I was so happy the day I finally learned what a Riemannian manifold was, so I could at least understand the entire line from Tom Lehrer's Lobechevsky song
@Balarka Here's a very nice question which I'll let you keep at the back of your head. Let $\Sigma$ be a smooth closed surface. What smooth functions $f: \Sigma \to \Bbb R$ can be the Gaussian curvature of some metric?
Fun problem. I can't immediately come up with a nontrivial characterizing property of the Gaussian curvature, but I'll think about it once I familiarize myself with it.
The angular velocity vector of a rigid body is defined as $\vec{\omega}=\frac{\vec{r}\times\vec{v}}{|\vec{r}|^2}$. But that's not how most people intuitively think about angular velocity.
Euler's theorem of rotation states that any rigid body motion with one point fixed is equivalent to a rotat...
I am asked to solve the following ODE involving constants $\alpha, L, V_0 > 0$ and $E < 0$:
$$-\psi'' - \alpha V_0\psi\cdot [\delta(x)+\delta(x-L)] = \alpha E\psi.$$
In particular, we want solutions $\psi:\mathbb{R}\to \mathbb{C}$ that are:
Continuous
$L^2$
We are given that the solutions l...
@keshav, this is a totally intuitionist answer, but my gut feel is that the reason a fixed point implies well-defined-ness of an axis of rotation is tautological
In geometry, Euler's rotation theorem states that, in three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point. It also means that the composition of two rotations is also a rotation. Therefore the set of rotations has a group structure, known as a rotation group.
The theorem is named after Leonhard Euler, who proved it in 1775 by an elementary geometric argument. The axis of rotation is known as an Euler axis, typically represented by a unit vector
...
But it sounds like rotation is defined in terms of Euler's notation theorem: "When a sphere is moved around its centre it is always possible to find a diameter whose direction in the displaced position is the same as in the initial position."
Oh wow, I did it for the unit sphere and it worked. K = +1.
Onto saddle.
@MikeMiller Without doing any calculations, I think nothing special should happen for radius r \neq 1. I and II are both multiplied by r, so they cancel off after taking det so you still get K = +1
@enthdegree If you rotate a sphere 90 degrees about the x-axis and then 90 degrees about the y-axis, it is not immediately obvious that there exists some rotation that will take you from the initial configuration of the sphere to the final configuration.
@enthdegree In any case, my question isn't about proving Euler's theorem of rotations, it's about relating Euler's theorem of rotations to the angular velocity vector.
@KeshavSrinivasan: The fastest way to see that visually seems to be to prove that a rotation can be decomposed into two reflections even if you can choose either the first or the second. This means that you can see that composing two rotations can be done by choosing the second and third reflections to cancel, leaving you only two reflections, which is a rotation.
Had to go away from keyboard. @0celo7 Can you elaborate what kind of thing you havei mind by quadratic approximation?
Also, $\kappa = -1/(1 + u^2 + v^2)^2$ makes sense for the saddle. As $u, v$ increases (so I go far away from the origin, on the surface of the saddle) $\kappa$ becomes $0$. At the origin $\kappa = -1$, so I suppose that's the local model for negatively curved surfaces.
Also the minus sign probably indicates that the Gauss map reverses orientation.
oh ok and then to show that the presentation given above for $Z_2 \times Z_4$ and $Z_2 \times Z_2 \times Z_2$ I have to show that they are isomorphic ?
For isomorphisms when you have a presentation, they are generally easy to prove, you just need to prove that whatever elements you are assigning to each of the generators obey all of the relationships in the presentation
so for the first, you'd show that $(1,0)=a,(0,1)=b$ obeys the 3 relationships, and that's your isomorphism.
Through a function name on there to be technical :)
@Alan can you find counter example of the following statement ? Suppose F is a free abelian group of finite rank n. I need counter examples for the following statements: i)Every linearly independent subset consisting of n elements is a basis of F. ii)Every linearly independent subset can be extended to a basis of F. iii)Every generating set of F contains a basis of F.
Hmm, double checking my defintions...free abelian of rank n juist means it has n generators that have infinite order and all commute, right? And what do we mean by linearly independant subset in a group? (I'm familiar with that for vector spaces, not groups). Likewise, basis?
Is the following true?
$G$ is a torsion free abelian group of rank $>n$.
Let $S$ be be a subgroup of $G$ generated by $s_1,s_2, \cdots ,s_n$.
