User pierre - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:48:31Z http://mathoverflow.net/feeds/user/12045 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127950/normal-subgroups-of-finite-index-in-free-groups Normal subgroups of finite index in free groups Pierre 2013-04-18T12:15:47Z 2013-05-07T14:22:45Z <p>Hi all,</p> <p>This is a question about the groups $H_{n,s}$ introduced by Völklein in his book "Groups as Galois groups", §7.1, and defined as follows: let $N$ be the intersection of all normal subgroups of index $\le n$ in $F_s$, the free group on $s$ generators, and define $H_{n,s} = F_s / N$.</p> <blockquote> <p>Have these groups been studied? Do they have a name? Is it possible to compute their orders, at least in some cases?</p> </blockquote> <p>For example for $s=1$, then $H_{n,1}$ is the cyclic group of order $lcm(1, 2, \ldots, n)$.</p> <p>In Völklein's book, these are introduced primarily to avoid talking about profinite groups (the inverse limits of the $H_{n,s}$, with fixed $s$, is the free profinite group of $s$ generators). </p> <p>Any information you may have on these will be great appreciated.</p> <p>Thanks!</p> <p>Pierre</p> http://mathoverflow.net/questions/125515/a-group-with-all-sylow-p-subgroups-cyclic/125516#125516 Answer by Pierre for a group with all sylow p subgroups cyclic Pierre 2013-03-25T09:25:46Z 2013-03-25T09:25:46Z <p>Take $p$ and $q$ two prime numbers with $q$ dividing $p-1$. Then there is a nonabelian semi-direct product $C_p \rtimes C_q$ which seems to be what you want, if i understand the question well. Here $C_n$ is the cyclic group of order $n$, and note that $p-1$ is the order of the automorphism group of $C_p$, when $p$ is an odd prime.</p> http://mathoverflow.net/questions/116365/terminology-question-for-real-k-theory Terminology question for real K-theory Pierre 2012-12-14T12:02:16Z 2012-12-14T15:18:57Z <p>This is a terminology question. Answers will help me satisfy a referee but I'm also genuinely interested. Consider the following two things that you could define for a topological space X:</p> <p>(1) The Grothendieck ring of the category of real vector bundles over X;</p> <p>(2) the set $[X, \mathbb{Z} \times BO]$ of homotopy classes of maps from $X$ to $\mathbb{Z} \times BO$ where $BO$ is the classifying space for the infinite orthogonal group.</p> <p>It is well known that (1) and (2) are in bijection when $X$ is, say, a finite CW-complex. This can fail when $X$ is not compact though, and I think it also fails when $X$ is a nested sequence of circles in the plane with the subspace topology. Anyway, the two are different in general.</p> <p>Now consider the following names and notation :</p> <p>(A) the real K-theory of $X$;</p> <p>(B) the real topological K-theory of $X$;</p> <p>(C) the K-theory of real vector bundles over $X$;</p> <p>(D) $K(X)$;</p> <p>(E) $KO(X)$;</p> <p>(F) $K_\mathbb{R}(X)$.</p> <blockquote> <p>Which would you pair to which? and what is your favorite for (1) and for (2)?</p> </blockquote> <p>For example I don't expect anyone to answer that C2 is reasonable. I'm curious whether E1 or E2 will appear. Some other notation may also be prefered, I think I saw $\mathbf{KO}(X)$ once, meaning (1), while $KO(X)$ was for (2). Or perhaps the other way around. There are also alternative names like "the Atiyah real K-theory", not sure meaning what.</p> <p>Thanks! There may not be any universal convention, but at least we can have a vote of sorts.