>Let $G$ be a reductive algebraic group over $\mathbb R$ and $K$ a maximal compact subgroup. Then we refer to the conjugacy class in $G$ of some $k \in K$ as an elliptic conjugacy class. </p> <p><strong>Question:</strong> Can one characterizes those conjugacy classes in $G$ which contain an elliptic conjugacy class in their closure?</p> <p>(For $G = GL_n(\mathbb R)$ they are characterized by the fact that all eigenvalues are of modulus one, if I a not mistaken.)</p> http://mathoverflow.net/questions/41689/why-symmetric-spaces/41695#41695 Answer by Tobias Hartnick for Why symmetric spaces? Tobias Hartnick 2010-10-10T16:48:48Z 2010-10-10T16:48:48Z <p>If one accepts curvature as a measure of complexity of a Riemannian manifold (which one might or might not agree with), then the "simplest" Riemannian manifolds are those of constant curvature. Unfortunately, there are not so many of these; besides Euclidean space the only simply-connected examples are the spheres (constant positive curvature) and the hyperbolic spaces (constant negative curvature). (Of course there is a rich theory of non-simply-connected spaces of constant negative curvature, but never mind.) So, how can one weaken the notion of constant curvature to obtain a larger class of interesting, but not "too complicated" spaces? Well, it seems natural to ask that the covariant derivative of the curvature should be 0. In this case, one has many more 1-connected examples, and these are precisely symmetric spaces. They are nice in various senses:</p> <ul> <li>They can be classified, so there are not too many of them (but still sufficiently many to be "interesting").</li> <li>They admit a description in terms of Lie groups. This allows for very explicit computations, e.g. of curvature and characteristic classes. </li> <li>They admit a natural duality (compact/non-neg. curved vs. non-compact/non-pos. curved) if one ignores flat factors. This allows for the transfer of ideas between two different worlds (see e.g. Hirzebruch proportionality).</li> </ul> <p>What more can one ask for? On the other hand, one should stress that they are really rare and special objects, just slightly less rare than manifolds of constant curvature.</p> http://mathoverflow.net/questions/52337/stiefelwhitney-classes-in-the-spirit-of-chern-weil Comment by Tobias Hartnick Tobias Hartnick 2011-04-11T14:41:57Z 2011-04-11T14:41:57Z @diverietti: I think the confusion is about real curves as opposed to complex curves ;-) http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration Comment by Tobias Hartnick Tobias Hartnick 2011-01-30T12:06:48Z 2011-01-30T12:06:48Z &quot;The trouble is that beginning math students are not nearly sophisticated enough to handle the technical baggage underlying the Lebesgue integral.&quot; Well, in the French/German/Swiss/Italian(?)/Israeli... system (i.e. all non-American systems I know), all math undergraduates are taught Lebesgue integral, and in many cases Lebesgue integral only (with only a weak version of Riemann integral, which is hardly more then an antiderivative &quot;Cauchy&quot; approach as preparation). Not all handle it equally well, but everyone is required to learn it. The above statement seems to be extremely culturally biased http://mathoverflow.net/questions/51085/subfields-of-mathbbc-isomorphic-to-mathbbr-that-have-baire-property Comment by Tobias Hartnick Tobias Hartnick 2011-01-04T21:51:53Z 2011-01-04T21:51:53Z @Zev: The fact that $\mathbb C$ has uncountably many field automorphisms does not seem so surprising to me. After all the only obstruction that keeps $\mathbb R$ from having many automorphisms is the fact that its order can be described algebraically, hence is preserved. Since there is no such obstruction on $\mathbb C$, there should be plenty of automorphisms, and in fact there are. http://mathoverflow.net/questions/48477/never-appeared-forthcoming-papers/48497#48497 Comment by Tobias Hartnick Tobias Hartnick 2010-12-08T19:56:00Z 2010-12-08T19:56:00Z @Victor: Burger, Sarnak: Ramanujan Duals II qualifies in so far as there is no published paper entitled Ramanujan Duals I, but I guess the point here is more that part I appeared under a slightly different title, so maybe it is not quite what you are looking for. http://mathoverflow.net/questions/2144/a-single-paper-everyone-should-read/43207#43207 Comment by Tobias Hartnick Tobias Hartnick 2010-11-17T13:44:46Z 2010-11-17T13:44:46Z Even if bashing Bourbaki seems to have become hip again (as it was actually during the Bourbaki era as well), and thus the myth of the &quot;wholesale rejection&quot; of