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Mar
29
comment Geometric intuition behind the Lie bracket of vector fields
@sifsa See drive.google.com/folderview?id=0B6x6GQ82vuH_YnE2STNKZXU1cUk page 19 at the bottom. The proof exploits a definition of the Lie bracket by its effect on a test function $\varphi$ via $L_v(L_w \varphi) - L_w(L_v \varphi) = L_{[v,w]}\varphi$. Then it is sufficient to know that $L_v \varphi$ transforms canonically...
Mar
23
comment Are there any objects which aren't sets?
The point of sets in set theory is that we can quantify (impredicatively) over sets, because they are assumed to exist as a completed whole. Even so you can explicitly name some classes (even using predicative quantification over other classes), they don't exist as a completed whole. So you cannot meaningfully quantify (impredicatively) over them. I explained the way they fail to exist here, but it is probably hard to understand despite the simple examples.
Feb
8
comment How Graph Isomorphism is used to determine Graph Automorphism?
It is good that you care about graph automorphisms. Any permutation is a product of transpositions, but it can happen that you have two transpositions which are not automorphisms of the graph, but their product might still be an automorphism of the graph. But it is fine for me, if you say that the transposition was just an example, and that the idea also works for permutations which are not transpositions. On an unrelated note: Did you notice that the text said: "continue till an isomorphism is found"? So it only computes a single representative. Can you work out why a single rep. is enough?
Feb
8
comment How Graph Isomorphism is used to determine Graph Automorphism?
"Now consider a transposition $\pi$ that moves $(i+1)$ th vertex to $j'$ th position where $i <j'\leq n$." Is it really sufficient to only look at transpositions $\pi$? What would happen, if you used this algorithm to compute the automorphism group of a directed circle (whose automorphism group doesn't contain any transposition)?
Jan
4
comment Isomorphism of Non-Symmetric Matrices
Your algorithm doesn't work as claimed. You basically only describe the divide phase of the algorithm, but use handwaving instead of a description for how to combine the results from the subproblems to form a global solution. You also rely on handwaving for your mathematical notation: $2^{\sum log_2(m)}$ is unclear (sum over what?), $E_2= \min(K_i,J_i)$ is unclear (minimum of two matrices?), $J_i < m_1/2$ is unclear (a matrix is smaller than some number?), ... I admit that I can guess what you want to say, but how can I help you understand where your algorithm fails, with such handwaving?
Dec
15
comment What does it mean if a free algebra has an unsolvable word problem?
@MattSamuel In the context of the word problem for groups, one talks about presentation of a group which means that a group is given by a set of generators and a set of relations among those generators. In this context, a relation is not allowed to contain free variables. So there might be a polynomial time algorithm for deciding the word problem for lattices (including relations), even so the word problem for free modular lattices is unsolvable.
Dec
15
comment What does it mean if a free algebra has an unsolvable word problem?
@MattSamuel I thought about it some more, and came to the conclusion that equational axioms and relations are completely different things. If $a,b,c$ are your generators, then a relation is an equation between $a,b,c$ containing no free variable (=universally quantified variable) at all. An equational axiom on the other hand is an equation between free variables, containing no generators at all.
Dec
15
comment What does it mean if a free algebra has an unsolvable word problem?
@MattSamuel Why do you mean that a free modular lattice is very far from free algebraically? It seems to fit exactly the definition of a free algebra. No sets of relations in addition to the equational axioms seem to be allowed, so it seems significantly more restricted than an algebra presented by a set of generators and relations. (In which case the relations would be allowed to explicitly name a specific generator, while the equational axioms only contain universal quantified variables without any reference to the generators).
Oct
10
comment Why do $f(x)= \frac {{x²-1}}{{x-1} } $ and $g(x)=x+1$ not have the same domain of definition, but are the same?
Voted to reopen, because this question has 6 upvoted and 1 downvoted answers. This is a pretty strong indication that the question itself is not really unclear. Partially undefined operations are indeed annoying, but it's no solution to pretend that questions about them are invalid or unclear. (The solution motivated by category theory to always specify domain and codomain and only consider total function is indeed quite powerful, but it is not the only possible solution. In the context of (complex) function theory, other solutions are more appropriate.)
Sep
25
comment SVD methods for minimal polynomial
It's not really a start point with respect to SVD. There is a close connection between Krylov subspace methods and minimal polynomials. It wasn't even a SVD, but just a general Eigen-decomposition. It should be clear how to get the minimal polynomial of a square matrix from its Eigen-decomposition. But they do use it for sequence extrapolation of transient (linear response) behavior when resonances are present, which could otherwise become annoying.
Sep
24
comment SVD methods for minimal polynomial
I'm pretty sure that an explanation of continuation by fast Krylov subspace Eigen-decomposition methods together with references to the original publications was in some introduction or appendix at least in the published book of ccrma.stanford.edu/~bilbao/master/goodcopy.html. I couldn't find it again at the moment, but why should I care?
Sep
1
comment Differential algebra and differential-algebraic equations
@Calle If you speak German, then section 3.2.1 in mod_geom_betr_klas_diffop.pdf gives a good idea for a relation between symmtries and structural properties, i.e. coupling or non-zero structure. Here are visualizations of non-zero structure of DAEs in integral form and non-zero structure of DAEs in semi-explicit form. See "Matrices and Matroids for System Analysis"
Apr
13
comment Does the “equality semigroup” have an accepted name?
If you introduce the name "subatom" for an element which is either the bottom or an "atom", then you can no longer pretend that you are using established terminology. If you slightly abuse existing terminology on the other hand, then you are just following established mathematical practice. (red herring principle...)
Apr
13
comment Does the “equality semigroup” have an accepted name?
In any partial ordered set with a bottom element, the meet-semilattice of atoms (en.wikipedia.org/wiki/Atom_(order_theory)) is a well-defined semilattice. So I would call it just this: "meet-semilattice of atoms".
Feb
21
comment Can every rational function be represented in barycentric form?
@JohnHughes You mean because $y_j$ is not just a coefficient, but also identical to $r(x_j)$? Maybe you have a good point, and it could be related to my troubles understanding the cited statement.
Feb
21
comment Can every rational function be represented in barycentric form?
In Barycentric Lagrange Interpolation. Jean-Paul Berrut, Lloyd N. Trefethen., the earliest references seem to be from 1997 of Jean-Paul Berrut. This is a case of an author citing another paper of himself as reference, so "it is known" might only refer to a very small set of people...
Dec
23
comment What's behind the Banach-Tarski paradox?
The physical world doesn't deal with sets at all!
Nov
29
comment Which associative and commutative operations are defined for any commutative ring?
@ThomasAndrews Yes, in case of addition the corresponding condition is that $b$ has to be in the range of $x\to ax$. What I like less is that $a$ and $b$ are currently not true parameters, but are restricted to the free commutative ring without generators, i.e. $\mathbb Z$. But I'm not sure how to extend Martin Brandenburg's "natural" argument to the case with parameters.
Nov
29
comment Which associative and commutative operations are defined for any commutative ring?
@ThomasAndrews It certainly won't be in the range of $\phi_{ab}$ if $b(b-1)/a$ is not defined, i.e. when $b(b-1)$ is not in the range of $x\to ax$. If on the other hand $b(b-1)$ is in the range of $x\to ax$, then the laws of commutative rings are sufficient to show that also the other expression will work.
Nov
29
comment How to show distributivity in a ring, and what is wrong with my algebra?
See also the appendix of math.stackexchange.com/questions/1001186/…