</p> http://mathoverflow.net/questions/109777/reference-for-representations-of-quaternion-group/109794#109794 Answer by Pierre for Reference for representations of quaternion group Pierre 2012-10-16T07:46:44Z 2012-10-16T07:46:44Z <p>Since $Q_8$ is an extraspecial group, if you want you can say that the result follows from Quillen's classification of their real representations in</p> <p>Quillen, The mod 2 cohomology rings of extra-special 2-groups and the spinor groups</p> <p>Mathematische Annalen, 1971</p> http://mathoverflow.net/questions/100910/homology-versus-cohomology-of-lie-groups/100918#100918 Answer by Pierre for Homology versus cohomology of Lie groups Pierre 2012-06-29T09:43:07Z 2012-06-29T09:43:07Z <p>With field coefficients, homology and cohomology are dual to each other. The cohomology of a space is an algebra, the homology of a space is a coalgebra. When the space is a group (or loop space or $H$-space...) both its homology and cohomology are Hopf algebras.</p> <p>"The literature" is full of computations of the cohomology of <em>the classifying space</em> $BG$. In some cases by the way, Hopf algebras help, for example if you compute with coefficients in $\mathbb{Q}$. Then graded Hopf algebras over the rational field are classified, which gives you a strong indication on what the cohomology of $G$ looks like, and a spectral sequence argument tells you that $H^*(BG, \mathbb{Q})$ is a polynomial ring, in the end. Details in McCleary's book <em>A user's guide to spectral sequences</em>.</p> http://mathoverflow.net/questions/93960/cryptography-and-iterations Cryptography and iterations Pierre 2012-04-13T14:30:22Z 2012-04-13T21:17:24Z <p>Hi,</p> <p>Here is a question in cryptography which is probably naive, and a reference request.</p> <p>I was wondering about the following key-exchange scheme, which is a variant on Diffie-Hellman. Consider a set $X$ (finite but very large) and a map $T : X \to X$, both made public together with a point $x \in X$. Now A chooses an integer $n$ secretly and publishes $T^n(x)$, while B does the same with $m$. A and B can both compute the key $T^{n+m}(x)$, and assuming that it is difficult to find $n$ from $x$ and $T^n(x)$, then noone else can.</p> <p>Traditionally one picks an element $g$ in a group $G$, then A publishes $g^n$, B publishes $g^m$, and A and B both know the key $g^{nm}$. For $G$ one picks $(\mathbf{Z}/p)^\times$, or an elliptic curve over a finite field, or a braid group, or what have you.</p> <p>It seems that with the above variant, it is easy to produce examples: for example take $X$ to be a vector space over $\mathbf{F}_2$, and let $T$ be some map which shuffles the bits around according to your fancy. My intuition is that it is easier to make the "log-problem" difficult in this way than by choosing the right group $G$. I may be so completely wrong!</p> <blockquote> <p>Is there an obvious weakness in this scheme? For example, is it very hard to prove that, for a given map $T$, the "log-problem" is indeed difficult?</p> </blockquote> <p>It may well be that I'm only describing something standard.</p> <blockquote> <p>What is a good reference, then?</p> </blockquote> <p>(Basic searches with "cryptography and dynamics" were not satisfactory.)</p> <p>Thanks for reading!</p> <p>Pierre</p> http://mathoverflow.net/questions/89396/indecomposable-modules-of-the-symmetric-group indecomposable modules of the symmetric group Pierre 2012-02-24T13:33:03Z 2012-02-24T14:31:28Z <p>Quite simply:</p> <blockquote> <p>What is known about the indecomposable modules of the symmetric group $S_n$ in positive characteristic?</p> </blockquote> <p>Since the answer is likely to be "very little", perhaps I should rather ask</p> <blockquote> <p>What is known about the indecomposable modules of $S_4$ in characteristic $2$ ?</p> </blockquote> <p>References appreciated.</p> <p>Thanks!</p> <p>Pierre</p> http://mathoverflow.net/questions/75967/filtered-ring-giving-rise-to-a-graded-commutative-ring Filtered ring giving rise to a graded-commutative ring Pierre 2011-09-20T16:02:58Z 2011-12-28T02:34:30Z <p>Hello,</p> <p>Given a ring $R$ with a filtration by two-sided ideals $F^0 \supset F^1 \supset F^2 \supset\cdots$, one can form the associated graded ring $gr R = F^0 / F^1 \oplus F^1/F^2 \oplus \cdots$. </p> <p>When $R$ is commutative, so is $gr R$. Now, the natural condition in the graded world is to be graded-commutative (that is $yx = (-1)^{pq} xy$ when $x$ resp $y$ has degree $p$ resp $q$). </p> <blockquote> <p>Are there natural/well-known/simple conditions on $R$ ensuring that $gr R$ is graded-commutative?