(1) If $m_1s_1+m_2s_2+ ...+m_ns_n$ is linearly independent modulo $p$ for every prime $p$, then $S$ is linearly independent. That is, if $m_1s_1+m_...
btw thinking about the earlier problem I would like to do it rigorously just want to see if I have the right steps so let us say we take $Z_8$ we have $\phi: F(X) \rightarrow Z_8$ where $X = \{a\}$ and the map $a \mapsto 1$
then it is easy too see that $R = \{a^8\} \subset ker\phi$
so $<R> \subset ker(\phi)$
it is also easy to see that the $\phi$ is surjective.
I would like now to prove that F(X) / <R> has at most 8 elements
In computer science, the iterated logarithm of n, written log* n (usually read "log star"), is the number of times the logarithm function must be iteratively applied before the result is less than or equal to 1. The simplest formal definition is the result of this recursive function:
log
∗
n
:=
{
0
if
...
Ahh, it's a measure of order of magnitude that's way, way more condenseced than normal logs. log massively shrinks a number. This says how many times do we go through that process before we get below 1. So for your example....what base log do you want to use? 2, 10, e?
Sure, but I'll warn you, my group theory is rusty :)
Wow, just read that it's only slower than the inverse ackermann function. The latter function actually came up in one of my PDE class work problems in a bound.
Well, let's see if this computation manages to finish in time before I lose my office and the computer it is running on. The calculation for type $B_6$ took about 60 hours, and I have just over two weeks for it to finish type $B_7$.
Caption: The convention of a cayley table is that the row element comes first followed by the column element. Therefore the table is basically the same as treating the row and columns as vectors and the entries of the table is given by the tensor product of Row and Column as $\textrm{Row}\otimes\textrm{Col}$
(If I recall correctly) The matrix product of a nx1 row vector times 1 x m column vector is a special case of a tensor product because it gives a nxm matrix . The reason I use the tensor product instead is because the next step in constructing the nmp array cannot be written by a matrix product
Perhaps more accurately, the row vector times column vector matrix multiplication is a Kronecker product, and that is the tensor product wrt to a certain basis
Now, after this is done, all entries of the algebraic structure that corresponds to the equation (ab)c is then given by the tensor product of the cayley table to the column of elements in the algebraic structure
The resulting entries will then be evaluated based on the rules of that algebraic structure
Similarly all entries that corresponds to a(bc) is given by the tensor product of the column of entries to the transpose of the cayley table
If these two arrays (which formally should be tensors) are equal, then the algebraic structure is associative wrt the binary operator under consideration
It's not very efficient as you are still computing all the terms unlike Light's associativity test, but it makes the organisation and perhaps automation easier in a computer as you effectively convert the abstract algebra problem into a matrix problem
As I recall, it is fairly common in Denmark at least, and I think I have seen one of the notations mentioned as belonging to some region, but I don't recall the details
@MAFIA36790 Munkres, corollary 6.4 illustrates this, because any attempt to plug an infinite set into Theorem 6.2, it breaks down thus can no longer conclude that the subset is not bijective for that set in question
anyone know of a proof that is really easy to check but hard to come up with? i post challenges on facebook. i'm thinking something along the lines of prove that there is a irrational number that can be raised to an irrational power to get a rational number but that one is pretty well seen
does anyone know if there exist a proof about a irrational number raised to an irrational algebraically independent number to get a rational result?
One thing I noticed is that its associativity table is quite trivial: given that we knew an associativity table of 3 elements is a 27 entry array. If we partition that into blocks of 3x1 columns, the result is always a,b,c
Btw, when reading wikipedia's article about semgroups, they said that absorbers are often interesting in semigroups. Why is that and how they are used in proving or investigating stuff in semigroups?
If a magma has both a left zero z and a right zero z', then it has a zero, since z=z\times z=z'z=zz'=z'. If a magma has a zero element, then the zero element is unique.
it also seems like its minimun sufficient information for somethings, like if we have x*y and we know that x is an absorb-er we have x, for example what is the minimum between x=(-infinity to 0) y=? we don't need to know y to note the minimum is -infinity
yeah but i gave an example to illustrate my point.
i should have used the words "for example" instead of "like"
it seems for operations there exist unique elements such that for operation , xy=y (identity) or x*y=x (absorb-er), the absorb-er then gains the interestingness of the identity in the groups where it exist @secret and thats why there looked for and are interesting
@BalarkaSen There's a trick about writing a surface locally as a quadratic, then the coefficients of the quadratic expansion can be used to easily compute the curvature
I have a question regarding the definition of homology with local coefficients defined using a module over the group ring. Is there someone who knows about this construction?