User james griffin - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:37:09Z http://mathoverflow.net/feeds/user/109 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83702/the-definition-of-a-cw-complex-and-related-notions The definition of a CW complex and related notions James Griffin 2011-12-17T13:19:33Z 2013-04-22T14:23:43Z <p>In the appendix of Allen Hatcher's book "Algebraic Topology", a CW complex is defined to be a space iteratively constructed by attaching $n$-cells onto an $(n-1)$ skeleton. There is a more general notion where a space can be built iteratively by attaching cells, however we pose no restriction on the order of attachment. For instance the endpoints of a $1$-cell could be glued onto the interior of a $2$-cell. This notion is discussed in Chris Schommer-Pries' answer to another MO question:</p> <p><a href="http://mathoverflow.net/questions/74863/what-does-actually-being-a-cw-complex-provide-in-algebraic-topology/74904#74904" rel="nofollow">http://mathoverflow.net/questions/74863/what-does-actually-being-a-cw-complex-provide-in-algebraic-topology/74904#74904</a></p> <p>Does this kind of complex have a name?</p> http://mathoverflow.net/questions/96840/q-deformation-of-the-permutation-group/96844#96844 Answer by James Griffin for q-deformation of the permutation group? James Griffin 2012-05-13T17:07:18Z 2012-05-13T17:07:18Z <p>The algebras you are looking for are called <em>Iwahori-Hecke algebras</em>. In the case of the symmetric groups the Iwahori-Hecke algebras are generated by transpositions' $T_i$ which satisfy the braid relations but don't square to zero; instead there is a relation which looks like $$T_i^2 = qT_i + (1-q)$$</p> <p>I'd recommend you read up on the monoidal category of modules for a Hopf algebra. The various properties of a Hopf algebra determine properties of its module category. For instance if the coproduct is cocommutative then the category of modules is symmetric monoidal. Many of these q-deformations aren't cocommutative but their module categories still have structure, they become braided monoidal.</p> <p>Once I understood this everything became much clearer, good luck!</p> http://mathoverflow.net/questions/5057/rational-group-cohomology Rational Group Cohomology James Griffin 2009-11-11T15:12:38Z 2012-02-16T09:07:57Z <p>This is a general question about group cohomology. I'm interested in the case when the coefficients are the rational numbers and hence I suppose when my groups are infinite. The question splits into two:</p> <p>1) Are there any favoured examples that you would recommend a look at? (Recommended references would be just as welcome.)</p> <p>And the main question:</p> <p>2) What sort of functors on the category of groups leave the rational cohomology unchanged? In particular is there a projection onto a special subcategory of groups that is in some way the right category to study?</p> <p>I have a feeling that someone with a good knowledge of rational homotopy theory would be able to answer this question with relative ease.</p> http://mathoverflow.net/questions/68312/posets-of-cosets-and-contractibility Posets of cosets and contractibility James Griffin 2011-06-20T19:08:03Z 2011-06-21T06:21:00Z <p>For this question let $G$ be a group, perhaps infinite, and let $H_i$ for $i\in I$ be a (finite) family of subgroups closed under taking intersections. I am interested in the coset poset $\mathcal{C}(G,{H_i})$ which is defined as the set of cosets $g H_i$ with ordering by inclusion. Note that $g H_i\subseteq g'H_j$ implies that $H_i\subseteq H_j$ and $g^{-1} g'\in H_j$.</p> <p>The family of subgroups $H_i$ defines a diagram of groups with maps given by inclusion inside $G$. Taking classifying spaces $B(H_i)$ gives a diagram of simplicial sets and I am interested in the colimit of this, denote it $B(G,H_i)$.</p> <p><strong>Question 1</strong> Where is there a reference for:</p> <hr> <p>The realisation $N(\mathcal{C}(G,H_i))$ is contractible if and only if $B(G,H_i)$ is a classifying space for $G$.</p> <hr> <p>I've got a sketch proof which takes the coset poset and applies the Borel construction using the category $\mathcal{E}G$ with object set $G$ and singleton homsets:</p> <p>$\mathcal{C}(G,H_i)\times\mathcal{E}(G)/G$</p> <p>This quotient category can be seen to be equivalent to a category with objects the family $H_i$ and homsets $Hom(H_i,H_j)\cong H_j$. The nerve of this is then seen to be both equivalent to the colimit $B(G,H_i)$ and the Borel construction applied to $N(\mathcal{C}(G,H_i))$.</p> <p><strong>Question 2</strong> What are some nice examples? I can think of right-angled Artin groups, with subgroups indexed by the simplices of the flag complex; I guess that the coset complex is CAT(0). Also if one of the subgroups is $G$ itself then the result holds.</p> <p>Of course a valid answer to this whole question could be that I've got everything wrong.</p> http://mathoverflow.net/questions/63923/shuffle-hopf-algebra-how-to-prove-its-properties-in-a-slick-way/66041#66041 Answer by James Griffin for Shuffle Hopf algebra: how to prove its properties in a slick way? James Griffin 2011-05-26T09:54:03Z 2011-05-26T09:54:03Z <p>I have come across a nice way to think about this Hopf algebra. Let $v_1\otimes\ldots\otimes v_n$ be a monomial in $TV$. Rather than thinking of it as just a word, think of it as a path of $n$ steps in $V$. Geometrically (if $V$ is defined over $\mathbb{R}$) it could be a piecewise linear path $\mu:[0,n]\rightarrow V$, where $\mu(k) = v_1+\ldots+v_k$. Ofcourse it's not really a geometric thing, it's a sequence of points, so we don't need anything to be over $\mathbb{R}$.</p> <p>The coproduct comes from splitting up the path at the integer values (including 0 and n). The coassociativity is obvious, both sides of the equation split a path into three. The counit sends any non-trivial path to 0, the trivial path to 1.