User richard borcherds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:29:42Z http://mathoverflow.net/feeds/user/51 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42929/suggestions-for-good-notation Suggestions for good notation Richard Borcherds 2010-10-20T19:58:32Z 2013-02-02T23:15:30Z <p>I occasionally come across a new piece of notation so good that it makes life easier by giving a better way to look at something. Some examples:</p> <ul> <li><p>Iverson introduced the notation [X] to mean 1 if X is true and 0 otherwise; so for example &Sigma;<sub>1&le;n&lt;x</sub> [n prime] is the number of primes less than x, and the unmemorable and confusing Kronecker delta function &delta;<sub>n</sub> becomes [n=0]. (A similar convention is used in the C programming language.) </p></li> <li><p>The function taking x to x sin(x) can be denoted by x ↦ x sin(x). This has the same meaning as the lambda calculus notation &lambda;x.x sin(x) but seems easier to understand and use, and is less confusing than the usual convention of just writing x sin(x), which is ambiguous: it could also stand for a number.</p></li> <li><p>I find calculations with Homs and &otimes; easier to follow if I write Hom(A,B) as A&rarr;B. Similarly writing A<sup>B</sup> for the set of functions from B to A is really confusing, and I find it much easier to write this set as B&rarr;A.</p></li> <li><p>Conway's notation for orbifolds almost trivializes the classification of wallpaper groups.</p></li> </ul> <p>Has anyone come across any more similar examples of good notation that should be better known? (Excluding standard well known examples such as commutative diagrams, Hindu-Arabic numerals, etc.)</p> http://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples/38893#38893 Answer by Richard Borcherds for Jokes in the sense of Littlewood: examples? Richard Borcherds 2010-09-15T23:47:29Z 2012-12-10T11:35:36Z <p>If $1-ab$ is invertible for $a$, $b$ in a (noncommutative) ring then so is $1-ba$.</p> <p>Proof: $$(1-ba)^{-1} = 1+ba +baba+\cdots = 1+b(1+ab+abab+\cdots)a = 1+b(1-ab)^{-1}a,$$<br> The meaningless infinite series give the right answer (which is hard to guess).</p> http://mathoverflow.net/questions/61315/lie-group-examples Lie group examples Richard Borcherds 2011-04-11T18:15:12Z 2012-05-20T10:24:13Z <p>I'm looking for interesting applications of Lie groups for an introductory Lie groups graduate course. In particular I'd like to hear of non-standard examples that at first sight do not seem to be related to Lie groups (so please don't suggest well-known things like Clifford algebras or triality that appear in standard Lie groups texts such as Fulton and Harris). Here are some examples of the sorts of things I'm looking for:</p> <p>*The cohomology of a compact Kaehler manifold is a representation of SL2, so the Hopf manifold cannot be Kaehler.</p> <p>*q-binomial coefficients are unimodal, as they are characters of representations of SL2</p> <p>*Hilbert's theorem on the finite generation of rings of invariants can be proved using invariant integration on compact Lie groups. </p> <p>*Holomorphic modular forms are really highest weight vectors of discrete series representations of certain Lie groups. </p> <p>*Most closed 3-manifolds are quotients of SL2(C) by discrete subgroups. </p> <p>*Bessel functions cannot be expressed using elementary functions and indefinite integration. (Differential Galois theory was one of Lie's original motivations, but seems to have been eliminated from texts on Lie theory.)