</p> </blockquote> <p>Thank you very much!</p> <p>Pierre</p> http://mathoverflow.net/questions/80272/a-question-about-connectedness-in-euclidean-space/80277#80277 Answer by Pierre for A question about connectedness in Euclidean space Pierre 2011-11-07T08:17:12Z 2011-11-07T11:49:20Z <p>Yes. Let $C$ be the closed complement of $U$, then by excision of $C$ we have $H_1(\mathbb{R}, \mathbb{R} - K) = H_1(U, U - K)$; since $H_1(\mathbb{R})=0$, you also have in fact $H_1(\mathbb{R}, \mathbb{R}-K)= H_1(U, U-K)= 0$ when $\mathbb{R}-K$ is connected. </p> <p>So $H_0(U-K)$ injects into $H_0(U)$ and $U-K$ must be connected.</p> http://mathoverflow.net/questions/76913/is-the-nc-torus-a-quantum-group/76918#76918 Answer by Pierre for Is the nc torus a quantum group? Pierre 2011-10-01T12:20:27Z 2011-10-05T07:15:03Z <p>I don't know about the $C^*$-algebra version, but I can tell you about the algebraic version (the algebra generated by $u$ and $v$, invertible, such that $uv = qvu$). </p> <p>It is not a Hopf algebra but a "braided group", that is, a Hopf algebra in some braided category (classical Hopf algebras being, in this parlance, "Hopf algebras in the category of vector spaces with the trivial twist"). Concretely, there is a map of algebras $A \to A \otimes A$ satisfying all the axioms you want, except that $A \otimes A$ is not made into an algebra in the way you think.</p> <p>If I were allowed a bit of self-advertising, I'd recommend §4 of</p> <p><a href="http://arxiv.org/abs/0911.5287" rel="nofollow">http://arxiv.org/abs/0911.5287</a></p> <p>Majid's book on quantum groups may have some formulae about the codiagonal in the quantum tori.</p> http://mathoverflow.net/questions/66019/what-is-the-cohomology-of-this-complex What is the cohomology of this complex? Pierre 2011-05-26T04:16:34Z 2011-05-26T12:07:41Z <p>I have a feeling that this may be a very easy question for some people on MO, but it isn't for me.</p> <p>Take a finite pointed set $X$, with $*$ the base-point. Build a cosimplicial set which in degree $n \ge 1$ is $X^n$ (the cartesian product of $n$ copies of $X$), and in degree $0$ is <code>$\{ * \}$</code>; the cofaces are:</p> <p>$d^0(x_1, \ldots, x_n) = (*, x_1, \ldots, x_n)$</p> <p>$d^i(x_1, \ldots, x_n) = (x_1, \ldots, x_i, x_i, x_{i+1}, \ldots, x_n)$</p> <p>$d^{n+1}(x_1, \ldots, x_n) = (x_1, \ldots, x_n, *)$</p> <p>Now apply the functor "free $k$-module on", where $k$ is your favorite ring. You get a cosimplicial $k$-module $A^*$, so you can build the associated cochain complex where the differential is the alternating sum of the cofaces. Note that $A^n = (A^1)^{\otimes n}$. </p> <blockquote> <p>What is the cohomology of this complex?</p> </blockquote> <p>Ideally someone will say something like "this is the cobar construction, it computes the cohomology of the loop space on the discrete space $X$, so the cohomology is $0$ in positive degrees", or something close. And it would be awesome. (The buzzword "cobar construction" seems to show up a lot among the papers I've skimmed through online.)</p> <p>Thank you so much for your help!</p> <p>Pierre</p> http://mathoverflow.net/questions/64326/metaplectic-group-does-not-split/64361#64361 Answer by Pierre for metaplectic group does not split Pierre 2011-05-09T07:40:05Z 2011-05-09T07:40:05Z <p>Have a look at</p> <p>Lion &amp; Vergne, The Weil representation, Maslov index, and Theta series.</p> <p>I haven't got it with me but I think I remember that it contains what you want.</p> <p>Also there is</p> <p>Teruji Thomas, The trace of the Weil representation (on arXiv).</p> <p>This paper is quite explicit but not "as explicit" as Rao (I know what you mean). It's got constructions of the metaplectic groups, and again I think I remember that what you're asking is an easy consequence of these.