</p> <p>The product is not the concatenation of paths of course, let $\mu$ and $\nu$ be paths of length $n$ and $m$ respectively. Their direct product $\mu\times\nu$ has domain $[0,n]\times[0,m]$ and codomain $V\times V$, <em>not</em> the tensor product. So how do we get a path from this direct product? Well we take paths in $[0,n]\times[0,m]$ where each step is a positive unit step in either the horizontal or vertical direction. The image of such a step is of the form $(v,0)$ or $(0,v)$, forget the zeros and we have a new path of the same type as we started with. But this required choice, we chose a path in $[0,n]\times[0,m]$, what possible choices were there? Well the set of such paths is easily seen as the set of shuffles! And so the product is given by summing all the possible paths.</p> <p>The associativity of the product is now easily seen as the product of three paths involves a summation over all possible paths in not a square but now a cube. The unit is the trivial path.</p> <p>So how about the bialgebra relation? Well this now has a simple combinatorial description. Take two paths $\mu$ and $\nu$, on one side of the bialgebra compatibility we have to take their product and then their coproduct. Their product is indexed by paths through a square, taking the coproduct splits such a path at a point, so taking the product then the coproduct gives a sum over paths in a square with a chosen point. Reindex: a sum over all integer points of the square with a path from $(0,0)$ to that point, followed by a path from the point to $(n,m)$ the other corner of the square. </p> <p>Now onto the other side of the equation, take the coproducts of each of path, this is indexed by an integer point in $[0,n]$ and an integer point in $[0,m]$, which taken together is an integer point in the square. Now take the products of the paths, left side with left side and right side with right side. We get precise what we hoped for, all possible paths from $(0,0)$ to the point in the square, followed by a path from the point to $(n,m)$.</p> <p>So we have a bialgebra, and the antipode just reverses the paths. Now I should stress that all I have done is to make the combinatorics apparent as paths in a square, if you want a quick proof, just do the combinatorics.</p> http://mathoverflow.net/questions/65901/glv-representation-theory-for-a-lie-bracket-kernel/65946#65946 Answer by James Griffin for GL(V)-representation theory for a Lie bracket kernel James Griffin 2011-05-25T10:29:23Z 2011-05-25T10:50:43Z <p>The representations are not in general irreducible. The second characterisation is the more useful to my mind because one can decompose $Lie((n+2))$ as an $Sym(n+2)$-module, which gives a decomposition of the $GL(V)$-module of interest. Theo already recited the magical words Schur-Weyl.</p> <p>The dimension of $Lie((n+2))$ as a vector space is $n!$. So the $Sym(6)$-module $Lie((6))$ has dimension 24. However the representation of maximal dimension is given by the partition $(3,2,1)$ which has dimension 16. Playing with the dimensions along with a little knowledge of the Lie operad (that it doesn't have any 1 dimensional modules) gives the decomposition of dimension 24 = 10 + 9 + 5. The 10 and 9 dimensional modules are unique up to transposition. For the 5 dimensional one there are (up to transposition) two possibilities (2,2,2) and (2,1,1,1,1), but we know that it must be the first because restricting (2,1,1,1,1) to $Sym(5)$ gives a one dimensional module in $Lie(5)$.</p> <p>I think that I've read about the general case somewhere but my preliminary google has drawn a blank. If I can find the reference I'll add it to my answer.</p> <h2>Update</h2> <p>The $Sym(n+2)$-module $Lie((n+2))$ is known as the Whitehouse module, there are some slides on Richard Stanley's <a href="http://www-math.mit.edu/~rstan/trans.html" rel="nofollow">website</a>.</p> http://mathoverflow.net/questions/65771/aspherical-amalgamations-without-injective-maps Aspherical amalgamations without injective maps James Griffin 2011-05-23T16:17:03Z 2011-05-23T17:36:20Z <p>The situation I find myself in is as follows: I have a CW complex $X$ which is covered by two subcomplexes $A$ and $B$ and I know that $A$, $B$ and $A \cap B$ are connected and aspherical. The term aspherical means that all higher homotopy groups $\pi_i(-)$ for $i\geq 2$ vanish. Recall that the Seifert van Kampen theorem implies that the fundamental group of $X$ is the amalgamation $A\ast_{A\cap B} B$.</p> <p>Now I had thought that the inclusions $\pi_1(A\cap B \rightarrow A)$ and $\pi_1(A\cap B\rightarrow B)$ were both injective. In this situation there is a Theorem which states that $X$ is itself aspherical, and so its homotopy type is described by its fundamental group $A\ast_{A\cap B} B$. However I have now learnt that my maps are not injective, however I still have strong reason to believe that my space $X$ is aspherical.</p> <p>I can prove that $\pi_1(A\cap B\rightarrow A\times B)$ is injective, so I want the following to hold: </p> <h2>Theorem(?)</h2> <p>Suppose that $X$, $A$, $B$ and $A\cap B$ are as above and that $\pi_1(A\cap B\rightarrow A\times B)$ is injective. Then $X$ is aspherical.</p> <p>I can sketch a proof: the main step is to show that any $\phi\in\pi_2(X;A\cap B)$ can be decomposed into a product of $\phi_A\in\pi_2(A;A\cap B)$ and $\phi_B\in\pi_2(B;A\cap B)$. Then in the long exact sequence of homotopy groups associated to $(X;A\cap B)$ the element $\phi$ comes from $\pi_2(X)$ only if the images of $\phi_A$ and $\phi_B$ in $\pi_1(A\cap B)$ are inverse. But these images must be in the kernels of the respective maps into $\pi_1(A)$ and $\pi_1(B)$, hence by the injectivity assumption are trivial. But now the long exact sequences for $(A;A\cap B)$ and $(B;A\cap B)$ imply that $\phi_A$ and $\phi_B$ are null homotopic and hence so is $\phi$.</p> <p>But I have not been able to find the result in the literature.</p> <h2>Questions</h2> <ol> <li>Is this theorem true?</li> <li>Is there a reference for this result?</li> <li>Are there any well known examples of this situation? Clearly any amalgamation of groups can be converted to one where the maps are injective, however the groups obtained in this way may not be themselves as nice as those with which we started. In my case the fundamental groups go from being finitely presented to only being finitely generated.