</p> <p>*Classifying manifolds up to cobordism uses orthogonal groups. </p> http://mathoverflow.net/questions/34938/when-does-iterating-z-z2-c-have-an-exact-solution When does iterating z -> z^2 + c have an exact solution? Richard Borcherds 2010-08-08T19:14:12Z 2012-04-14T04:29:00Z <p>If one iterates the map z -> z^2 + c there is obviously a simple formula for the sequence one gets if c=0. Less obviously, there is also a simple formula when c = -2 (use the identity 2 cos(2x) = (2cos(x))^2 - 2). Are there any other values of c for which one can solve this recurrence explicitly? (For all initial values of course: there are many trivial explicit solutions for special initial values, such as fixed points.)</p> <p>Related links:<br> <a href="http://en.wikipedia.org/wiki/Mandelbrot_set" rel="nofollow">http://en.wikipedia.org/wiki/Mandelbrot_set</a> (the points c where 0 remains bounded under iteration of this map: this strongly suggests that there is no simple exact solution for general c).<br> <a href="http://en.wikipedia.org/wiki/Logistic_map" rel="nofollow">http://en.wikipedia.org/wiki/Logistic_map</a> (gives the explicit solutions above, after a change of variable)</p> <p>Motivation: I once used the map with c=-2 in a lecture to show that one could prove limits exist even without a formula for the exact solution. A first year calculus student pointed out the non-obvious exact solution above, and I don't want to be caught out like this again. </p> http://mathoverflow.net/questions/88141/varieties-cannot-be-isomorphic-to-proper-open-subsets/88143#88143 Answer by Richard Borcherds for Varieties cannot be isomorphic to proper open subsets Richard Borcherds 2012-02-10T21:04:36Z 2012-02-10T21:04:36Z <p><a href="http://en.wikipedia.org/wiki/Ax-Grothendieck_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Ax-Grothendieck_theorem</a></p> http://mathoverflow.net/questions/82408/non-uniqueness-of-solutions-of-the-heat-equation/82410#82410 Answer by Richard Borcherds for Non-uniqueness of solutions of the heat equation Richard Borcherds 2011-12-01T21:14:16Z 2011-12-01T21:14:16Z <p>Tychonoff in his 1935 paper <a href="http://mi.mathnet.ru/eng/msb/v42/i2/p199" rel="nofollow">Théorèmes d'unicité pour l'équation de la chaleur</a> proved uniqueness if the solutions are not too large, and gave an example to show that the solution is not unique in general. His counterexample grows extremely rapidly for large x. </p> http://mathoverflow.net/questions/76580/famous-mathematicians-with-background-in-arts-humanities-law-etc/76688#76688 Answer by Richard Borcherds for Famous mathematicians with background in arts/humanities/law etc Richard Borcherds 2011-09-28T20:25:44Z 2011-09-28T20:25:44Z <p>Karl Marx. So you didn't know he was a mathematician? A book of his collected mathematical papers is in our math library, which is more than most mathematicians can claim. (They are mostly attempts to understand the definition of a derivative if I recall correctly.) They were quite popular during the cultural revolution, Chinese mathematicians presumably figuring that the study of dialectical calculus was better then a one-way trip to one of Mao's holiday resorts. </p> http://mathoverflow.net/questions/34424/number-of-finite-simple-groups-of-given-order-is-at-most-2-is-a-classification/34432#34432 Answer by Richard Borcherds for Number of finite simple groups of given order is at most 2 - is a classification-free proof possible? Richard Borcherds 2010-08-03T20:06:45Z 2011-09-19T22:20:30Z <p>It is usually extraordinarily difficult to prove uniqueness of a simple group given its order, or even given its order and complete character table. In particular one of the last and hardest steps in the classification of finite simple groups was proving uniqueness of the Ree groups of type $^2G_2$ of order $q^3(q^3+1)(q-1)$, (for $q$ of the form $3^{2n+1}$) which was finally solved in a series of notoriously difficult papers by Thompson and Bombieri. Although they were trying to prove the group was unique, proving that there were at most 2 would have been no easier. </p> <p>Another