If I take the cross product between the angular rotation of earth and a train travelling north in the northern hemisphere the result is $|\omega||v|\sin 90- \lambda$ where lambda is the latitude right?
Alright, I have an attempt at a proof that every element of a finite group has a finite order. I'll copy and paste it in here. If anyone has a moment could they tell me if there are any holes in the argument?
Let us first establish a base case of a group with only a single element. Since this group must have an identity element, it must be this single element, and so every element in this group has finite order. Now, assume that every finite group with up to $n$ elements has been shown to only have elements of finite order and add a single new element $a$ to such a group.
We know that $a$ must have an inverse, so either we assign it to be it's own inverse, in which case it automatically has finite order, or we rearrange relations in the group such that $a$ has an inverse which is not itself. Let us assume, for ease of argument, that the latter situation is possible, and consider $a^2$. If $a^2\neq a$, then we know that $a^2$ is an element with finite order, and so $a$ also has finite order. So, allow that $a^2=a$.
Then we have that, for any element $h=ah_{0}=h_{1}a$ in the group, $h=ah=ha$, and so $a$ must be the identity, which we know is unique and already defined, and so $a^2$ cannot equal $a$, and every element of this group has finite order.
I'm skeptical that induction makes sense for this kind of argument. It's not obvious to me how, say, $S_3$ can be thought of as $\Bbb Z/5\Bbb Z$ with an element adjoined.
Indeed. $\Bbb Z/5\Bbb Z$ is the integers mod 5---$\{0,1,2,3,4\}$ with addition modulo 5.
My point is that Z/5Z is the only group of order 5 (up to isomorphism), and $S_3$ is a group of order 6, but the relations change so much that there's not really any correspondence between the elements of one and the elements of the other.
I'd argue that, since the relations are arbitrary to the fact that elements must have finite order, we can rearrange the relation until it's unrecognizable and the argument would hold, because $a^2$ has to go somewhere. (Also, making note that that is for p-adics.)
Though, if induction doesn't hold, what would be a more suitable approach?
Right, but my point is that you don't know whether the elements other than $a$ have finite order anymore, because they don't exist in the same group anymore. Order is entirely relative to the overlying group.
I'd try a proof by contradiction.
Although, I think there is a way to do this by strong induction, reasoning on subgroups.
So, start off by perhaps assuming that for every element $a$, $a^{-1}\neq a$, and then give the logical statement that if $a^n= a^{-m}$ for any $m$ or $n$, then $a$ has finite order?
Is anyone in this room simply explain the difference between a conditional probability, and a stochastic kernel? If not, then I'll just ask it as an SE question.
Well, I've already convinced myself that $a^2=a\implies a=1$, and so $a^2\neq a$, and now I'm thinking about the path that the $a$ traces out in the group as you increase the exponent. If it has infinite order it has to trace out a loop which never encounters $1$.
Well, we only have finite elements in the group. By closure, $a^n$ has to be in the group for every $n\in\Bbb{N}$. So, $a^n=a$ for some $n$. (Is this correct?)
$a^n$ doesn't have to equal $a$ ever, but you're right---if we're assuming we're in a finite group, then the set I stated is a subgroup, and so must be finite.
And that would require that $a^n = a^m$ for some integers $n,m$.
I'm having a little bit of confusion. So, I'm looking at the statement that "any group generated by two elements of order two is dihedral," which I'm fine with, but that should only allow for one such group, correct? "The dihedral group for the line segment," in a sense.
Oh, I think I see what I'm missing. If $a^2=h^2=1$, then we don't necessarily know that $(ah)^2=1$. All we know is that the order of $ah$ is finite.
It's also a way of getting around the fact that there's not "really" only one of a group. You can label the elements or operations in any way that you like, and those different labelings are taken to be different sets.
As an example, there are lots of infinite cyclic groups (Z, 2Z, 3Z, ...).
But there's only one infinite cyclic group up to isomorphism---all the groups I just listed are "the same" in a well-defined sense.
So what this is saying is that "any group generated by two elements of order 2 must specifically be of this form", and in this case that gives enough info to say "there's only one group generated by two elements of order 2: namely, the one of this form".
You move out a layer in abstraction and you can see that the different structures are just one and the same. I think I'm going to enjoy studying isomorphisms when we get to that point in the semester.
The thing that tells you that $(ah)^2 = 1$ is the combination of the facts that: - $a$ and $h$ generate the group - $a^2 = h^2 = 1$. If both these things were not true you could not necessarily conclude the above.
Hmmm, so $(ah)^2=1$ for sure? Okay, then I don't immediately see the way to arrive at that conclusion. Do we know that a group generated by elements $a,h$ with $a^2=h^2=1$ is necessarily abelian?
Yeah. Now work backwards from this reasoning and see why two order 2 elements always give another generating set whose relations are the dihedral ones.
I guess finiteness of the group in question is being assumed?
In fact it's a three part question, and the first part is asking "Assuming that the order of $G$ is finite, what can you say about the order of the element $xy\in G$?"
Which kind of confuses me because there was already an exercise which said "Let $G$ be a finite group. Show that any element of $G$ has finite order," and I don't just want to say "refer to the above work."