</li> </ol> http://mathoverflow.net/questions/63923/shuffle-hopf-algebra-how-to-prove-its-properties-in-a-slick-way/64367#64367 Answer by James Griffin for Shuffle Hopf algebra: how to prove its properties in a slick way? James Griffin 2011-05-09T10:25:03Z 2011-05-09T10:25:03Z <p>I believe the right way to consider this algebra is to view it as the free zinbiel algebra. A zinbiel algebra has a single operation o which must satisfy</p> <p>(x o y) o z = x o (y o z + z o y)</p> <p>The zinbiel operation o in your algebra is the sum over all of the (p,q) shuffles which in the notation of the question have </p> <p>${\sigma^{-1}(1)} &lt; {\sigma^{-1}(i+1)}$</p> <p>So the commutative product defined in the question is a.b = a o b + b o a. The answer to your question now comes from a result which will state that the free zinbiel algebra viewed as a commutative algebra is itself free, then you just need to check that your shuffle coproduct is defined on the generators. This will show that it is a bialgebra.</p> <p>An analogous result is that the free associative algebra is a free Lie algebra with the associated Lie bracket. </p> <p>One way to prove this result would be to decompose the Zinbiel operad as a left module for the commutative operad. But I imagine that the result is already in the literature somewhere. I guess that there are other names for zinbiel algebras, perhaps shuffle algebras or something similar. </p> <p>They do occur naturally, for instance if you want to decompose the direct product of two simplices (which isn't a simplex) into simplices in a natural way; see p278 of Allen Hatcher's Algebraic Topology.</p> http://mathoverflow.net/questions/59563/configuration-space-of-little-disks-inside-a-big-disk/59584#59584 Answer by James Griffin for Configuration space of little disks inside a big disk James Griffin 2011-03-25T17:16:29Z 2011-03-25T17:16:29Z <p>I would very much appreciate a good answer to this question, perhaps a follow up to Igor's answer as this is something that I have thought about before, but have not come across in the literature. </p> <p>I quickly (perhaps too quickly) abandoned the discs model in favour of the little cubes model, or perhaps I should say the hard cubes model.</p> <p>For hard 2-cubes and k=3 I think the homology groups are: </p> <p>for r > 1/2: clearly 0</p> <p>for 1/2 >= r > 1/3: H_0 = Z^6, H_1=Z^6 and H_i=0 for i>1</p> <p>the three squares are effectively arranged in a circle which can be rotated, the order (and not just the cyclic order!) parametrises the 6 connected components.</p> <p>for 1/3 >= r we get the usual configuration space: H_0 = Z, H_1 = Z^3, H_2 = Z^2 and H_i=0 for i>2.</p> http://mathoverflow.net/questions/45651/is-there-a-q-analog-to-the-braid-group/45685#45685 Answer by James Griffin for Is there a q-analog to the braid group? James Griffin 2010-11-11T11:47:16Z 2010-11-11T11:47:16Z <p>I suggest you look at the Iwahori-Hecke algebras of type A. These deform the symmetric group algebras with relations that look like</p> <p>$T_i^2 = q + (1-q)T_i$</p> <p>for generating elements $T_i$. The braid relations</p> <p>$T_iT_{i+1}T_i = T_{i+1}T_iT_{i+1}$</p> <p>still hold though so you get a (surjective) morphism</p> <p>$kB_n\rightarrow \mathcal{H}_n\rightarrow 0$</p> <p>(<strong>From here on I'm less sure of the details</strong>) giving you a short exact sequence'</p> <p>$0\rightarrow K_n\rightarrow kB_n \rightarrow \mathcal{H}_n\rightarrow 0$.</p> <p>To define the kernel you should look at the coinvariants of $kB_n$ w.r.t. the coalgebra map of $\mathcal{H}_n$. This makes $K_n$ an algebra but not necessarily a Hopf algebra (although it may be a braided Hopf algebra in a suitable category).</p> <p>The Hecke algebra may be the algebra that you want because the q parameter counts the way the Borel double cosets in some $GL_n(k)$ multiply (recall the Bruhat decomposition). When the Borels become trivial then $q$ becomes one.</p> <p>But notice also that this deformation does not require a deformation of the braid group. I have no idea what the algebra $K_n$ looks like and if indeed it is well defined, it may still turn out to be $kP_n$.</p> <p>To offer an answer to your final question: there may very well be q-analogues of the braid groups, but they may not be what you should be looking for.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/40999#40999 Answer by James Griffin for Compelling evidence that two basepoints are better than one James Griffin 2010-10-04T09:22:50Z 2010-10-04T09:22:50Z <p>A short example.</p> <p>The family of pure braid groups does not possess a symmetric operad structure.</p> <p>But the fundamental groupoid of the little 2-discs operad is naturally a symmetric operad.</p> <p>Although the fundamental groups of the little 2-discs operad are the pure braid groups, there is no way to choose basepoints consistent with the operad structure.</p> <p>The moral is that groupoids are not naturally pointed, whilst groups are. If you're working with fundamental groups you should really be working with pointed spaces. Ofcourse you can ignore this and you'll only run into trouble if your mathematics doesn't work with pointed spaces, see the example above.</p> http://mathoverflow.net/questions/40291/what-are-operad-automorphisms/40311#40311 Answer by James Griffin for What are operad automorphisms? James Griffin 2010-09-28T13:15:59Z 2010-09-29T13:18:42Z <p>I'm not sure what the question you're <strong>trying</strong> to ask is, but the answer to the question that you have asked is that an operad automorphism is an invertible operad endomorphism.</p> <p>[EDIT] (just restating Ryan's comment on the original post) An operad endomorphism is an operad morphism where the source and target operads are the same. An operad morphism is a collection of maps $\mathcal{O}(n)\rightarrow\mathcal{P}(n)$, one for each arity, such that the obvious squares involving the operad structure maps commute. [/EDIT]</p> <p>Perhaps you want to know about operad automorphisms in a homotopy category. In that case you want to understand what an operad quasi-isomorphism is. This is an operad morphism which is a quasi-isomorphism on each underlying space of operations.