Eivind Dahl
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 Apr21 revised Closure of $A= (0,1) \cup (1,2)$ vs. Closure of $A = [0,1] \cup \{2\}$ added 11 characters in body Apr21 answered Closure of $A= (0,1) \cup (1,2)$ vs. Closure of $A = [0,1] \cup \{2\}$ Oct14 revised Guide to mathematical physics? added 368 characters in body Oct14 revised topological properties of a given set added 199 characters in body Oct14 answered topological properties of a given set Oct8 comment Guide to mathematical physics? I couldn't really say. I mean, if you're interested in 1 dimensional field theory as described in Faria-Melo one could say that you don't really ever need to say "category," it's just a convenient way to think. Though if you're interested in d dimensional field theories, you should probably know about d (or at least d-1) categories. I couldn't tell you what mathematical physicists usually do because I'm not one of them. Maybe I'll be able to send one your way but I can't promise anything, heh. Oct7 comment Guide to mathematical physics? Yeah well, principal bundles with their connections, vector bundles, lie groups and their representations, operator algebras for infinite dimensional stuff, Lagrangian and Hamiltonian mechanics for the dynamics of states (or the unifying frame work of symplectic geometry), and as always category theory. Cohomology becomes useful for understanding existence of spin structures etc. I'll let someone with more experience write a more comprehensive answer :) Oct7 answered Guide to mathematical physics? Oct7 comment Guide to mathematical physics? My favourite book so far on gauge theory is "Mathematical Aspects of Quantum Field Theory" by Faria-Melo -- you might find it useful. Oct5 comment In how many ways can you arrange all letters in the word MISSISSIPPI so that For 1, the symmetric group on 11 letters acts on the set of arrangements of these 11 letters. The stabilizer of any four letters is a copy of the symmetric group on 7 letters. A consecutive arrangement of four I's is determined by the location of the first I (there are 8 possible choices) and by a permutation on 4 letters. Hence there should be about $(11!/7!)\cdot 8\cdot 4!=1520640$ such arrangments. This is assuming two copies of a single letter are different; if not, scrap the copy of $\Sigma_4$ and some of the stuff in $\Sigma_7$. Oct3 comment Why is one “$\infty$” number enough for complex numbers? By the way, you could consider a closed 2-dimensional disc a compactification of $\mathbf{C}$ with a full circle worth of infinities. Oct3 revised Why is one “$\infty$” number enough for complex numbers? deleted 1 character in body Sep30 comment A smooth non-stably trivial smooth vector bundle Yes so it would appear. I have no idea why I had convinced myself that it didn't. Sep30 accepted A smooth non-stably trivial smooth vector bundle Sep30 asked A smooth non-stably trivial smooth vector bundle Sep24 awarded Autobiographer Jul28 awarded Yearling Jul2 awarded Curious Jul28 awarded Yearling Jul25 awarded Good Answer