User jake - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:51:50Z http://mathoverflow.net/feeds/user/14352 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43148/basic-results-with-three-or-more-hypotheses/130715#130715 Answer by Jake for Basic results with three or more hypotheses Jake 2013-05-15T13:11:05Z 2013-05-15T13:11:05Z <p>A basic example from undergraduate topology that comes to my mind is the theorem on the existence of universal covers.</p> <p><strong>Theorem.</strong> Let $X$ be a topological space. Then, there exists a universal covering space $\pi\colon \tilde{X}\rightarrow X$ if $X$ is connected, locally path connected and semi-locally simply connected.</p> http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product/109891#109891 Answer by Jake for References for Eilenberg-Zilber shuffle product Jake 2012-10-17T10:21:51Z 2012-10-17T10:21:51Z <p>A complete reference for this with full proofs can be found in Bredon's <em>Topology and Geometry</em> (Springer GTM). More specifically, this is written up in chapter IV, part 16 (p220-223) and chapter VI, part 1 (p315-318). </p> http://mathoverflow.net/questions/97139/what-is-the-difference-between-up-to-conjugacy-and-up-to-conjugation/97149#97149 Answer by Jake for What is the difference between "up to conjugacy" and "up to conjugation" ? Jake 2012-05-16T18:43:15Z 2012-05-16T18:43:15Z <p>To expand upon Gerhard's comment, I would add the following:</p> <p>In English, the -acy suffix tends to denote the nounification of an <em>adjective</em> and the -ation suffix tends to denote the nounification of a <em>verb</em>. </p> <p>Obviously, 'conjugate' can act as both an adjective and a verb e.g. '$x$ and $y$ are conjugate' or 'One may conjugate $x$ by some element to get $y$'</p> <p>Since these phrases are mathematically the same, there is thus no mathematical difference between '$x$ and $y$ are equal up to conjugacy' and '$x$ and $y$ are equal up to conjugation' </p> http://mathoverflow.net/questions/96563/origin-of-the-sign-convention-in-the-tensor-product-of-graded-vector-spaces/96569#96569 Answer by Jake for Origin of the sign convention in the Tensor product of graded vector spaces Jake 2012-05-10T12:59:25Z 2012-05-10T12:59:25Z <p><em>Is this the only one or are there others defined by other sign conventions?</em></p> <p>As has already been mentioned - there are essentially two but the Koszul one is amongst other things, the only one which makes the tensor product of complexes a complex.</p> <p>Since I have been interested in this topic of late, I would point you towards the following internal (within MathOverflow) references:</p> <p>For a recent discussion see MathOverflow question: <a href="http://mathoverflow.net/questions/93464/tensor-product-of-linear-mappings-versus-chain-complexes" rel="nofollow">http://mathoverflow.net/questions/93464/tensor-product-of-linear-mappings-versus-chain-complexes</a></p> <p>In particular Simon Letner and Qiaochu Yauns' answers provide a nice conceptual backdrop.</p> <p>I also found Theo Johnson-Freyd's answer in this MathOverflow question quite illuminating: <a href="http://mathoverflow.net/questions/53315/references-for-sign-conventions-in-homological-algebra" rel="nofollow">http://mathoverflow.net/questions/53315/references-for-sign-conventions-in-homological-algebra</a></p> <p>Theo Johnson-Freyd's answer in this thread is also instructive: <a href="http://mathoverflow.net/questions/95795/sign-convention-for-derivations-in-cdgas" rel="nofollow">http://mathoverflow.net/questions/95795/sign-convention-for-derivations-in-cdgas</a></p> http://mathoverflow.net/questions/2533/homological-algebra-texts/95765#95765 Answer by Jake for Homological Algebra texts Jake 2012-05-02T13:12:39Z 2012-05-02T13:12:39Z <p><em>Basic Homological Algebra</em> by Scott Osbourne is a nice beginners text. It is very thorough and detailed yet well motivated and conversational with a particularly engaging style.</p> <p>Although old fashioned and outdated in many respects; I would have to say that Cartan-Eilenberg is still of great value as a reference.</p> http://mathoverflow.net/questions/93464/tensor-product-of-linear-mappings-versus-chain-complexes/93846#93846 Answer by Jake for Tensor product of linear mappings versus chain complexes Jake 2012-04-12T10:31:54Z 2012-04-12T16:27:35Z <p>This is just a naive comment on the second to last paragraph in Qiaochu's answer:</p> <p><em>'So what you've observed is that the category of chain complexes is morally closer to the second construction than the first. Why this is the case is something that I don't have a completely satisfactory answer for, but in any case the point I want to make is that the second comultiplication is just as natural as the first from the appropriate perspective.'</em></p> <p>Surely the motivation comes from thinking of the associated double complex and the desire to 'embed' both complexes into the tensor product complex? From this point of view the choice $x \mapsto x \otimes 1 + 1 \otimes x$ is very natural. </p> <p>The signs in the formula are then intepreted as arising from the braiding i.e. Given complexes $(C,\partial)$ and $(D,\delta)$, we evaluate $\partial \otimes 1 + 1 \otimes \delta$ on tensors $\sum_i c_i \otimes d_i$ via the pairing</p> <p>$$\mathrm{End}(C) \otimes \mathrm{End}(D) \otimes C \otimes D \xrightarrow{\Psi} \mathrm{End}_k(C) \otimes C \otimes \mathrm{End}(D) \otimes D \to C \otimes D$$</p> <p>Where $\Psi$ is the braiding morphism. We see that taking $\Psi$ to be the plain tensor flip doesn't work and then modify this so that on homogeneous elements $c$ and $d$ we have $\Psi\colon c \otimes d \mapsto (-1)^{|c||d|} d \otimes c$ which <em>does</em> work.</p> <p>To me this is main unsatisfactory part of the story: why the Koszul braiding? The only explanation I have seen so far is: <em>'It works! Just accept it.'</em> I guess if I could see that the Koszul braiding was the <em>only</em> braiding that worked then I would be happy but I can't see that off the top of my head...</p> http://mathoverflow.net/questions/93464/tensor-product-of-linear-mappings-versus-chain-complexes/93846#93846 Comment by Jake Jake 2012-04-12T13:41:01Z 2012-04-12T13:41:01Z Thanks for this Simon - I think I understand now. We just consider the braiding on a one dimensional vector space where we see that there is but one choice and then uniqueness of the Koszul braiding follows from there via naturality. It seems that the difference between how you are stating things (with the action of the Taft algebra providing the signs) vs. the signs coming from a braiding is somewhat like the Taft Algebra acting as some kind of 'bosonization' a la Majid.