example is given in the paper by Higman in the book "finite simple groups" where he tries to characterize Janko's first group given not just its order 175560, but its entire character table. Even this takes several pages of complicated arguments. </p> <p>In other words, there is no easy way to bound the number of simple groups of given order, unless a lot of very smart people have overlooked something easy. </p> http://mathoverflow.net/questions/74889/what-is-the-modern-understanding-of-the-order-of-a-mock-theta-function/74911#74911 Answer by Richard Borcherds for What is the modern understanding of the order of a mock theta function? Richard Borcherds 2011-09-08T17:29:54Z 2011-09-08T17:29:54Z <p>This is not a characterization of the order: it is a definition of the order. You cant really ask if it is the same as Ramanujan's definition of the order, because Ramanujan never defined the order in general. All he did was say what the order was in a few examples, so you can't really do any more than check it is the same on these examples. </p> http://mathoverflow.net/questions/71816/w-completion-of-a-c-algebra/71833#71833 Answer by Richard Borcherds for W*-completion of a C*-algebra? Richard Borcherds 2011-08-01T21:35:33Z 2011-08-01T21:35:33Z <p>The universal enveloping W* algebra of a C* algebra is discussed in detail in chapter III.2 of volume 1 of Takesaki's work on "Theory of operator algebra". It is universal in the sense that any map to an W* algebra factors through it (modulo some mumbling about topologies), and as mentioned above is given by the double dual of the C* algebra. This is called the Sherman-Takeda theorem, and was announced by Sherman in 1950 and proved by Takeda in 1954.</p> http://mathoverflow.net/questions/71732/bimonster-and-heterotic-string-theory/71751#71751 Answer by Richard Borcherds for Bimonster and Heterotic String Theory Richard Borcherds 2011-07-31T19:05:44Z 2011-07-31T19:05:44Z <p>The connection between the bimonster and moonshine and the 26 dimensions of string theory is still mysterious (at least to me), though there are several intriguing hints that there is something going on. </p> <p>Some papers discussing this are as follows: </p> <p>The paper by Miyamoto "21 involutions acting on the Moonshine module" J . Algebra 175 (1995), no. 3, 941–965. gives a relation between the 26 involutions generating the bimonster and the natural module of the monster. His construction involves the 26 dimensional even unimodular Lorentzian lattice that also appears in string theory and moonshine. </p> <p>Basak "The complex Lorentzian Leech lattice and the Bimonster" J. Algebra 309 (2007), no. 1, 32–56 gives a complex reflection group generated by 26 complex reflections similar to the bimonster, except that the generators have order 3 rather than 2. </p> http://mathoverflow.net/questions/65957/general-integer-solution-for-x2y2-z2-1/67874#67874 Answer by Richard Borcherds for General integer solution for x^2+y^2-z^2=+/-1 Richard Borcherds 2011-06-15T15:52:28Z 2011-06-15T15:52:28Z <p>The question is the same as looking for points of norm 1 or -1 in the unimodular Lorentzian lattice $Z^{1,2}$. This has an infinite group of automorphisms, with an index 2 subgroup that is a Coxeter group generated by 3 reflections. This group acts transitively on the vectors of norm 1 and -1 if I remember correctly, so all solutions can be obtained from 1 particular solution by acting with this group. </p> http://mathoverflow.net/questions/67595/why-was-it-reasonable-to-ask-what-the-higher-k-groups-are/67609#67609 Answer by Richard Borcherds for Why was it reasonable to ask what the higher K-groups are? Richard Borcherds 2011-06-12T22:00:16Z 2011-06-12T22:00:16Z <p>The idea of considering higher K-groups comes from topology, and is due to Atiyah, Bott, and