</p> <p>Some random comments now. The Maslov index gives you an extension of $Sp_{2n}(k)$ by the Witt group $W(k)$. For $k=\mathbb{R}$ the connected component of 1 in this extension is the simply-connected extension of $Sp_{2n}(\mathbb{R})$ (this fact is in Lion-Vergne), so it's certainly non-trivial. The metaplectic group is a quotient of this. Note that for $n=1$ you get an extension of $SL_2(\mathbb{R})$, and the inverse image of $SL_2(\mathbb{Z})$ is the braid group $B_3$, proving again that the extension is non-trivial (see Kassel &amp; Turaev, Braid groups, appendix, and Milnor's book on algebraic K-theory).</p> http://mathoverflow.net/questions/56351/should-there-be-a-specified-standard-knowledge-of-mathematicians Should there be a specified standard knowledge of mathematicians? Pierre 2011-02-23T01:43:25Z 2011-03-24T05:21:40Z <p>(Feel free to close this if it is too vague/chatty/soft/etc, I won't be offended!)</p> <p>Some very quick background. I am a visitor this year at some university, and they have very kindly organized a number of events for all "temporary" people (so postdocs and visitors such as myself mainly). Some of these events are social, but I want to talk about the "Colloquium" which we run (mostly between us though some extra people show up). I want to talk about the difficulty we have to understand each other.</p> <p>Apart from being the "temporary" people, on a mathematical level we have little in common. I am deeply frustrated by how little we are able to share in the colloquium.</p> <p>The first talk was about number theory, modular forms in particular. The first sentence was "as you know the Galois group is profinite, that is, compact and totally disconnected". The PDE people in the audience rolled their eyes, as you can imagine, and pretty much stopped listening after this one sentence. </p> <p>We talked about this, and the next speaker decided he'd keep things very basic. He gave a talk on PDEs, and I was only able to follow about 10 minutes. Which is terrible. </p> <p>When it was my turn, I tried to do something about the quadratic reciprocity law (in fact based on a MO question!). It's not for me to say how it went, but I remember being in shock at some point: I was saying "this set $G$ is in fact a group, it has $n$ elements, so if $g \in G$ we have $g^n = 1$". Somebody (a respectable specialist in probability theory) asked "why is that?" Having learned from this experience, I know that it's worth saying that it's called Lagrange theorem, and believe me, whoever you are, you've learned it within the first two years of university. It blew my mind to find out that some mathematicans have forgotten this one -- but of course the things I have myself forgotten must be equally fundamental to others.</p> <p>Hence my question:</p> <blockquote> <p>Would it be useful to write down a document specifying what ALL working mathematicians can be expected to know? People giving a colloquium talk (as opposed to a seminar talk) could be required to adjust their presentation so that it is understandable by anyone knowing what's in that document.</p> </blockquote> <p>I'm thinking of something similar to what the Word Wide Web Consortium (W3C) has achieved: a standard, a protocol.</p> <p>I want to stress a difficult point: it's not only about known results, but also standard habits with notations. Let me give you an example of the things which confused me greatly during the aforementioned PDE talk. There was a map $(x, t) \mapsto f(x, t)$ and at some point the speaker wrote $\hat{f}(t)$. I was unable to decide if he meant (i) fix $t$, take $x\mapsto f(x,t)$, take the Fourier transform of that, call it $\hat{f}(t)$, it is a function; or (ii) fix $x$, consider $t\mapsto f(x, t)$, take the Fourier transform and evaluate it at $t$, call it $\hat{f}(t)$, it is a number; or (iii) something else. </p> <p>I was uncannily reassured to find out that next to me, some other specialist in PDE said "that's funny, I would have written $\hat{f}(x)$ for the same thing". But then I was more confused than ever when they agreed that the notation did not matter since the meaning was obvious anyway (!). I'm not more anal than the next person, and I'm certainly no Bourbaki fanatic, but I found the need for clarification. (I tried to ask a question but they thought, again, that I was argueing against the notation, not that I was utterly confused as to its meaning.)