</p> <p>Would you like to refine the question?</p> http://mathoverflow.net/questions/39441/topologizing-a-free-product-gh-of-discrete-groups/39480#39480 Answer by James Griffin for Topologizing a free product G*H of discrete groups? James Griffin 2010-09-21T09:59:10Z 2010-09-21T15:00:34Z <p>I wouldn't agree with your interpretation of free groups, I think that your understanding comes from the free group on 2 letters with a particular action on a fractal. That's not to say that there isn't much of interest to say about free products and actions on particular spaces.</p> <p>But if you're just interested in free groups and how they are related to free products then I'd recommend reading up on Bass-Serre theory: <a href="http://en.wikipedia.org/wiki/Bass%E2%80%93Serre_theory" rel="nofollow">http://en.wikipedia.org/wiki/Bass%E2%80%93Serre_theory</a></p> <p>This studies how groups act on simplicial trees.</p> <p>Current research is focused on the automorphisms of free groups, if you want to learn about this you should search for articles on "Outer space", possibly by Karen Vogtmann.</p> <p>[EDIT: As pointed out the last paragraph points to just one topic of active research, which by no means describes fully the current research environment. ]</p> http://mathoverflow.net/questions/29636/operad-terminology-operads-with-and-without-o0/34643#34643 Answer by James Griffin for Operad terminology - Operads with and without O(0). James Griffin 2010-08-05T15:42:19Z 2010-08-05T15:42:19Z <p>I can second Jeffrey's comment, reduced is used to say that O(1) is just the monoidal unit (it allows us to use the Boardman Vogt resolution in homotopy theory). It's my opinion that this terminology will probably stick.</p> <p>I would also say that a $\mu$ in O(n) had arity n.</p> <p>That the O(0) part of an operad is referred to as the 'constants' of the operad makes a lot of sense, every algebra for O must contain O(0) and the composition of those must behave in a certain way.</p> <p>Calling O(0) the point also makes sense, because in the category of algebras O(0) will be the initial object.</p> <p><em>Here my comment has become too long, just as I've got to the point of my comment:</em></p> <p>The comments to the question tend to prefer terminology that relates to the behaviour of the operad (eg "reduction", because a unit lowers the arity). My personal preference (and I think the literature follows it), is that terminology should have more of a relation to the category of algebras than to the operad itself.</p> <p>So my vote is that you call O(0) the initial of O. And you call an operad without O(0) initial-less or uninitiated.</p> http://mathoverflow.net/questions/32889/a-few-questions-about-kontsevich-formality/34002#34002 Answer by James Griffin for A few questions about Kontsevich formality James Griffin 2010-07-31T10:54:06Z 2010-07-31T10:54:06Z <p>To (1): Daniel is right, there is a map of homotopy Gerstenhaber algebras between the two algebras. However the full story is quite complicated and to show that the hochschild cochains form a homotopy Gerstenhaber algebra is hard, it's known as the Deligne conjecture. I don't know the details of the proof.</p> <p>Recall that a Poisson algebra is a commutative algebra with a Lie bracket and these two products satisfy a Leibniz identity. A Gerstenhaber algebra is a bit like a Poisson algebra, except the Lie bracket is of degree 1 not 0. The bracket satisfies a graded Leibniz identity wrt to the commutative algebra structure.</p> <p>The formality morphism as homotopy Gerstenhaber algebras restricts to a formality morphism as homotopy Lie algebras and to a formality morphism as homotopy commutative algebras.</p> <p>In my view the simplest proof of the formality of the Hochschild cochains of a nice enough algebra as a homotopy Gerstenhaber algebra is contained in</p> <p><a href="http://arxiv.org/abs/math.KT/0605141" rel="nofollow">http://arxiv.org/abs/math.KT/0605141</a></p> http://mathoverflow.net/questions/31851/decomposing-a-large-colimit-as-a-pushout-of-smaller-colimits Decomposing a large colimit as a pushout of smaller colimits James Griffin 2010-07-14T13:20:44Z 2010-07-20T15:11:37Z <p>I would like to find a reference in the literature for the following result. I have it on high authority that it isn't in 'Categories for the Working Mathematician' and I can't find it in Borceux's handbook. It's a result that I'm confident is true (at least when stated correctly) and is probably second nature to category theorists. I however am writing for group theorists and so want to reference results thoroughly.</p> <p>I have a functor $F:\mathcal{C} \rightarrow \mathcal{D}$. The target category $\mathcal{D}$ is cocomplete. The source category $\mathcal{C}$ is finite and can be decomposed as the pushout of smaller categories $\mathcal{C}_1\leftarrow\mathcal{C}_0\rightarrow\mathcal{C}_2$. The functors from these into $\mathcal{D}$ are denoted $F_1,F_0$ and $F_2$ respectively.</p> <p>I need to take the colimit of $F$ and I think that it can be taken to be the pushout of</p> <p>$\text{colim}F_1\leftarrow\text{colim}F_0\rightarrow\text{colim}F_2$.</p> <p>Obviously if $\mathcal{C}$ were constructed from a different colimit rather than a pushout one might expect an analogous result.</p> <p>Giving a proof is an option, but would be out of context with the rest of the paper and probably consigned to an unread appendix. Or I could just quote it without proof. Help, or just opinions would be very welcome.</p> http://mathoverflow.net/questions/23782/a-positive-formula-for-the-dimensions-of-homogeneous-components-of-free-lie-algeb/25982#25982 Answer by James Griffin for A positive formula for the dimensions of homogeneous components of free Lie algebras James Griffin 2010-05-26T09:45:40Z 2010-05-26T09:45:40Z <p>This doesn't answer the question, but might still be of interest to you. Let $V$ be the $n$-dimensional vector space spanned by your $n$ letters.