Sure:The goal of this problem is to show that any finite group generated by two elements of order two is dihedral (with $\mathbb{Z}_2\times\mathbb{Z}_2$ being considered a "degenerate" dihedral group). Suppose that $G$ is generated by the elements $x,y\in G$, both of order $2$. a) (3 pt) Assuming that the order of $G$ is finite, what can you say about the order of the element $xy\in G$? b) (3 pt) Show that the group generated by $x$ and $y$ is the same as the group generated by $xy$ and $y$ c) (3 pt) Show that the group generated by $x$ and $y$ is dihedral.
I understand that how to convert left module to right module by involution.But what is meaning of we always interpret dual module M* as right module in this sense? Also what is meaning of scalar multiplication is twisted by involution?
Definition of Concrete Categories
Let $\mathbf{X}$ be a category. A concrete category over $\mathbf{X}$ is a pair $(\mathbf{A}, U)$, where $\mathbf{A}$ is a category and $U : \mathbf{A} \to \mathbf{X}$ is a faithful functor. Sometimes $U$ is called the
forgetful (or underlying) functor ...
(From what I can tell, gyroid infill doesn't have any clear advantages against other types of infill, but it doesn't have any disadvantages either and it looks cool)
Question: If $p : E \to B$ is a covering map and $B$ is locally compact, does it follow that $E$ is locally compact? I know it follows if we further assume that every compact set of $B$ has a open set containing it that is evenly covered by $p$. Is it possible to drop this assumption?
Interesting. I believe if $B$ is locally compact AND Hausdorff, then $E$ must also be locally compact Hausdorff. Are we able to do away with the Hausdorff assumption?
What is the difference between substituting and changing variables suggested by the Wikipedia quote, "Change of variables is an operation that is related to substitution. However these are different operations, as can be seen when considering differentiation (chain rule) or integration (integration by substitution)"?
I'll probably end up taking 30 credits just between set theory and logic next semester... I signed up for the graduate set theory seminar just earlier today :P
I don't know about Chern's classes but I know he has nice glasses.
@AlessandroCodenotti I think there are quite a number of physics courses too in a math degree in UK, but maybe we shouldn't count UK as part of Europe.
Are you asking how you induce a measure on $G/H$ when $H$ is normal and you have compatible measures on $G$ and $H$? This is Fubini's Theorem/integration over the fiber.
Today I saw a second cool application of the fact that modules are the colimit of their f.g. submodules, I'm starting to appreciate some category theory when used sparingly and appropriately
Suppose the gaussian curvature is always positive that means that the eigenvalues of the shape operator are of the same sign which means the min max of the second fundamental form are of the same sign which means that the shape operator which is the derivative of the gauss map N is always positive or negative . How i am supposed to continue in order to prove that the sueface is orientable . What i nees to prove is the det of the deriavte of composition of patches is positive
I don't know how much you know. If $K>0$ everywhere, then the surface is convex, and so it lies on one side of each tangent plane. That gives you an orientation (make the normal point outward).
I guess that if you choose coordinate patches where the principal curvatures are always positive, you can also deduce that the coordinate change will have positive determinant.
@Manolis: I guess the point is that the unit normals glue compatibly when you keep the sign consistent for the principal curvatures, and so the fundamental cross products being in the same direction will tell you that $(u,v)\mapsto (u',v')$ has positive determinant. You have to do that calculation.
@Manolis: Yes, but the sign of the principal curvatures is determined by the sign of $N$. So getting $N$ pointing in the same direction is equivalent ... and I'm saying that's equivalent to your chart-overlap jacobian.
You need to write out the chain rule for $\partial\vec x/\partial u'$, etc., in terms of $\partial\vec x/\partial u$ and $\partial\vec x/\partial v$.
Pullback looks like $f^{-1}$ so it has an upper star
Until you get to AG where $f^{-1}$ is pullback of sheaves and $f^\ast$ is pullback of $\mathcal O_x$-modules since you can't just pull them back as sheaves
@Leaky: Yes, except you mixed $F$ and $f$. It should be the companion of the pullback formula (usual change of variables) $\int_X f^*(g\,dy) = \int_Y g\,dy$.
Of course, that only works with the same dimension. The pushforward allows you to collapse dimensions.
@CaptainAmerica: That doesn't seem a very interesting exercise.
Am I correct in saying that $S^n$ with cell structure consisting of one $0$-cell and one $n$-cell can't be built up inductively (like start off with a discrete set $X^0$ attach some $1$-cells to get $X^1$ attach some $2$-cells to $X^1$ and get $X^2$ and so on)? Because in this case the $n-1$ skeleton of $S^n$ is just an empty set
Thanks @Ted I'm sure I have to use the other definition of a CW complex (as a Hausdorff space together with a cell decomposition that satisfies blah blah blah) to recognize $S^n$ as a CW complex in that case
I understand that how to convert left module to right module by involution.But what is meaning of we always interpret dual module M* as right module in this sense? Also what is meaning of scalar multiplication is twisted by involution?
Definition of Concrete Categories
Let $\mathbf{X}$ be a category. A concrete category over $\mathbf{X}$ is a pair $(\mathbf{A}, U)$, where $\mathbf{A}$ is a category and $U : \mathbf{A} \to \mathbf{X}$ is a faithful functor. Sometimes $U$ is called the
forgetful (or underlying) functor ...
