By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$\langle a, y\mid y^2, y^{-1}a^2=a^{-2}y\rangle,$$
where $y=ab$. This presentation is equivalent to
$$\langle a, y\mid y^2, y^{-1}a^2y=a^{-2}\rangle,$$
which defines the gr...
By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$\langle a, y\mid y^2, y^{-1}a^2=a^{-2}y\rangle,$$
where $y=ab$. This presentation is equivalent to
$$\langle a, y\mid y^2, y^{-1}a^2y=a^{-2}\rangle,\tag{P}$$
which defines...
@user330477: Holomorphic functions are always locally bounded unless they have a (non-removable) singularity. But we're talking about a neighborhood of a zero, not a singularity.
@AkivaWeinberger The 2-color pics reminds me of those molecular dynamics simulation results of my school's research groups where two different liquids get segregated into a manner similar to a 3D version of that. I wonder if the mathematics of Truchet tiles will be useful to further characterise those intermolecular interactions....
Aha. The function $f$ couldn't have a pole or essential singularity (or else $f^8$ would too). But there could be branch points. Continuity on the region rules that out.
How do we find $$S=\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$$
I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is the Euler–Mascheroni constant.
But I could not manipulate this ...
If $f(z)$ is an entire function which is not a polynomial, then what kind of singularity can it have at $\infty$? The hint at the back of Gamelin says that $f(z)$ is infinitely many negative powers of $\frac{1}{z}$, so the singularity is essential. I think this is incomplete as it ignores the case that $f(z)$ may some positive powers of $\frac{1}{z}$, even if it is not a polynomial.
@MikeMiller @Ted Couldn't sleep, so posted an answer anyway
Thanks for the helpful discussion. I hadn't previously thought about the fact that $H^1_{dR}(G) \cong (\mathfrak{g}^{ab})^{*}$ (I didn't do it in full generality in the answer but still)
Does anybody know how on page $8$, the author is getting that on $C_2$ the integral is equal to $e^{i2\pi/3}I$? Why $e^{i2\pi/3}$? I'm not following https://math.mit.edu/~jorloff/18.04/notes/topic9.pdf
Let $A \subseteq X$; suppose $r : X \to A$ is a continuous map such that $r(a) = a$ for each $a \in A$. If $a_0 \in A$, show that $r_* : \pi_1(X,a_0) \to \pi_1(A,a_0)$ is surjective...This seems easy. Given $[f] \in \pi_1(A,a_0)$, wouldn't $r_*$ send $[ \iota * f]$ to $[f]$, where $\iota : A \to X$ is the inclusion map? QED?
I was at least able to follow what the speaker was saying, though there were obviously a lot of points where I had to trust what they were claiming made sense
"Recall that the totally nonnegative Grassmannian Gr≥0(k, N) is the subset of the real Grassmannian Gr(k, N) where all Plucker coordinates are nonnegative"
Ugh that must be a weird real semialgebraic subvariety of the Grassmannian
If I have a topological manifold $X$ and an embedding $f\colon X\to\Bbb R^n$ what can I say about $\dim_H(f(X))$ where $\dim_H$ denotes the Hausdorff dimension?
There's a Jordan curve passign through any bounded and totally disconnected subset of $\Bbb R^2$ so I suspect Hausdorff dimension of an embedded curve can already be very terrible
The reason why I feel like nothing should happen for submanifolds is
that fractals always seem to rely on ideas of "infinitely pointy" which would make it hard to find little open balls around points such that the intersection with the submanifold is homeo to a Euclidean space
Well, remember that by Jordan-Schoenflies theorem there's a homeomorphism of $\Bbb R^2$ taking any Jordan curve to the circle. So if I have a Jordan curve $C$ in $\Bbb R^2$ and a point $p$ on it there's a ball $B$ around $p$ such that $(B, B \cap C)$ is homeomorphic as pairs to $(\Bbb R^2, \Bbb R^1)$
@Danu What is the precise question? Let $C$ be the Osgood curve and $p$ a point on it. I am claiming there is a ball $B_\delta(p)$ around $p$ in $\Bbb R^2$ such that there's a homeomorphism $B_\delta(p) \to \Bbb R^2$ taking $B_\delta(p) \cap C$ to $\Bbb R^1$
In the sense that topological submanifolds need not have normal neighborhoods as in the smooth case; i.e. don't have product neighborhoods even locally
@Danu An example of a topological submanifold which does not have normal neighborhood, by the way, is the Alexander horned sphere in $\Bbb R^3$. At the "bad points" there is no ball $B$ such that $(B, B \cap \text{Horned Sphere})$ is homeomorphic to $(\Bbb R^3, \Bbb R^2)$.