</p> <p>The vector space $V^{\otimes k}$ has a natural $S_k$ action. There exists an $S_k$ module, which I will denote $\text{Lie}(k)$, such that the $k$th homogenous component of the free Lie algebra on $V$ is isomorphic to</p> <p>$V^{\otimes k} \otimes_{S_k} \text{Lie}(k)$.</p> <p>And this module has dimension $(k-1)!$. This wont help you with the dimensions you want, but I think that it's interesting.</p> <p>If you want to read more then you need to learn about operads, and in particular the Lie operad.</p> <p>If you just want to know the $S_k$-module structure on $\text{Lie}(k)$ then it can be given as follows: Let $C_k$ be a subgroup of $S_k$ generated by a $k$-cycle. Let $W$ be a 'primitive' representation of $C_k$. (this requires a primitive $k$th root of unity in your field). Then the module we are looking for is $W$ induced up to $S_k$.</p> <p>This last bit is a bit mysterious to me.</p> http://mathoverflow.net/questions/17742/automorphism-group-objects/17891#17891 Answer by James Griffin for Automorphism group objects James Griffin 2010-03-11T18:55:09Z 2010-03-11T18:55:09Z <p>I think that it might be worth refining your conditions to allow for some other interesting examples. Rather than have a diagonal map defined for each object, instead have the diagonal map as part of the group object. This would include vector spaces in your list of examples. The group objects would then by Hopf algebras.</p> <p>The following example may be of interest: Let H be a Hopf algebra, this may act on an algebra A, we just ask that the algebra multiplication map of A is a morphism of H-modules. Equivalently A is an algebra object in the monoidal category of H-modules; A is called a H-module algebra.</p> <p><strong>Example</strong>: Let V be a representation of a Lie algebra g. Let A=SV be the free commutative algebra on a vs V. Then the universal enveloping algebra Ug acts on SV with g acting as derivations.</p> <p>One may define a category of Hopf algebras acting on a fixed algebra A. This category has a terminal object, PA. This is very much reminiscent of what you might call Sym(A). PA contains the automorphism group algebra of A.</p> <p>I'm afraid that I can't relate this example directly to your questions, but it's a good generalisation of the automorphism group, perhaps it might help.</p> http://mathoverflow.net/questions/9146/noncommutative-rational-homotopy-type/9266#9266 Answer by James Griffin for Noncommutative rational homotopy type James Griffin 2009-12-18T12:27:37Z 2009-12-18T12:27:37Z <p>This is an attempt to rephrase some of Joey's answer and also to add in the higher operations required.</p> <p>Suppose we have two cdgas A and C and that they're connected by an $A_\infty$ morphism. This is encoded as a map from $BA$, the associative bar construction on A to C. A $C_\infty$ morphism is encoded by a morphism from the commutative bar construction, $B_cA$, this takes the form of a quasi-free colie algebra on A. There is a natural transformation from the free coassociative algebra functor to the free colie algebra functor and this coincides with a natural transformation from the associative to commutative bar functor. This is ofcourse the wrong direction, the natural transformation says "every $C_\infty$ morphism is an $A_\infty$ morphism".</p> <p>To go the other way we observe that the maps $BA\rightarrow B_cA$ split. I wont offer a proof here but the proof goes by constructing certain idempotents, it's known as the Hodge decomposition of the bar complex of a commutative algebra. We now have the composition</p> <p>$B_cA\rightarrow BA \rightarrow C$</p> <p>which is our $C_\infty$ morphism. <strong>Warning</strong> The $A_\infty$ morphism we get from this $C_\infty$ morphism is not necessarily the one we started off with, we have just stripped away all the "stuff" coming from other parts of a Hodge filtration.</p> <p>Finally I'd like to add that your intuition was broadly right, there will be examples of non char 0 cases where the answer to your question will be "no".</p> http://mathoverflow.net/questions/5161/groupoid-of-moves-on-trivalent-fatgraph/7468#7468 Answer by James Griffin for Groupoid of moves on trivalent fatgraph James Griffin 2009-12-01T18:10:32Z 2009-12-01T18:10:32Z <p>This is a nice question and I find it very hard to make such things understandable so I'm impressed.</p> <p>The relationship to Outer Space of course looks very strong, but you're just looking at the simplicies of maximal dimension (trivalent graphs), so to understand the relationship my first step would be to try to lift your moves to the whole of outer space. </p> <p>Your edge moves get factored into edge expansion and edge contraction, these are well understood and studied, see any reference on outer space. As for the vertices this means having to alter fat vertices of higher degree and looking at the trivalent case I think that means that you have to allow moves between any two cyclic orderings. A first glance seems to indicate that there is some kind of compatibility between edge and vertex moves: essentially it does look to me that you can lift your concept of moves to the whole of "fattened outer space".</p> <p>Now we've done this we can study your object in the formalism of Kontsevich's graph complexes (well the associated (semi-)simplicial structures). Fattened outer space becomes the associative graph complex but with some extra simplices (your vertex moves). We can characterise these as certain simplices coming from the kernel of the map from the associative graph complex to the commutative graph complex.</p> <p>As such I don't think that your groupoid will mean much beyond outer space, but it may encode some information from the associative complex, although I'm guessing only something along the lines of enumeration of the kernel.</p> <p>I've run out of time and need to leave, hopefully what I've said makes some sense and the details are fill-in-able. I'll address any comments tomorrow.