Reputation
3,908
Top tag
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
1 10 40
Impact
~87k people reached

9h
comment Does a $\Pi_2^0$ sentence becomes equivalent to a $\Pi_1^0$ sentence after it has been proven?
@DavidC.Ullrich If that were true, then the answer to my question should be a resounding YES! But the example of Goodstein's theorem given in the question should make in clear that a $\Pi_2^0$ sentence can be true, without being equivalent to every other true sentence. (I actually hoped that Goodstein's theorem would be equivalent to (or at least imply) the consistency of PA, but I only found such a statement for the related Paris-Harrington theorem.) The case of the model existence theorem is more problematic, because it isn't really a statement about natural numbers.
17h
asked Does a $\Pi_2^0$ sentence becomes equivalent to a $\Pi_1^0$ sentence after it has been proven?
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
19
awarded  Popular Question
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
25
revised Bound on the size of Permutation Set for Isomorphism
question has been modified, wrote an extended comment on my attempts to help Jim!
Jan
19
accepted Graphs embeddable into tree like simplicial 2-complexes
Jan
18
answered Bound on the size of Permutation Set for Isomorphism
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?
Jan
4
answered Isomorphism of Non-Symmetric Matrices
Dec
27
awarded  Popular Question
Dec
18
asked Graphs embeddable into tree like simplicial 2-complexes
Dec
16
reviewed Leave Open Can we always multiply some function that is not differentiable everywhere with function that is to obtain differentiable product?
Dec
16
reviewed Leave Open Solution about a Michael Aschbacher exercise [edit]
Dec
16
reviewed Leave Open Is there a non-trivial cyclic quotient group of a non-cyclic group?
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
revised What does it mean if a free algebra has an unsolvable word problem?
added links to definitions and the "terminology" tag
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.