@Sophie Ah, I see. A group action of $S_3$ on $\mathbb{Z}/p\mathbb{Z}$ is the same data as a homomorphism $S_3\rightarrow\operatorname{Aut}(\mathbb{Z}/p\mathbb{Z})$, but the latter group is cyclic. A subgroup of a cyclic group is cyclic and $S_3$ isn't cyclic, so that homomorphism necessarily has non-trivial kernel. But that means there's a non-identity element of $S_3$ acting like the identity and that gives what we want (for all $a$ simultaneously, even).
Let $h$ be a smooth complex-valued function. Show that $h$ is harmonic if and only if $\partial h / \partial \overline{z}$ is conjugate-analytic.
Recall that $\displaystyle \frac{\partial f}{\partial \overline{z}} = \overline{\frac{\partial \overline{f}}{\partial z}}$ and $\displaystyle \frac{\...
@sheltonBenjamin yes P is the circumcenter because the angle of a segment of a circle seen from the center is twice that seen from another point in the circle
if literally every short exact sequence splits then every R-module is projective by definition (if your definition is every exact sequence of the form blah splits)
so for every nonzero $x$, $R \to R/(x)$ has a section that sends $1 \in R/(x)$ to say $y \in R$, then $y=1 \pmod x$ (since it is a section) and $xy=0$ (since the section is well-defined). from the latter we have $y=0$, so from the former $x$ is a unit
@Sayan quick question: need to parse a definition! Suppose I have a representation V of a group G and let K be a subgroup of G. Suppose I have a full decomposition of V into irreducible subreps. Some of those might be isomorphic to each other, so group these together into equivalence classes and sum together the subspaces of V fixed by the action of K wrt these equivalence classes. The sum of those guys is called the K-isotypic decomposition of V, right?
In the finite dimensional case all you're doing here is taking the sum of all $K$ submodules right? And then you're original representation decomposes into these sums
I am wondering if anyone here is familiar with the setup of the Leray spectral sequence. I am having problems with the differentials because I am getting that many of them are always zero and that ought to be wrong...
the problem basically boils down to the following: I have a map f:X-->Y and a sheaf S on X, which I push forward to Y and resolve with injective sheaves f_S --> I^.
Sorry, f_* S --> I* maybe better notation here in chat.
Ok, cool. So I forget how the story is phrased; you take an injective resolution $S \to I$, apply $f_*$, and take hypercohomology of this complex of sheaves?
And the spectral sequence you speak of is a corollary of the general hypercohomology spectral sequence, so you're basically asking how to set that one up. OK, let me read the rest of the post.
From there on I want to use John McCleary's Book on spectral sequences. The Leray spectral sequence is the one obtained from this double complex as I understand it.
@ChristianSander Yes, that's right. I typically don't think about the details with the Cartan-Eilenberg resolution, but you can take any injective resolution of the complex $f_* S \to f_* I$, apply $\Gamma$, and compute spectral sequence of the double complex.
The differentials on the $r$-th page of a spectral sequence of a double complex is given by going along an $r$-step ladder in the double complex by snake lemma.
You have to solve a system of the form $a_0 = da_1$, $d' a_2 = d a_3$, etc
Okay, I want to take the filtration that cuts threw the collums and keeps the rows of the double complex intact (I refer to that as the vertical filtration)
Now the $r$-th differential in the spectral sequence is coming from the map $Z_r^{p,q}\xrightarrow{d} Z_r^{p+r,q-r+1}$ where $d$ is the differential of the double complex.
$d = d_h + d_v$
Okay, I am getting to the point :).
I think of $Z_r^{p,q}$ as elements on the diagonal with total degree $p+q$ that map to something nontrivial only above the $p+r$-th row under my differential.
So in order to find these I (visually) start in the bottom right with an element $a_0$ sitting in the spot $(p,q)$ and apply $d_h$ and $d_v$. $d_h$ must be $0$ in this case leaving me with one element $d_v(a_1)$ which I have to kill with an element sitting in the $(p-1,q+1)$-st spot.
I gotta continue this process up to the top left making sure that I kill all elements along the way that are below the $p+r$-th line on the diagonal with total degree $p+q+1$, right?
$B_{r-1}^{p+r,q-r+1}$ is the image of $d$ of all elements on the diagonal with total degree $p+q$ above the $p+1$-st row that also land above the $p+r$-th row, right?
@BalarkaSen Or let me rephrase the last question: In order to understand $B_{r-1}^{p+r,q-r+1}$ I need to do that "going up the latter" just as it was done for $Z_r^{p,q}$ but instead of starting in the spot $(q,p)$ I start at the next (up and left) spot $(q-1, p+1)$.
Sorry, I was away from keyboard. It's hard for me to follow symbols, but maybe you could skip it because I know the description you're trying to recall. Could you point out why your differentials are becoming zero?
So far, everything in the question you linked seems correct. So if there is some inconsistency it's in this part -- certainly in the Leray spectral sequence not all the differentials beyond page 2 are zero in general so we know something you're doing is wrong.
I'll try to keep it as simple as possible: Take an element in $E_r^{p,q}$. This is represented by an element in $Z_r^{p,q}$.
To compute the image of that under the differential $d_r$ we have to do that "going up the ladder".
Let's say the module at the bottom right on our diagonal with total degree $p+q$ is $A_0 \oplus B_0\oplus C_0$.
this is where we start that ladder process. Since we want to land above the $p+r$-th row, we cannot take something in $C_0$, right?