Hirzebruch. Atiyah and Hirzebruch defined topological K theory and observed that Bott periodicity says that $K(X)$ is more or less the same as $K(S^2X)$. This suggested to them defining a generalized cohomology theory of period 2 by using all the groups $K(S^nX)$ (this was the first example of a generalized cohomology theory). Once one realizes that topological $K^0$ can be extended to topological $K^n$, it does not take much imagination to suggest that algebraic $K^0$ also has an extension to algebraic $K^n$. (Of course, finding this extension was much harder than guessing it existed.)</p> http://mathoverflow.net/questions/66793/reference-on-the-veblen-young-characterization-of-projective-spaces/66815#66815 Answer by Richard Borcherds for Reference on the Veblen-Young characterization of projective spaces Richard Borcherds 2011-06-03T12:12:47Z 2011-06-03T12:12:47Z <p>The 2-volume book by Veblen and Young is an obvious reference for the original proof. A modern reformulation of their result is the work of Tits classifying buildings of rank at least 3 in terms of algebraic groups over fields. </p> http://mathoverflow.net/questions/66551/structure-on-the-set-of-elliptic-curves-via-j-invariant/66560#66560 Answer by Richard Borcherds for Structure on the set of elliptic curves via $j$-invariant Richard Borcherds 2011-05-31T15:23:41Z 2011-05-31T15:23:41Z <p>The normalization of the j-invariant is a historical accident: it was chosen to have a zero of order 3 at a cube root of 1. There is no really good reason for this: one could equally well normalize it to have a double zero at i, or a zero at some integer of an imaginary quadratic field of class number 1. So it is unlikely that the ring structure on isomorphism classes of elliptic curves has any meaning. In any case j is not really a bijection: there is really only 1/2 or 1/3 of an elliptic curve with j invariant 1728 or 0. One can mumble something about stacks at this point.</p> http://mathoverflow.net/questions/66127/number-of-triples-of-roots-of-a-simply-laced-root-system-which-sum-to-zero/66239#66239 Answer by Richard Borcherds for Number of triples of roots (of a simply-laced root system) which sum to zero Richard Borcherds 2011-05-27T22:29:37Z 2011-05-27T22:29:37Z <p>Assuming all roots have norm 2, this is essentially the same as showing that the number of roots having inner product 1 with a fixed root $\beta$ is 2h-4, which in turn follows from the property that $\sum_\alpha(\alpha,\beta)^2/(\alpha,\alpha)(\beta,\beta)=h$. This equality is one of many standard properties of h, given in Bourbaki ch V no 6.2 corollary to theorem 1. </p> http://mathoverflow.net/questions/65904/kummer-and-fermats-equation/65913#65913 Answer by Richard Borcherds for Kummer and Fermat's Equation Richard Borcherds 2011-05-24T23:09:03Z 2011-05-24T23:09:03Z <p>Washington's book has a table of irregular primes together with the power of p dividing the first factor of the class number. There are plenty for which the first factor is divisible by $p^2$; the first is p=157. So for these Kummer's first condition fails. </p> http://mathoverflow.net/questions/65858/series-whose-convergence-is-not-known/65891#65891 Answer by Richard Borcherds for Series whose convergence is not known Richard Borcherds 2011-05-24T19:21:31Z 2011-05-24T19:27:27Z <p>$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ where $\mu$ is the Moebius function. This series is known to converge for $s\ge 1$ and diverge for $s\le 1/2$. Its convergence is unknown if $1/2&lt; s&lt; 1$ (convergence in this interval is essentially the Riemann hypothesis).