@EricSilva Consider $M$ a compact Riemannian manifold, and take $p\in M$, $\delta>0$ small. Consider $B_\delta(p)$. Given $\varepsilon>0$, is there a diffeomorphism $\phi$ of $M$ such that $\mu(M\setminus \phi(B_\delta(p)))<\varepsilon$?
That being said, the basic idea is that you want to write $$\frac{-24}{(x+5)(x-3)} = \frac{p(x)}{x+5} + \frac{q(x)}{x-3},$$ where $p$ and $q$ are polynomials
and the functions on the right-hand side are "proper fractions"
which means that $p$ and $q$ are each of degree 1 (in this case)
the fact that the constant on the left is $-24$ has no bearing on how you choose to represent the constants on the right---I mean, if you wanted, you could write $-A$ and $-B$ in the numerators, but then you have extra negative signs running around that you will have to deal with later
again, I am looking only at the expression $$\frac{-24}{(x+5)(x-3)}.$$ This is an expression that can be decomposed into partial fractions. Do that, then plug it back into the more complicated expression that you are interested in.
@PVAL-inactive Is there an easy formula for $H_1 \Sigma(p,q,r)$ when $p,q,r$ are not necessarily pairwise coprime? It seems like I need to compute Seifert invariants and then take the determinant of the resulting matrix, which sucks. In particular $H_1 \Sigma(2,4,5)$ seems to be $\Bbb Z/5$ which I for whatever reason did not expect.
@MikeMiller At some point I wrote a mathematica script which gave the intersection form of M(2,a,b) where M(2,a,b) is the smoothing of the (2,a,b) Milnor fiber. I think it's pretty easy to generalize to M(p,q,r), but I don't think I ever made that. There should be a way to compute H_1 directly, but I don't recall one.
I was particularly interested in computing the htpy type of the plane field coming from the canonical contact structure on \Sigma(p,q,r). That's why I did all this.
There's better formulas for the signature without presenting the matrix apparently, but I didn't know them at the time.
I see. I wanted some examples with nontrivial homology and an irreducible $SU(2)$-representation to run an instanton spectral sequence on but that's probably too hard in any single example and there are too many irreducible representations.
This probably isn't the type of solution you are looking for, but If you think about it, the circle in the corner satisfies the parametric equation $r \cos(\theta) + r, r \sin(\theta) + r$ if the bottom left of the triangle is the origin. Because you know the diagonal you know one point on this circle, which gives you two equations and two unknowns. You could solve it that way.
Hello! I'd like somebody to hint at my mistakes. I rewrote the first expression as $a_03^0+a_13^1+a_23^2+a_33^3$ but I can't figure out what $a_0$ means here, I supposed tht it is -1, then added them up like this: $a_03^0+a_13^1+a_23^2+a_33^3=62$. But in the method of solution of the problem there is written: 'Note that: $46=1201_3$'. What does the index 3 mean here?
Oh god yeah, but they're rather unrelated objects. $K(X)$ is defined using vector bundles on a space $X$, which can often be determined in terms of cohomological information like characteristic classes, and is a cohomology theory itself so can be computed using cut-and-paste operations via Mayer-Vietoris. Then $K^{-1}(X)$ is defined as $K(\Sigma X)$, and all other values of $K^{-n}$ are determined by these.
$K_n(R)$ in algebraic K-theory is $\pi_n\left(BGL(R)^+\right)$, the homotopy groups of a complicated space arising from the classifying space of the group of automorphisms of $R^\infty$ (more or less) by a procedure called the Quillen plus construction which kills its fundamental group but doesn't change its homology. On the other hand, it wildly changes homotopy groups.
Homotopy groups of spaces are hard to calculate, period. There are methods nowadays (that I don't know much at all about) to calculate algberaic K-theory but they're much more complicated than asking to do the sort of cohomological things you need for $K(X)$.
the calculation of $K_n(\Bbb F_q)$, the algebraic K-theory groups of a finite field, was a 40-page paper by Quillen; $K_n(\Bbb Z)$ will probably always be an open question
Ah, I can't right now and I don't think I have the time to commit to it. But feel free to send me a copy when it's finished and I can leisurely find time to read it :)
Two people, $P$ and $Q$, decide to independently roll two identical dice, each with $6$ faces, numbered $1$ to $6$. The person with the lower number wins, In case of a tie, the roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by $P$ and $Q$. Assume that all...
