When talking about a deformation retraction of a space $X$ onto a subspace $A$ being a homotopy from the identity map of $X$ to a retraction of $X$ onto $A$, a map $\gamma:X\to X$ such that $\gamma(X)=A$ and $\gamma\mid A=\mathbb{1}$ (the identity map)
In Isaacs Algegra, in an exercise in the group theory part he uses the notation $x^\sigma$ for $x\in G, \sigma\in$Aut(G). Is that supposed to mean application of $\sigma$ to $x$?
@BenjaLim I see. So basically, sketches and extrapolations.
I had a book on analysis by protter morrey, that had a similar style. However, there, they had explicitly mentioned as such," to allow the student to fill in the proof." and just guidiing him in the right direction.
My current potential research topic is on how to use Witt vectors to characterize degree p^2 extensions of p-adic fields. Other topics may come up in the future.
Pretty good. I don't like having to get up in the morning for a lecture I don't even listen to, but...
Now, I need to count the matrices with entries in $\mathbb{Z}_+$ such that the rows add to the vector $\underline{r}$ and the columns add to the vector $\underline{c}^t$. All my vectors are row vectors, unless specified otherwise.
@KannappanSampath See, consider a single row. Then the set of solutions for your main problem is a subset of the set of solutions of the partition problem for that single row.
His problem is one of counting, not enumeration (the latter of which costs much more computationally). Obviously if you wanted to enumerate all $\Bbb Z_{\ge0}$ solutions you could enumerate integer partitions of each row and column component individually and then form two families of matrices by putting together permutations of the row and column partitions respectively and find the intersection of these two families, but that's way infeasible.
Suppose the vectors and columns of sums are $\underline{r}$ and $\underline{c}$ respectively. And suppose the number of solutions is $S(\underline{r},\underline{c})$
Suppose the sum in the first row is $r_{1}$ and the first column is $c_{1}$. Let $m_{1,1} = \operatorname{min}(r_{1},c_{1})$. Then the element $a_{1,1}$ has the range ${0,1,...,m_{1,1}}$
@KannappanSampath If you want a recursive algorithm you might consider thinking about filling the top row and leftmost column in by generating things $\le$ the given sums, and then calling the algorithm for the left-over $(n-1)\times(n-1)$ minor (via excising said row and column) in each instance, but that seems like a lot of recursion.
@anon I tried this; but the problem, there are several ways a column and a row can be filled. So we'll have to multiply at each stage by that number and so on... But to find that factor itself seems very hard.
I'm unsure of what multiplication you're referring to. You iteratively generate the left corner of the matrix (corner = leftmost column + topmost row) (which corners are possible given the row/column sum vectors is an easy task), and given each corner of the matrix you decrease the row/column sums by the relevant amount and the apply the algorithm to the $(n-1)\times(n-1)$ minor left behind (obviously creates a lot of recursion).
I think I know how to show that given any $X \in \mathfrak{g}, Y \in \mathfrak{h}$ that $e^Xe^Ye^{-X} \in H$. Now
$$\begin{eqnarray*} e^{X}e^Ye^{-X} &=& 1 + e^XYe^{-X} + e^{X}\frac{Y^2}{2!}e^{-X} + \ldots\\
&=& 1 + e^{\textrm{ad}_X}(Y) + \frac{1}{2!}(e^XYe^{-X})(e^{X}Ye^{X}) + \...
@KannappanSampath I guess my method is good. Start with the problem of the sum of all row vectors zero and column vectors zero. For that we have one solution, namely the zero matrix. Now let only one row and one column have the sum 1. Count the number of such matrices. It will be $m \times n$ where $m$ and $n$ are the number of rows. Next, consider with two rows of sum one and two columns of sum 1. Calculate this, we will get $mC2 \times nC2$. Continue this, till we are left with a matrix with all columns 1 and all rows 1. If rows are not equal to columns, then obviously, one of them will …
Show that $\tau\sigma_{\mathfrak{P}}\tau^{-1}$ satisfies the defining property of $\sigma_{\tau(\mathfrak{P})}$, and by uniqueness of the latter they are one and the same
that's supposed to be a subscript, but $\mathfrak{P}$ is a big letter
In other words, show that for all $\alpha$, we have $$(\tau\sigma\tau^{-1})(\alpha)\equiv\alpha^{N(\tau(\mathfrak{P}))}\pmod{\mathfrak{\tau(P)}}$$ given that you know that $$\sigma(\alpha)\equiv\alpha^{N(\mathfrak{P})}\pmod{\mathfrak{P}}$$ for all $\alpha$. You can do this by noting that $N(\tau(\mathfrak{P}))=N(\mathfrak{P})$ and $\alpha^{N}=\tau\left(\big(\tau^{-1}(\alpha)\big)^{N}\right)$ and $x\in X\iff f(x)\in f(X)$ if $f$ is a bijection.
Honestly I do not see the point of the base field $k$ or the lower prime ideal $\mathfrak{p}$, are they even relevant? (Or is it part of the setup to more parts of an overall problem?)
@FortuonPaendrag I'm not trying to fight umbral calculus - I mean, for understanding it. I'm in search of: 1 - A simple explanation on some aspects of it; 2 - To make people know that it exists, it may be useful for someone.
$$(y+x)^n=\sum_{k=0}^n{n\choose k}y^{n-k} x^k$$
What's the meaning of the parentheses with N and K?
I understand. But I feel like you would have a better feel for these ideas once you take more classes. So, I am ashamed to ask, as you told me once.. but what is your math background?
It is the number of ways to choose $k$ distinct objects from $n$ objects.
Thanks for the information on the N and K in the parentheses.
user19161
4:14 AM
@GustavoBandeira Do you mean the n choose k part? Well, that is just the number of ways to choose k objects out of n objects, where distinct arrangements of the k objects constitute the same way. You may read up more on "combinations" and "permutations".
@MarianoSuárez-Alvarez It is quiet possible though. It was that way at my univ too. We had access to only articles, no books. Confirmed with authorities.
I think I know how to show that given any $X \in \mathfrak{g}, Y \in \mathfrak{h}$ that $e^Xe^Ye^{-X} \in H$. Now
$$\begin{eqnarray*} e^{X}e^Ye^{-X} &=& 1 + e^XYe^{-X} + e^{X}\frac{Y^2}{2!}e^{-X} + \ldots\\
&=& 1 + e^{\textrm{ad}_X}(Y) + \frac{1}{2!}(e^XYe^{-X})(e^{X}Ye^{X}) + \...