</p> http://mathoverflow.net/questions/6175/cohomology-of-associative-algebras/6261#6261 Answer by James Griffin for Cohomology of associative algebras James Griffin 2009-11-20T13:17:10Z 2009-11-20T13:17:10Z <p>In my view Hochschild cohomology is the most interesting cohomology on associative (and I dare say commutative algebras). So far all that has been said is about different methods of computation. But there are also many applications and ways of viewing it. </p> <p>Skip the following paragraph if you want, it's just a side point.</p> <blockquote> <p>The one that sticks in my mind is the application to deformation theory. The Hochschild cochain complex is actually the object of interest in deformation theory, its homology is just one invariant of it and captures the infinitesimal deformations. The cochain complex carries a specific algebraic structure; it's an algebra for the braces operad. And then there's the celebrated (and many times proved ;-)) Deligne conjecture which says that it may be viewed as a homotopy Gerstenhaber algebra. Finally there's Kontsevich's formality result which says that for smooth commutative algebras that looking at homology and its Gerstenhaber algebra structure actually does capture all information about the Hochschild cochains and hence the deformation theory of the algebra.</p> </blockquote> <p>Anyway I didn't mean the write that, but just got overexcited, my point in writing this answer was to say that there are other homology theories. </p> <p>For example there's the bar homology. This homology is little known which is a big pity because it's actually rather special! There's a very good reason why it's not studied though and that's because for a unital algebra its homology is always zero, but it is still interesting because it the chain complex a coalgebra and we're not interested in its homotopy type as a complex and so shouldn't be taking its homology at all! The coalgebra actually gives generators and relations for the algebra, it's the derived functor of $A \mapsto A/(A.A)$ from the category of associative algebras to vector spaces.</p> <p>But you guys like taking homology, so I should give you a better reason for studying the bar homology. Suppose you have an augmented algebra, so we can split the identity off and write</p> <p>$A = k\oplus A'$</p> <p>Then the bar homology of $A'$ is not necessarily zero and gives interesting invariants of the algebra. In the char 0 commutative case this is well studied, you guys might know it as part of rational homotopy theory. The commutative bar homology of the cohomology ring of a nice space is the rational homotopy of the space.</p> http://mathoverflow.net/questions/5833/symmetric-powers-tableau-and-wreath-products Symmetric Powers, Tableau and Wreath Products James Griffin 2009-11-17T16:23:47Z 2009-11-18T10:11:46Z <p>Let V and W be irreducible representations of $S_n$ and $S_m$ over a field of characteristic 0. Then the Littlewood-Richardson coefficients allow us to compute the isomorphism type of the induced $S_{n+m}$-module V&otimes;W&uarr;. This induction comes from the inclusion</p> <p>$S_n\times S_m \rightarrow S_{n+m}$.</p> <p>Now suppose V=W. Then V&otimes;V&uarr; is a $S_{2n}$-module. But actually there's a symmetry coming from the symmetric monoidal category structure, so there is another induction up to an $S_{2n}$-module structure:</p> <p>Extend the action of $S_n\times S_n$ on V&otimes;V by including the symmetry $c_V$, this naturally extends the group to the wreath product $S_n\sim S_2$. Induction along</p> <p>$S_n\sim S_2 \rightarrow S_{2n}$</p> <p>gives the representation that I want:</p> <p>$(V\otimes V)_{S_n\sim S_2}\uparrow^{S_{2n}} \hookrightarrow V\otimes V\uparrow^{S_{2n}}$.</p> <p>Using the Littlewood-Richardson rules we know the structure of the last term in terms of semi-standard skew tableau. My question is, how do we characterise the inclusion?</p> http://mathoverflow.net/questions/5031/model-structure-of-commutative-dg-algebras-inside-all-dg-algebras/5200#5200 Answer by James Griffin for Model structure of commutative dg-algebras inside all dg-algebras James Griffin 2009-11-12T16:12:04Z 2009-11-12T16:19:23Z <p>I can second Mark's example, however I would point out that the cocycle z lies in degree 8 and not 6. There's another one, w in degree 11 whose purpose is to kill y^2, which is 0 in the abelianisation. Not to mention xz in degree 10. The resultant commutative algebra seems to be of the form k[x]&otimes;k[V], where V has some kind of algebraic structure, I'm guessing a (co)lie-module after a suspension, but I can't work out what it should be.</p> <p>I also have an interpretation of all of this. In working out an associative quasi-free presentation (TW, &delta;) we're actually working out the bar homology W of the algebra (actually including the A-infinity coalgebra structure as well). But there's a decomposition of the bar homology of a commutative algebra known as the &lambda;-decomposition (see for instance Loday's Cyclic Homology book). If we were to take the "middle" piece W' of this decomposition we would get the commutative bar homology (and it's associated L-infinity coalgebra structure). And this is just what we need to get the resolution of the original algebra as a quasi-free commutative algebra (SW', &delta;'). All the extra bits (the V of the example) come from other pieces of the &lambda;-decomposition. I think that this should impose strong structural conditions on (SW, &delta;'').</p> <p>Going back the example we can work out that z is in the second part of the decomposition, although I'm not sure about w.</p> <p>So what this means for the various homotopy functors between the homotopy categories in question I'm not sure, the following is speculative: It seems to me that it indicates that they're not full. There are homotopy morphisms from k[x]/x^2 to another commutative algebra that are not homotopy commutative algebra morphisms. They should fit into the &lambda;-decomposition and their position there should indicate just how "not commutative" they really are.</p> <p>Final note on "not commutativity": this notion can be made more rigorous using operads, but in a hand-wavey way it just means to what level the higher homotopies of the commutativity condition hold.