Please don't confuse modular arithmetic with division by zero, which has a very precise meaning
> Macieks300 1 week ago These kind of videos that are year 1 level are what you should've started with. Work your way towards the more complex stuff first. Also, the title is a little misleading; equivalence class of integers congruent to 0 is not really something most people picture when thinking about division by 0.
$(\sqrt{2} - 1)(\sqrt{2} +1) = 1$. Does the notion of conjugation exists for field extensions?
Other thoughts:
$$\begin{matrix} \cdot & a & b & c \\ a & a^2 & ab & ac \\ b & ba & b^2 & bc \\ c & ca & cb & c^1 \end{matrix}$$
off diagonal elements are entisymmetric if $xy=-yx$ for all $x,y \in \{a,b,c\}$
so that gives 3 equations that relates $a,b,c$
$ab=-ba$ $bc=-cb$ $ac=-ca$
While that basically means the anticommutator of all pairs must vanish, this seemed to be not efficient enough to determine whether given a polynomial that is squared, only the diagonal terms survive
However, if this is a field, then we quickly obtain the conditions: $ab=bc=ac=0$ which basically means one cannot have off diagonal terms in a multiplication all antisymmetric for a field (since a field has no zero divisors)
I am not sure however how to generalise the observation of $(a+b)(a-b) = a^2-b^2$ for all alternating polynomials with n terms, though
I need to figure out a good way on determining the condition when two polynomials of degree n are multiplied together, where the cross terms will be cancelled without computing all $n^2$ terms in the multiplication
When $P(0)=0$ there is no constant term. Therefore, the polynomial must be of the form $P(x)=xQ(x)$. Now plug $x \in \Bbb{Z}$, therefore there must be a factor $n!$ and hence the proposition is true for $x=0$
Let $P(x)=A(x)+B(x)$ where $A(x)$ are odd powers of $x$ and $B(x)$ are even powers of $x$. Then $P(-x)=A(-x)+B(-x)=-A(x)+B(x)$. When $P(-x)=0$, $A(x)=B(x)$ for that particular $x$. therefore...
Here's a proof that algebraists can draw pictures, too. Question: Does every Noetherian local commutative reduced ring have a unique minimal prime ideal? Answer: No, consider:
Why do we have
$$ \ln(z) = \frac{z-1}{e -1} \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {z^{2^{-n}} + 1} $$
Is it possible to show that the derivative of the product is $z^{-1}$ without showing that the product is the logarithm ?
Does that imply there exists a solution set $a_n$ such that :...
This goddamn language… so apparently 消す (to put out, to extinguish) is pronounced "kesu", but 消える (to disappear) is pronounced "kieru". The first is the transitive version of the second, but in the first one, 消 is "ke", while in the second one, 消 is "ki".
(Japanese)
The kanji represent ideas, not sounds... but still
(す, え, and る are kana, meaning that they represent syllables and not ideas.)
Saying something like "Fermat's results were published posthumously by his son Samuel de Fermat in an appendix to a restored edition of Arithmetica" would mean that Samuel published Fermat's results after Fermat's death, right?
I suppose if $\Omega'$ is not measurable, you can still construct an outer measure on $2^\Omega$ from $\mu$, restrict that to $2^{\Omega'}$ and then take the subsets of $\Omega'$ which satisfy the Carathéodory condition as your sigma-algebra
Terminology question, does the term vector field only apply to the tangent bundle, or would you also refer to a section of any vector bundle on a manifold as a vector field?
But a vector field itself is a section of the tangent sheaf. Why in the smooth affine variety setting is a vector field given by $A:\mathcal{O}_X\to \mathcal{O}_X$ such that $A(fg)=A(f)g+fA(g)$?
Or in the analytic setting, why is it $C^\infty(M)\to C^\infty(M)$ with that property, rather than a section $s:M\to TM$
I imagine this is induced in algebraic and analytic setting (or equivalent?)
Two people, $P$ and $Q$, decide to independently roll two identical dice, each with $6$ faces, numbered $1$ to $6$. The person with the lower number wins, In case of a tie, the roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by $P$ and $Q$. Assume that all...
Why do we have
$$ \ln(z) = \frac{z-1}{e -1} \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {z^{2^{-n}} + 1} $$
Is it possible to show that the derivative of the product is $z^{-1}$ without showing that the product is the logarithm ?
Does that imply there exists a solution set $a_n$ such that :...
