Hugh Thomas
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Registered User
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Apr 2 |
answered | Expected distance of a random point to the convex hull of N other points |
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Mar 28 |
accepted | Number of Permutations with k-inversions and with a single clamped value |
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Mar 16 |
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Random walk on the hypercube "Go on like that" is not entirely clear to me. Do we pick a new random $k$ at each step? Or do we proceed $k$, $k+1$, ..., which is consistent with the example you gave? Also, I suggest that it will probably help to visualize the problem to think of these 0,1 strings as lattice paths from $(0,0)$ to $(t,n-t)$, where we read 1's as horizontal steps and 0's as vertical steps. The basic swap move which you describe looks at two adjacent steps. If they are both horizontal or both vertical, nothing happens; otherwise, a single box is added or subtracted from the region under the path. |
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Mar 14 |
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Is always a Crepant birational map between smooth varieties a small modification I don't see why you're removing a point. As stated, the Lemma applies even before you remove the point. |
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Mar 14 |
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How to prove the NP-hardness of this problem I don't think you have stated the problem in a sufficiently precise form in order for it to have a solution. Also, I still cannot understand the sentence "Now for each sequence $S_i$, there is an element that is used to replace only one element in $S_i$." |
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Mar 14 |
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A counterexample to a conjecture of Nash-Williams about hamiltonicity of digraphs? From what I can see it's a counter-example. Though I would go back and check the statement of the conjecture as given by Nash-Williams (the source of which doesn't seem to be online). The check of non-Hamiltonicity is pretty easy to do by hand: a Hamiltonian cycle has to alternate between vertices in {0,1,2} and {3,4,5} since there are no edges among {0,1,2}. Now it's clear that 1 has to fall between 4 and 5, so the two possibilities are 3,$x$,5,1,4,$y$ and 3,$x$,4,1,5,$y$ and they can both be ruled out. |
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Mar 14 |
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How to prove the NP-hardness of this problem I cannot understand your question. In particular, I don't know what the sentence that starts "Now for each sequence..." means. If we know nothing about $f$, how can we minimize it except by checking all the possible inputs? |
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Mar 12 |
answered | Number of Permutations with k-inversions and with a single clamped value |
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Feb 5 |
awarded | ● Nice Answer |
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Jan 14 |
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Tensor product of quivers Thanks for the additional explicit description. I think I now understand your definition (on a superficial level). Am I right that the tensor product of an oriented path with m arrows and one with n arrows gives you an oriented path with n+m-1 arrows (where this works properly if one of m,n is zero, but not if both are)? |
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Jan 12 |
awarded | ● Guru |
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Jan 10 |
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Not isomorphic varieties with isomorphic tilting algebras Related to your last paragraph: there are examples of non-isomorphic varieties which are derived equivalent. The derived equivalences are then called Fourier-Mukai transforms. For example, in dimension 3, all crepant resolutions of a variety with terminal singularities are derived equivalent. There is lots of information in the chapter "Derived categories of coherent sheaves on algebraic varieties" by Yukinobu Toda in Triangulated Categories, edited by Holm, Joergensen, and Rouquier, including some discussion of tilting objects, but an answer to your question didn't obviously follow for me. |
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Jan 10 |
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Tensor product of quivers Another tensor product that has been somewhat studied is the pointwise tensor product, see Ryan Kinser (arXiv:0711.1135) and Martin Herschend ("Tensor products on quiver representations", Journal of Pure and Applied Algebra 212 (2008) 452-469. This is different from yours, though. In Martin's paper he also discusses some other notions of tensor product (but I didn't read carefully). Also, I don't understand exactly what you've written: for e in E and e' in E', what is s(e\otimes e')? |
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Jan 9 |
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Why is there a unique increasing maximal path in any Bruhat interval under any reflection order? I agree that if W is not finite, the argument I gave potentially runs into trouble, since nothing guarantees that the process I described will terminate (and thus you might never obtain an increasing path). However, I think an argument like the following should work: among all reflections t such that u < vt <. v (where <. means "is covered by") take the first possible t (wrt the reflection order). Keep doing this. I bet if this wasn't increasing, the existence of an earlier rank 2 replacement would give a contradiction. |
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Jan 3 |
revised |
Why is there a unique increasing maximal path in any Bruhat interval under any reflection order? Edited to add argument for uniqueness |
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Jan 2 |
answered | Why is there a unique increasing maximal path in any Bruhat interval under any reflection order? |
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Dec 19 |
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Reference/ elementary proof of a result about projective dimension in group rings Thanks. So the equivalent statement is that a kG-module M admits a resolution of that form iff M is projective as a kG-module. Is that right? Further up in the question, you say that O(M) will be 1 or infinity. I'm confused. I would expect the projective dimension of a projective to be zero, so I would expect the projective dimension of O(M) to be zero or one. Can you clarify this? I should apologize in advance that, even once I understand the question, I may well not be able to help a lot in solving it. |
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Dec 17 |
accepted | Why is the representation dimension of an Artin algebra never equal to 1? |
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Dec 17 |
accepted | Representation dimension of a special algebra |
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Dec 17 |
answered | Why is the representation dimension of an Artin algebra never equal to 1? |
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Dec 16 |
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Proving indecomposability of special modules I think the best approach is staring at the module and convincing yourself it can't be written non-trivially as a direct sum. |
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Dec 16 |
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Representation dimension of a special algebra I think mainly what's going on is that a vector space of morphisms is implicitly being replaced with a suitable basis for the vector space. One basis element is the identity map, and the other basis elements (i.e., all the basis elements "except for" the first one) are chosen so that they do factor as desired. |
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Dec 15 |
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Reference/ elementary proof of a result about projective dimension in group rings I do not understand your equivalent reformulation. What are the morphisms in the sequence P -> P -> M? Or what do we know about this sequence? We must know something (which is not clear to me from what you've written) if there is to be any hope that a property exclusively of M will tell us whether or not the sequence is exact. Also, are all the P's in that paragraph isomorphic? |
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Dec 15 |
answered | Representation dimension of a special algebra |
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Dec 14 |
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The BCH series in terms of Lyndon words Let me mention that the Lyndon words of length n obtained by prepending an x, form a basis of the subspace of FL_n of the form [x,FL_{n-1}]. This seems as if it could be helpful for showing that only terms of this form appear in the n-even case. |
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Dec 14 |
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The BCH series in terms of Lyndon words Okay, I can establish what is probably the least interesting of my observations. It's straightforward to check that the terms in BCH of the form yx^n are exactly yx/(e^x-1). Now x/(e^x-1) is a familiar power series, and its only odd-degree term is -x/2 (subtract this off and what is left is an even function). This shows that the Lyndon word <x^{n-1}y> doesn't appear for even n at least 4. (It also points out that my statement wasn't quite correct for n=2: <xy> does appear.) |
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Dec 12 |
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The BCH series in terms of Lyndon words It looks to me (from the data in the linked pdf) like there are two different perhaps not-so-related things that are happening: at n odd, BCH gets all possible Lyndon words except one of them when n=6k+3 (k>0), in which case it misses x^{4k+1}yx^{2k}y, and at n=2k, the Lyndon words that appear in BCH are exactly those obtainable from the Lyndon words of length 2k-1 by prepending x, except that x^{2k-1}y does not appear. I have no explanations, though. |
