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 Dec18 comment Using quaternions to represent an affine transformation? I know I read about some parts of the above on wikipedia and in Hatcher's book on Vector Bundles and K-theory (there are many other fantastic insights in that book). But I got much of my intuition from Bott and Tu's Differential Forms in Algebraic Topology. That one is a gem. Dec18 comment Using quaternions to represent an affine transformation? hm, not sure where I have the idea from. One source is definitely reading about K-theory. To classify rank $k$ vector bundles on $S^n$, you can consider maps from the equator (which is $S^{n-1}$) to $GL(k)$. This tells you how to glue the north and south portions of the bundle (each of which is a trivial bundle because disk is contractible). So you need to find the homotopy type of $GL(n)$ and it turns out it's $O(n)$ resp. $U(n)$ for reals resp. complexes (by Gramm-Schmidt). So in the end one is studying homotopy groups of classical groups. Dec18 comment Using quaternions to represent an affine transformation? thank you for your kind words. Although I'm not as fluent as I'd like to be, I'm beginning to see the light here and there :) By the way, one piece of analysis was missing: we always stay in the positive-determinant component of $GL(2)$. That's why it's enough to work with $SO(2)$. Otherwise there would be a further ambiguity of adding a reflection. Dec18 comment Using quaternions to represent an affine transformation? As $GL(2)$ has the homotopy type of $O(2)$, this is the only ambiguity we get. The remaining degrees of freedom (scaling and skewing) can always be divided in half uniquely. Dec18 comment Using quaternions to represent an affine transformation? Nice answer. As for when the method works: basically what you're doing is picking a point on the group of transformations, connecting it to the origin using the geodesic and picking a point in the middle. So the method works when the geodesic is unique (this happens almost always). Your statement that it works for rigid rotations is not quite correct. If you pick rotation by $\pi$ then there are two solution $\pi/2$ and $3\pi/2$. Your equation $(c + I_2)$ also degenerates in this case since $c=-I_2$. So there is inherent ambiguity of left vs. right. Dec18 comment Using quaternions to represent an affine transformation? Quaternions are not used to represent any 3d transformations. They are used to represent rotations. The group of rotations is $SO(3)$ and unit quaternions with quaternion multiplication form a double cover of that group. That's the sole reason they are useful. If you are doing anything other than 3d rotations (in particular arbitrary 2d transformations) chances are quaternions won't help you one bit. Dec18 comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots @Goos: we can agree to disagree. It largely depends on context and audience when you should conflate two problems into one (for example when teaching limits for the first time, you should definitely not conflate the two). I for one prefer to be explicit and don't consider making myself clear a pain. Rather I find it painful if someone doesn't take the effort in presentation and assume everybody thinks the same way they do and knows the same facts as well. If you want to continue the discussion let's move it into chat, I think we spammed this thread enough :) Dec18 comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots @Goos: if it's a standard usage then be it (I don't think it is) but I still need not like it. Same thing with that multiple of $2$. It's really two distinct questions, one for integrality and one for divisibility by $2$. Conflating them into one question doesn't really do anyone a service. You save about three words at the expense of possibly confusing anyone who doesn't follow exact same conventions you do. Dec18 comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots @Goos: there is nothing really interesting in this question as I said. One can asks dozen similar questions. But the accidental connection to Legendre polynomial pointed out in Ragib's answer is neat, so that's the only interesting thing going on here. Dec18 comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots @DanielV: thank you Captain Obvious. Since the question talks about negativity of roots it obviously assumes they are real (since there is no notion of negativity on complex roots, there being no order on complex numbers). If this is not so then the question's title is misleading and should state that OP wants to know whether the roots are real. Dec18 comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots @Goos: fair enough. But I don't think the interesting part of the question is in proving the reality of the roots. Dec18 comment Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots Minor remark: the fact that the roots must be negative is obvious since the coefficients are all positive. A second minor remark: almost all polynomials have distinct roots and so does probably yours as well. It might be very hard to prove it but unless something special is going on (like the binomial formula for $(1+x)^n$) it's really not surprising, or even interesting. You could ask similar question for a polynomial having coefficients given by any function of binomial numbers whatsoever (except the trivial one I just mentioned). Dec18 revised Aplanar covering of $S^1 \vee S^1$? Found all aplanar finite degree coverings Dec18 answered Aplanar covering of $S^1 \vee S^1$? Dec18 answered Are these functions isomorphisms of the first binary structure with the second? - Fraleigh p.34 3.13, 3.15 Dec18 comment Are these functions isomorphisms of the first binary structure with the second? - Fraleigh p.34 3.13, 3.15 You cannot suspect it, you just have to figure it out. Either find a proof or counterexample. A map can fail to be an isomorphism because of only three reason: it's not injective or not surjective or not homomorphism. So check each property in turn. Dec17 comment Aplanar covering of $S^1 \vee S^1$? Universal covering is a tree, so certainly planar. But I'm not sure about finite degree coverings. They seem intricate. Dec17 comment Irreducible and regular representation @James: there are many ways to decompose representations into the irreducibles. It's hard to help you without knowing what you know and what you're supposed to use. E.g., are you required to use pure linear algebra to block-diagonalize the matrices? Or can you use standard character theory or other tools from rep. theory? Dec14 comment Cohomology groups @Daniel, thanks for pointing this out. I was implicitly assuming finite generation. Dec13 answered How to identify surfaces of revolution