How in the world does one even begin to solve for $k$ when given $\int_k^1 \left(\sqrt{1-x^2}-\frac{1}{3}\right) dx=\int_{-k}^k \left(\sqrt{1-x^2}-\frac{1}{3}\right) dx$ ?
I mean I know the solution, I just don't remember how to do an integral with a variable in the boundary things.
Oh yeah, it was because I wanted to cut a circular cake into 9 equally sized slices using 2 horizontal and 2 vertical cuts, with each cut being equidistant from the center of the cake.
@LeakyNun @loch If $\pi_1(B)$ doesn't act trivially on $H(F)$ you still have a spectral sequence but now with $E^2_{p, q} = H_p(B; \underline{H_q(F)})$ where $\underline{H(F)}$ is the local system on $B$ coming from the representation $\pi_1(B) \to \text{Aut}(H(F))$.
@LeakyNun Which is alternately degree $2$ and degree $0$. For example, the 2-cell of $\Bbb{RP}^2$ is attached to 1-skeleton/0-skeleton = $S^1$ by a degree 2 map $S^1 \to S^1$. The 3-cell of $\Bbb{RP}^3$ is attached to 2-skeleton/1-skeleton = $S^2$ by a degree $0$ map $S^2 \to S^2$.
It's tricky to see that it has degree 0. The 3-cell is attached to the 2-skeleton by the double cover $S^2 \to \Bbb{RP}^2$. So that might lead you to think $S^2 \to \Bbb{RP}^2 \to \Bbb{RP}^2/\Bbb{RP}^1 = S^2$ is degree 2, but $\Bbb{RP}^2$ is nonorientable so degree doesn't work that way. Think instead of the map $S^2 \setminus S^1 \to \Bbb{RP}^2 \setminus \Bbb{RP}^1$ by throwing away the equator. This is a map $D^2 \sqcup D^2 \to D^2$ which is a homeomorphism on each factor.
The first factor is degree $1$, the second is degree $-1$! One is obtained from another by the antipodal map.
So $1 + (-1) = 0$ is the degree.
This in particular implies $\Bbb{RP}^3/\Bbb{RP}^1 \cong S^3 \vee S^2$ if I'm not mistaken.
In general $\Bbb{RP}^n/\Bbb{RP}^m$ are called stunted projective spaces.
@MikeMiller Are the homotopy types of these spaces completely understood?
Let $H$ be a separable Hilbert space and let $K(H)$ be the algebra of compact operators on $H$. Is there an easy way to see that $K(H)$ has no nontrivial characters (*-morphisms into $\Bbb C$)? I can show this but taking a huge detour
Because it's an algebra homomorphism, so in particular linear, I suppose
So it is determined on the subset of diagonalizable operators in K(H), which means it's determined upto real scalars by the image of diagonal operators...
you would have to explain more carefully why that's a proof, normally I expect to see a more complicated diagram showing how they fit together
but yeah pretty much that, the proof should be about exactly as enlightening as the definition
if this proof is unenlightening then that probably means you're unsatisfied with the definition of the cellular chain complex using differential given by the composite of those two maps
another argument is that $C_*(X)$ is a filtered chain complex, with $F_k C_*(X) = C_*(X^k)$. Filtrations give spectral sequences, and the cellular chain complex is the $E^2$ page of this spectral sequence, and the $E^2$ page has its own differential; because the $E^2$ page here is concentrated in a single line, there are no further differentials
If I know that $\Lambda_1, \Lambda_2 \subset A$ are two orders in a $\mathbb{Q}$-algebra $A$ with $\Lambda_1 \subset \Lambda_2$, does it follow that $\Lambda_1^\times$ has finite index in $\Lambda_2^\times$?
Also does the same hold for, say, the subgroup $SL_n(\Lambda_1)$ of $SL_n(\Lambda_2)$?
To be precise the snake map $\partial : H_n(X, A) \to H_{n-1}(A)$ comes from taking a relative class $[\xi]$ and sending it to $[\partial \xi]$ since by definition $\xi \in Z_n(X, A)$ means $\partial \xi \in Z_{n-1}(A)$
In light of this, interpret the cellular boundary maps and why they are "obviously" the right maps
So I have an upcoming course on Global Analysis, in the winter semester. The book that they are going to follow is S Ramanan's Global Calculus. Considering I have functional analysis now and that I am attending this school on Operator theory and NC Geometry, what else is necessary for Global Analysis?
I'm looking for an example of a bounded operator on an Hilbert space that has a closed invariant subspace $\mathcal N$ such that $\mathcal N^\perp$ is not invariant, any idea?