Because the horizontal differential is sending that not to zero.
So th eprocess starts with an element like $(a,b,0)$. We apply the vertical differential and get $(d_va,d_vb,0)$ and this we can only kill with an element next to it on the left if $d_vb=0$. In this case an element that kills it via the horizontal differential looks like $(x,y,d_va)$.
And here I could freely choose $x$ and $y$. Am I talking sense up to here?
Let's just do $r = 2$. $E_2^{p, q}$ is represented by an element $\alpha \in K^{p, q}$ such that $d_v \alpha = 0$, $d_h \alpha = - d_v \alpha'$ where $\alpha' \in K^{p-1, q-1}$. Am I correct?
In en.wikipedia.org/wiki/Horseshoe_lemma indeed they do not say anything about that. But in Weibel's book he says that it acts componentwise if I understand it correctly.
Yeah, at least for finite direct sums I don't see a problem either.
Hmm I admit that I am thinking about this for two weeks now. I couldn't find anything but I agree: It is probably just a stupid mistake that I made :-/.
Judging from his statement I would say that it is a direct sum of complexes because of him writing "...the right hand column lifts to an exact sequence of complexes $0\to P'\xrightarrow{i} P\xrightarrow{\pi} P''\to 0$.
It's all good, I learnt something as well. Good to see that I wasn't 100% off as to where the issue was.
I never remember this horseshoe lemma because it's an immediate corollary of the definition of injectivity/projectivity. So I never remember the differentials either
@Christian Here's the takeaway, which is what was surprising me about your confusion. You can never expect to know everything about a spectral sequence; if you know $d_h$, you must not know what $d_v$ is, at least very explicitly. And that's right, the differentials in the horseshoe lemma aren't super-trivial - the off-diagonal term you need to compute.
If you know everything about a spectral sequence you either did a easy computation or you're wrong.
hi, i read this def in my notes: $ f \in \mathbb{C}[x,y]$ has no repeated factors over $\mathbb{C}$ iff it is not a product of hte form $g^2 h$ where $g$ is non-constant. Equivalently, $f = P_1 ... P_k$ where $P_i$ are distinct irreducible polynomials. I don't think this is quite right, I think what they mean to say is 'Equivalently, $f = P_1...P_k$ where $P_i$ are mutually non-associate irreducible polynomials'. Which one is correct?
@ChristianSander for any given quadratic of the form $ax^{2} + bx + c$, what is the point along the parabola at some arbitrary distance along the parabola and away from the vertex of the parabola towards either negative or positive infinity, and consequently, what are the functions of x and y that describe this parabola parametrically as $(\operatorname{cosp}(\theta), \operatorname{sinp}(\theta))$?
The parametric equation
$$\begin{align*}
x(t) &= \cos t\\
y(t) &= \sin t
\end{align*}$$
traces the unit circle centered at the origin ($x^2+y^2=1$). Similarly,
$$\begin{align*}
x(t) &= \cosh t\\
y(t) &= \sinh t
\end{align*}$$
draws the right part of a regular hyperbola ($x^2-y^2=1$). The ...
@LeakyNun No, but since what I'm doing requires that all things be waves, and norms, it makes sense to find a wave function that describes a parabola and work up to all polynomials eventually as a generalization of sinp and cosp.
Also the infinite series shows this is surprisingly close to $\arctan(x)$. Hm. Interesting. (Now how to remove the coefficients in the numerator I wonder?)
In the question that LeakyNun posted here which I was also looking at, it would appear that maybe $\operatorname{sinp}(t)=t$ and $\operatorname{cosp}(t)=1+\frac{1}{2}t^{2}$ in the accepted answer, however, he doesn't state that as the definition, but more as a guess.
Indeed there are, but they're not usually called that. What ordinary trigonometric and hyperbolic functions have in common is that they are solutions to the differential equation $$f''(t) = af(t)$$
When $a$ is negative, the solutions are ordinary sines and cosines, scaled horizontally by a factor...
One of the limitations I have put on myself as well in this area that I am researching (I don't know what it would be called) is no calculus of infinitesimals, so I can't get something normally via integration or differentiation. It has to be through composition or applying some other function, and in a closed-form expression.
The benefits will be enormous. I find that while it is easy to compose functions using elementary particles of addition and subtraction fundamentally, it is difficult to decompose functions into their elementary parts and describe them all the same using a closed-form expression.
But I also don't want the fruits of my labor to be a long "list of identities for common functions" that you have to memorize, but rather something very terse that can be universally applied and easily remembered to "split" a function into parts which can be manipulated and rearranged.
The simple questions I'm still contemplating the answers to are "What is every $x$?", "What is every function?", and "What are all axes?" They'll be the substance of my study.
Hyperbolic functions I consider to have closed-form definitions that can be trivially computed since they are real functions of x and do not strictly require imaginary numbers.
If I could trivially compute complex-valued functions, well then that would be a different story.
I've been looking for a cheaply computable form of the circular functions, you see, but I would say that is only a catalyst rather than the motivation for what I'm now studying, as this is actually relevant to a project of mine which is non-trivial in computer science and applicable to all sciences once complete.
The parametric equation
$$\begin{align*}
x(t) &= \cos t\\
y(t) &= \sin t
\end{align*}$$
traces the unit circle centered at the origin ($x^2+y^2=1$). Similarly,
$$\begin{align*}
x(t) &= \cosh t\\
y(t) &= \sinh t
\end{align*}$$
draws the right part of a regular hyperbola ($x^2-y^2=1$). The ...
