By that technique, you get a feel for the (true) fact that the genus of "the" Siefert surface of the trefoil is 1. That's an invariant of knots in general (genus of the trivial knot is 0 so now you know these two knots are inequivalent: the trefoil cannot be unknotted).
A simple closed disconnecting curve on a torus gives a connected sum description of the torus, hence there's a disc on one side, hence it's the unknot.
@TedShifrin oh also, i'm writing a little animation thing that shows how the intersection of two constantly rotating lines forms a conic section in $\Bbb R^2$
nice, that looks much better than my drawings :P so if I remove a trefoil from the torus and give some width to the removed part I get a surface whose boundary looks like 2 trefoils
@trilolil I don't think Ted is asking for that information necessarily, I think he is asking what you're trying to learn about the system $H$ represents. At least, I'm fuzzy about what you're asking...
I have no idea, @trilolil. You haven't given me any information except that you told me the $1.12$ came from the Pythagorean Theorem with $\sqrt{1^2+(1/2)^2}$.
@TedShifrin Well by converting the imaginary numbers j and 0.5 to polar coordinates one can construct a triangle and use the pythagorean theorem to calculate the hyppothenusa which here equals 1.12.
However I would like to know whether there is another way which doesn t depend on geomettry and conversions to polar coordinates in order to get the value 1.12 (which was the the length of the hypothenusa)
I decided I could do with some random background reading regarding fiber bundles, so I looked into some books. I found that Novikov wrote three books on geometry/topology, and that the second half of the last one looks quite interesting. So now I'm reading a bit more about obstruction theory, building towards the gneeral concept of cohomology operations
But the paper McCrory and I wrote on cusps of the projective Gauss map, we spent pages proving carefully the sort of "genericity" claims that Arnol'd just asserted with no justification whatsoever.
There's at least a few math books I have that are written with a typewriter font (which I'm pretty sure means "written with a typewriter".) They're hard to read, especially the formulae
i don't find those hard to read at all, other than sometimes I can't tell that some symbol is $\mathscr N$ or whatever (but I can still identify the symbol as "it's that symbolf rom the previous page")
When I wrote my Ph.D., I hired the math secretary to type the official copy. She used an IBM Selectric with a symbol ball that had to go in every time there was a symbol.
@Alessandro: Regardless, it's just a 2x2 determinant.
@MikeMiller Understanding individual symbols is not so much the problem---the flow of reading is just continuously disrupted by bad spacing/slightly weird looking sequences of symbols.
There was a thing Knuth wrote that I read a while ago (about an alternate approach to teaching calculus) that I couldn't find a PDF of, just the LaTeX code
Let X be a normed space. Let $B(X)$ denote the set of bounded linear operators from X to itself. Let $K(X)$ denote the set of compact operators from X to itself.
Suppose that $dim(X) < \infty$ then $B_x$ is compact(the open ball centered at x. Let $T \in B(X)$ then it is continuos. So, $T(B_x)$ is compact as it is the continous image of compact set. Compact sets in haussdorff space are closed so $Clos(T(B_x)) = T(B_x)$ so $T \in K(X)$.
@MikeMiller Suppose that $K(X) = B(X)$ WTS $dim(X) < \infty$. This is equivalent to proving that given $B_x$ in X it is compact. Suppose $T \in B(X)$ since $T \in K(X)$, so this means that $closure(T(B_x))$ is compact. Since T is continous so the preimage is compact but $B_x$ is contained in the pre-image. Since it is closed subset of the preimage so $B_x$ is compact.
@MikeMiller so it is exactly as you said. Suppose that X is infinite dimensional then we will prove that $K(X)$ is proper subset of $B(X)$. Take the identity operator and let it act on $B_x$ the ball centered at x of some radius. Then the closed ball can't be compact as X is infinite dimensional.