I was mostly just confused because normal neighborhood means something slightly different in Riemannian geometry (geodesics and stuff, though I guess it's closely connected)
In general if $M^k$ is a topological submanifold of $\Bbb R^n$ such that at a point $p \in M^k$ there is a ball $B$ such that $(B, B \cap M) \cong (\Bbb R^n, \Bbb R^k)$, you'd envision that has having sort of "topological normals" to $M$ around a neighborhood of $p$ (pull the $\Bbb R^{n-k}$ copies perpendicular to $\Bbb R^k \subset \Bbb R^n$ back)
But a-priori there is no reason to have those, and indeed you don't. In 2D miraculously you do
I have ecnountered the following statement: If $T$ is a compact operator on a Hilbert space, then for $\lambda\in \Bbb C$, $T-\lambda\operatorname{id}$ is either invertible or has non-trivial kernel.
Am I correct to rephrase this as: The spectrum of a compact operator consists only of eigenvalues?
The bullshit thing that happens is even though the Alexander gored ball is not homeomorphic to $B^3$, if you double it (glue two copies of the gored ball along their horned boundaries), you get $S^3$. This was proved by Bing, the madman
Speaking of, I was reading Gromov's Abel prize interview yesterday. He just says everything he discovered, except perhaps pseudoholomorphic curves, is obvious
@AlessandroCodenotti if you are summoned by func ana I saw an interesting thing today: C(X) is a Banach algebra when X is compact va the sup-norm. a) is there another norm on C(X) that makes it into a Banach algebra? (Easy) b) is there another norm on C(X) that makes it into a normed algebra? (Hard)
@anakhro dimension of local killing fields cannot decrease locally, rank of maps into vector space cannot increase locally are some statements ive just read
I mean, the statement is that for any point $x$ you've got a neighbourhood so that $\dim kill^{loc}(x)≤\dim kill^{loc}(y)$ for all $y$ in the nieghbourhood is what the statement means
if I have a path between $x$ and $y$ so that the rank changes, suppose the starting point has higher rank. Now choose points along the path and open sets containg those points so that the rank on the set is always greater than the rank on the associated point.
do it with finitely many and why dont i have a contradiction here
no, we're looking at a map on some space to the space of matrices, the map is continuous so the the rank cannot decrease locally
if we cover the path with finitely many neighbourhoods so that the rank does not decrease relative to the "based point" of the neighbourhood, we could jump from base point to base point along the path, rank never decreasing, so the rank shouldnt decrease on the entire path
ah ok, nvm. There is no way to ensure that the next base point is contained in the neighbourhood of hte previous one, just that the two neighbourhoods intersect
That's not what the statement means. Let $\Sigma_{< k+1}$ denote the space of matrices of rank $< k+1$ and $\Sigma_{<k}$ the space of matrices of rank $<k$. You can easily join a point of $\Sigma_{< k}$ to an interior point of $\Sigma_{< k+1}$ by a continuous path.
It means for every point $p$ in the interior of $\Sigma_{< k+1}$ there is some ball $B$ around $p$ that lies entirely inside $\Sigma_{< k+1}$.
Even if $p$ is very very very close to $\Sigma_{< k}$, you can find a very very very small ball around it that fits in $\Sigma_{< k}$.
I was confused about this a long while ago; I'd say the point is, if you use the terminology in this answer, "stratums of $M_{m \times n}(\Bbb R)$ are open bounded submanifolds"
Hmm. Suppose I have three non-collinear points in R^2. Then the convex hull of these points is a triangle, with every point on the triangle corresponding to a unique convex combination of the vertices.
By contrast, if I chose at least four such vertices in R^2 (all of which are assumed to be non-collinear) then the resulting hull is still convex but now the points in the interior can be written as infinitely many such combinations.
Trying to see how to classify that extra freedom in that case...
@Semiclassical a convex combination of $n$-points can be viewed as a point on an $n-1$ dimensional simplex. As such if you have the convex hull $K$ of $n$ points in $\Bbb R^d$ and you want to see the degeneracy of how many combinations describe the same point, what you can do is look at the subset of $K\times \Delta$ given by $\{ (x, (t_1,..,t_n)) \mid \sum_n t_n x_n = x\}$
at each point you've got as a fibre basically this degeneracy, and you can look at quantifiers like the dimension or the volume of this fiber to describe how degenerate one point is
for example with your triangle every fiber is just a point, if you are looking at a square then you've got as a fiber a line over every point except the center point
@SharathZotis You see that from the beginning by not thinking about elements at all. There are 2 disjoint right cosets which partition $G$, and 2 disjoint left cosets which partition $G$, right?
But then both of them has to be $G - H$, because if you have a partition of $G$ into two disjoint sets, and one of them is $H$, the other has to be $G - H$.
@Daminark Ok, here is a question for you. Let X and Y be measure spaces. Is the natural map $L^\infty(X \times Y) \to L^\infty(X, L^\infty(Y))$ an isomorphism?
By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$\langle a, y\mid y^2, y^{-1}a^2=a^{-2}y\rangle,$$
where $y=ab$. This presentation is equivalent to
$$\langle a, y\mid y^2, y^{-1}a^2y=a^{-2}\rangle,$$
which defines the gr...
How do we find $$S=\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$$
I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is the Euler–Mascheroni constant.
But I could not manipulate this ...