</p> <p>I would be interested in reading your thoughts about this. Of course a subsidiary question is, who would write the document, and what authority would it have?</p> <p>(anecdotal stories of complete confusion during a talk can also provide comic relief, by the way)</p> <p>Thanks for reading, Pierre</p> <p>EDIT: based on one answer below, I want to add the following: I'm not looking for advice on how to improve my skills in exposition, I think I'm doing fine, thank you... The suggestion I'm making, should it be efficient at all, would be rather to improve the <em>average</em> quality of exposition in talks. And specifically, when a speaker adresses an audience of non-specialists. Of course whenever a speaker truly cares (and I think I do, for example) s/he will be doing fine. The point is to make them care.</p> <p>EDIT: turned into community wiki.</p> http://mathoverflow.net/questions/52099/cohomology-theory-for-algebraic-groups/52102#52102 Answer by Pierre for cohomology theory for algebraic groups Pierre 2011-01-14T18:38:16Z 2011-01-14T18:38:16Z <p>Yes, it's called "rational cohomology" -- not to be confused with cohomology with rational coefficients... see eg "Rational and Generic cohomology" by Cline, Parshall, Scott and van der Kallen, Inventiones.</p> <p>By using google I have even found a link, make sure it is legal for you to download this file:</p> <p><a href="http://www.digizeitschriften.de/main/dms/gcs-wrapper/?gcsurl=http%253A%252F%252Flocalhost%253A8086%252Fgcs%252Fgcs%253F%2526%2526%2526%2526%2526%2526%2526%2526%2526action%253Dpdf%2526metsFile%253DPPN356556735_0039%2526divID%253Dlog12%2526pdftitlepage%253Dhttp%25253A%25252F%25252Fwww.digizeitschriften.de%25252Fmain%25252Fdms%25252Fpdf-titlepage%25252F%25253FmetsFile%25253DPPN356556735_0039%252526divID%25253Dlog12%2526targetFileName%253D_log12.pdf" rel="nofollow">http://www.digizeitschriften.de/main/dms/gcs-wrapper/?gcsurl=http%253A%252F%252Flocalhost%253A8086%252Fgcs%252Fgcs%253F%2526%2526%2526%2526%2526%2526%2526%2526%2526action%253Dpdf%2526metsFile%253DPPN356556735_0039%2526divID%253Dlog12%2526pdftitlepage%253Dhttp%25253A%25252F%25252Fwww.digizeitschriften.de%25252Fmain%25252Fdms%25252Fpdf-titlepage%25252F%25253FmetsFile%25253DPPN356556735_0039%252526divID%25253Dlog12%2526targetFileName%253D_log12.pdf</a></p> http://mathoverflow.net/questions/51259/are-there-homology-groups-for-cosimplicial-groups Are there homology groups for cosimplicial groups? Pierre 2011-01-05T22:46:31Z 2011-01-06T03:05:05Z <p>Hi,</p> <p>Assume you have a cosimplicial group $G$, so that for each $n \ge 0$ there is a group $G_n$, and you have the usual cofaces and codegeneracies. </p> <blockquote> <p>Is there a known way to associate to this a collection of homology/homotopy groups in a sensible way? </p> </blockquote> <p>"Sensible" means at least that it should provide a generalisation of the following two particular cases:</p> <p>(1) When each $G_n$ is abelian, one can get a cochain complex (I believe this is called the Dold-Kan construction), and one can consider its cohomology groups.