</p> http://mathoverflow.net/questions/65282/are-any-natural-examples-of-godel-speed-up-known/65291#65291 Answer by Richard Borcherds for Are any natural examples of Gödel speed-up known? Richard Borcherds 2011-05-18T04:16:23Z 2011-05-18T04:16:23Z <p>Friedman has given many examples of such speedups. One well known one is his finite version of <a href="http://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem" rel="nofollow">Kruskal's tree theorem</a>. In particular he gave examples of reasonably natural statements that have very short proofs in 2nd order arithmetic, and can be proved in Peano arithmetic, but the shortest proof in Peano arithmetic is ridiculously long. </p> http://mathoverflow.net/questions/65041/adem-wu-relations-from-bullett-macdonald-identities/65048#65048 Answer by Richard Borcherds for Adem-Wu relations from Bullett-Macdonald identities Richard Borcherds 2011-05-15T16:25:25Z 2011-05-15T19:34:00Z <p>Bullet and Macdonald gave an algebraic proof of this in their paper <a href="http://dx.doi.org/10.1016/0040-9383(82)90015-5" rel="nofollow">On the Adem relations</a></p> http://mathoverflow.net/questions/64881/irreducible-decomposition-of-tensor-product-of-irreducible-s-n-representations/64928#64928 Answer by Richard Borcherds for Irreducible decomposition of tensor product of irreducible $S_n$ representations Richard Borcherds 2011-05-13T20:11:56Z 2011-05-13T20:11:56Z <p>The numbers you want are called Kronecker coefficients. Bürgisser and Ikenmeyer "<a href="http://www.dmtcs.org/dmtcs-ojs/index.php/proceedings/article/viewFile/dmAJ0131/2487" rel="nofollow">The complexity of computing Kronecker coefficients</a>" showed that they are hard to compute in general, so in particular there are no "easy" formulas for them. (There are some explicit formulas for simple special cases.) </p> http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/63824#63824 Answer by Richard Borcherds for Examples of common false beliefs in mathematics. Richard Borcherds 2011-05-03T15:52:33Z 2011-05-03T15:52:33Z <p>"The universal cover of $SL_2(R)$ is a universal central extension" (which I believed until recently...)</p> http://mathoverflow.net/questions/63529/universal-cover-of-sl2r-does-it-admit-central-extensions/63564#63564 Answer by Richard Borcherds for universal cover of SL2(R): does it admit central extensions? Richard Borcherds 2011-04-30T22:10:32Z 2011-05-01T22:19:53Z <p>(This is incorrect: see below)</p> <p>A proof of this can be extracted from Steinberg's paper "Générateurs, relations et revêtements de groupes algébriques". If Ive understood it correctly, he shows that a Cartan subgroup of the universal central extension of $SL_2(K)$ (for a field $K$ with at least 4 elements) is generated by elements $h(t)$ for $t$ a nonzero element of the field, that in particular satisfy $h(tu^2)=h(t)h(u^2)$, and the center of the universal central extension is the kernel of the map from the Cartan subgroup to $K^*/\pm 1$ taking $h(t)$ to $t$. In the case when $K$ is the reals, this kernel is generated by $h(-1)$ as every element is a square or a square times $-1$. So the center of the universal central extension is generated by $h(-1)$ and in particular is cyclic. The center is known be be at least $Z$, so the center of the universal central extension is exactly $Z$, and the universal central extension is therefore the same as the universal cover. </p> <p>Added later: this answer is wrong. I overlooked that the proof of $h(tu^2)=h(t)h(u^2)$ in Steinberg's paper requires that $t$ and $u$ generate a cyclic subgroup of $K^*$, which holds in the finite field case he was considering but not over the reals. </p> http://mathoverflow.net/questions/63434/avatars-of-the-ring-of-symmetric-polynomials Avatars of the ring of symmetric polynomials Richard Borcherds 2011-04-29T16:10:59Z 2011-04-30T13:43:12Z <p>I'm collecting different apparently unrelated ways in which the ring (or rather Hopf algebra with $\langle,\rangle$) of symmetric functions $Z[e_1,e_2,\ldots]$ turns up (for a Lie groups course I will be giving next year). So far I have: </p> <p>*The ring of symmetric functions</p> <p>*Irreducible representations of symmetric groups =Schur functions</p> <p>*Irreducible representations of general linear groups = Schur functions</p> <p>*The homology of $BU$, the classifying space of the infinite unitary group. (It also turns up in several other related generalized homology rings of spectra.)