</p> http://mathoverflow.net/questions/4682/how-much-morse-theory-can-be-accomplished-given-only-a-continuous-transformatio/4723#4723 Answer by James Griffin for How much "Morse theory" can be accomplished given only a continuous transformation of a space? James Griffin 2009-11-09T12:22:16Z 2009-11-09T12:22:16Z <p>Morse theory can be generalised in many directions. The generalisation of which you speak would cover Morse-Novikov homology; this is where there is a function which is locally Morse, but it might not integrate to a full Morse function (there may be loops of 'flow').</p> <p>Another generalisation it could cover is Morse-Bott homology and this will cover the situation where you don't have isolated fixed points, but the fixed points form nice enough shapes.</p> <p>Both of these generalisations may be studied by taking cellular decompositions (the 'flow' out of a fixed point), although in the Novikov case things are more complicated as you have to take the limit of relative decompositions, but I wont try to make that precise here. These cellular decompositions give a filtration (by index) of the cell complexes of the underlying space. All of the Morse conditions are carefully chosen to make the associated spectral sequences nice (converging to the homology in the nicest case!). So an approach to generalisation would normally involve relaxing these conditions and seeing what happens to the spectral sequence.</p> http://mathoverflow.net/questions/3207/what-kind-of-geometric-operations-scale-up-cohomology/3252#3252 Answer by James Griffin for What kind of geometric operations "scale up" cohomology? James Griffin 2009-10-29T12:32:49Z 2009-10-29T12:32:49Z <p>To get from spaces to graded rings you go via dg-rings. That is,</p> <p>Spaces &rarr; dg-Rings &rarr; graded Rings.</p> <p>where the first arrow is the functor taking simplicial cochains, and the second is the functor taking cohomology. So any "scaling up" of spaces should exist in dg-Rings. But how do you scale up a cochain complex? You can't because because that would break the degree of the codifferential.</p> <p>So I don't think that there is a sensible "scaling up" for spaces.</p> <p>Nice idea though, I don't know how you could classify functors that act on graded rings that can be lifted to spaces. An obvious example might be that of setting all groups of degree greater than i say to 0. That would correspond to filling in things in the space to kill the cohomology groups (I don't know the correct terminology off the top of my head).</p> http://mathoverflow.net/questions/1951/does-the-cohomology-ring-of-a-simply-connected-space-x-determine-the-cohomology-g/2066#2066 Answer by James Griffin for Does the cohomology ring of a simply-connected space X determine the cohomology groups of ΩX? James Griffin 2009-10-23T09:39:06Z 2009-10-23T11:07:07Z <p>My feeling is that Charles is on the right track with the answer above. But rather than looking for a counterexample, I think we should have a go at correcting the original question. Now I'm not quite sure over which rings the next statements work, possibly only over rings over a field of char 0. Perhaps someone knows the details better than I, but to make it work will probably require working with simplicial algebras as these carry a model structure over any ring.</p> <p>The cochains of X carry a dg-algebra structure A. Since &Omega;X is the homotopy pullback of &bull; &rarr; X &larr; &bull; and taking cochains should preserve the relevant (co)limits (can someone help me here), then the cochains ring of &Omega;X is the homotopy pushout of k &larr; A &rarr; k, that is, the derived tensor product. We can then take cohomology.</p> <p>For the next bit we probably do need characteristic 0. The cochains ring will be rather large, so to keep track of things we could take the cohomology, but remember the higher operations. Then as an infinity ring the cohomology H(A) will be quasi-isomorphic to A (which isn't necessarily true if we don't remember the higher operations). Then with that in mind we can calculate the derived form of k &otimes;_H(A) k. Its cohomology should be the cohomology of the loop space.</p> <p>It would be nice to have a counterexample though, how about complements of links, the cohomology rings aren't so bad to calculate (only depending on the number of links over the rationals at least). What about the loop spaces?</p> http://mathoverflow.net/questions/1247/mystery-of-the-monstrous-moonshine/1404#1404 Answer by James Griffin for Mystery of the Monstrous Moonshine James Griffin 2009-10-20T09:45:24Z 2009-10-20T09:45:24Z <p>Sorry, this isn't an answer but just a piece of related information for people who might be in the Cambridge (UK) area:</p> <p>Marcus du Sautoy (Simonyi Professor for the Public Understanding of Science) is giving a public talk tonight entitled "Monstrous Moonshine". So if you want to see how someone might go about explaining the monster group to the general public then it might be a worthy talk. Perhaps I could ask Ilya's question at the end. :)</p> <p>I think that it's worthwhile going just for some inspiration on how to present maths. He's a good communicator.</p> <p>Details on talks.cam.ac.uk</p> <p>(on another note, if anyone with special powers could fix my account I'd be very grateful, this could have been a comment!)