applications of mathematics &quot;during the Bourbaki era&quot; has become generally accepted in certain circles, it still remains a myth, which like all myths contains a germ of truth surrounded by a lot of prejudices, misunderstandings and plainly wrong statements. http://mathoverflow.net/questions/2144/a-single-paper-everyone-should-read/2256#2256 Comment by Tobias Hartnick Tobias Hartnick 2010-11-17T12:41:39Z 2010-11-17T12:41:39Z A very nice case study indeed, but I found the general remarks surrounding it superfluous and besides the point (and even insulting in parts). Not to talk about the horrible title, which raises expectations that the paper cannot keep. Why not call it &quot;On Szemeredi's theorem&quot;, skip Sections 1 and 3 and leave it to the reader which conclusions to draw? I was very disappointed to see that a great mathematician like Tao felt the need to write such a strange convolute of nice insights (in the case study) and complete trivialities (in Section 1). But then, my position seems to be an isolated one. http://mathoverflow.net/questions/45700/cohomology-of-the-unitary-group/45702#45702 Comment by Tobias Hartnick Tobias Hartnick 2010-11-12T01:02:32Z 2010-11-12T01:02:32Z Of course there is a striking similarity to the formula for Chern classes in terms of curvature here, which is no coincidence, since the latter arise from the above classes via transgression along the universal bundle. http://mathoverflow.net/questions/45730/transitive-semigroups-of-2-times-2-matrices/45740#45740 Comment by Tobias Hartnick Tobias Hartnick 2010-11-12T00:55:17Z 2010-11-12T00:55:17Z I agree. But then, the non-asked question seemed so much more interesting to me. http://mathoverflow.net/questions/45730/transitive-semigroups-of-2-times-2-matrices/45740#45740 Comment by Tobias Hartnick Tobias Hartnick 2010-11-11T23:19:17Z 2010-11-11T23:19:17Z I think he wants to assume all matrices to be invertible. http://mathoverflow.net/questions/45748/what-are-the-compact-symmetric-kahler-algebraic-varieties/45753#45753 Comment by Tobias Hartnick Tobias Hartnick 2010-11-11T23:14:51Z 2010-11-11T23:14:51Z No. Among the symmetric spaces, the compact K&#228;hler ones are the same as the complex projective ones. But not every complex projective variety is a symmetric space! http://mathoverflow.net/questions/45748/what-are-the-compact-symmetric-kahler-algebraic-varieties Comment by Tobias Hartnick Tobias Hartnick 2010-11-11T22:32:17Z 2010-11-11T22:32:17Z Of course the right context to study symmetric spaces is real varieties, and these are in general not K\&quot;ahler. http://mathoverflow.net/questions/45748/what-are-the-compact-symmetric-kahler-algebraic-varieties/45753#45753 Comment by Tobias Hartnick Tobias Hartnick 2010-11-11T22:31:17Z 2010-11-11T22:31:17Z But projective varieties are not affine. You asked for affine varieties... Otherwise, the answer is, that compact symmetric spaces are K\&quot;ahler if and only if they are complex projective. All compact symmetric spaces are real projective, but this does not imply K\&quot;ahler. http://mathoverflow.net/questions/44501/what-are-the-canonical-and-earliest-references-to-trivial-symmetries-in-gauge-sys/44535#44535 Comment by Tobias Hartnick Tobias Hartnick 2010-11-02T12:08:08Z 2010-11-02T12:08:08Z Well, a functional is also a function, and the notation $S(q, \dot q, ...)$ seems to indicate that it depends only on input from the configuration space. http://mathoverflow.net/questions/44018/why-is-symmetric-group-not-matrix/44020#44020 Comment by Tobias Hartnick Tobias Hartnick 2010-10-28T22:37:10Z 2010-10-28T22:37:10Z So can we agree that the main point is the following: If $S_\infty$ had a finite-dimensional faithful representation over some field $k$, then the dimension of minimal faithful representations of finite groups over $k$ was uniformly bounded, which is &quot;clearly&quot; wrong, and there are many ways to make that clearly really clear. http://mathoverflow.net/questions/41689/why-symmetric-spaces/41695#41695 Comment by Tobias Hartnick Tobias Hartnick 2010-10-11T10:34:25Z 2010-10-11T10:34:25Z Concerning Suresh's question, one should emphasize that to the best of my knowledge there is no such thing as a general duality between non-pos and non-neg curved spaces. This is something very special to symmetric spaces (and some of there generalizations). In this context, instances of duality can be found in every book on symmetric spaces (e.g. Helgason). As far as duality of char. classes is concerned, the above cite article of Kobayashi-Ono is state of the art, as far as I know.