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Regular. To start off: The regular representation of a group $G$ over a field $k$ is the action on $k[G]$ given by group multiplication. A topology is regular if a closed set and a point not in ...

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One of the things I like to mention, since I study topology, is the Brouwer fixed point theorem. The idea to explain is that if you pick up a piece of paper, DON'T RIP IT, but crumple it, turn it ...

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I was recently amazed at a quick connection between two facts I've known since high school. The Euler characteristic of a sphere, thought of as #vertices + #faces - #edges on a polyhedron, buckyball, ...

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I'm not sure if this is what you're looking for. But to me, a Hermitian metric is just $g+i\omega$, where $g$ is a real inner product, and $\omega$ is a symplectic form (alternating, but still non-...

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Simon's answer points out that the Euclidean algorithm shows that gcd(A,B) divides C is necessary, but the lack of large container makes the problem more difficult, because obviously you can't get $C$ ...

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I second the references to Hatcher and to Mosher &amp; Tangora, though you can also find Steenrod's original paper. At least the first two of those start out by listing the various axioms of Steenrod ...

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In addition to the above answers, This question about understanding Steenrod squares may help. The thing that Hatcher does that is different from, say, Steenrod and Epstein as Charles mentioned, is ...

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Brian Conrad has a handout (pdf) in which he talks about tensorial maps. In it he notes that one should construct maps independently of bases, but that in order to prove the properties of such maps ...

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One place for h-principles, and where PDEs come up is jet bundles. (There seems to be a MathOverflow question about PDEs and jet bundles, here). For example, consider the 1-jet space of maps $\... View answer 6 answers votes 3k views 7 votes I find it easiest to get$\mathbb RP^n$from$S^n$with two cells in every dimension. That is to say, you start with two points for$S^0$, attach two one-cells oriented opposite ways for$S^1$, ... View answer 15 answers votes 18k views 6 votes One thing I've learned recently (moving into symplectic geometry from topology) is that people often underestimate the value of regarding manifolds locally as graphs of functions, or submanifolds of ... View answer 3 answers votes 2k views 6 votes Ari's answer is good because you can see how even flowing along a symplectic vector field is not enough, but you could add a nice touch to the picture. Because a Hamiltonian diffeomorphism is exact, ... View answer 4 answers votes 612 views 5 votes While working on my thesis, I (by accident) constructed a Lagrangian Klein bottle in$(S^2\times S^2, \omega\oplus\omega)$. The construction works in$S^2\times D^2$as long as the area of$D^2$is ... View answer 37 answers votes 195k views 5 votes I think the myth of being "too old" is a stereotype threat like others that tend to push people out of mathematics at various stages, at least in my experience in the USA. But it's not just about ... View answer 13 answers votes 3k views 3 votes I think the use of calculators at an early level is a great thing. For one thing, calculators give kids a sense that math actually works, a solid thing that can be checked and thus grasped without ... View answer 3 answers votes 3k views 3 votes I think so: it looks like the local degree according to Hatcher's definition measures whether$f$preserves orientation or reverses it on the neighborhood of$x$. On page 233 he begins discussion of ... View answer 2 answers votes 578 views 2 votes I found a useful reference on Doug Ravenel's website that first explains Browder's definition and then relates it to Kervaire's. Here it is (pp. 142-143; the file's a whopping 5MB, but just because ... View answer 2 answers votes 488 views 2 votes Edit: This is meant to answer the question of why we can't have an embedding$\mathbb S^{n-1}\times I\hookrightarrow\mathbb R^n$such that the boundary is two side-by-side spheres rather than two ... View answer 2 answers votes 598 views 1 votes Here's a thought too long for a comment, but prompted by Gerhard's comment and Robert's answer: It seems to me that when$m = 3$(or$3k$), and$p_n, p_{n-1}$are both not small, we cannot have the ... View answer 1 answers vote 860 views 1 votes I was going to suggest that all the connectivity properties were either preserved or sometimes acquired by completion: e.g. a totally disconnected$X$may become a path-connected$\bar X$, and the ... View answer 3 answers votes 3k views 1 votes EDIT: The two spaces I made are not n-connected; the homotopy groups up to n are trivial EXCEPT for the fundamental group. The idea is to use the fact that a covering space for n-connected spaces ($...