</p> <p>(2) (I apologize for the vagueness of this one.) In low degrees, one can sometimes use a couple of tricks. I have, for example, come across the following situation: the map $x\mapsto d^0(x) d^2(x)$ was a homomorphism $G_2 \to G_3$, as was the map $x\mapsto d^3(x)d^1(x)$; so I could consider their equalizer. Moreover the map $x\mapsto d^0(x)d^1(x)^{-1}d^2(x)$ was a homomorphism $G_1 \to G_2$, whose image was normal in the preceding equalizer. Taking the quotient gave a generalisation of $H^2$ in this lucky situation. The "general" definition which I'm asking for, should it exist, would hopefully coincide with this equalizer trick whenever it makes sense.</p> <p>Let me also point out that, in the example above, I had originally started with some bigger cosimplicial group $(\Gamma_n)$ and then decided to restrict to the smaller $(G_n)$ precisely so that I could use this little trick. I do believe that the details of this example are completely irrelevant to the general discussion; I mention it because someone might know how to go from a cosimplicial group to a "nicer" one somehow.</p> <p>Thank you for reading this.</p> <p>Pierre</p> http://mathoverflow.net/questions/127950/normal-subgroups-of-finite-index-in-free-groups/127973#127973 Comment by Pierre Pierre 2013-04-19T09:50:11Z 2013-04-19T09:50:11Z Nice! I'm eager to read that paper. http://mathoverflow.net/questions/116516/orientation-of-manifold-in-topological-sense Comment by Pierre Pierre 2012-12-16T14:06:07Z 2012-12-16T14:06:07Z for a manifold which is not assumed to be smooth, orientability is a more delicate question, which requires an understanding of homology AFAIK. See Bredon, &quot;Geometry and Topology&quot;. http://mathoverflow.net/questions/116365/terminology-question-for-real-k-theory/116381#116381 Comment by Pierre Pierre 2012-12-14T17:34:40Z 2012-12-14T17:34:40Z Thanks! I'm starting to guess what happened in my referee's mind. He or she suggested the use of (E) to mean (1), which confused me; I see now that it would have confused other people. My guess at the moment is that s/he did not really bother to distinguish between (1) and (2) at all but rather, worries that a confusion with Atiyah's theory was possible. It's making more sense. I'll leave this open for a while to see if other people have something to add. http://mathoverflow.net/questions/116365/terminology-question-for-real-k-theory Comment by Pierre Pierre 2012-12-14T13:02:37Z 2012-12-14T13:02:37Z @Tom: I think there's a nice paper by Bob Oliver about (1) for classifying spaces of compact Lie groups. I seem to recall that the answer is interesting and quite different from the Atiyah completion theorem which describes (2) in this case. http://mathoverflow.net/questions/114508/is-there-any-good-heuristic-for-what-we-should-expect-to-happen-when-we-take-alge Comment by Pierre Pierre 2012-11-26T10:46:19Z 2012-11-26T10:46:19Z read: $1$ or $2$ or $+\infty$ (not 'of'). http://mathoverflow.net/questions/114508/is-there-any-good-heuristic-for-what-we-should-expect-to-happen-when-we-take-alge Comment by Pierre Pierre 2012-11-26T10:45:20Z 2012-11-26T10:45:20Z Perhaps he's asking about the dimension of extensions of the form $L/K$ with $L$ algebraically closed. This dimension may well be $1$ of course. It cannot be $3$, for the group of order 3 is not an absolute Galois group (its mod 3 cohomology is not generated in degree 1). In fact a similar argument shows that this number can only be 1, or 2, of $+\infty$. http://mathoverflow.net/questions/111480/lambda-operations-on-stable-homotopy-groups-of-spheres/111486#111486 Comment by Pierre Pierre 2012-11-05T09:32:30Z 2012-11-05T09:32:30Z Let me add that, for some strange reason, when I uploaded this to the arxiv there was a bug with the latex compilation, and all the lambda's appear as L's. (On one other occasion I had the O in O_n, the orthogonal group, come out as the empty set. I think some commands such as \O and \l should not be redefined...). If you send me an email I can get you a copy with proper lambdas ! http://mathoverflow.net/questions/109777/reference-for-representations-of-quaternion-group/109794#109794 Comment by Pierre Pierre 2012-10-17T11:58:02Z 2012-10-17T11:58:02Z @darij grinberg: this is Proposition 5.8 in