</p> <p>*The universal commutative $\lambda$-ring on one generator $e_1$</p> <p>*The coordinate ring of the group scheme of power series $1+e_1x+e_2x^2+\cdots$ under multiplication</p> <p>What other examples have I missed? </p> http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63454#63454 Answer by Richard Borcherds for Checkmate in $\omega$ moves? Richard Borcherds 2011-04-29T18:09:40Z 2011-04-29T18:09:40Z <p>See <a href="http://arxiv.org/abs/math/9905198" rel="nofollow"> On numbers and endgames: Combinatorial game theory in chess endgames</a> by Elkies for some chess positions with non-integer values. </p> http://mathoverflow.net/questions/63033/what-would-you-want-on-a-lie-theory-cheat-poster/63193#63193 Answer by Richard Borcherds for What would you want on a Lie theory cheat poster? Richard Borcherds 2011-04-27T16:12:59Z 2011-04-27T16:12:59Z <p>Affine Dynkin diagrams (with the linear combination of simple roots that has norm 0)</p> http://mathoverflow.net/questions/63033/what-would-you-want-on-a-lie-theory-cheat-poster/63044#63044 Answer by Richard Borcherds for What would you want on a Lie theory cheat poster? Richard Borcherds 2011-04-26T14:36:40Z 2011-04-26T14:36:40Z <p>Maximal subgroups</p> http://mathoverflow.net/questions/63033/what-would-you-want-on-a-lie-theory-cheat-poster/63042#63042 Answer by Richard Borcherds for What would you want on a Lie theory cheat poster? Richard Borcherds 2011-04-26T14:33:05Z 2011-04-26T14:33:05Z <p>Finite subgroups</p> http://mathoverflow.net/questions/63033/what-would-you-want-on-a-lie-theory-cheat-poster/63041#63041 Answer by Richard Borcherds for What would you want on a Lie theory cheat poster? Richard Borcherds 2011-04-26T14:30:11Z 2011-04-26T14:30:11Z <p>Homology and homotopy groups</p> http://mathoverflow.net/questions/63033/what-would-you-want-on-a-lie-theory-cheat-poster/63040#63040 Answer by Richard Borcherds for What would you want on a Lie theory cheat poster? Richard Borcherds 2011-04-26T14:29:11Z 2011-04-26T14:29:11Z <p>Maximal compact subgroups</p> http://mathoverflow.net/questions/90317/can-you-name-these-orthogonal-polynomials Comment by Richard Borcherds Richard Borcherds 2012-03-06T21:47:30Z 2012-03-06T21:47:30Z There is a good chance that they are special cases of Macdonald polynomials of type $A_n$. Check Macdonald's book on symmetric functions and Hall polynomials. http://mathoverflow.net/questions/79093/an-identity-involving-an-infinite-integral-with-a-sinh-in-the-denominator/79107#79107 Comment by Richard Borcherds Richard Borcherds 2011-10-25T21:51:33Z 2011-10-25T21:51:33Z This integral is due to Barnes: see Barnes, E. W. (1908), &quot;A New Development of the Theory of the Hypergeometric Functions&quot;, Proc. London Math. Soc. s2-6: 141–177, doi:10.1112/plms/s2-6.1.141 http://mathoverflow.net/questions/63529/universal-cover-of-sl2r-does-it-admit-central-extensions/63599#63599 Comment by Richard Borcherds Richard Borcherds 2011-05-02T00:04:16Z 2011-05-02T00:04:16Z Just to confirm that my answer was indeed wrong and this one is correct. http://mathoverflow.net/questions/63529/universal-cover-of-sl2r-does-it-admit-central-extensions/63564#63564 Comment by Richard Borcherds Richard Borcherds 2011-05-01T16:35:34Z 2011-05-01T16:35:34Z You can find the rule $h(tu^2)=h(t)h(u^2)$ on page 121 of Steinberg's paper, as the special case of 7.3 (e) when r=s and