I have seen two different definitions for an analytic function, and I am wondering whether they are equivalent. Here is one definition: $f$ is analytic on some region $R$ in $\Bbb{C}$ if $f$ has a derivative at each point of $R$. The other definition something like "$f$ is analytic if it can be locally represented by a power series."
@BalarkaSen Here's a simple question: I have seen several people writing the following: $SO(6)/U(3)=SU(4)/S(U(3)\times U(1))=\Bbb C\mathrm P^3$. The first equality sign is what I don't understand: The denominator is unchanged because $U(3)\cong S(U(3)\times U(1))$, with the isomorphism of Lie groups given by $A\mapsto (A,\det A^{-1})$, but the top is weird. $SU(4)$ is the double (universal) covering of $SO(6)$, so I don't understand how this can work. Do you understand what might be going on?
i'm trying to figure something out and i'm really confused which way to interpret it
im trying to figure out how to compute the surface normals at the vertices of a polygonal surface. I've heard that it can be done by computing a weighted average of the various normals on the faces on the surface
the problem is im not sure if I need all the faces sharing some point or just the faces touching the face im examining.
:/
it seems like the latter should work
in the sense that i doubt an algorithm to find all the faces even exists
@LeakyNun that's because you can construct the field with $p^n$ elements as the roots of that polynomial and the algebraic closure is the union of all those fields
the way to compute the "soft normals" in that diagram is some kind of "weighted average" of the normal vectors on each plane (not at all ever explained in detail). My question is that (for instance) with something like the icosohedron do you actually need all 5 normal vectors to compute that weighted average or would using only 3 adjacent triangle work the same.
tbh, that's probably not true but it seems like the icosahedron is a prime example where computing the normal vectors are becoming harder and harder tocompute
Guys. Let $G$ be a group, and $A\subset G$ a finite subset, with $n=\vert A\vert$. Let $H=\{g\in G\mid\text{for all }a\in A,gag^{-1}\in A\}$. I have to show that $H$ is a subgroup of $G$. It’s clear that $e\in H$ and I can also show that if $g,h\in H$, then $gh\in H$.
However, I’m having difficulty showing that if $g\in H$, then $g^{-1}\in H$. So say $g\in H$. Then we know that $gag^{-1}\in A$ for all $a\in A$. We need to show that $g^{-1}ag\in A$ for all $a\in A$. The closest I could get to showing this is: $g^{-1}ag=g^{-1}agag^{-1}ga^{-1}$. Now unfortunately, the order here isn't right, for we need $a^{-1}g$ istead of $ga^{-1}$.
Now I'm kind of stuck here. Could someone give me a hint or the answer? Maybe I should use the fact that $A$ is finite, but I wouldn't know how.
Proof of Fermat little theorem: $\Bbb Z_p$ is a field whose multiplicative group has order $p-1$. Therefore, for $0 \ne a \in \Bbb Z_p$, $a^{p-1} = 1$.
Guys. Let $G$ be a group, and $A\subset G$ a finite subset, with $n=\vert A\vert$. Let $H=\{g\in G\mid\text{for all }a\in A,gag^{-1}\in A\}$ and let $N=\{g\in G\mid\text{for all }a\in A,ga=ag\}$. I’ve already shown that $H$ is a subgroup of $G$, and that $N$ is a normal subgroup of $H$. I need to find a homomorphism $f\colon H\to S_n$ such that $N=\ker(f)$. I am quite clueless at this point. I'm not going to try out an example, because that would take forever.
It seems to me that we want the kernel to constitute of 'commutative elements', in the sense that those elements are in $N$. So as soon as they commute with elements in $A$, we want to get $(1)$. But I'm not really sure how I could relate elements of $H$ with permutations. Any help would be appreciated.
@RandomVariable In that question, yes, but I think the same functions should work for $\int_{-\infty}^\infty f(x-1/x)\,\mathrm{d}x =\int_{-\infty}^\infty f(x+\tan(x))\,\mathrm{d}x =\int_{-\infty}^\infty f(x)\,\mathrm{d}x$
We had a homework problem earlier to show that commuting matrices over C are simultaneously triangularizable
His idea was to do it via quotient spaces
(This was second quarter analysis)
Now, our first quarter prof had us work through this linalg book that proved more generally that a matrix is triangularizable iff its minimal polynomial factors completely
If $T:V\to V$ is an operator, $W$ is an invariant subspace, and $v\in V$ is some vector, the $T$-conductor of $v$ into $W$ is the unique monic generator of the ideal of polynomials $p$ such that $p(T)v \in W$
@Salt you got the joke, our prof saw a few people who used that to answer the midterm problem and said "Don't give me the book Souganidis told you to read and start talking to me about conductors. This is linear algebra, not trains, no need to bring in Amtrak to solve your problems"
hi is the direct sum of rings different than direct product of rings. I think it is because in direct sum, we have multiple representation of 1 which is (0+0+..1+0+0+..) and in direct product if we a different but unique representation of 1 which is (1,1,...,1,,,)
i remember having a calculus instructor that drew the most beautiful diagrams on the blackboard, but passing him on campus he would always be covered in chalk