Quillen's paper. @Mariano: it is my understanding that the only point is to satisfy a referee :) And yes, even understanding the notation takes longer than proving the result for oneself, but hey, somebody asked for a reference. http://mathoverflow.net/questions/109777/reference-for-representations-of-quaternion-group Comment by Pierre Pierre 2012-10-16T16:30:14Z 2012-10-16T16:30:14Z Gill! I didn't know you were on MO. Well everybody is, I suppose. Hope you're well. http://mathoverflow.net/questions/108046/is-it-common-practice-to-publish-parts-of-a-phd-thesis-in-advance Comment by Pierre Pierre 2012-09-25T12:24:17Z 2012-09-25T12:24:17Z It is certainly quite common to publish papers during your PhD. Then your PhD thesis consists of your papers with some re-writing, an introduction to the state-of-the-art, some concluding words... and it can be considered a fairly different document from your articles. Lots of people have done this (myself included) and I'm not aware of any subsequent issues. This doesn't mean that it's 100% legal though. http://mathoverflow.net/questions/105920/advantages-of-the-sequence-definition-of-limits Comment by Pierre Pierre 2012-08-30T12:02:27Z 2012-08-30T12:02:27Z little motivation to give the alternative definition. Nevertheless, you could argue that it gives a &quot;dynamical system&quot; feel to the concept of limit, and that can be pretty visual and intuitive with some students. http://mathoverflow.net/questions/105920/advantages-of-the-sequence-definition-of-limits Comment by Pierre Pierre 2012-08-30T12:00:11Z 2012-08-30T12:00:11Z A common approach, as pointed out in an answer below, is to teach limits of sequences first and then to define limits of functions using sequences, in an attempt to use as few $\epsilon - \delta$ as possible (basically, just enough to prove that $1/n$ converges to $0$ and to prove what you call the &quot;algebra of limits&quot; -- after that, you can get away without ever seeing an $\epsilon$ again in the first year, if you so wish). In your case though, if the students have seen epislons and deltas and do not hate them (as many students do, I have found), there is little motivation to (to be continued) http://mathoverflow.net/questions/100910/homology-versus-cohomology-of-lie-groups/101069#101069 Comment by Pierre Pierre 2012-07-01T15:48:29Z 2012-07-01T15:48:29Z while using the Hopf algebra structure a little. It was worth providing complements to what I said. I object to the way it was phrased. PS I should make the following point clearer: the simple (and very nice IMHO) approach in McCleary's book does <i>not</i> work for other fields, unless you throw more hypotheses in. http://mathoverflow.net/questions/100910/homology-versus-cohomology-of-lie-groups/101069#101069 Comment by Pierre Pierre 2012-07-01T15:46:03Z 2012-07-01T15:46:03Z If we want to be precise (and I think we do), this spectral sequence is indeed irrelevant if one wants to prove the result I mentioned, to the effect that $H^*(BG,\mathbb{Q})$ is a polynomial ring (see theorem 6.38 in McCleary's book as well as th 3.27). I never said one had to restrict to $\mathbb{Q}$. As should be apparent now, I was trying to state something as elementary as possible, while... (to be continued) http://mathoverflow.net/questions/100910/homology-versus-cohomology-of-lie-groups/100996#100996 Comment by Pierre Pierre 2012-07-01T08:56:55Z 2012-07-01T08:56:55Z Dear Peter, pointing out that a sentence such as &quot;the graded Hopf algebras over the rational field are classified&quot; lacks precision is indeed harsh, if fair. The context being cohomology rings, graded commutativity seems implicit to me, and the fact that the Hopf-Borel theorem does not classify the possible diagonal maps seems irrelevant for the application I'm describing. I would have been more careful if (i) I had not given a reference for the full details and (ii) the original question had been more advanced.