d=2. http://mathoverflow.net/questions/63529/universal-cover-of-sl2r-does-it-admit-central-extensions/63573#63573 Comment by Richard Borcherds Richard Borcherds 2011-05-01T04:46:35Z 2011-05-01T04:46:35Z There is a more complete version available at Steinberg's home page <a href="http://www.math.ucla.edu/~rst/" rel="nofollow">math.ucla.edu/~rst</a> http://mathoverflow.net/questions/63434/avatars-of-the-ring-of-symmetric-polynomials/63486#63486 Comment by Richard Borcherds Richard Borcherds 2011-04-30T12:33:19Z 2011-04-30T12:33:19Z Grassmannians are more or less the same as BU, though this is not trivial so I guess it counts as another independent example. I'm slightly surprised Hall knew this as I've seen no sign in his papers that he knew anything about topology. http://mathoverflow.net/questions/63434/avatars-of-the-ring-of-symmetric-polynomials Comment by Richard Borcherds Richard Borcherds 2011-04-29T17:55:37Z 2011-04-29T17:55:37Z Schur proved something like this, except using finite dimensional vector spaces rather than finite sets. http://mathoverflow.net/questions/63434/avatars-of-the-ring-of-symmetric-polynomials/63449#63449 Comment by Richard Borcherds Richard Borcherds 2011-04-29T17:53:45Z 2011-04-29T17:53:45Z There is no need to work over the complex numbers: the integral ring of symmetric functions gives an integral form of the vertex algebra. Though the harmonic oscillator modes do not generate this integral form. http://mathoverflow.net/questions/63434/avatars-of-the-ring-of-symmetric-polynomials/63442#63442 Comment by Richard Borcherds Richard Borcherds 2011-04-29T17:49:46Z 2011-04-29T17:49:46Z Macdonald's book is a good reference for these. If I remember correctly you need to work with the symmetric functions over polynomials in q or something like that, and if you set q=1 you recover the usual symmetric functions. http://mathoverflow.net/questions/63033/what-would-you-want-on-a-lie-theory-cheat-poster/63041#63041 Comment by Richard Borcherds Richard Borcherds 2011-04-26T22:59:04Z 2011-04-26T22:59:04Z If you don't know everything, it does not mean you know nothing. The homotopy groups of Lie groups are known up to dimension higher than most people are likely to need. And in the stable range they are all known by Bott periodicity. http://mathoverflow.net/questions/5372/dimension-leaps/62167#62167 Comment by Richard Borcherds Richard Borcherds 2011-04-19T16:00:22Z 2011-04-19T16:00:22Z It's doable in 26 dimensions, just MUCH harder than in 25 dimensions. The main obstruction is that there seems no point in doing it; all one would get for a lot of effort would be a boring list of a couple of thousand lattices. (Their root systems have already been found by King by finding the mass of each root system.) http://mathoverflow.net/questions/53515/decidability-of-tiling-r2/53547#53547 Comment by Richard Borcherds Richard Borcherds 2011-01-28T01:05:12Z 2011-01-28T01:05:12Z <a href="http://arxiv.org/abs/1003.4279" rel="nofollow">arxiv.org/abs/1003.4279</a> http://mathoverflow.net/questions/50610/what-are-classical-groups/50629#50629 Comment by Richard Borcherds Richard Borcherds 2010-12-29T18:26:21Z 2010-12-29T18:26:21Z I don't understand this comment, as my answer did not mention Lie algebras. http://mathoverflow.net/questions/42929/suggestions-for-good-notation/42936#42936 Comment by Richard Borcherds Richard Borcherds 2010-10-20T20:25:39Z 2010-10-20T20:25:39Z According to Knuth, this notation and these names were introduced by Iverson in his book &quot;A programming language&quot; in 1962. http://mathoverflow.net/questions/42439/how-can-one-express-the-dedekind-eta-function-as-a-sum-over-the-lattice/42444#42444 Comment by Richard Borcherds Richard Borcherds 2010-10-18T05:02:32Z 2010-10-18T05:02:32Z There was a typo in my post: logarithm should have been logarithmic derivative.