@AlessandroCodenotti let your Hilbert space be $\Bbb R^2$ and your operator be $\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$ and your invariant subspace be $\langle (1,0)\rangle$
Just compute the cellular homology with $\Bbb Z/2$ coefficients. You get $\cdots \to \Bbb Z/2 \to \Bbb Z/2 \to \Bbb Z/2$ with all zero homomorphisms between them
@LeakyNun Learn what goes wrong when you do universal coefficient with your coefficient ring not a PID and then teach it to me. The point should be that if $M$ is an $R$-module and $R$ is a PID then there's a 3-term resolution $0 \to F_1 \to F_0 \to M \to 0$ so the whole abelian group proof pushes through. There's a universal coefficient spectral sequence which says there's a spectral sequence with $E^2_{p, q} = \text{Tor}^q_R(H_p(X;R),M)$ that converges to $H_*(X;M)$ or some such.
Namely, the higher Tor terms won't vanish
I guess it makes sense. If you have a $R$-module chain complex $C_*$ and you take a projective resolution of $C*$, tensor with $M$, that's a double complex. You filter the totalization in the obvious fashion and take it's cohomology. The $E^1$ page takes cohomology in vertical direction giving rise to resolutions for $H_p(X; R)$ tensored with $M$. Then the $E^2$ page takes cohomology in the horizontal direction giving rise to $\text{Tor}^q_R(H_p(X; R), M)$.
Simple enough.
I don't suppose this spectral sequence is computable without effort in general. Who knows what the higher differentials do
If $\gamma : [a.b] \to \Bbb{C}$ is a function of bounded variation and $f : [a,b] \to \Bbb{C}$ is continuous, how does one define $\int_{a}^{b} f(t) d \gamma(t)$? Is it defined as $\sup \{\sum_{k=1}^m f(\tau_k)(\gamma (t_k)-\gamma(t_{k-1})) \mid \text {etc.} \}$?
In mathematics, the Riemann–Stieltjes integral is a generalization of the Riemann integral, named after Bernhard Riemann and Thomas Joannes Stieltjes. The definition of this integral was first published in 1894 by Stieltjes. It serves as an instructive and useful precursor of the Lebesgue integral, and an invaluable tool in unifying equivalent forms of statistical theorems that apply to discrete and continuous probability.
== Formal definition ==
The Riemann–Stieltjes integral of a real-valued function
f
{\displaystyle f}
of a real variable with respect...
@BalarkaSen Hatcher says that if $0 \to A \to B \to C \to 0$ is exact then there is a corresponding fibration $K(A,1) \to K(B,1) \to K(C,1)$
Take any contractible space $X$ on which $\Bbb Z$ acts freely properly discontinuouslyand any contractible space $Y$ on which $\Bbb Z_2$ acts freely properly discontinuously. Consider $(X \times Y)/\Bbb Z$ where $\Bbb Z$ acts on the first factor usually and on the second factor by factoring through $\Bbb Z \to \Bbb Z_2$.
There's a map $(X \times Y)/\Bbb Z \to Y/\Bbb Z = Y/\Bbb Z_2$ by forgetting about the first factor.
$(X \times Y)/\Bbb Z$ is weak homotopy equivalent to $S^1$ and $Y/\Bbb Z_2$ is weak homotopy equivalent to $\Bbb{RP}^\infty$ :)
"Weak" can be removed if the spaces in question are CW complexes and the action is cellular etc
So for an explicit example, consider $(\Bbb R \times S^\infty)/\Bbb Z \to S^\infty/\Bbb Z_2$.
This is your homotopy fibration map $S^1 \to \Bbb{RP}^\infty$
@LeakyNun $S^1 \to \Bbb{RP}^1 \subset \Bbb{RP}^\infty$ is not a fibration, but it can be made into one. See this.
Meh nevermind the last comment. It is a double cover, so whatever.
Anyway the point is you shouldn't think of $K(A, 1)$ simply by it's homotopy type. They are by construction massive spaces. They just shrink a lot sometimes.
OK, no, my comment was correct, actually. If you take $S^1 \to \Bbb{RP}^\infty$ by just naively sending $S^1$ to the 1-skeleton $\Bbb{RP}^1$, then make it a homotopy fiber sequence using the wikipedia page I linked, it's homotopy fiber is going to be $S^1$.
Because suppose it is $F$. Then you have a homotopy fiber sequence $F \to S^1 \to \Bbb{RP}^\infty$, where the middle $S^1$ is not really $S^1$ but some homotopy-enlarged space.
The map $S^1 \to \Bbb{RP}^\infty$ induces $\Bbb Z \to \Bbb Z_2$ in $\pi_1$. So by running the homotopy fibration sequence, you get $1 \to \pi_1 F \to \Bbb Z \to \Bbb Z_2 \to 0$ and $\pi_n F \cong 0$ for all $n \geq 2$.
That means $\pi_1 F = \Bbb Z$ and $F$ is a $K(\Bbb Z, 1)$ space. A circle.
write $S^\infty$ for the unit sphere in $\Bbb C^\infty$, and let $S^1$ act on itself with weight two; it is convenient to rename this second circle, maybe call it $C$. then the projection of your fiber sequence can be modeled as $$S^\infty \times C \to (S^\infty \times C)/S^1 = \Bbb{RP}^\infty$$