</p> http://mathoverflow.net/questions/96840/q-deformation-of-the-permutation-group/96844#96844 Comment by James Griffin James Griffin 2012-05-17T09:53:29Z 2012-05-17T09:53:29Z Having said that, it is still a good comment. But the deformation theoretic view sort of misses the point. The Hopf algebra is there to represent a monoidal category in some sense, in a related sense it's a decategorification of something. Both viewpoints are lost when you view it in terms of formal deformations. I suppose the `moral' is that q-deformations are somewhat different from formal deformations. http://mathoverflow.net/questions/96840/q-deformation-of-the-permutation-group/96844#96844 Comment by James Griffin James Griffin 2012-05-17T09:49:17Z 2012-05-17T09:49:17Z Alexander, in reply to your first comment. Don't you have to be careful with this? The Hecke algebra isn't a formal deformation in the sense that it isn't defined over the power series ring, but rather over polynomials in q and perhaps q^-1. Enlarging the base ring using something like q=e^t you do have this isomorphism, but that's a necessary step. Over C you also have to be careful, if q were a root of unity then your statement wouldn't be true. http://mathoverflow.net/questions/5057/rational-group-cohomology/88613#88613 Comment by James Griffin James Griffin 2012-02-17T09:43:58Z 2012-02-17T09:43:58Z Thank you Mark, interestingly enough I've been reading recently about the homology of the group of upper uni-triangular matrices with integer coefficients. http://mathoverflow.net/questions/87831/the-coproduct-on-the-cohomology-of-a-hopf-algebra Comment by James Griffin James Griffin 2012-02-08T15:01:25Z 2012-02-08T15:01:25Z My intuition is that Theo is right. Does the construction you have in mind work for any coproduct, or does it have to be cocommutative? http://mathoverflow.net/questions/86818/a-flag-complex-is-contractible-iff-the-underlying-graph-is Comment by James Griffin James Griffin 2012-01-27T12:21:38Z 2012-01-27T12:21:38Z Suppose you have a graph G and we know that the flag complex C(G) is contractible. Pick a clique K inside G and define H to be the graph given by 'taking the cone' over K, so adding an extra vertex v and adding edges (v,k) for each k in K. Then C(H) is still contractible. My guess is that any G with C(G) contractible can always be built by repeatedly taking cones over subcliques. I think this counts as a graph theoretic condition. Can anyone verify my guess? http://mathoverflow.net/questions/86494/software-for-combinatorial-algebra-sought Comment by James Griffin James Griffin 2012-01-25T19:23:52Z 2012-01-25T19:23:52Z You can try sage without installing it via their online notebook site at www.sagenb.org http://mathoverflow.net/questions/83702/the-definition-of-a-cw-complex-and-related-notions Comment by James Griffin James Griffin 2011-12-19T10:48:45Z 2011-12-19T10:48:45Z Thank you Dmitri, I am aware of this. I think that I'm for a more friendly notion for a broader audience. http://mathoverflow.net/questions/83702/the-definition-of-a-cw-complex-and-related-notions/83709#83709 Comment by James Griffin James Griffin 2011-12-19T10:46:29Z 2011-12-19T10:46:29Z Can you persuade me that this is standard language? For instance J.H.C. Whitehead in Combinatorial Homotopy I, describes a cell complex as a space with a CW-like decomposition but without the topological conditions. In particular he asks that the boundary of an n-cell is in the (n-1)-skeleton. http://mathoverflow.net/questions/32889/a-few-questions-about-kontsevich-formality/34002#34002 Comment by James Griffin James Griffin 2011-12-13T11:38:44Z 2011-12-13T11:38:44Z I realise that I never got back to writing an explanation of that proof. Sorry Kevin, maybe one day, I do plan to return to formality style questions sometime in the future. With regards to the char 0 point, the general ideas in that paper may work in any characteristic. However I don't recall all the steps so much care is required, in particular the little discs operad isn't formal as a symmetric operad, only as a shuffle operad (I think). http://mathoverflow.net/questions/68312/posets-of-cosets-and-contractibility/68344#68344 Comment by James Griffin James Griffin 2011-06-23T09:52:32Z 2011-06-23T09:52:32Z Thanks for the reference, their covering is just what I want. You're right that this does deserve to be more widely known. I'll try to find the time to have a proper look at the paper on global actions, I think that I would learn a lot. http://mathoverflow.net/questions/63923/shuffle-hopf-algebra-how-to-prove-its-properties-in-a-slick-way/66041#66041 Comment by James Griffin James Griffin 2011-05-30T09:27:35Z 2011-05-30T09:27:35Z I think there is a nice way to prove the antipode identity, but I've not quite worked it out. You can view the convolution of the antipode and the identity as a kind of summation over foldings of the paths; for each point of the path you look at the two paths leaving that point, then you shuffle them together, now when you compare the 'foldings' arising from the two neighbouring points you find that they cancel out. Perhaps. http://mathoverflow.net/questions/63923/shuffle-hopf-algebra-how-to-prove-its-properties-in-a-slick-way/64367#64367 Comment by James Griffin James Griffin 2011-05-26T15:55:57Z 2011-05-26T15:55:57Z So I think that Zinbiel is a Hopf operad, maybe that's enough to finish the proof this way. But showing that the operad is Hopf is probably just the same as showing that the algebra is a Hopf algebra. http://mathoverflow.net/questions/63923/shuffle-hopf-algebra-how-to-prove-its-properties-in-a-slick-way/66041#66041 Comment by James Griffin James Griffin 2011-05-26T15:52:25Z 2011-05-26T15:52:25Z Well a path is a series of steps, the antipode applies those steps in the reverse order with a sign added for each step. At least that's what the formula in the question does :-). However actually showing that this reversal defines a good antipode seems to be much harder than I had thought. I think I have a proof, but it's really just checking the combinatorics by hand and certainly isn't enlightening. I'll try to think of a better proof involving paths. http://mathoverflow.net/questions/65901/glv-representation-theory-for-a-lie-bracket-kernel/65946#65946 Comment by James Griffin James Griffin 2011-05-25T12:55:57Z 2011-05-25T12:55:57Z Whoops, yes that would be unfortunate. http://mathoverflow.net/questions/65901/glv-representation-theory-for-a-lie-bracket-kernel/65946#65946 Comment by James Griffin James Griffin 2011-05-25T11:42:55Z 2011-05-25T11:42:55Z Oh, and finding that webpage was pure luck. In searching for a reference for this I got distracted and started reading a paper on PreLie algebras, which just happened to mention that $Lie((n+2))$ is known as the Whitehead module.